Guest Post by Willis Eschenbach
Since at least the days of Da Vinci, people have been fascinated by perpetual motion machines. One such “perpetuum mobile” designed around the time of the civil war is shown below. It wasn’t until the development of the science of thermodynamics that it could be proven that all such mechanisms are impossible. For such machines to work, they’d have to create energy, and energy cannot be either created or destroyed, only transformed.
I bring this up for a curious reason. I was reading the Jelbring hypothesis this afternoon, which claims that greenhouse gases (GHGs) are not the cause of the warming of the earth above the theoretical temperature it would have without an atmosphere. Jelbring’s hypothesis is one of several “gravito-thermal” theories which say the heating of the planet comes from gravity rather than (or in some theories in addition to) the greenhouse effect. His thought experiment is a planet with an atmosphere. The planet is isolated from the universe by an impervious thermally insulating shell that completely surrounds it, and which prevents any energy exchange with the universe outside. Inside the shell, Jelbring says that gravity makes the upper atmosphere colder and the lower atmosphere warmer. Back around 2004, I had a long discussion on the “climateskeptics” mailing list with Hans Jelbring. I said then that his theory was nothing but a perpetual motion machine, but at the time I didn’t understand why his theory was wrong. Now I do.
Dr. Robert Brown has an fascinating post on WUWT called “Earth’s baseline black-body model – a damn hard problem“. On that thread, I had said that I thought that if there was air in a tall container in a gravity field, the temperature of the air would be highest at the bottom, and lowest at the top. I said that I thought it would follow the “dry adiabatic lapse rate”, the rate at which the temperature of dry air drops with altitude in the earth’s atmosphere.
Dr. Brown said no. He said that at equilibrium, a tall container of air in a gravity field would be the same temperature everywhere—in other words, isothermal.
I couldn’t understand why. I asked Dr. Brown the following question:
Thanks, Robert, With great trepidation, I must disagree with you.
Consider a gas in a kilometre-tall sealed container. You say it will have no lapse rate, so suppose (per your assumption) that it starts out at an even temperature top to bottom.
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
I say no. I say after a million collisions the molecules will sort themselves so that the TOTAL energy at the top and bottom of the container will be the same. In other words, it is the action of gravity on the molecules themselves that creates the lapse rate.
Dr. Brown gave an answer that I couldn’t wrap my head around, and he recommended that I study the excellent paper of Caballero for further insight. Caballero discusses the question in Section 2.17. Thanks to Dr. Browns answer plus Caballero, I finally got the answer to my question. I wrote to Dr. Brown on his thread as follows:
Dr. Brown, thank you so much. After following your suggestion and after much beating of my head against Caballero, I finally got it.
At equilibrium, as you stated, the temperature is indeed uniform. I was totally wrong to state it followed the dry adiabatic lapse rate.
I had asked the following question:
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
What I failed to consider is that there are fewer molecules at altitude because the pressure is lower. When the temperature is uniform from top to bottom, the individual molecules at the top have more total energy (KE + PE) than those at the bottom. I said that led to an uneven distribution in the total energy.
But by exactly the same measure, there are fewer molecules at the top than at the bottom. As a result, the isothermal situation does in fact have the energy evenly distributed. More total energy per molecules times fewer molecules at the top exactly equals less energy per molecule times more molecules at the bottom. Very neat.
Finally, before I posted my reply, Dr. Brown had answered a second time and I hadn’t seen it. His answer follows a very different (and interesting) logical argument to arrive at the same answer. He said in part:
Imagine a plane surface in the gas. In a thin slice of the gas right above the surface, the molecules have some temperature. Right below it, they have some other temperature. Let’s imagine the gas to be monoatomic (no loss of generality) and ideal (ditto). In each layer, the gravitational potential energy is constant. Bear in mind that only changes in potential energy are associated with changes in kinetic energy (work energy theorem), and that temperature only describes the average internal kinetic energy in the gas.
Here’s the tricky part. In equilibrium, the density of the upper and lower layers, while not equal, cannot vary. Right? Which means that however many molecules move from the lower slice to the upper slice, exactly the same number of molecules must move from the upper slice to the lower slice. They have to have exactly the same velocity distribution moving in either direction. If the molecules below had a higher temperature, they’d have a different MB [Maxwell-Boltzmann] distribution, with more molecules moving faster. Some of those faster moving molecules would have the right trajectory to rise to the interface (slowing, sure) and carry energy from the lower slice to the upper. The upper slice (lower temperature) has fewer molecules moving faster — the entire MB distribution is shifted to the left a bit. There are therefore fewer molecules that move the other way at the speeds that the molecules from the lower slice deliver (allowing for gravity). This increases the number of fast moving molecules in the upper slice and decreases it in the lower slice until the MB distributions are the same in the two slices and one accomplishes detailed balance across the interface. On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down, with zero energy transport, zero mass transport, and zero alteration of the MB profiles above and below, only when the two slices have the same temperature. Otherwise heat will flow from the hotter (right-shifted MB distribution) to the colder (left-shifted MB distribution) slice until the temperatures are equal.
It’s an interesting argument. Here’s my elevator speech version.
• Suppose we have an isolated container of air which is warmer at the bottom and cooler at the top. Any random movement of air from above to below a horizontal slice through the container must be matched by an equal amount going the other way.
• On average, that exchange equalizes temperature, moving slightly warmer air up and slightly cooler air down.
• Eventually this gradual exchange must lead to an isothermal condition.
I encourage people to read the rest of his comment.
Now, I see where I went wrong. Following the logic of my question to Dr. Brown, I incorrectly thought the final equilibrium arrangement would be where the average energy per molecule was evenly spread out from top to bottom, with the molecules having the same average total energy everywhere. This leads to warmer temperature at the bottom and colder temperature at elevation. Instead, at thermal equilibrium, the average energy per volume is the same from top to bottom, with every cubic metre having the same total energy. To do that, the gas needs to be isothermal, with the same temperature in every part.
Yesterday, I read the Jelbring hypothesis again. As I was reading it, I wondered by what logic Jelbring had come to the conclusion that the atmosphere would not be isothermal. I noticed the following sentence in Section 2.2 C (emphasis mine):
The energy content in the model atmosphere is fixed and constant since no energy can enter or leave the closed space. Nature will redistribute the contained atmospheric energy (using both convective and radiative processes) until each molecule, in an average sense, will have the same total energy. In this situation the atmosphere has reached energetic equilibrium.
He goes on to describe the atmosphere in that situation as taking up the dry adiabatic lapse rate temperature profile, warm on the bottom, cold on top. I had to laugh. Jelbring made the exact same dang mistake I made. He thinks total energy evenly distributed per molecule is the final state of energetic equilibrium, whereas the equilibrium state is when the energy is evenly distributed per volume and not per molecule. This is the isothermal state. In Jelbrings thought experiment, contrary to what he claims, the entire atmosphere of the planet would end up at the same temperature.
In any case, there’s another way to show that the Jelbring hypothesis violates conservation of energy. Again it is a proof by contradiction, and it is the same argument that I presented to Jelbring years ago. At that time, I couldn’t say why his “gravito-thermal” hypothesis didn’t work … but I knew that it couldn’t work. Now, I can see why, for the reasons adduced above. In addition, in his thread Dr. Brown independently used the same argument in his discussion of the Jelbring hypothesis. The proof by contradiction goes like this:
Suppose Jelbring is right, and the temperature in the atmosphere inside the shell is warmer at the bottom and cooler at the top. Then the people living in the stygian darkness inside that impervious shell could use that temperature difference to drive a heat engine. Power from the heat engine could light up the dark, and provide electricity for cities and farms. The good news for perpetual motion fans is that as fast as the operation of the heat engine would warm the upper atmosphere and cool the lower atmosphere, gravity would re-arrange the molecules once again so the prior temperature profile would be restored, warm on the bottom and cold on the top, and the machine would produce light for the good citizens of Stygia … forever.
As this is a clear violation of conservation of energy, the proof by contradiction that the Jelbring hypothesis violates the conservation of energy is complete.
Let me close by giving my elevator speech about the Jelbring hypothesis. Hans vigorously argues that no such speech is possible, saying
There certainly are no “Elevator version” of my paper which is based on first principal physics. It means that what I have written is either true or false. There is nothing inbetween.
Another “gravito-thermal” theorist, Ned Nikolov, says the same thing:
About the ‘elevator speech’ – that was given in our first paper! However, you apparently did not get it. So, it will take far more explanation to convey the basic idea, which we will try to do in Part 2 of our reply.
I don’t have an elevator speech for the Nikolov & Zeller theory (here, rebuttal here) yet, because I can’t understand it. My elevator speech for the Jelbring hypothesis, however, goes like this:
• If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest air at the top and the warmest air at the bottom.
• This also is happening with the Earth’s atmosphere.
• Since the top of the atmosphere cannot be below a certain temperature, and the lower atmosphere must be a certain amount warmer than the upper, this warms the lower atmosphere and thus the planetary surface to a much higher temperature than it would be in the absence of the atmosphere.
• This is the cause of what we erroneously refer to as the “greenhouse effect”
Now, was that so hard? It may not be the best, I’m happy to have someone improve on it, but it covers all the main points. The claim that “gravito-thermal” theories are too complex for a simple “elevator speech” explanation doesn’t hold water.
But you can see why such an elevator speech is like garlic to a vampire, it is anathema to the “gravito-thermal” theorists—it makes spotting their mistakes far too easy.
w.
However, we do have a real heat source in the earth’s core with the fission of heavy nuclei. Not perhaps a lot, but some. I never have gotten any really good estimates of how large the effect is, and I am not enough of a geologist to derive it. Anyone around have an idea??
I have a partner who has been taken in by the Thrive Movement and especially the Free energy -torus machine I have tried to explain about nergy to no avail. The Thrive promoters are linking their philosphy to climate change and saving the planet by promoting their device as a saviour.
I think Hans uses a definition which is ‘per unit area’
There is more room at the top of the atmosphere for more thinly spread molecules. More area per isobar as altitude increases. Needs thinking about.
I’ll sleep on it.
ShrNfr:
http://en.wikipedia.org/wiki/Geothermal_gradient
Per Wikipedia “Heat flows constantly from its sources within the Earth to the surface. Total heat loss from the earth is 44.2 TW (4.42 × 1013 watts).[12] Mean heat flow is 65 mW/m2 over continental crust and 101 mW/m2 over oceanic crust.[12] This is approximately 1/10 watt/square meter on average, (about 1/10,000 of solar irradiation,)”
“More total energy per molecules times fewer molecules at the top exactly equals less energy per molecule times more molecules at the bottom. Very neat.”
Except that more of the total energy of the molecules at the top is locked up in gravitational potential as opposed to being available as kinetic energy capable of generating heat in collisions.
The isothermal/adiabatic distribution for an isolated ideal gas in a gravitational field has long been debated.
For the isothermal distribution we have Maxwell, Boltzmann and Clausius.
For the adiabatic distribution we have Loschmidt, Laplace and Lagrange.
The smart money must be with the isothermal advocates but I would not regard this as a debate of which was settled and of historical interest only.
Clausius clincher argument of the perpetual motion machine being possible for the adiabatic distribution turns out to be very hard to prove with real components given 9.8K/km scale.
Perhaps Willis will suggest a real experiment with real materials to test the alternative conjectures.
Say with a thermoelectric device to make use of the temperature difference.
A computer simulation program would not be any kind of proof
I think he will find with real materials that this is beyond him
Perhaps this is why there has never been an experiment to settle the matter!
If the adiabatic conjecture turned out to be correct I’m sure there would be an explanation that did not conflict with the second law.
Here for instance is a member of the physics department of the University of California making a very up to date case for the adiabatic distribution.
http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4990v3.pdf
Some Qs:
1. can the same result be found using gas laws, i.e., as P drops, so drops n thus leaving T unchanged? Seems reasonable.
2. do P and n necessarily change 1:1? I do not know this answer.
3. in light of 1., what if the total volume of the cylinder is not fixed?
4. in light of 2., what if they do not change 1:1?
Both 3. and 4. seem like complications beyond my reach.
Mark
Final thought for the night. Here’s the reply I gave Robert Brown on my site Earlier today:
Robert Brown says:
Second, my comment about egregious violation of the laws of thermodynamics were specific to Jelbring, who (IIRC, I’m not looking at his article again as I type this) explicitly asserted a column of fluid with no energy inputs, and then claimed that in equilibrium it would exhibit a thermal gradient. No, it wouldn’t.
Hi Robert,
I think the laws of thermodynamics talk about energy, rather than temperature or heat, but there are several formulations of them, so maybe we’d better discover who is using which definitions. We’d better do this, because in the application of classical mechanics to energy distribution in the model atmosphere, as defined by Hans Jelbring, there will indeed be a thermal gradient, as confirmed by Graeff’s empirical experimental data (Which should be replicated by an accredited laboratory).
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
http://www.emc.maricopa.edu/faculty/farabee/biobk/biobookener1.html
Laws of Thermodynamics
Energy exists in many forms, such as heat, light, chemical energy, and electrical energy. Energy is the ability to bring about change or to do work. Thermodynamics is the study of energy.
First Law of Thermodynamics: Energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe remains constant, merely changing from one form to another. The First Law of Thermodynamics (Conservation) states that energy is always conserved, it cannot be created or destroyed. In essence, energy can be converted from one form into another. Click here for another page (developed by Dr. John Pratte, Clayton State Univ., GA) covering thermodynamics.
The Second Law of Thermodynamics states that “in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state.” This is also commonly referred to as entropy. A watchspring-driven watch will run until the potential energy in the spring is converted, and not again until energy is reapplied to the spring to rewind it.
———————-
I’m not seeing the words ‘heat’ or ‘temperature’ in these definitions, so please could you clarify. Thanks.
I’m not looking at his article again as I type this
Maybe you should. This is one of Jelbring’s chief complaints. People answer what they think he said, instead of answering what he actually said.
I think there is still a bit of a fudge factor in that elevator speech.
“Since the top of the atmosphere cannot be below a certain temperature…”
Why? It can’t be below absolute zero but that isn’t relevant here.
Willis Eschenbach wrote:
Yes, it was a long discussion and my, how time flies when you’re having fun. It’s even more fun when you finally “get it”, whatever the “it” is that’s been bugging you.
And thanks again for your help around that time and continuing pedagogy. Yer blood’s worth bottlin’ Willis.
I think I need to hear an elevator speech for why “same total energy per unit volume is a physical necessity. That is not obvious at all.
Here is another elevator speech for why in the constant insolation example you get an Isothermal atmosphere.
You have a fixed amount of energy per unit (Eg) area received by the ground. This is a experiment in spherical symmetry. Things like temperature, pressure, potential energy only vary by r (radius). So at the ground, at equilibrium, you have some temperature Tg = T(r=ground) that we will constrain by the SB law of a black-body ground. The catch is that in order to radiate the ground with Eg, you must have a uniform shell at r=(very large) also radiating Eg uniformly over its entire area. When you dissipate the energy per area by 1/r^2, you also increase the area of the shell radiating the ground by r^2. But if the shell is radiating at Eg, as a black body, then the temperature of the shell is Tg, the same temperature as the ground. So T(r) = Tg at r=ground and Tg at r=(very large). But (very large) is arbitrary, so it can be any value between ground and infinity. Therefore, T(r) = Tg at all r. Isothermal regardless of pressure and gravitational potential energy.
So I can get to Isothermal with the completely artificial initial conditions of uniform insolation. If and only If. Does that tell us anything useful about the real world? I’m skeptical. We need day and night.
ShrNfr says:
January 19, 2012 at 4:03 pm
ShrNfr, thanks for a good question. I ran the numbers in the past. I don’t have them in front of me, but it is very small, less than a tenth of a watt per square metre from memory.
There are a variety of geothermal regions, and hot vents under the sea, and hot springs and pools on land. But you have to consider—for every hot springs you know of, there are thousands of square miles of land with no hot springs, where if you go to sleep in the morning, you wake up very cold.
So yes, you are right, there is a heat source down under. But it is very small, even if we have greatly underestimated it. That’s why sleeping on the bare ground is dang chilly.
All the best,
w.
> If left undisturbed in a gravity field, a tall container of air will stratify vertically,
> with the coolest air at the top and the warmest air at the bottom.
My thermodynamics is a bit rusty, but I am fairly sure that warm air always rises to the top.
It is called the “adiabatic” lapse rate for a reason. It arises from adiabatic processes. What you are describing is the exact opposite of adiabatic. Your atmosphere is static and you allow it to reach a static thermal equilibrium over a very long period of time. Under those conditions you will indeed see equal temperatures everywhere. Hey – it isn’t at all surprising that you can get rid of the adiabatic lapse rate if your model eliminates all possibility of adiabatic processes.
But add in some adiabatic action – a nice little bit of vigorous vertical mixing – and the temperature gradient reappears. The elevator speech goes thus: “When air moves in a vertical airflow from the top to the bottom it is compressed and thus heats. When air moves in a vertical airflow from the bottom to the top it is decompressed and thus cools. In an atmosphere with a lot of vertical mixing you therefore will see a temperature gradient.”
Our atmosphere has such a temperature gradient.
I am skeptical of both the gravitaional model and the GHG model.
With respect to the former, i think that there are a number of factors overlooked, the extent to which they may be material is moot.
First, gravity is not a constant force acting on the atmosphere in the sense that the atmosphere is not subject to only the force resulting from the mass of the Earth. The atmosphere is constantly being flexed by the Sun and the Moon (and even to a small extent by the Gas Giants). The diurnal bulge/atmospheric bulge is the consequence of this and is well known and this means that work is constantly being inputted into the atmosphere and as one knows a by product of work is heat. One can see the effect (an extreme example admittedly) of gravitaional pull on Io which is the most geologically active body in the solar system and this is due to the gravitational pull imposed by Jupiter and the other Galean moons. Thus there is a top down force (gravity from the Sun, Moon etc) in addition to the bottom up force of the gravity from the Earth all working on the atmosphere.
Second, and this is a factor of the first point, the atmosphere is constantly being displaced at the bottom with the ebb and flow of the tides. Again, although this is obviously weak, this too results in work being exerted on the atmosphere, the by product of which is heat,
The result of these two factors is that the atmosphere is being squeezed much like the walls of a car tyre and any motorsport fan will know that this flexing generates heat in the tyre. It is very effective at heating up the air in a tyre or at any rate maintaining the heat generated by inflating the tyre to the desired pressure.
Third, the Earth itself is a heat source and imports heat into the atmosphere. The Earth is geologically active such that the ground is well above absolute zero. Indeed, even if the sun was to stop shinning, unlike the moon, the temperature of the ground would take a long time to cool to levels seen on the dark side of the Moon. Further, we know little of the deep ocean and there is every likelihood that the amount of thermal energy being inputted into the oceans is considerably under-assessed.
Fourth, the sun warms the atmosphere irrespective of GHGs simply because of aerosol particulate matter in the atmosphere which then warms surrounding gases by conduction/thermalisation
;
May be all of this does not add up to all that much. However, the atompsphere was obviously born warm (being the left over from what was in effect a condensing fireball and the out pourings of volcanoes etc) and one only needs it to add up to the amount of energy that the system is net losing to space to maintain an equalibrium balance.
I think that there may be more to the gravitational theory than you presently give it credit. .
I must modify my isothermal elevator speech with a slight complication. I said that to have Eg insolation at the ground, the black body shell at r=(very large) must also be at Eg. That would only be true if the index of refraction of the ideal gas is 1.000 at all pressures (r). I don’t think it is reasonable to assume index of refraction for a compressing ideal gas does not increase with pressure. Assuming that Index of refraction (Ir(r=ground) is higher near the ground than at high altitude Ir(r=large), then there is a focusing of energy. Therefore, the energy radiated per unit area by the shell must be Es < Eg. therefore T(r=very large) < Tg. But we still have an arbibrary "very large", so I'm talking myself into a slight decrease in T(r), as r increased from "ground" to "very large". Likewise, Es(r) also must = Eg at r=ground, but decrease as r increases as the Index of refraction decreases as r increases.
tallbloke says:
January 19, 2012 at 4:34 pm
Hans Jelbring, on the other hand, proposes a model earth with an impenetrable shell around it that does not allow energy to enter or leave the system, op. cit.:
In other words, no energy can get into or out of Jelbring’s system.
It seems to me that you just proved that Jelbring’s hypothesis can’t work, Tallbloke. If the potential energy of the state inside the shell can only move to where it is less than the initial state, as your quote clearly states … then how can the gravity possibly separate a low-energy, isothermal atmosphere into a higher energy state of cold at the top and warm at the bottom?
My best regards to you,
w.
Many people added comments, presumably physicists and engineers, indicating they had a huge problem with the Nikolov & Zeller theory. I think the issue is dead. There is no such thing as gravity creating higher temperatures.
Deep in the earth, radioactive decay (fission) of higher elements converts mass into energy. In extreme cases, larger bodies like our Sun, the gravitational pressures and heat can result in fusion, which creates even more energy by converting even more mass. The reality is that the center of our planet is well insulated from the near -273 C of space and it therefore remains hot as it is unable to dissipate the heat generated from radioactive decay quick enough to cool down. It has little to do with gravity and everything to do with fission heat and insulating properties of hundreds of miles of rock. What ultimately happens is what we call “steady state” (not the same as equilibrium).
It is the same for the atmosphere – it has energy sources from above (Sun) and from below (Earth black body & some reflected Sun) and ultimately our atmosphere has also reached a quasi-steady state which only fluctuates in response to the changes in energy gained and energy lost. Water being a huge stabilizer of our atmospheric temperature by virtue that it can store energy as it converts from water to a vapor and vice-versa. Anyone can see that after the sun’s radiative energy, water is the single biggest factor in the highly stable behavior of our atmosphere and finally orbital parameters and albedo play a small role to.
This atmospheric gravity thing is complete and utter codswallop.
Here’s a shorter elevator speech:
If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.
Give us time Willis.
I’ve now read and understood N&Z, and their first reply at TT to commenters from WUWT and TT. I now think their work is a paradigm-shifting cracker, but I also realize that paradigm-shifters, while ultimately incredibly simple, need to do a lot of background detail work to answer all significant details AND eliminate all possible scientific stupidities of one’s own AND cope with the psychological challenge of answering classy experts without letting one’s scientific immaturity, feyness or hippiedom lose one the necessary credibility, AND provide an FAQ-type approach to those who’ve found something to doubt and given up on the spot. Like I was sure your elevator speech last time was flawed but my answer, though still IMHO denting your thesis, didn’t really deal with it properly. And I just haven’t yet read Jellbring at all. I might find I agree with you, re Jellbring, for all I know.
Give us time, and we will repay our debts, I mean explain the new paradigm in acceptable ways and with sufficient evidence from data in the public domain. There really is a lot of stunning high quality evidence now available.
The problem is that GHGs and back radiation does not explain the vertical temperature of the atmosphere. The inescapable conclusion of this is that the GHG model is not capable of explaining our atmosphere and that there is more at ‘play’ than the GHG model would suggest.
The vertiacal temperature profile of Earth’s atmosphere is not fully explained by the gravitational model but there does appear to be, for the main part some, causal connection. Ditto, other celestral bodies that we know of.
Presently, we do not know enough, or understand enough to fully evaluate either model. When you neither know or understand enough thought experiments invariably lead to flawed conclusions. Testing and the accumulation of observational data is required to take the matter forward/
Your elevator speech:
Here’s my elevator speech version.
• Suppose we have an isolated container of air which is warmer at the bottom and cooler at the top. Any random movement of air from above to below a horizontal slice through the container must be matched by an equal amount going the other way.
• On average, that exchange equalizes temperature, moving slightly warmer air up and slightly cooler air down.
• Eventually this gradual exchange must lead to an isothermal condition.
The air moving up and down exchanges potential energy (PE) for kinetic energy (KE). The air moving down loses PE but gains KE, and vice versa for the air moving up. A higher KE means a higher temperature, a lower KE means a lower temperature. So the air moving down increases in temperature (KE), while the air moving up decreases in temperature (KE). This will maintain the adiabatic lapse rate, warmer air at the bottom of the column and cooler air at the top.
Your second point is wrong, there is no equalisation of temperature. Therefore, your conclusion in the third point of your elevator speech is also wrong.
@Josh C Thanks.
To test any theory i like to look at the extremes and see if they work. So with this theory in mind, what temperature would the planet be left at if the sun were to switch off?
Should i now blame air pressure for my sunburn?
More briefly, one might refute the gravitational theory of atmospheric warming by reference to the empirical fact that at the top of the Earth’s atmosphere, outgoing radiant flux is closely similar to incoming radiant flux, whereas, if gravity contributed significantly to warming at the surface, Earth would be luminous, i.e., have a positive net outgoing radiant flux.
The only time gravity causes a net increase in the thermal energy content of the atmosphere is during the process of atmosphere formation, e.g., when an airless planet passes through a gas cloud. Then, gravitational compression of the gas will cause heating, the greatest effect being at the surface. However, the heat added to the atmosphere during gravitational compression will warm the surface, the added energy being then radiated to space until equilibrium is reached.
If we wanted gravity derived concepts for heat, we should look at the temperature profiles of planets like Jupiter to see if there is a corresponding gravity induced effect. A quick look at the temperature profile here:
http://en.wikipedia.org/wiki/File:Structure_of_Jovian_atmosphere.png
If the elevator goes down far enough, there is a similar curve found in most planets with a solid cloud cover.
Thanks to papers brought to our attention by Paul Dennis at tallbloke’s blog, namely:
Coombes, Ch. A. and Laue, H., 1985, Am. J. Phys, v53, 272-273
Velasco, S., Roman, F.L. and White, J.A., 1995, Eur. J. Phys. v17, 43-44
I have become convinced that the isothermal hypothesis, although correct as an approximation, is theoretically true only in the limit. The former paper purports to demonstrate the strictly isothermal result, but, if my reading (with which it is not clear that Paul Dennis agrees) of the latter, Velasco et al. paper is correct, its Equation 8, a result of statistical mechanics, dictates that average kinetic energy decreases with height even at equilibrium.
A layman’s-eye view of what’s going on in a paper by Román et al., on which Velasco et al. rely, is found at tallbloke’s here: http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/#comment-13608
Understand that ‘back radiation‘ is as if innumerable miniature dipole antennas tuned to specific wavelengths (as per the CO2 and H2O resonant frequencies) were present in the atmosphere and they ‘catch’ (the forward or out-to-space-bound EM energy) and re-radiate (in ALL directions including back-to-earth so-called ‘back radiation) that same EM energy AT the specific frequencies/wavelengths where they are resonant (and this includes spectral ‘line broadening’ due to molecular collisions at higher pressures e.g. at low altitudes) …
It is as simple as that.
(You do understand, too, that temperature is a measure of molecular vibration, and that gas molecules have different characteristics when ‘vibrating’ than do solids?)
.
What about the gas giant planets? Jupiter at 99% hydrogen and helium has had about 4.6 billion years to become isothermic. Yet the lowest gas/liquified gas/metallic hydrogen temperatures are 35,700K and the top of the atmosphere is only 165K.
There are lots of perpetual motion machines. But only one really works.
Well, he does have a point. IMHO gravity has never been fully explained.
Lucy Skywalker says:
January 19, 2012 at 5:10 pm
Lucy, thank you for your comment, but time is what you don’t have. The clock is running, the elevator speech for N&Z is way overdue. No one has done it to date, they either say nothing or like you and Ned Nikolov himself they give us Wimpy’s line from Popeye, “I’ll gladly pay you Tuesday for a hamburger today”.
Look, you describe N&Z as “ultimately incredibly simple” … so how about you share that simplicity with us? I mean if it’s all that simple … why the wait?
You say you have “read and understood” N&Z. There is an easy test to determine if that statement is true—explain the incredibly simple N&Z theory that you say you understand.
So no, I fear you don’t have time, Lucy. You say you understand the N&Z theory now, this moment. You say now, this moment, you know how “incredibly simple” it is … so no need to wait ’til next week, how about you give us the elevator speech now, this moment, so you can see if you do understand it? Don’t put it off until you collect the “stunning high-quality evidence”, that only does any good after we understand the theory. Hit us with the elevator speech, we’re tough, we can take it.
The elevator speech is only partly so we can see if you understand it, Lucy. Mostly, it’s so you can see if you understand it …
My very best to you,
w.
“As this is a clear violation of conservation of energy, the proof by contradiction that the Jelbring hypothesis violates the conservation of energy is complete”.
Can somebody please conceive this fact; Energy cabn be also employed, not just conserved.
Add kinetic energy to matter and the E does not equal mc/2.
Dr. Brown says that your “column of gas” will have the same temperature at the top as at the bottom, due to equal numbers of molecules with identical kinetic energy moving each way between adjacent thin “slices”. So those adjacent “slices” have equal overall kinetic energy and MB distribution, and therefore equal temperature. In that case they must also have equal numbers of molecules, and so the two slices must have equal pressure. If two adjacent thin slices have the same pressure, then the whole column has the same pressure. Extend the column or cylinder to TOA and you have a mechanism to lose the top layer to space fairly rapidly, and by logical extension, the whole atmosphere.. Some disconnect with reality here?
Bill Illis says:
January 19, 2012 at 5:50 pm
Bill, the isothermal condition only obtains when there is no energy entering or leaving the system, as is the case with Jelbrings thought experiment about his theory. Obviously, this is not the case for any of the planets. As a result, we would not expect Jupiter to be isothermal.
w.
We know that gravity exists but not one acedemic has managed to explain with any confidence why our boots are attracted to the Earth’s core.
“”The explain the incredibly simple N&Z theory that you say you understand””.
The kinetic energy of mass is the mechanism that enhances its employment of energy.
But it is explained Willis, so fully by N&K, it turned my lights on, and what a brilliant sight it is to see.
All sounds like my first year university Physics lectures.
Perhaps a few more Physicists need to read and understand and comment.
Where’s Richard Feynman when we really need him?
@Josh C and ShrNfr
Ref Earth’s Heat
Don’t forget that we still quite a bit left from the accretion of the planet.
Willis Eschenbach wrote:
“at thermal equilibrium, the average energy per volume is the same from top to bottom, with every cubic metre having the same total energy. To do that, the gas needs to be isothermal, with the same temperature in every part.”
No it is not, equipartition states that temperature is proportional to the internal energy per molecule. Energy is an extensive property, temperature intensive.
“Zac says:
January 19, 2012 at 6:01 pm
We know that gravity exists but not one acedemic has managed to explain with any confidence why our boots are attracted to the Earth’s core”.
They can now. Wrong way up we are.
Its not that your boots are attracted to the earths core, it’s that your boots are attracted less to the earths core, because you have more kinetic (employed) energy. As is matter with more kinetic energy above you, like gas.
Quite simple really.
WRT crustal heat flux: surface. http://www.solid-earth.net/1/5/2010/se-1-5-2010.pdf
I had made the following argument above:
I realized after posting this that some people might say “well, maybe the isothermal case is the high energy state”. There’s an easy way to settle that.
If an energetically isolated system is in its lowest energy state, it cannot perform work.
If the isolated atmosphere in Jelbring’s thought experiment is warm at the bottom and cold at the top, I can stick a thermocouple into it and use the temperature differential to generate electricity to perform work.
Therefore, the isothermal state (same temperature everywhere) is the lower of the two energy states, since I cannot use it to do work.
As Tallbloke points out, the second law says an isolated system can only move towards a lower energy state. That means Jelbring’s thought experiment must inexorably move towards the isothermal condition as its equilibrium state.
Since Jelbring claims an adiabatic state will obtain at equilibrium, his hypothesis is falsified.
w.
100 kg of gas 100 kms from Earth will have 98,000,000 less joules of energy than 100 kg of gas at the surface.
Gravitation potential energy actually turns into real thermal energy for a mass that is falling through a gravity field or objects with mass that are closer/farther from the centre of the gravity field.
GPE = Mass * Gravity * Height
Without this (albeit very unnatural effect to us but nevertheless real effect in the real universe) there would be no stars or galaxies anywhere and there would no elements beyond hydrogen, helium and tiny amount of Lithium and there would no us. The universe is made up of many different types of energy beyond photonic EM radiation. There is the strong force, the weak force, gravity and dark energy in addition to the electro-magnetic force.
If we are going to just accept every “settled scientific fact” about radiation theory, then we might as well all be running climate models. If they don’t work, then there is a reason. One is, they do not consider everything that is actually happening (in the quantum world, in the real universe).
Willis Eschenbach says:
January 19, 2012 at 5:54 pm
Lucy, thank you for your comment, but time is what you don’t have.
That is the same argument used by the cap and trade shills to try and force people to agree without due taking time for due diligence. Buy now before it is too late.
Temperatures have leveled. Sea levels have stabilized. CO2 has not. These events contradict GHG theory predictions, which in science is a strong indication the GHG theory is wrong.
It is as though Einstein predicted that gravity would bend light, and when light was measured it was found not to bend. So, then Einstein proposed a new aerosol particle to explain why gravity did not bend light as predicted, but would bend light in the future.
There is plenty of time to review the science and find out why the GHG predictions failed.
Willis Eschenbach says:
January 19, 2012 at 4:58 pm
“…then how can the gravity possibly separate a low-energy, isothermal atmosphere into a higher energy state of cold at the top and warm at the bottom?
How do you know that the “cold at the top and warm at the bottom” state has a higher energy?
Isothermal is just that isothermal, no change in temperature with altitude. The energy contained per unit volume would be greater at altitude than at the surface if fewer molecules occupied that volume. So here is an interesting thought experiment.
To be isothermal, the atmosphere would require perfect insulation. The number of collisions of molecules per unit volume at the surface would be the same as the number of collisions per unit volume at the top of the atmosphere. There would be no lapse rate. The density of the atmosphere would be greater, but there would still be a top of the atmosphere. How high would that be?
Now, if gravity increased, the height of the TOA would decrease. If the atmosphere was perfectly insulated, the temperature of the volume would increase due to compression. There is still no lapse rate, the atmosphere is still isothermal. The number of collisions per unit volume increase.
Now let’s let energy flow from the surface out of the TOA. Unless the atmosphere is perfectly conductive or perfectly transparent to radiant flow, energy will be lost to the atmosphere. With the additional energy, the atmosphere expands, creating a lapse rate, temperature decreases with altitude. Since energy must be conserved, the total energy of the atmosphere must remain the same if the average energy of the atmosphere is to be maintained, unless we add energy. Now the “potential temperature” of the parcel of air at altitude would equal the true temperature of the parcel of air at the surface.
Now that we have a lapse rate and the total energy of the atmosphere fixed since we have not added energy. If the surface layer of the atmosphere warms by 33C then the TOA decreases by 33C. There is a 66C difference in temperature between the surface and the TOA, which is the tropopause. Add sunlight and we have the stratosphere and a new TOA.
Manabe knows that and his estimate for CO2 forcing is half of Hansen’s. Trenberth doesn’t know that, or at least won’t admit that, so his cartoons are meaningless.
CO2 has an impact, just not as much as estimated by people that confuse the tropopause with the surface.
richard
“The problem is that GHGs and back radiation does not explain the vertical temperature of the atmosphere. ”
The vertical temperature differential is a requirement for the GHG effect. You’ve not understood why GHGs cause the surface to cool less rapidily than it would otherwise.
Simply. If the upper altitudes were not colder, then GHGs wouldnt have the effect they do.
More GHGs means the earth radiates from a higher elevation.
If that higher elevation is cooler, then the surface must “warm” or cool less rapidily.
Lapse rate is a requirement for the GHG effect to take place.
Hi,
I have some questions and some observations but as I am not as smart as some please excuse my mistakes.
Due to the spherical nature of our planet, any column of air will not be parallel sided but will be trumpet shaped and the volume will not be linear. Does this effect your equilibrium of energy?
Your statement that a column of air will stratify in a gravity field is observed on earth and I believe that this temperature gradient is in part due to the effect of gravity. Does not the ideal gas law have some application here? However, as we know, the atmosphere is very complicated and probably cannot be described so simply.
The heat engine Dr Brown describes in his proof by contradiction sounds to me like the description of a thunder storm and I personally have observed many of those. If we could harness the power of thunder storms and hurricanes we could indeed power our cities but of course our atmosphere also has other sources of energy, namely the sun. Does Dr Brown disprove the theory or add weight to it?
I am not trying to suggest that anything is right or wrong but the idea that the lower atmosphere is warmed by gravitational pressure from the air above it makes sense to me.
ShrNfr says:
January 19, 2012 at 4:03 pm
There are actually at least 3 intrinsic heat sources for the earth:
1) fission of radionuclides
2) primordial heat from earth’s formation, and the impact that created the moon, or other impacts
3) flexion friction from gravity interactions, especially from the moon
Chemical oxidation heat is mostly externally provided and driven (photosynthetic oxidation of carbon, oxygen being provided by conversion of nitrogen by cosmic and solar radiation, etc.)
Gas molecules don’t travel in straight lines very far. They collide. Energy in these molecules is spread in a distribution described by temperature, across various degrees of freedom (1/2 kT).
Thermal conduction in the Earth seems to be very slow. There are temperature gradients below ground. Caves near the surface seem to have a temperature roughly equal to the average yearly temperature. Temperatures under ground increase with greater depth. Could the heat of radioactive decay make much difference at the surface? Perhaps it’s the old chicken and egg dilemma.
It seems a significant contribution to the Earth having an elevated temperature compared to a black body is the delay in reradiating absorbed light from the sun. The surface warms and some of that energy works its way deep into the ground, or into water below the surface. There is heat capacity, heat conduction, and fluid mixing in the oceans all moving absorbed energy to and from the surface where it might be radiated back to space. The Earth is a heat sink with poor thermal conduction characteristics. If the Earth absorbs light more efficiently than it radiates, then the surface temperature must rise before equilibrium is reached. That higher temperature is thermal energy that can be transported deeper into the Earth. It will take time to penetrate, and it will take time to rise back up to the surface after the sun sets.
If some of the daily energy is retained and not radiated at night, then the Earth’s temperature rises. A glacier has similar properties. Eventually all of the snow that falls on a mountain melts, but if some of the snow never melts during the summer, a glacier will form. The length depends on various parameters like snowfall, but eventually all of the snow melts. The glacier length stabilizes when the rate of snowfall balances the rate of melting. If some of the heat absorbed during the day is not reradiated, then the surface the next day will start out warmer. The average surface temperature will rise until the total amount of energy radiated over 24 hours equals the total amount absorbed during the day.
The atmosphere can pick up thermal energy by conduction and move it to cooler places. The atmosphere also has its own heat capacity. However, there is no way in hell that our atmosphere retained excess heat due to some hypothetical compression event over 4 billion years ago.
If you reduce the number of molecules to a fairly small number, you can see that Willis’s argument, beguiling as it is, is wrong. Consider a single-molecule system, for example. Does anyone doubt that the molecule’s translational kinetic energy is greater when it is lower in the gravitational field than when it is higher? Does that reverse relationship disappear when a second molecule is added?
As the number of molecules gets large, the system approaches isothermic. But a (very small) lapse rate persists.
DAMMIT, WILLIS:
PLEASE ADDRESS THE EMPIRICAL EVIDENCE, WHICH WILL ULTIMATELY RESOLVE THIS ARGUMENT! WHY DO YOU REFUSE TO DO THIS?
[Moderator’s suggestion: If you didn’t YELL at him maybe more would be accomplished? Maybe? -REP]
Willis, do you have a rebuttal or reply to my post http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-870067?
I think that you had it right to begin with – the molecules move faster down low. That means temperature is higher down low. The fact that energy density is constant is a red herring.
I am a semanticist at heart. I read with interest these posts and wonder about two little words: “at equilibrium”. I suspect these theories are thinking about something that doesn’t actually exist in the “Real World”, like so many thought experiments I see from so many very learned and intelligent people.This begs me to ask several questions:
1) Is the atmosphere of the Earth “at equilibrium”?
2)If the atmosphere of the Earth is not “at equilibrium”, does this discussion have any meaning, other than as a diverting thought experiment?
3)If the atmosphere of the Earth is not “at equilibrium”, how will it behave?
Like much climatic, I suspect we are undone by our “Human” view of things.
IMHO,
When, Dr. Robert Brown speaks, people should shut up and listen.
A gentle reminder of the core issue raised by Willis: It seems to me that many posts are injecting unwanted complexities. While it is true that the Earth includes a large body of water, rotates in a 24 hour period, has a core that produces a (small) amount of energy, and so on and so forth, the question to be resolved is the behaviour of a column of “air” – actually any gas – in a gravity field, particularly its change in temperature (if any) with altitude.
To resolve this requires the question to be formulated as simply as possible. So to understand how gravity affects temperature distribution we ignore – for the time being – anything extraneous. No sun, no rotation of the Earth, no surface or sub-surface effects. Simply a column of gas in a gravity field. Nothing more.
I appreciate the effort many have put into their posts, and many are very interesting. But first let us understand how this works in the simplest manner possible.
David says:
January 19, 2012 at 5:21 pm
“The air moving up and down exchanges potential energy (PE) for kinetic energy (KE). The air moving down loses PE but gains KE, and vice versa for the air moving up. A higher KE means a higher temperature, a lower KE means a lower temperature. So the air moving down increases in temperature (KE), while the air moving up decreases in temperature (KE). This will maintain the adiabatic lapse rate, warmer air at the bottom of the column and cooler air at the top.”
I agree with this. Respectfully, Willis is wrong.
Gravity fields are not acceleration fields. In uniform acceleration fields the temperatures are indeed uniform. For a column of material In a gravity field to be of uniform temperature the column would need to be walled radially from the gravity point source.
Dear Mr Willis Eschenbach;
As you have [SNIP: Markus, sorry, but this sort of stuff contributes nothing to the discussion. You can make this sort of comment AFTER a cogent debunking. Please. -REP]
If we had a single pane of glass and the temperature outside was 10 degrees and it was 20 degrees inside, then the temperature within the pane would vary linearly from 10 to 20 degrees across the width of the pane. Right? Now let us pretend we had a 10 kilometre long solid copper tube in the form of a vertical pole that was totally insulated except that the bottom 10 metres was in asphalt at 30 degrees C and the top 10 metres and was exposed to air at -50 degrees C. Then you would automatically have a temperature gradient of 8 degrees C per kilometre. How is this fundamentally different from what really happens with air? If the sun heats the surface and outer space is cold, you automatically get a temperature gradient without resorting to work done by increasing pressures.
I take your kilometer’s tall cylinder of atmosphere in thermal equilibrium and flip it over, like flipping an hour glass. It involved no input of work as its height did not change, so the column’s energy remains constant. But now the gas at temp T that was at the top has been wildly compressed, making it much, much hotter, while the gas at temp T that was at the bottom has been expanded, making it much, much colder. If you let it get anywhere close to thermal equilibrium I’m going to flip it over again, and since I’m not putting in any work, I can do this all day, continuously forcing your air column back to the dry adiabatic lapse rate as I sip a margarita.
I love doing jobs that don’t involve an input of work and constantly overturn an idealized thermal equilibrium, because it’s just that easy.
Willis, thanks, that is a beautiful picture of a perpetual motion machine.
When I was a lad with a crude workshop I did play for a while with the PM notion. My version was powering an air turbine with the exhaust from a compressed gas source. Of course the the turbine was going to compress even MORE gas and the turbine would produce lots of free energy.
Then I went to engineering school…..
One thing I learned in school was that if something works there will be tens or hundreds of people using it. For example the IC engine, the airplane, the integrated circuit, and (not for much longer sadly, see the recent news about Kodak) photographic emulsions on flexible roll film.
So I ask myself, if the Greenhouse Gas Effect is REAL (still in question it seems after these many decades) WHY is it that NOBODY has figured out how to apply it to any practical problem and solve, or at least ameliorate said problem ?????
Think about it for a second, there are lots of very obscure physical effects that have practical applications. For example, the Bernoulli effect is what makes a plane fly (although there is still some debate; does the effect “suck” the plane upwards, or does it “push” the plane upwards, this one seems like tomato .vs. thamato to me). The Peltier effect has been used for decades to cool electronic devices and has lately been applied to drink coolers.
It sure seems that once a physical effect is observed and characterized some sharp engineers find a way to apply it to solve a problem.
So why is it that after decades no sharp engineer has figured out how to apply the Greenhouse Effect to solve any problem, like maybe using the effect to create “net energy gains” in the insulation surrounding a building, that would sure reduce energy usage ???
I see a few possibilities;
1) Engineers are dumber than climate scientists, seems unlikely, some are probably dumber, but I bet at least a few are smarter.
2) There are a dearth of practical problems to be solved, seems unlikely, there always seems to be lots of problems to be solved, like how do we feed everybody and keep them warm. Or how do we allow folks to have fun on a floating cruise ship without letting an incompetent skipper rip a hole in the side by driving it over rocks and “catching” one in the hull, OK we have some more work to do there….
3) Maybe the Greenhouse Effect does not exist, or it exists, but does not provide any “net energy gains” or produce a “higher equilibrium temperature” as claimed ?
Personally, I’m betting on choice #3 and giving odds of several Million to One.
I have not yet evaluated any of the “gravity caused” theories others have suggested and you are discussing, and they may in fact have some merit. But I am convinced that the ”Greenhouse Effect” hypothesis is a modern version of the Perpetual Motion Machine. BTW you are not supposed to be granted a patent on PM machines, and I see nobody trying to patent the “Greenhouse Effect” or applications of if, surely somebody would have tried to lock up all the potential business opportunities if the “effect” really existed as promised ???
Cheers, Kevin.
Somewhat off topic, has anyone built up an error budget for the AGW hypothesis? Not just how skill(less) the models are, but from a mearsurement perspective.
Our understanding of the historic record has error bars that dwarf the analysis. Our recent data has significant error. The spatial error is enormous.
What is all of this hysteria actually saying? We think a 2° rise is too fast, but the historic record can’t be resolved to that fine a point; and that 2° rise is based on models that don’t replicate the record fed with data widely dispersed, inconsistent measurements of fluctuating weather.
Gravity has NO AFFECT ON TEMPERATURE.
How many times must it be said.
You people are reading science fiction.
You have to do WORK to create a change in temperature – this is basic thermodynamics!!!!
If an object falls in a gravitational field then potential energy will be converted to kinetic energy which will create heat. However a stable column of air in equilibrium does not create any energy or heat.
This is so so so basic that I am afraid I may have to give up this website altogether in disgust.
Imagine you have a tube of air 1km long. Now instead of being a tube, you basically turn it onto a cone. Lets say the bottom of the tube represents 1/10,000 of the surface of the Earth, and the top represents 1/10,000 of the “surface” of an imaginary sphere at 1km altitude. This is why I said I would model the atmosphere as a series of concentric spheres representing conditions at the altitude of each sphere. Now, temperature and heat. At molecule at the ground might be 100F and a molecule at 40km might be 100F but there are far fewer of them at 40km. So if you stick your thermometer out the window of the bazillionth floor at 40km altitude, far fewer molecules will strike your thermometer and transfer heat to it. So your thermometer will cool until it reaches an equilibrium where the heat it is radiating is equal to the heat it is receiving. So the temperature of the molecules can be the same (100F) but there is less heat per given amount of space because there are fewer molecules.
Equilibrium of the atmosphere doesn’t exist. As can be seen in this thread, just trying to understand the climate in even the simplest terms is a daunting task.
Topologists long ago proved that the wind will always blow somewhere…
right after they mistook their donuts for their coffee cups.
@ David says:
January 19, 2012 at 7:20 pm
From what I’m understanding, I agree.
However, I can’t help but wondering what would happen, keeping the extraneous items out as you said – “no sun, no rotation of the Earth, no surface or sub-surface effects. Simply a column of gas in a gravity field. Nothing more” out, except in a situation where the gravity isn’t constant. Wouldn’t, in that case, we have a continuously moving gas due to the changing gravity? Would we then see heat, energy, or work being “created” by the fluctuating gravity?
Just wondering.
Willis,
You keep asking for an elevator speach supporting the N&Z analysis.
Here is my attempt.
(1) any worthwhile theory should describe reality or it is worthless and should be discarded.
(2) When you make due allowance for differences in their distances from the sun, the temperature of Mercury, relative to Venus, is too low to be explained by the greenhouse theory.
(3) However this is very closely explained by their different atmospheric pressures at the surface.
(4) That is also true for other bodies in the solar system.
(5) That in turn suggests that the various laws on physics mentioned in this thread, while themselves highly likely to be true, do not interact in the manner that has been outlined by people critical of the two unpublished N&Z (20110, 2012) papers.
I really do not think that this arguement will be settled using theoretical “thought” experiments. These remind me in so many ways of the best work of the IPCC.
We need a theory to explain why the surface temperature of the various solar bodies can be derived as a function of distance from the sun (solar radiance) plus near surface atmospheric pressure.
N&Z have provided their theory.
The task of critics is now to come up with better, more economical (Occham’s Razor) theories.
As Lucy Skywalker has said – this is a game changer.
the game HAS changed.
We must now all respond to the new paradigm.
Jeremy says:
January 19, 2012 at 7:52 pm
///////////////////
Jeremy
Just three quick questions.
1. How much work is involved in the creation of the diurnal/atmospheric bulge?
2. How much work is involved in the moving of the tides?
3. Are not the same processes that are involved in moving the tides also at work on the atmosphere but not so readily apparent to an Earth boud observer since he cannot see the ebb and flow of the atmosphere as it is sublected to the gravitaional forces of the celestrial bodies?
The gravitational forces being exerted on the atmosphere are not constant. The atmosphere is never in equalibrium.
Perhaps you will answer my questions before departing this web site.
In the absence of an external energy source the column would indeed become isothermal.
Temperature at both top and bottom would be the same despite the higher energy content per unit volume at the bottom.
Mass is simply a form of energy so a denser mass per unit volume contains more energy but it does not follow that it has a higher temperature than a less dense unit of volumre.
Temperature is simply a measure of kinetic or vibrational energy and molecules can have the same averaged kinetic energy in both a dense and a less dense unit of volume.
Gravity just primes the system by placing greater density of molecules at the bottom of the column. It does not provide any heat or kinetic energy in itself.
If an external energy source is then switched on then the kinetic response to that energy input is density dependent and so the temperature gradient with density then appears.
More molecules per unit volume will convert a larger proportion of the incoming radiative energy into kinetic form and it is kinetic energy that is refected in a higher temperature.
Furthermore higher density involves more collisional activity due to closer packing of the molecules so that kinetic energy stays in kinetic form for longer whilst it is bounced to and fro between molecules before eventually being released in the form of outgoing longwave.
The more incoming radiation that is converted to kinetic energy per unit volume AND the longer it stays in kinetic form the higher the temperature will become.
The adiabatic temperature gradient is therefore a consequence of gravity induced pressure PLUS uneven energy distribution (more molecules per unit volume) PLUS incoming radiation.
ALL the components must be in place at the same time to produce the temperature gradient.
THEN the entire structure of the planetary atmosphere is effectively forced to adjust itself to provide the most efficient mix of energy transfer mechanisms both radiative and non radiative so as to maintain that adiabatic temperature gradient.
Radiative processes only perform a mopping up function. In so far as non radiative processes fail to return the system to that adiabatic lapse rate then radiative processes step in to make up the difference.
The S – B equations do not deal with the non radiative processes.
The final oucome in terms of atmospheric structure can become highly complex and that is where composition becomes relevant and why no planet with an atmosphere matches the S – B equations.
Imagine a glass tube 100 miles tall reaching from the surface of the earth to outer space. Insert the base of the tube in concrete and then fill it with Coca-Cola or some other carbonated beverage. Now add peanuts at the top and watch what happens. At first the peanuts sink but as the fall, they gather bubbles which slows their descent until …..
LOL.
Sorry. But Jeremy is right. This discussion is almost totally nuts.
KevinK says:
January 19, 2012 at 7:42 pm
////////////////////
Kevin
I have made this point to Willis on a number of occassions and it has always been met with deadly silence.
Why are we seeking to exploit solar energy by way of PV cells when this has obvious drawbacks such as sunlight is not received at night or when cloudy. Trenberth suggests that on average solar power is just 184 w per sq m. Why are we not exploiting backradiation or at any rate researching its use when it is claimed to be in the order of 333 w per sq m come rain or shine 24/7? If this was truly a souce of energy capable of doing sensible work, we would be exploiting it since it would cure over night the world’s energy needs. Something is amiss here and it is probably that physists employed by energy companies or at the cutting edge of research do not consider it a source of sensible work.
Richard, yes I also get silence from Willis regarding any of my comments. I think that this is his loss…
I still think that if the “greenhouse effect” had any merit a whole bunch of engineers would have JUMPED on it a long time ago. BTW I am an engineer and I always JUMP on any “effect” that helps me solve a problem.
Cheers, Kevin.
ferd berple says:
January 19, 2012 at 6:34 pm
Thanks, ferd. I come to a very different conclusion from the same facts. I say that the greenhouse effect has done the heavy lifting of bringing us to our warm current temperature. At the current equilibrium condition, however, the various governing systems have come fully into effect, keeping the earth from warming any further. These mechanisms include the daily cycle of thunderstorms in the tropics, the annual swing of clouds which warm the planet when it is cold, and cool it when it is warm, and the multi-year swing of the Nino/Nina cycles.
This does not mean that GHGs have no effects. It means that at the equilibrium condition, the minor changes in GHGs are totally swamped by the effects of the clouds and other governing mechanisms.
All the best,
w.
Willis,
I was in the same boat as you, thinking that the equilibrium temperature profile of an insulated column of air would be warmer at the bottom than the top. Like you, I changed my mind.
I discusses the topic with a few bright physicists. We bandied back and forth several ideas. There are a few convincing (but wrong) arguments that a lapse rate is the expected result. Eventually we concluded that the equilibrium profile must indeed to isothermal, because the arguments for that were more convincing (and right).
The simplest and most convincing argument ended up being your same perpetual motion approach. You could run an insulated copper bar from bottom to top. This bar does not have the lapse rate effect. (Or use another gas that would have a different lapse rate because it has a different heat capacity). The copper should have the same temperature at top and bottom. If the gas had a different temperature, would could use this temperature difference to continually run the sort of heat engine you suggested.
And that is how theoretical science is supposed to work. You come up with a couple different ideas about the some new situation. You apply fundamental, laboratory-tested principles and discuss it with others. When several different people who understand the science agree, then you have your conclusion.
PS A “back of the envelope” calculation suggests the “time constant” for this process would be weeks. Reaching something close to uniform temperature could takes months. But in the real world, there would be other processes to remix the air long before that.
PPS. This whole gedanken experiment was, of course, predicated on perfect insulation and a constant temperature at the bottom. The same “back of the envelope” calculation suggests that heat flows involved in conduction are less and 1 mW/m^2. Even a tiny amount of GHG at the top of the atmosphere would radiate a 1,000 times more energy, meaning that conduction would never actually achieve a uniform profile.
jae says:
January 19, 2012 at 7:05 pm
jae, I fear I don’t have a clue which “empirical evidence” you are speaking of. If it has to do with the N&Z hypothesis, I don’t understand the hypothesis so “evidence” means nothing. Let me know which evidence and which theory you’re talking about.
More light and less heat would help here.
w.
David says:
January 19, 2012 at 7:07 pm
David, your rebuttal is to the argument made by Dr. Brown, and I don’t want to explain his argument more than I’ve done. For an answer, you should post it on his thread, the link is at the top.
Many thanks,
w.
Dougmanxx says:
January 19, 2012 at 7:10 pm
To the contrary. Thermal equilibrium simply means that the objects have stopped exchanging energy because they are at the same temperature. This happens all the time.
w.
“• If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest air at the top and the warmest air at the bottom.
• This also is happening with the Earth’s atmosphere.
• Since the top of the atmosphere cannot be below a certain temperature, and the lower atmosphere must be a certain amount warmer than the upper, this warms the lower atmosphere and thus the planetary surface to a much higher temperature than it would be in the absence of the atmosphere.”
Strange as seems I would say it’s it’s true. We are assuming the gas that we talking about can’t freeze or liquify.
Nitrogen in near vacuum pressure can be close to absolute zero. And Helium:
“All known liquids, except liquid helium, freeze when the temperature is lowered enough. Liquid helium remains liquid at atmospheric pressure even at absolute zero, and can be solidified only under pressure.”
Liquid helium: Boiling point at 1 atm: 4.2 K 3.2 K
http://en.wikipedia.org/wiki/Liquid_helium
So certainly true if you are talking about a helium atmosphere.
Of course to have gravity you need mass- something which is giving 1/10th of gee will have internal heat.
Question is would work with artificial gravity.
I think there some thought experiment which says under certain you can’t tell the differences between acceleration and gravity, perhaps this would be a way to tell the difference, and maybe not.
• This is the cause of what we erroneously refer to as the “greenhouse effect””
Hmm, don’t think this explains everything in regards to the “greenhouse effect”.
If include the sun’s energy, the heat capacity of atmosphere, water vapor, clouds, ocean temperature, and land temperature, then yeah, most of the greenhouse effect.
And I don’t think it explain stratosphere and higher.
George Turner says:
January 19, 2012 at 7:34 pm
You have not allowed for the fact that the atmosphere in the cylinder is mostly at the bottom. As a result, you have to move much more air up than down when you flip it. So it does involve a large input of work, despite the fact that its height did not change.
w.
A question about the diameter of the column of air extending upwards 1km. If the column were small diameter (say 5m across at ground level – expanding as it rises) would we not get different conditions/events than if it were, say, 1 km diameter? At some point, won’t we see convective effects happening? With hot air rising as it would in a hot air balloon? As Willis has so convincingly explained with his thunderstorm arguments (or did I miss something somewhere?).
And, further demonstrating my lack of knowledge, if warmer air rises due to convective effects, how come we notice that air temperatures are (generally) warmer at lower altitudes? Is that to do with the relatively greater air density at surface than at altitude.
Clearly I must do more reading to keep up.
the universe is a perpetual motion system, expanding and contracting simultaneously in patterned ways, ad infinitum.
the universe is matter and space. temperature is associated with matter. it seems that a critical amount of matter determins black hole versus supernova outcomes for astronomical bodies.
what is the trigger for a big bang ? an accumulation of matter.
given that time is a measure, we could describe the life cycle of big bang as how many years ?
my point is that entropy is the normal condition.
more specific phenomena are contextual. our entropy is affected by local conditions, but in the end physical constructions return to matter and space.
my miniscule understanding of the topic under question favours the idea of gravity increasing temperature. where are the hottest places in our environments ? the centres of our astronomical bodies ?
ergo gravity wins up to the point of bigbang initiation, which may well be another gravitational effect.
of course I’m brain-sailing, surmising, and thinking aloud.
I have a follow-up thought experiment to my previous comment.
If flipping the air column upside down doesn’t require an input of work and returns the column to the dry adiabatic lapse rate, then the dry adiabatic lapse rate must be the column’s thermal equilibrium because a system cannot be shifted from thermal equilibrium without an input of work.
If the adiabatic lapse rate is the thermal equilibrium then it should be impossible to devise a heat engine driven by a column of air whose temperature varies with the adiabatic lapse rate. If the isothermal condition is the thermal equilibrium then the same should apply, because a system in thermal equilibrium can’t drive a heat engine.
If I take a tall column of isothermal air in a gravitational field and exchange a pair of parcels from top to bottom, the descending parsel warms up and the ascending parcel cools down, with no change in energy and no net work. Yet in their new positions both these parcels have a large delta T relative to the surrounding air, and that delta T can drive a heat engine such as a Stirling cycle.
If I try moving parcels within a column whose temperature follows the adiabatic lapse rate, each parcel always stays at the same temperature as the surrounding air, so I cannot drive a heat engine. From that I conclude that an isothermal column of air in a gravitational field is not in thermal equilibrium, and a column of air at the adiabatic lapse rate is.
So if the column of air is isothermal, I can drive lots of little heat engines until it reaches the adiabatic lapse rate, after which I can extract no more energy from the system.
On caveat is that if the system is at the adiabatic lapse rate then the bottom air is warmer, and thus less dense, raising the center of mass of the entire column in slightly increasing its potential energy, so perhaps flipping the column does involve an input of work. Perhaps a more detailed analysis would take this into account and provide a slightly shifted equilibrium point.
Jeremy says:
January 19, 2012 at 7:52 pm
Thanks, Jeremy. You are a hundred percent correct, gravity can’t do ongoing work to change the temperature. And yes, there’s a heap of folks that are not clear about that, I wasn’t until recently.
One of the things that happens here is that we try to provide information and informed discussion about basic science, for me (as you can see from my story above) as well as for others. Every blog has lots of folks who are in need of this. Most don’t even deal with it.
Now, since the proponents of these “gravito-thermal” theories have been gaining some prominence and some traction on the blogosphere, I’m trying to provide a counterweight of science. When you go out to battle the forces of scientific fantasy, there will be (as you have observed) lots of folks that don’t get the basics.
If you are fighting basic ignorance of science, you will be deluged with ignorant people. Not much I can do but just keep putting the facts out there.
Certainly there are a host of much more sophisticated threads, and those tend to attract a more scientifically literate commenter. But when you are discussing “gravito-thermal” theories …
So you could stick to the posts that don’t even mention gravity. Or you could stick around and try to explain it. If so, prepare for frustration. In any case, I can only agree with your statements, thanks for your post.
w.
AusieDan says:
January 19, 2012 at 8:06 pm
Thanks, Dan. The problem with your elevator speech is that nowhere does it say how the N&Z effect actually works, which is the point of the elevator speech.
w.
Quark says:
January 19, 2012 at 8:35 pm
If you don’t like it, Quark, then lead, follow, or get out of the way … but just complaining does nothing.
Many questions have been settled or elucidated by thought experiments. Einstein was very fond of them. Your idea that they mean nothing ignores a long history of their use in science.
w.
“Temperature at both top and bottom would be the same despite the higher energy content per unit volume at the bottom.”
The “higher energy content” seems to mean it’s hotter.
It seems the gas molecules would all have same average velocity- with exception that faster molecules would tend to be higher and slow molecules would be lower. This tendency- depends upon the amount of gravity- 10 gees would more of tendency than 1 gee. But this means more fast molecules could found and average speed isn’t different, the gravity sort them more. If follow a molecule it tends to stay lower, and spend less time higher.
Or the gas molecules are always varying velocities, the velocity is random, but is averaged by the zillion of molecules. If heads is faster, molecules with 10 heads in a row are tend to be higher, those with 10 tails in row tend to be lower.
In gravity field the average molecule speed will lower in the higher density.
Put 1 cubic meter of 1 atm gas, into another 1 cubic meter of 1 atm gas- doubles pressure to 2 atm, and is hotter, when cools to “room temperature” the 2 atm gas will have more density and less velocity.
“Energy content” to me suggests density and/or higher velocity of gas molecules- *either one or both” are higher temperature.
Willis says:
But I get all that work back when it passes the tipping point and the air rushes back to the bottom, just like flipping over a half-full bottle. The kinetic and potential energy in the final state is the same as in the initial state (except, in the case of a gas, for that thermally induced change in the center of mass I mentioned above), so there is no net input of work. You could drive the bottle flip with a spring and make a cute perpetual motion machine whose only flaw would be internal friction.
David says “To resolve this requires the question to be formulated as simply as possible. So to understand how gravity affects temperature distribution we [should] ignore – for the time being – anything extraneous. No sun, no rotation of the Earth, no surface or sub-surface effects. Simply a column of gas in a gravity field. Nothing more.”
But this then considers only the local effect of gravity on the planet itself – not the effect of the gravitational fields of other bodies. The second law (and gravity) applies universally – you must take into account that no known planet simply wanders about the universe as an orphan; all known planets (and moons) are under the control of some greater external gravitational force (unless perhaps caught in a supernova explosion – but that would be a special case!). This implies rotation (Keplerian orbits, axial rotation by conservation of momentum or gravitational/ tidal coupling), therefore nights and days, atmospheric mixing, heating, and radiation by the atmosphere itself.
Part of the problem with Jelbring is the assumptions applying to the model planet. Interestingly the same applied to Willis’s orphan planet with the non-GHG atmosphere (in his original Some Gravity “trap” thread) – perhaps in the hypothetical world we will find some form of paradox applying until it is accepted that all known planets rotate.
The actual atmosphere temperature distribution cannot be modeled in terms of random molecular motions alone. The atmosphere is heated at the base by the absorption of UV and visible light. It warms to the point of hydrodynamic instability in the day and there is bulk flow energy transport from the warm base to higher elevations. Even without water vapor, CO2, methane or other greenhouse gases, ozone would absorb some of the outgoing IR. Without attributing either validity or falsity to any of the theories of atmosphere heat transport in discussion here, this is a complicated problem that is not going to be settled by simple arguments. Do the diurnal fluid mechanics problem along with atmospheric circulation and pole-equator insolation differences and then try to explain it in simple terms if you can.
Willis – at some point I got lost and it was at the molecular replacement part. A highly energized molecule takes up more space than a lesser energized molecule. For there to be a one-to-one replacement of a displaced (convected) molecule, the molecule replacing it has to consume the same volume. Meaning it has to be at the same energy level. What compels molecules at the same energy level to swap chairs? Describe what happens to a lesser energized molecule when it drops into the hole left by a more energized molecule. And I know you know.
Then we will need to talk about gradients where all gradient elements are very close to every other gradient element. This gets to a very earthly feature known as long runout earth slides such as that which buried Pompeii. But first things first. Please give me the elevator speech description of why molecules of identical energy levels would swap chairs. I will snip your posts if you go off topic.
Willis, It is hard to argue with your thought experiment due to your requirement of equalibrium which excludes convection. Once convection is allowed then adiabatic temperature differences will result.
Temperature vs Energy of a gas.
What is the temperature of a gas? It is proportional to the *average* kinetic energy of the gas molelules whose temperature you are trying to measure but then we must always define which molecules are and are not included for the temperature measurement. We can do this by defining a region in space – a volume with x,y and z dimensions. The gas molecules in our region of interest don’t all have to have the same kinetic energy – as long as the average kinetic energy of the molecules in a given region of gas is the same it has the same temperature.
The total Energy (kinetic plus potential energy) can be very different in a cubic meter of gas at sea level to the same volume of a gas at the same temperature at altitude. The density (and mass) of gas in a given unit of volume also changes with altitude. It is simplistic to think of one molecule of gas as it turns potential energy into kinetic as it falls, when you consider a given volume of gas as it falls the number of gas molecules per unit volume increases this means that the potential energy of 1 unit of volume of a gas does not decrease simply as a function of altitude but also of density (mass per unit voume).
At altitude the number of gas molecules per unit volume is less than at the planets surface so that means the total kinetic energy (sum of the kinetic energy for each gas molecule) is declining with altitude even if the gases have the same temperature (average kinetic energy of the molecules). Again the Potential energy of the unit of volume of gas at altitude is higher per molecule but there are less molecules per unit volume so the sum total of the potential energy for a unit of volume of the atmosphere is also a function not only of gravity but also of the number of molecules in a unit of volume.
My hat goes off to those who delve into this further! Good luck to all of the smart people thinking about these matters.
richard verney says:
January 19, 2012 at 8:37 pm
Richard, I’m sorry you think I’ve avoided the question. I receive dozens and dozens of questions daily and I cannot answer them all.
Several things. First, it’s not clear but it sounds like you think that downwelling longwave radiation (DLR) does not exist. I assure you that scientists routinely measure it daily around the world.
Second, photons of ultraviolet light have plenty of energy to stimulate band-gap jumps, creating the photoelectric effect. Blue light somewhat less. Green light and yellow light, only slightly. Red light, almost none. Thermal infrared, AKA “longwave”? Sorry. See here for a discussion. We can’t use it photoelectrically with our current technology, it doesn’t have enough energy to knock things loose. That means we’d have to use it as heat.
Here’s the problem with that. Bear in mind that do work with that heat, you need more than the heat. You also need to have a colder region to which you can reject the waste heat from the heat engine. In other words, you need a hot end and a cold end for a heat engine. It runs off of the temperature differential, not the temperature.
Finally, consider that wherever that IR is radiating from on high, it is coming from someplace colder than the surface. The global average upwelling radiation is on the order of 390 W/m2, or about 14°C. Downwelling global average radiation, on the other hand, is only on the order of about 320 W/m2, a few degrees above freezing.
So you can build your heat engine to use the downwelling radiation, but because everything around you is warmer than that radiation … where are you going to reject the heat? So you can’t use it to drive a heat engine.
Curiously, in certain circumstances you can pull your trick off, but only in in the backwards direction. By that I mean in areas where there are very few GHGs, you can utilize the lack of downwelling radiation. This was done in the old days in the southwestern US. The desert air is bone dry so GHG radiation is low. The folks there would put out shallow flat trays of water on the roof. This allowed the trays of water to reject their heat to the infinite coldness of space with little back radiation.
And as a result, the trays of water will freeze even though the ground temperature and the local air temperature never go below freezing at any point during the night.
At dawn, the people would take the trays of ice indoors before the sun came up, and put them in their ice storage, typically a hole in the ground filled with sawdust or other insulating material.
So in theory: You make ice during the night, then during the day you use the both solar and IR to heat the hot end of your heat engine and you use the ice to cool the cool end. But I digress.
That’s why we can’t pull work out of the downwelling longwave. Not enough energy for photoelectric, no cool place for rejecting heat.
All the best,
w.
PS—it strikes me that a satellite based heat engine (think thermocouple) could extract work from the upwelling planetary longwave, because it could reject the heat to outer space …
Willis-san: Here’s your logic, above, with the symbolic logic thereof:
“If an energetically isolated system is in its lowest energy state, it cannot perform work”
If EIS = LES THEN -W
“If the isolated atmosphere in Jelbring’s thought experiment is warm at the bottom and cold at the top, I can stick a thermocouple into it and use the temperature differential to generate electricity to perform work.”
If EIS = HBCT THEN W
“Therefore, the isothermal state…is the lower of the two energy states, since I cannot use it to do work.”
-W THEREFORE EIS = LES
Do you see what you’ve done, Willis?
If Roger is a goose, I can’t ride him like a bicycle.
If Roger is a Schwinn, I can ride him like a bicycle.
I can’t ride Roger like a bicycle, therefore he is a goose.
We seem to be in the same boat Willis.
* We both thought that the lapse rate might be the equilibrium condition in this thought experiment.
* We both discussed it with other smart, informed people and decided the equilibrium condition is isothermal.
* We both realized a perpetual motion machine was the simplest argument against the permanent lapse rate situation.
That is how science should be done. I hate to admit I was wrong about physics, but there is no other conclusion possible here about the answer to this question. In our defense, there are arguments that sound very convincing that the temperature should drop as it go up. You REALLY have to know thermodynamics to avoid getting sucked in by those alluring arguments.
PS. The thought experiment is a very specific situation, not likely to be seen in the real world. Thermal conduction through the atmosphere is less than 1 mW/m^2 if the lapse rate is the maximum stable amount of 10 K/km. This number is SO much smaller that other energies involved (convection, incoming solar, GHG radiation, evaporation) that such an isothermal condition would never be realized in real life.
For those of you that believe empiricism trumps thought experiment, consider the following, attributed to Galileo, though it was also recorded ~1,000 years before Galileo by John Philoponus and also Oresme (IIRC) more than a century before.
Galileo asked us to consider what would happen if two iron balls were tied together as one by an iron rod. The smaller and lighter ball, according to Aristotelian physics, would slow down the ascent of the larger, heavier ball. Yet the combined weight, being greater than either ball alone, meant that they would fall faster when tied together, as well as slower. Since a contradiction was (and remains) not allowed, the answer to the problem was that objects necessarily fall at the same rate, regardless of their weight. Galileo had successfully demonstrated that Aristotle had been wrong about falling weights.
While Galileo is widely (and incorrectly) believed to have performed the experiment of dropping two cannonballs of differing weight from the tower at Pisa, this cannot be the case. In his record of the experiment, Galileo refers to the height from which a wooden and iron cannonball were dropped: 300 feet. This would make his assistant the tallest man in the world — ever!
In the event, Galileo recorded that the wooden cannonball initially fell faster than the iron cannonball, and that the iron cannonball overtook the wooden cannonball, beating it to the ground by a measurable margin.
Think about what we are doing when we attempt to reconcile the empirical result withe the deductive result.
dp says:
January 19, 2012 at 9:58 pm
dp, you should ask the author, Dr. Robert Brown, that question, as the description is his. I have only quoted part of it here. There is a link to his thread in the head post.
w.
Willis:
The problem is that the atmosphere Jelbring described cannot exist, and thus no meaningful statements can be made about it.
Suppose we have an atmosphere with a near surface temperature of 288K, and translucent in infrared frequencies. Then it must radiate to space at about the black body rate. Thus it will cool to the temperature of space, about 3K. In order to keep the atmosphere at a temperature above 3K, energy must be added. Indeed, to keep the atmosphere at 3K, energy must be added, via the cosmic background radiation. Once energy is added, whether at the top or the bottom of the atmosphere, the conditions for an isothermal atmosphere no longer apply.
It’s like dividing by zero. Once we let that slip in, whether by Jelbring or Eschenbach or Mann & Jones, we can get to any conclusion we want by following one chain of logic while ignoring another.
okay Willis, I take your point re gravity not affecting temperature. however I would argue that gravity is associated with temperature – I assume that temperature is highest in the core of astronomical bodies, as written.
yes temperature dissipates by convection and radiation.
but what is kinetic energy other than mainly a gravitational effect, a pull or a push ?
whilst this topic is ‘restricted’ to movements of air and temperature in a tube, and the subject of a planetary body minus an atmosphere, the real life said body of air operates in relation to a wider context – wind, height, temperature of surrounding air.
equilibrium is a temporary phenomenae. perhaps slowest rate of change is a near definition.
I have to go out. I’ll return to this later.
“It’s an interesting argument. Here’s my elevator speech version.
• Suppose we have an isolated container of air which is warmer at the bottom and cooler at the top. Any random movement of air from above to below a horizontal slice through the container must be matched by an equal amount going the other way.
• On average, that exchange equalizes temperature, moving slightly warmer air up and slightly cooler air down.
• Eventually this gradual exchange must lead to an isothermal condition.
I encourage people to read the rest of his comment.
Now, I see where I went wrong. Following the logic of my question to Dr. Brown, I incorrectly thought the final equilibrium arrangement would be where the average energy per molecule was evenly spread out from top to bottom, with the molecules having the same average total energy everywhere. This leads to warmer temperature at the bottom and colder temperature at elevation. Instead, at thermal equilibrium, the average energy per volume is the same from top to bottom, with every cubic metre having the same total energy. To do that, the gas needs to be isothermal, with the same temperature in every part.”
[-Sounds wrong.-]
“Yesterday, I read the Jelbring hypothesis again. As I was reading it, I wondered by what logic Jelbring had come to the conclusion that the atmosphere would not be isothermal. I noticed the following sentence in Section 2.2 C (emphasis mine):
The energy content in the model atmosphere is fixed and constant since no energy can enter or leave the closed space. Nature will redistribute the contained atmospheric energy (using both convective and radiative processes) until each molecule, in an average sense, will have the same total energy. In this situation the atmosphere has reached energetic equilibrium.”
So, it’s the “average molecule” which has same energy [velocity] and in less density the amount energy per cubic meter is less [and is therefore is colder- doesn’t warm as much something like thermometer].
But no molecule stays at same velocity. Or make it simple, maybe, molecules don’t go [normally] in any vector for more nanosecond.
In convection one can have a group “averaged molecules” going in one direction or vector or have something one calls velocity- but none of these actual molecules is actually going in one direction- it’s more like sound wave. it’s possible for a gas molecule travel say a meter or more in one direction, but chances are low of this happening. With zillion and zillion them, one probably goes mile fairly commonly:)
Example of the theory as I understand it.
Have two flat areas of real estate. Have them at different elevation. On at sea level and one at 10 km in elevation. And they are on same planet. They will have different density of air 1 meter above their surfaces.
On the land at 10 km in elevation should receive more sunlight. But let’s say the both receive same amount of sunlight. These areas of real estate are hundreds of square km and flat.
The one at higher elevation will not warm the air to as high a temperature as the land at sea level. The higher elevation land will increase the gas molecules speed to same velocity as the sea level land, but there is low density, so it’s cooler.
And even with more sunlight in higher elevation it would make the gas molecules travel faster, but it still would be cooler air temperature.
If this is true, than lower elevation has a “greenhouse effect”.
And the land at higher elevation should have it’s dirt at higher temperature [if has more sunlight]- and less greenhouse effect in this sense.
jorgekafkazar says:
January 19, 2012 at 10:14 pm
Not a clue, my friend. Your pseudo-algebra is meaningless. Try putting units on it to see what I mean. Unless you think that is symbolic logic, but it’s not any kind that I recognize.
jorge, i seriously have no clue what you are on about. It seems you think I’ve made a logical mistake. Let me recap the logic.
I stick a thermocouple in an isothermal atmosphere. I cannot extract any work.
I stick a thermocouple in an atmosphere that is cold at the top and warm at the bottom. I can extract work.
Which one is the lower energy state?
w.
Gravitational heating comes from contraction of a gas. Potential energy is converted to kinetic energy.
The sun when it originally formed would have had enough kinetic energy to shine for about 18,000,000 years. However – after about 18,000,000 years it would be cold and dead without nuclear fusion.
The atmosphere of earth may have contracted at some point but that energy dissipated several billion years ago.
The gravito-thermallers are kind of right – it may have happened, once. But that train has left the station and doesn’t impact current atmospheric conditions. The potential energy was converted and radiated away a long time ago.
Willis,
I am beginning to think that there may be another effect at work to raise the temp due to gravity, but not for the reasons suggested. Because of the small angle subtended by the earth vis a vis the sun, the energy comes in a collimated beam. Essentially, most of the photons, whatever their energy, have a pretty good shot at the surface of the earth. However, once the surface or the gas
near (first couple miles let’s say) the surface reradiate this energy, it goes off it all directions. Due to density differences in the atmosphere caused by gravity, I would expect some of this energy to be reflected back at lower angles of incidence to the horizon due to refractive effects in the atmosphere, similar to the mirror like effect seen on a hot highway at low look angles. The compressed gas of the atmosphere creates a diode effect for inbound radiant energy. No clouds required.
Thomas L says:
January 19, 2012 at 10:22 pm
First, this is a thought experiment. Einstein famously explained relativity using the thought experiment of an elevator in space. Of course, something like that cannot exist … but it can still be very useful in understanding the real world.
Second. suppose we find a planet with an atmosphere with no nearby sun. We put a foam shell around it, so it is almost entirely energetically isolated from the universe.
That is the atmosphere Jelbring describes … so why can’t that atmosphere exist?
w.
George Turner says:
January 19, 2012 at 7:34 pm
Nice try. Before you rotate the cylinder, the air at the top of the cylinder has lower pressure than the air at the bottom of the cylinder, and therefore the density at the top is less than the density at the bottom. So when you turn the cylinder over, you are doing work, as moving the bottom up takes more energy in a gravity field than you get from moving the top down. And the air in what was the top (now the bottom) gets compressed and heated, while the air in what was the bottom (now the top) expands and cools. But you knew this.
Bryan says:
January 19, 2012 at 4:30 pm
Perhaps Willis will suggest a real experiment with real materials to test the alternative conjectures.
Say with a thermoelectric device to make use of the temperature difference.
A computer simulation program would not be any kind of proof
I think he will find with real materials that this is beyond him
Perhaps this is why there has never been an experiment to settle the matter!
Yes, the distance from textbook to rigorously formulated physical hypotheses is huge. In fact, it is so large that it contains all the science, except for checking the predictions. To the Warmists it is a vast desert that they dare not enter.
JeffT says: Jan 19, 5:00 pm If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part.
/headslap! Thank you, Jeff, it is that simple. Total energy, chemical Potential energy, mechanical potential energy do nothing to prevent temperature equilibrium.
Take a pressure cylinder. Fill it with iron springs and Freon. Put in the piston. Place it is a bath or air, water, or any other fluid. Press the piston to its max. The contents of the cylinder heats, the iron springs compress, the Freon compresses then liquefies releasing more heat. Wait a bit. The Cylinder system contains a lot of potential energy, but it and its bath will come to the same temperature in time.
Release the piston and the temperature of the cylinder will drop and then be warmed by the bath back to a lower temperature. This example does not employ gravitational potential energy, but I’m convinced. Total energy is irrelevant.
So on a uniformly heated, dead planet at equilibrium, the atmosphere will have a zero lapse rate, isothermal atmosphere. But bring on a day and night cycle, then differential heating will force a lapse rate to become established by adiabatic processes regardless of the atmospheric composition. All we need to know is the specific heat of the atmosphere and the surface gravity: Cp/g. http://wiki.answers.com/Q/What_is_the_dry_adiabatic_lapse_rate_on_Venus.
Could one do the experiment using water as the fluid?
A long insulated tube (thermos like?) with temperature sensors along and a small tube inside for air to get out. Pour water at fixed temperature. Measure along tube. It would not have to be very long with good temperature sensors, a tower would do.
1. The elevator soliloquy:
The mass of the atmosphere, solar insolation, and the temperature of the upper atmosphere determine the temperature at the surface of a planet of sufficiently thick atmosphere. The exact makeup of the gasses of the atmosphere is not important, as it can be assumed that essentially all outgoing long wave radiation is absorbed and re-radiated in the lower atmosphere, finally being radiated to space from the upper atmosphere. Surface temperature is simply a function of the temperature of the effective black body temperature of the upper atmosphere required to radiate heat equal to the net insolation of the total planet, the mass of the atmosphere, and the gravity of the planet.
Climate change and excursions from average can be attributed mainly to changes in the mass of the atmosphere, albedo, and insolation, and are largely independent of minor changes in LW absorption in the atmosphere.
2. I must disagree with Dr. Brown. In a tall column of gas acted on by gravity, and at equilibrium, in a perfectly insulated container, the top of the column will be colder than the bottom of the column. All gas will be at an equal energy state (by mass, not volume), and since gas at the top has more potential energy, it must be at a lower thermal energy state in order to have the same entropy value. In other words, the adiabatic lapse rate still applies. This is essentially the situation that occurs in the atmosphere, on a very large scale.
Stephen Rasey says: 4:54 pm: Assuming that Index of refraction (Ir(r=ground) is higher near the ground than at high altitude Ir(r=large), then there is a focusing of energy.
Nope! Effects cancel out. Index of refraction might make it possible for someone on the surface of a planet to see more than 180 degrees of the outer shell. But each point on the outer shell is illuminating an equally larger area of the planet, thus diluting its contribution at any one point. Therefore, there is no net focusing, no net increase in energy, and an isothermal atmosphere even with index of refraction that increases with pressure.
Willis Eschenbach says:
January 19, 2012 at 10:32 pm
You have a planet not near a star and heat the atmosphere to 288K. You construct an insulating sphere to hold the heat in. If the atmosphere is isothermal, the TOA is at 288K. This will heat the insulation to 288K. The insulation will then radiate to space at 288K. Of course the radiation rate could be very slow, but we are talking about equilibrium here. If we make the foam slightly radioactive, then we get near equilibrium, but only until enough half-lives have passed.
If there is a material that is 100% reflective at microwave through visible frequencies, that would keep the atmosphere either isothermal or iso-energetic. Iso-energetic implies a lapse rate related to the gravity gradient. My memory of thermodynamics is not enough to tell me which way the energy gets partitioned, but I think it’s iso-energetic.
Or we could surround the planet with insulation in the form of turtles. Then it’s turtles, all the way up.
Thus spake Willis (re thermonuclear heating): “I ran the numbers in the past. I don’t have them in front of me, but it is very small, less than a tenth of a watt per square metre from memory. There are a variety of geothermal regions, and hot vents under the sea, and hot springs and pools on land. But you have to consider—for every hot springs you know of, there are thousands of square miles of land with no hot springs, where if you go to sleep in the morning, you wake up very cold. So yes, you are right, there is a heat source down under. But it is very small, even if we have greatly underestimated it.”
Right, Willis. I did the calculation a couple of years ago, and the conducted core heat is minuscule. If we look at mass transfer of hot material (hot springs, volcanoes, etc.), the numbers may be more significant. It is estimated that there are on the order of three million undersea volcanic vents. Most of these are small, but we don’t really have a handle on their total heat load, nor do we know exactly where it goes or how fast. (These seeps also emit CO2, but that’s another story,)
There is no equilibrium of a column of air in the gravitational field of a spherical planet unless there is a heat sink drawing energy away from the atmosphere. The heat equation does not allow a spherically distributed atmosphere to attain thermal equilibrium.
Without radiating gasses to draw the heat energy away, the atmosphere of a planet is like an ideal electrical capacitor hooked up to a constant voltage source, or an ideal inductor hooked up to a constant voltage source. It has nowhere to dissipate the energy, so the energy just keeps accumulating.
Eventually, the capacitor pops, or the inductor melts. In the real world, stray conductance or resistance, respectively, if they are large enough, prevent destruction. In the same way, a planet with insufficient radiation boils away its atmosphere, or achieves an equilibrium when sufficient radiation takes place from unaccounted emitting particles.
I have fleshed this all out in the previous thread starting about here.
So-called greenhouse gasses do not heat the surface – they prevent it from overheating. They are like a resistance put in parallel with that capacitor, or in series with the inductor.
In steady state, there is no difference between my description and the standard greenhouse hypothesis. On Earth, IR emitting gasses arrest the runaway heat accumulation that would otherwise exist. Backradiation equilibrates with the net overage in steady state Stefan-Boltzmann emissions from the planet’s surface.
So, there is no discriminator in the steady state, which is why people have been led astray for so long. They have been using Stefan-Boltzmann in ways it was not intended to be used – Stefan-Boltzmann only holds in conditions of equilibrium or, at least, quasi-equilibrium. The atmosphere of a planet without significant emissions is never in equilibrium, or even quasi-equilibrium.
The solution of the heat equation in spherical coordinates tips the balance in determining the real process going on. Without atmospheric emissions, there is always a temperature gradient leading into the atmosphere, and the atmosphere therefore continually accumulates energy. There is no adiabatic lapse rate possible without a heat sink in the atmosphere above. But, there is a lapse rate.
Look at the earlier thread. I’ve got it nailed. There is no doubt about it.
Erratum:
Without radiating gasses to draw the heat energy away, the atmosphere of a planet is like an ideal electrical capacitor hooked up to a constant current source…
“On average, that exchange equalizes temperature, moving slightly warmer air up and slightly cooler air down.”
Willis, I’m not sure whether that summarizes what Dr Brown said, but it’s wrong, and for a fundamental reason.
When the lapse rate is below the dry adiabat (toward isothermal) it is referred to as convectively stable. Above the adiabat, it is unstable. At the adiabat, it is neutrally stable.
At the adiabat, rising air cools by expansion, at exactly the rate at which the nearby air is becoming cooler (by lapse rate). And falling air warms. There is no buoyancy issue created. Moving air retains the same density as the environment. And no heat is transferred.
But in the stable regime, falling air warms faster than the change in ambient. It becomes less dense, so there is a force opposing its rise. That is why the air is stable. This motion both takes kinetic energy from the air and moves heat downwards (contra your statement). It is a heat pump which works to maintain the lapse rate.
Rising air does the same. It cools faster, and so rises against a buoyancy force. It takes KE from the air and moves “coolness upwards”, ie heat down. It pumps heat just as falling air does.
That is why air in motion tends to the adiabat lapse rate. A heat pump requires energy. Where from?
The atmosphere is famously a heat engine, driven by temperature differences, most notably from equator to pole, but also innumerable local differences, eg land/sea. This provides the kinetic energy that maintains the lapse rate, and it is hard to imagine any planetary atmosphere where the energy would not be available.
The effectiveness of the heat pump tapers as the adiabat lapse rate is approached. Beyond, in the unstable region, everything is reversed. The pump becomes an engine, with heat moving upward creating KE. This of course quickly diminishes the temperature gradient.
I blogged about this here.
It’s true that with no motion at all, conduction will render the air isothermal.
“a force opposing its rise”
I meant, opposing its fall.
A few posts from the other thread to help the discussion:
Solution of the Heat Equation in Spherical Coordinates
Bart says:
January 18, 2012 at 12:11 pm
In an atmosphere with no convection (the thought experiment world), the PDE governing the temperature is
dT/dt = alpha*del^2(T)
where alpha is the conductivity parameter and “d” is actually a partial differential operator and del^2 is the Laplacian. To solve this equation, we set T = T1(t)*T2(r), where T1 is wholly a function of t, and T2 is wholly a function of r. This leads to
(dT1/dt)/T1 = alpha*del^2(T2)/T2
where the “d’s” are now total differential operators. Since the left side is wholly a function of time, and the right wholly a function of r, they must both equal a constant, call it lambda.
Then, the solution of dT1/dt = lambda*T1 is elementary, T1 = T1(0)*exp(lambda*t). The solution of
del^2(T2) = (lambda/alpha)*T2
is a modified and adjusted zeroth order Bessel function of the second kind, which is qualitatively similar to a constant divided by radius.
Thus… the full solution is
T = T(0,0)*exp(lambda*t)*F(r)
There is always a gradient downhill in the radial direction. Thus, there is always heat flow into the higher altitudes. This heat flow will continue until it stops, either by high energy radiation if the atmosphere allows, or boiling away of the atmosphere.
Solution is a little more complicated than that, but same general morphology
Bart says:
January 18, 2012 at 1:56 pm
The alpha parameter is actually called “thermal diffusivity”, and it is inversely proportional to density. So, as the altitude increases, alpha will grow larger. So, even the Bessel function solution is not precise, and will only hold approximately in the lower atmosphere and before significant thermal expansion has taken place. A precise global solution would have to take all of these factors into account.
But, there is no chance of steady state because the steady state solution is T = B/r for a constant B, and since there is a gradient, there cannot be a steady state, and that creates a contradiction. So, the conclusion remains: there is always a thermal gradient pushing heat continuously into the atmosphere, and it will not stop until either there is some kind of radiative release, or the atmosphere flees.
Why the solution cannot be a constant, part I
Bart says:
January 19, 2012 at 1:52 pm
…Starting from some temperature at the base, in any finite time, the temperature is going to be less out farther than it is nearer, and it is going to go to zero at infinity. So, there must be a gradient over any finite time.
What is the form of that gradient? Well, through separation of variables, the temperature function is the product of a time varying part, and a spatial varying part. The spatial varying part is a function of spatial coordinates only, so it does not change with time. Therefore, since for any finite time, the spatial solution is of the form A + f(r), where A is a constant and f(r) is a monotonically decreasing function of r, and A must be zero at that time, then it must always be zero, and f(r) must go to zero as r approaches infinity.
QED.
Why the solution cannot be a constant, part II
Bart says:
January 19, 2012 at 8:14 pm
…So, if the “mode shape” of this thing is a constant independent of radius, that means the entire spherical shell has to heat uniformly. It starts at zero everywhere. It is at 1K everywhere at the same time. It is at 100K everywhere at the same time. This is physically impossible based solely on the fact that the rate of heat conductance is finite.
Willis,
If I understand what you are asking in this post, here is my elevator speech. If I have misunderstod, my apologies.
Imagine a 10km perfectlyl insulated column of gas.
Outside the tube connect one end of a suitably sensitive, perfectly insulated thermocouple to the top and bottom of the tube.
1. If a continual current flow is detected from the thermocouple, then the temperatures at the top and bottom of the column are different and one has a perpetual motion machine. *
* Series-parallel enough of these structures and our energy problems are solved – until gravity runs out, anyway.
2. If, after a suitable period of time no current flow is detected then the top and bottom of the column are at the same temperature.
I’m betting on #2.
I disagree with idea that the temperature on each layer is the same or that the energy per volume is the same, both are IMO wrong.
Let’s imagine an ideal gas container in gravitational field. In our gas particles don’t interact at all. As long as the gas is in equilibrium, each of its particles has the same total energy at every moment (but the important point is that total energy is sum of kinetic and potential energy). These particles travel on parabollic trajectory, bouncing off the bottom. This gives us overview of the energy distribution over the container volume – at the very top, the speed of particles is zero and the temperature is zero as well. At the very bottom the particle temperature is proportional to square of the height of the trajectory. The equilibrium condition is held too because on each horizontal layer boundary you get exactly the same number of particles traveling up and down at exactly the same speed.
This model I believe gives us very good idea about temperature, pressure and energy distribution over the container even for normal gases with interacting particles.
Another important thing is, your thermal perpetuum mobile cannot work in this temperature gradient as the medium you use in it follows the same rules the gas does. To raise your medium to the height where you intend to let it cool down you use more energy than you gain by letting it cool down there.
_______________
Now, I may be completely wrong somewhere in the above (transition from ideal non-interacting gas to normal interacting one feels to be the weakest point) so my main points are:
– potential energy is completely omitted in the article arguments
– the argument with thermal perpetuum mobile is invalid because the medium must follow the same rules the atmosphere does. If gravity created temperature gradient, it cannot be used by thermal engine.
One needs to ask, why is our air at different temperatures on earth at different altitudes? For one thing, we do not have an impervious membrane around us, and we have this thing called the sun shining at us. Result, the ground and especially water is heated or energized, and conducts heat to the lower air by direct contact, evaporation, etc. Result, the air at the bottom is warmer than air higher up. However, if you go very high, you reach ozone, which can and does absorb UV radiation, making it hot (and heating the non ozone around it) and causing it to expand way out into space. This will have some effect heating from the top down, although since this air is very thin the total amount of energy it can add to air lower down isn’t much. There is also some heating caused by the various GHG’s (especially water vapor) being heated by radiation and heating the air around them. Conclusion, there are reasons why our air is at different temperatures at different altitudes and it has nothing to do with any gravito thermal effect.
Second, if earth were as described, with an energy impervious barrier around it (and it had no fissionables), it would be at absolute zero. Even if gravito thermal worked, any air that contacted the earth would contact a very cold surface. The much more dense earth would quickly absorb all the heat from the air, which would freeze out rapidly, goodby gravito thermal effect.
However, let us assume, for arguments sake, that the surface of the earth and the air start out at the same temperatures as our surface and air. If gravito thermal works, and the people make huge heat engines, they will have to produce a lot of energy to make up for the fact that the cold ground will absorb and keep on absorbing the energy. They could, I suppose, surround the entire earth with an insulating material to slow this heat loss. If they did, they could be comfortable assuming the heat engines can create enough heat and light for them and to make up for the slight loss to ground. An impervious insulator below them would result in the energy and heat from their engine building up rapidly in the space between the insulator above and below them, which would continue until it melted the engines and stopped their operation. A not quite perfect insulator below them could keep them comfortable until the ground below the insulator had warmed up all the way through, at which point the people would find it slowly getting warmer, continuing until once again it gets warm enough that the heat engines melt and no more heat is, apparently, being produced from nothing.
Third, let us assume that the earth, as ours does, has fissionables in it’s core, and that is why the surface is warm. However, since there is a barrier to energy above them, this heat will slowly build up, since it cannot escape, until all the fissionables are used up. If it has a lot of them, as our earth seems to, it will get very hot before that happens, and once all the fissionables are gone, it will stay at that hot uniform temperature forever. However, lets us say, for arguments sake, that the last fissionables are used up just as it reaches a nice comfortable temperature (one where air does not freeze out). Now let us say that the people there, somehow, in the dark, invent heat engines to take advantage of this gravito thermal effect to make light and power. This will slowly add heat to the earth (the solid planet will absorb most of it) and very slowly heat things up. Eventually, it will get too hot, all the people will die, the engines will stop, and it will stay at this new, somehow elevated temperature forever. If you could somehow create indestructible heat engines that never melted and didn’t need tending, the heat would slowly build up forever, eventually the inside of the membrane would be hotter than any sun (which, I suppose, since it melted the earth, would stop the gravito thermal effect). As we can see, even under ideal starting conditions, if we assume that this membrane surrounds the earth, and we assume ideal starting temperatures for gravito thermal to work, it always results in energy being created from nothing, and the space inside the membrane getting hotter and hotter.
I think all this shows that under any circumstance we start with, even absolutely ideal (and impossible) starting conditions, gravito thermal results in an impossible outcome, with energy inside the membrane building up from nowhere. It should also be noted that the whole thought experiment is ridiculous anyway, such and energy impervious barrier cannot exist in our universe with it’s laws anyway. Thus we see that the only way gravito thermal can work is in a universe that is entirely imaginary, and has different natural laws that ours does. In other words, the authors of this idea are living in a world all their own, literally. They would realize that if they thought through all the implications of all this, as I have above.
Bart says:
January 19, 2012 at 11:06 pm
So in the Jelbring thought experiment you are saying the atmosphere never, ever achieves thermal equilibrium? If not … why can’t we pull work out of it with a thermocouple?
Regards,
w.
Nick, first you say:
Nick Stokes says:
January 19, 2012 at 11:15 pm
Then you say:
Since Dr. Brown is talking about air in a tall container, we’re talking about still air. The same thing is true in the Jelbring thought experiment, air with no motion in both cases.
As a result, I don’t see why you say Dr. Brown is wrong.
w.
F. Ross says:
January 19, 2012 at 11:31 pm
I’m betting the same way. I note that Bart wants to drag in a thousand and one complex equations. I say no matter what Bart claims about the equations, we can’t get an ongoing current from the tall columns. Period. Which means that they are isothermal.
w.
Let us forget gases for a moment and consider golf ball size particles made from lead, as our atmosphere, closely packed near the surface and twenty miles above loosely packed.
The potential energy of the balls at twenty miles up is huge. Consider these as cold molecules falling under gravity and impacting other molecules. Other molecules heated by conduction from the Earth by the sun are rising to meet them, and meet them they do head on. Thus slowing the transfer of Sol’s heat, graciously given to the world back into space. This is the major part of the misnamed green house effect. The rest of the green house nonsense can be explained by compression of the atmosphere close to the surface, where the molecules are elbowing for room.
The modulation to all this is water in it’s various phases, that gives us a livable planet, chaotic as it is.
For such machines to work, they’d have to create energy, and energy cannot be either created or destroyed, only transformed.
So the earth can’t orbit around the sun because that would break the laws of thermodynamics?
I hate to break it to you Willis but you have gone off the deep end.
The reason that gasses in an atmosphere are of higher temperature then if they were floating around in a vacuum are the same reasons why the earth can continue to orbit the sun without constantly being pushed.
i will restate my thought experiment again.
Imagine gas floating around in a cloud in the vacuum of space. it is spread out and the interactions between the atoms in the cloud are sparse…now imagine that cloud being pulled down a gravity well of a planet. the gas gains no energy yet it is now under pressure….now the atoms are all close together and colliding and doing what atoms do when they are forced to be in close proximity to one another. Again the atoms energy has not changed. yet their temperature is higher.
see how that works?
And it can be proven. Take the temperature of any old room….now compress air from that room into a container…now measure the temperature of the compressed air in that container…..guess what the air inside the container is hotter then the air in the room.
now i know what you are going to say “you used energy to compress that air” you are correct i did…but when atoms are compressed by gravity as in an atmosphere then no energy is used to compress it.
anyway i just thought of how this can balance thermodynamics….think of school teacher holding a ball in the air then he drops it. the teacher then says the ball while he held it had potential energy and when he released it it had kinetic energy. Think of the gas floating around in space as potential energy and when they are pulled down a gravity well of a planet as kinetic energy….you see they had that potential energy from when the universe was created and now that they are compressed by the planet’s gravity they have kinetic energy.
I really cannot explain it simpler then this…..if you cannot grasp it i have to suspect it is from willful ignorance.
“If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest air at the top and the warmest air at the bottom.”
Note that the atoms at the top have the highest energy and the lower ones the lowest.
The reason the temperatures are reversed is simply because at the top of the column you have less atoms hitting the side of your thermometer then you do at the bottom.
I think part of the problem is that although you understand that hot air moves to cold air i think you either are forgetting or don’t understand why atoms do that.
Willis replied to Ferds comment;
‘Temperatures have leveled. Sea levels have stabilized. CO2 has not. These events contradict GHG theory predictions, which in science is a strong indication the GHG theory is wrong.’
By remarking;
‘Thanks, ferd. I come to a very different conclusion from the same facts. I say that the greenhouse effect has done the heavy lifting of bringing us to our warm current temperature….’
Temperatures have been slightly higher than today and sea levels also higher on several ocassions during the last 5000 years. Self evidently that wasn’t due to raised co2 levels (according to the stated records)
What did the ‘heavy lifting on those occasions? Natural variability? if so couldn’t that be the cause again today?
tonyb
Willis Eschenbach says: January 19, 2012 at 11:52 pm
“I don’t see why you say Dr. Brown is wrong.”
In the quote that I cited, it explicitly says that motion of the air equalizes temperature. And it doesn’t, for the reasons that I gave.
But I was more concerned to say what it does do. It provides the heat pump which counters conduction and radiative transfer down the gradient.
And still air is impossible. You need a heat source to prevent liquefaction (from radiative loss), and that in practice will create temperature differences and motion, which in turn pushes toward the adiabatic lapse rate.
What the guys makea big tohuwabohu out of a very simple problem?
What is all this poetic story about?
Willi, it’s trivial as a nutshell.
1. There is a big difference between a tall air “container” and an atmosphere: the atmosphere is MIXING (troposphere = mixing sphere), your “tall container” is not.
2. If there is mxing, the only equilibium is very simple: ds/dz=0, where s is the entropy density. You have the thermal source at the bottom – the Earth surface heated by the Sun, and the thermal sink at the top – radiation to the space by the GHGs or clouds. In between you automatically get ds/dz=0. This is the adiabate that gives you the temperature lapse.
3. In your non-mixing “tall container”, there is no energy sink at the top. Thus, the temperature is constant. T=const. Very simple.
4. The temperature at the Earth’s surface is NOT defined by the GHG concentration, if the GHGs screen the Earth’s IR radiation completely. In this case, you must calculate the energy balance AT THE RADIATING SURFACE that is special for every wavelength: it is the altitude, where the atmosphere becomes transparent for the radiation. In this way you obtain the temperature up there.
5. Now you start your adiabate from this radiation surface at the top of the troposphere (the energy sink) and go down to the planet surface. Here you use the adiabate with its temperature lapse rate. The higher the atmospheric pressure at the bottom, the higher the temperature. Has nothing to do with “perpetuum mobile”.
6. For the reason discused above, Venus is hot as a hell due to the high atmospheric pressure (NOT GHG “concentration”), Mars with its low atmospheric pressure (although 20x more CO2 than the Earth!) is freesing and we have the comfortable climate.
Alex.
5.
Thomas L said:
But that’s also true if the air was already at the adiabatic lapse rate. In that case, when I flip the cylinder, the end state will be exactly the same as the initial state, so no net work was performed. I lift the air, I lower the air. I do work lifting, then have work done to me as I set it down. I do work tilting the cylinder 90 degrees, then have work done to me as I resist the further rotation as the dense air pours to the other end to try and complete the flip. It’s the same as pendular motion.
Back to Willis’s original thought experiment (and perhaps its key point) I’m not sure you can pull work out of the adiabatic atmosphere with a thermocouple, as I’m not sure anyone has dug that deeply into heat transfer in a solid in a gravitational field. For example, the thermocouple has to conduct the energy (as heat) along its length. If we pick the best standard material for heat conduction, silver, and work at the Earth’s adiabatic lapse rate of about 2 degrees C per thousand feet, we can only conduct about 2.4 Watts per square meter of silver conductor, even horizontally. Such low levels of heat transfer over such vertical distances might require the inclusion of effects that are irrelevant in most applications. putting us below the round-off error in the equations that work acceptably well in other applications. I’m pretty sure nobody has tried to use 2 degrees C to move only 2.4 watts upward through a silver or copper slab a thousand feet tall and a square meter in cross section, which is the same as using 160,000 tons of silver, the weight of almost four Iowa class battleships, to cool a 200 Watt CPU by just 2 degrees.
We could sidestep the question by positing that solid masses are allowed to reach thermal equilibrium at the top and bottom of the atmosphere, then are insulated and moved into contact with the thermocouple. I don’t know if this would raise questions about KE and PE of the solid bodies, though, or whether moving them that distance without losses would itself be a perpetual motion machine.
joshua Corning says:
January 20, 2012 at 12:08 am
The Jelbring thought experiment concerns a closed system, with no energy going either in or out. I don’t recall saying the earth could not orbit the sun, you’ll have to cite that. In any case, that’s an open system.
Thanks,
w.
Joe Born (January 19, 2012 at 5:42 pm) wrote:
[If the] … Velasco et al. paper is correct, its Equation 8, a result of statistical mechanics, dictates that average kinetic energy decreases with height even at equilibrium.
I’m not an atmospheric scientist, just an electrical engineer who had a single thermodynamics class 40+ years ago, but the comment above matches my reservations. Let’s postulate a sealed container filled with a perfect gas (defined as an ideal gas which additionally is completely transparent to radiation), similar to that proposed by Willis. The container begins at ground level, where the pressure is 1,000 millibars, and extends to the altitude at which the pressure would be 500 millibars if the gas were isothermal. Let’s grab a small amount of gas at ground level and raise it (without disturbing any other molecules) to the top. When we do that, we add potential energy of position to the gas molecules that rode the elevator to the top.
Where did that energy come from? Unless there’s another source of energy that can be tapped (a violation of the assumed equilibrium conditions in our sealed container), the energy has to come from the kinetic energy of the molecules that were raised from bottom to top. If it comes from anywhere else, we’ve added extra energy to the molecules that were raised that can then be used to perform work, creating yet another perpetual motion machine!
If we take the sealed container postulated above and evacuate it until there’s only a single gas molecule left, it’s clearly true that the total energy of that lonely gas molecule is the sum of its potential energy of position and its kinetic energy. When the molecule is at ground level, it will have the highest possible kinetic energy (implying a higher temperature), and zero potential energy of position. If it travels straight upward until it reaches the top of the sealed container, it will have the highest possible potential energy of position for that system, and correspondingly lower kinetic energy (implying a lower temperature).
When the sealed container has the normal complement of perfect gas within it, the mean free path of any given gas molecule will be very short (Wikipedia says 68 nanometers for air at 1013 millibars). However, between collisions, the vertical component of each molecule’s velocity will be affected by gravity, just as in the case where only a single molecule is present. If the vertical component is upward, it will trade kinetic energy for energy of position, cooling slightly in the process. If the vertical component is downward, it will trade energy of position for kinetic energy, warming slightly in the process.
So, the heresy, spoken aloud: A perfect gas within a sealed container will exhibit a temperature gradient in a gravitational field. This is not, however, a substrate for a perpetual motion machine. The total energy (kinetic plus potential energy of position) of each molecule is the same. Any attempt to extract energy from the difference in temperature between the top and bottom will founder on the reality that entropy has already been maximized for the gas within the sealed container. There’s nothing left to get.
You’ll note that at the beginning, I specified that the sealed container had a pressure of 1,000 millibars at the bottom and 500 millibars at the top. It’s been too many years since I took physics to allow me to easily perform the calculations, but someone who performs such calculations regularly should be able to calculate the energy required to raise a molecule from the bottom to the top, subtract it from the assumed kinetic energy of a molecule at 1000 millibars, and determine the remaining kinetic energy for that molecule. This would allow calculation of the actual pressure at the top (nominally 500 millibars), and the temperature gradient that would exist in the sealed container because of gravity.
I have never seen so much ignorance in one blog
Does no one understand the dry adiabat for the atmosphere on Earth (DALR=9 degrees per Km)?
This is entirely controlled by gravity and by the mass of the atmosphere– nothing else.
(hence it varies for other planets)
It applys to the larger atmosphere in exactly the same way it it does in an insulated cylinder of dry air
Does no one here understand the TePhigram which plots equal lines of equal entropy against air pressure (altitude)?
Why not take 5 minutes and get out of this ignorance
Neil
joshua Corning says:
January 20, 2012 at 12:08 am
Thanks, joshua. I do see how it works. And you are right, that will heat it … once. But you cannot get continuous work out of that.
You can’t get work out of gravity. I understand the lure of perpetual motion, of a wheel endlessly turned by gravity, but it can’t happen. Gravity is a force. Either you are moving against it or with it. But it is a zero-sum game (actually it’s a loss). If I start at my house and walk down the hill and walk back up, the energy gained on the walk downhill I lose on the way back up. Plus friction loss, to add insult to injury. The three laws of thermodynamics say you can’t win, you can’t break even, you can only lose.
Sure, as in your example, things pick up kinetic energy when they fall in a gravitational field, which in a gas is temperature. But if you want it to happen twice, you have to move them against the force of gravity back up high so they can fall again. No way to come out ahead doing that.
Regards,
w.
Tim Folkerts says
“The simplest and most convincing argument ended up being your same perpetual motion approach. You could run an insulated copper bar from bottom to top. This bar does not have the lapse rate effect. (Or use another gas that would have a different lapse rate because it has a different heat capacity). The copper should have the same temperature at top and bottom. If the gas had a different temperature, would could use this temperature difference to continually run the sort of heat engine you suggested.”
Tim go one stage further and source any real heat engine connected by real copper connections of one kilometer in length for a 9.8K temperature difference.
Factor in the resistance of the copper.
Who’s to say that one kilometer of copper will not have its own gravitational/thermal effect with its own value?
Indeed perhaps all this is in accordance with the second law
Maybe you can now see why an experiment to settle the isothermal/adiabatic distribution has never been attempted.
Its interesting that Claudius (who supported the isothermal conjecture) dropped the second law proof later on in life
Willis: Sorry to have confused you. WordPress doesn’t allow the use of symbolic logic symbols here. I thought you’d immediately see the problem. In essence, you state:
1. “If [the isolated system is in its lowest energy state]{B} THEN [it cannot perform work]{not W}”
2. “If [the isolated system…is warm at the bottom and cold at the top]{D} THEN [it can perform work]{W}”
3. “[The isolated system cannot perform work]”{not W}
4. “therefore [the isothermal state…is the lower of the two energy states,]{J}”
Simplified even more, your argument boils down to:
1, if B THEN not W
2. if D THEN W
3. not W,
4. therefore J
But what you’ve proved, so far, is simply “not D.” (I can think of another state which can’t perform work.) For the argument to be valid, you must establish
1a. if not D THEN B, as well as
1b. if B THEN J, and so forth, 2., 3., 4., as you state.
Sorry for the confusion. I looked for the symbols for quite a while before giving up.
“The Jelbring thought experiment concerns a closed system, with no energy going either in or out. I don’t recall saying the earth could not orbit the sun, you’ll have to cite that. In any case, that’s an open system.”
Atoms of high energy will be able to climb higher against gravity in the column then atoms of low energy. Energy is not going in and out but atoms still run into one another within the system right? and at any given time some atoms will whack into each other harder, simply by virtue of angle and velocity, then other atoms right?
If low energy atoms can only be at the bottom and high energy atoms can be anywhere in the column simple addition should tell you there will be more atoms at the bottom then on the top.
If there are more atoms on the bottom then the likelihood of those atoms (even though on average they are of a lower energy then the top) hitting your thermometer at the bottom then there is of them hitting the thermometer at the top. ie the temperature will be higher on the bottom then on the top.
You are confusing temperature with heat.
Hi Willis,
I think the thermocouple analogy is broken. In order to extract energy from a thermocouple, it is true there has to be a temperature difference, but there has to be heat to operate the thermocouple pair. The top thermo has no ‘sink’ and as a result would simply get warm by conduction, from the lower thermocouple.
It your hypothetical isolated planet the atmosphere would become Isothermic. My reading of N&Z is that an input of energy is required to maintain the gradient. The mechanism is unclear, however ignoring KE, PE and convection is as foolish as expecting gravity to do all the work (pun intended)
Cheers
Robert
Willis says
“To the contrary. Thermal equilibrium simply means that the objects have stopped exchanging energy because they are at the same temperature. This happens all the time.”
No at equilibrium they exchange equal quantities of energy
“But you cannot get continuous work out of that. ”
yes you can. An object in motion will to stay in motion. Atoms don’t stop…they don’t have wind resistance, they don’t get tired, they just keep going and going and going, and as you stated no energy comes in and no energy goes out. So those atoms cannot transfer energy out…ever. they will bounce around in your column until the end of the universe and they will segregate out into a pressure gradient simply by virtue of the fact that at any given time some atoms will have low energy and some will have high energy…the low energy atoms stay close to the bottom while higher energy ones will be everywhere….thus you get high pressure at the bottom and low pressure near the top….a thermometer will measure this gravity induced pressure gradient as hot below and cold on top simply because more atoms will hit the thermometer on the bottom then will hit it on top.
Willis,
Firstly, thank you for the courtesy of your response. It is encouraging that we are able to set aside non-scientific issues that hang between us and conduct scientific debate rationally and reasonably.
Some preamble, and then some science.
You said to Lucy that
“time is what you don’t have. The clock is running, the elevator speech for N&Z is way overdue.”
Nobody is king of the clocks. Lucy and I are not sales people with obligations to meet targets within timeframes or elevators. Paradigms don’t change overnight. Resistance to the theory of plate tectonics continued for as long as the old guard were in tenure at their institutions. Time must be spent in evaluating new theories properly, not concertina’d into a gish gallop of instant rebuttal and ‘counterproof’.
A more relevant example than plate tectonics is the Loschmidt vs Maxwell and Boltzmann debate regarding thermal gradients matching the theoretical dry adiabatic lapse rate in equilibrium atmospheres subject to a gravitational field. It’s been going on for over a hundred years without resolution and we don’t need to force a conclusion within the next few days just because it has been thrust to the centre of the stage at the moment.
OK thanks for reading that, lets address some science.
Willis Eschenbach says:
January 19, 2012 at 6:26 pm
If an energetically isolated system is in its lowest energy state, it cannot perform work.
Agreed
If the isolated atmosphere in Jelbring’s thought experiment is warm at the bottom and cold at the top, I can stick a thermocouple into it and use the temperature differential to generate electricity to perform work.
Excellent, a proposed experiment. Let us know the result. I think you’ll find that even as a thought experiment it doesn’t work out though. Peter Berenyi emailed me that argument and I sent him my disproof. He hasn’t got back to me in the two days since. I’ll post it in a separate comment if you are interested in defining your setup.
As Tallbloke points out, the second law says an isolated system can only move towards a lower energy state. That means Jelbring’s thought experiment must inexorably move towards the isothermal condition as its equilibrium state.
No, as we’ve been saying all along, as have other people on this thread, at the lowest energy state, molecules at the top of the atmosphere have the same total energy as those at the bottom, but less of the total is available as kinetic energy which manifests as heat via collisions because more of the total energy is locked up as gravitational potential energy.
Since Jelbring claims an adiabatic state will obtain at equilibrium, his hypothesis is falsified.
If my statement above is correct, or if the equivalent macroscopic arguments provided on this thread and by Jelbrings 2003 paper are correct, then this statement is false.
Willis Eschenbach says:
January 19, 2012 at 9:21 pm
Jeremy says:
Gravity has NO AFFECT ON TEMPERATURE.
How many times must it be said.
You people are reading science fiction.
You have to do WORK to create a change in temperature – this is basic thermodynamics!!!!
If an object falls in a gravitational field then potential energy will be converted to kinetic energy which will create heat. However a stable column of air in equilibrium does not create any energy or heat.
Thanks, Jeremy. You are a hundred percent correct, gravity can’t do ongoing work to change the temperature
Jeremy and you are 100% wrong. Work is done by energy. Gravity is not a type of energy, it is a force. It cannot and does not need to “do ongoing work”. Nor is a change in temperature under discussion. Gravity, via the pressure profile it induces in an atmosphere, and considering the compressibility of the medium, causes the denser per unit area of the atmosphere near the surface to be warmer than it is at higher altitudes. g/Cp
If you are fighting basic ignorance of science, you will be deluged with ignorant people. Not much I can do but just keep putting the facts out there.
Certainly there are a host of much more sophisticated threads, and those tend to attract a more scientifically literate commenter. But when you are discussing “gravito-thermal” theories …
This is an ad hominem attack which has no place in scientific discourse.
“joshua Corning says:
January 20, 2012 at 12:08 am
but when atoms are compressed by gravity as in an atmosphere then no energy is used to compress it”.
Probably be better to say, “then no energy is used to compress it, other than the potential energy of its mass.”
IR is (forced) employed into potential energy viz conduction of paired electrons around a molecules covalent bond, heating its chemicals, and adding kinetic energy to its energy budget
.
The adding of kinetic energy to mass causes heat, because of the collisions and transfer of the Atmospherically Thermally Enhanced kinetic energy that has been temporarily employed by the energy of mass.This heat is enhanced at higher pressures.
The energy budget of atmospheric trace chemicals should equal the square of its mass plus its employed kinetic energy.
And all the UW IR, DW IR, IR from enhanced mass, are just mixing gases in a pot up to re-radiation when kinetic energy of mass returns the employed energy to space
I have confidence in the N&K equation describing this phenomenon.
..
Fascinating stuff in all these comments ..As a reader I especially liked the ones posted by ed_b , Ian H and I found the intriguing question why there are no PV-cells exploiting the energy supposedly radiated from the GHG:s during night by Richard Vernay very reveiling.
Nick Stokes say
“When the lapse rate is below the dry adiabat (toward isothermal) it is referred to as convectively stable. Above the adiabat, it is unstable. At the adiabat, it is neutrally stable. ”
At the adiabat the parcel of air has no unbalanced force.
It can move up or down at constant speed or stay stationary.
This condition is called the neutral atmosphere and it can be surprisingly stable at night.
In dry conditions the gradient will be – 9.8K/km.
So its not correct to say that in OUR troposphere, the air will become isothermal if still.
That’s an awful lot of talking people. So? What did you come up with?
Let’s try this again.
1) Sun heats surface. Some heat goes into land and sea, below the surface.
2) If an atmosphere is present, hotter areas are cooled and the heat is transported elsewhere and not radiated away as quickly.
3) After the sun goes down, the latent heat begins to radiate away. However a lot of energy remains below the surface and takes time to reach the surface.
4) If not all of the energy absorbed during the day is radiated away, then the surface will start off warmer the next day.
5) The average surface temperature continues to rise until the surface is able to radiate in 24 hours the same energy it absorbs during the day.
At least on land, the key difference between the Earth and the Moon is the atmosphere. On Earth the hot surface is cooled by air and therefore is unable to radiate as efficiently (because it was cooled). Cooler (shaded) areas are warmed by the heat transferred to the air by conduction from the warmer locations. On the Moon, shadowed areas radiate to 3K and are very cold. The Sun can heat surface rocks and these will get hotter until they are able to radiate efficiently the same amount of energy they absorb. No heat is transfered from one location to other places on the Moon. It should be no surprise the Moon’s surface cools rapidly after sunset. Although water is interesting, it seems not the driving factor given the data from planets and moons we saw a few weeks ago.
A recent BBC TV programme on the Earth’s core featured a claim by some scientist that the strength of the planet’s magnetic field has been falling significantly for the past 170 years. If true, might this tie in with the Svensmark hypothesis that changes in cosmic rays hitting the atmosphere affect cloud cover? Might one of WUWT’s resident brainboxes look into this and see if it has legs?
Same answer as I gave early in the “Matter of Gravity” post on January 14, 2012 at 2:45 am
Your argument about a heat engine using heat difference not working has been countered by the Norwegians who had a power station off coast in the Atlantic using the heat difference between water layers. It produced 60Mw, until the first severe winter storm when it sunk without trace.
The gravitic heat is better called adiabatic heat by compression which does exist and is the principle used by many machines used daily by everybody. It would happen in a planetary atmosphere because that is open to space and has free movement to all sides. there would be vertical movement due to convection and atmospheric pumping between day and night sides due to temperature difference/heat loss rates. Also your column, despite being at the same temperature, would not have the same heat content at the top compared to the bottom because gravity would introduce a density differential, A Km high enclosed column of air does NOT represent an atmospheric column by any stretch of the imagination.
I also ask my Jupiter question again. Why does this gas giant radiate more heat than it receives from the sun. your argument above makes this impossible.
My reply to all the questions raised above on both sides is WWES…….What Would Einstein Say? I would genuinely like to know. I wonder if he would have enjoyed these blogs. Anyone channel him lately?
“George Turner says:
January 19, 2012 at 9:20 pm
I have a follow-up thought experiment to my previous comment.
If flipping the air column upside down doesn’t require an input of work and returns the column to the dry adiabatic lapse rate, then the dry adiabatic lapse rate must be the column’s thermal equilibrium because a system cannot be shifted from thermal equilibrium without an input of work.”
I am sure what would happen. It seems you going heat the top when reaches bottom- and that heat isn’t going to get back to the new top.
Let’s move to orbit, where you could spin it [forever], and you have gravity gradient- the length column will make lower part travel slower than upper part. You will have gravity- gravity gradient in “free fall”.
Of course spinning will also add artificial gravity.
As I said you will heat the lower end- and therefore cool the gas.
So how a way to cool gas.
Now, let’s insulate it somehow.
Hmm. Use water instead of gas.
So without spinning water will cover entire inside [water tension] and will end up at lower end, mostly. Start slow rotation, and water will run to to middle then new bottom.
It seems that will affect the orbit somehow:).
But question is whether cools.
It doesn’t seem to. Back to air.
A new question is how fast would air move.
It’s going to fall, so it’s going to move fairly fast.
So back to surface again [or be some orbit and have respectable amount gravity gradient-big column]
The vacuum end when rotates to ground will “suck” the heavier air
and heavier air will fall- and this is going to be violent- as in supersonic.
So, there will be good mixing of the air- because the hot air going go straight up,
or break your machine.
You going to cause the air to become more directed.
It seems is you made air more balance in it’s direction- every molecule stopped another other molecule you get colder temperature.
But seems that if you start with cold air you could get hot air- and the more vacuum the better:).
One problem is normal atmospheric air isn’t balanced in terms of weight- the bottom is a lot heavier than the top.
So work is being done by rotating it.
So question is how small can make this thing and have do something. 1 km long seems challenging. But if used colder air and have vacuum at top, it seems something smaller is possible.
If it was 5 km in length falling velocity is somewhere around 500 mph, which somewhat interesting- but smaller are going make a little bit of wind. So entire atmosphere excitement, and 5 km maybe somewhat interesting, but probably something even bigger in space environment and using gravity gradient- might actually do something useful..
Your first thought experiment sounded like Mawells Demon in which a gas can be partitioned into a hot side and a cold side by a force (a demon) that only allows high kinetic energy molecules to move in one direction and excluded the slow ones. In this case gravity is the demon.
I still don’t like your “perpetuum mobile” argument and a claim it violates the law of conservation.
I am not a physicist, but instead a logician. As I understand it physics claims energy cannot be destroyed. Thus it cannot be used up just converted and lost.
Jelbrings model according to my reading encapsulates energy and does not allow any to exit the atmosphere. So while you talk about light forever, if that energy is manifested as light and its always light where does it go when it cannot go anywhere?
Logically it seems to me the only issue is whether the atmosphere would be isothermal or not.
If isothermal then Jelbring is wrong and gravity does not cause the lapse rate and if the lapse rate exists at internal equilibrium he is most likely right.
I tend to think, but could be wrong, that for molecules under less pressure with the same temperature need to contain more energy.
So the question is what equilibrium is, is it energy or is it temperature. Personally I like the basal concept that its energy and not temperature but I have not wrapped my mind, or perhaps warped it, with Caballero.
There are some terrific quotes by Richard Feynman in his Nobel Lecture The Development of the Space-Time View of Quantum Electrodynamics, that bear on these issues of thermodynamics:
What we teach our engineering students is that you don’t really understand a phenomenon until (1) you can analyze it in more than one framework (formalism), and (2) within each framework, you can present it as an picture, an equation, and a numerical computation, and (3) you are able to unite all these frameworks within an encompassing narrative, that dovetails with other folks’ time-tested narratives.
With regard to thermodynamics and transport theory (which is broadly what this WUWT topic is about), an historically recent and very broadly applicable framework regards thermodynamics and transport theory as (essentially) the study of the geometry of flow on manifolds, specifically the study of Hamiltonian dynamical flows on manifolds that are equipped with a symplectic structure.
One great virtue of the symplectic approach to thermodynamics and transport is that it naturally encompasses both classical and quantum dynamics. A pretty significant downside, though, is that it takes a full year to learn the basic ideas and notations of differential geometry, in terms of which the whole framework is given.
What the resulting geometric dynamical frameworks predicts, though, is simple. If we computationally model the atmosphere (by a brute-force numerical calculation of molecular dynamics) as a cloud of particles in a gravitational potential, such that the particles bounce off each other elastically (with each collision conserving both energy and the number of particles), and the distance between collisions is small compared to the height of the atmosphere, and we do not “stir” the atmosphere with radiation heating at the bottom (or allow any other external influence to do work on the particles), then we will find (from the brute-force numerical codes) that the kinetic temperature of the particles is constant from top-to-bottom (note: the Wikipedia page the kinetic theory of gases has a picture of this kind of simulation, which are described in-depth in the well-respected textbook by Frenkel and Smit, titled Understanding Molecular Simulation: from Algorithms to Applications).
The dynamical geometer will argue as follows:
Summary: The folks here on WUWT who conceive that the gravitational potential gradient of the earth’s atmosphere has to induce some kind of thermodynamic gradient are 100% correct—but that thermodynamic gradient is a gradient of pressure (or equivalently, of chemical potential), not of temperature.
If these ideas sound kind of complicated and subtle … well, they are! Mistakes are very easy to make, and that is why it is prudent to evolve multiple independent explanations (pictures, analysis, numerics) as a cross-check on any given calculation.
A final thought when you have an equilibrium you need energy to change it. Thus our Stygian friends cannot tap into the equilibrium and generate light, they need energy to throw it out of equilibrium or if that energy is manifested as light to change the lighting arrangement concentrate it, focus it, or do any work with it.
re perpetual motion machines, we can say that there is ‘perpetual’ motion in our universe, such as orbits, bigband expansions and the like. how many motions are we undergoing at the moment ? 5 ?
If it’s energy for work you are seeking, then the best perpetual producer of energy on our planet might be it’s magnetic field. the findings of Nikola Tesla are to be considered. wiki has a reasonable summary. I am led to believe that I can access the electrical differential between 2 points separated by height, enough for a domestic supply.
my view is that there is much to be said for domestic solutions to energy requirements, passive solutions in the main.
getting back to the flow of energy, matter, temperature and gas atmospheres, bottled and otherwise.
heat is associated with increased molecular activity, ditto for pressure and molecular activity. we have bonding processes, chemical aggregations, there’s an evolution of structure. certainly our systems are extremely dynamic (large energy packages) with a myriad of structures – guess a number.
entropy – the eventual breakdown of structure and atoms. our little atom no longer has the energy to maintain it’s structure. pop ! it stops, electrons touch protons and neutrons, probably losing their shape. the molecular structure loses it’s space, and the volume of the matter reduces by a massive amount. in our astronomy class we would marvel at how a battleship could fit into a matchbox if you removed the space between the atoms. so we can ‘witness’ the reduction of volume at the earth’s core, I don’t know the pressure or gravitational measurements at that place, nor electrical discharge. obviously the process depletes the fuel source and imploding occurs. so we are supported from below as well as from above.
I do believe in using a multivariable approach to solving problems, rather than isolating certain variables.
I think we can generally expect more of the same; cosmic rays, sun bursts, volcanic eruptions, convection currents, and the odd meteorite.
cheers !
anna v says:
January 19, 2012 at 10:36 pm
“Could one do the experiment using water as the fluid?”
Yes, you could, but there is no need. We already have some; they are called oceans and lakes. The water at the bottom of the ocean is under great pressure. Dos this make it warmer? No, we all know that warm water rises. The hotter water is at the top. You can verify that in your bathtub. So pressure does not cause warming. QED.
So what is the difference between water and the atmosphere? We all know that warm air rises too, just like warm water rises. But in the atmosphere the cold temperatures are at the top ( up to the tropopause). So what is different? I could indicate this in one word! – but it is best left as an exercise to the reader. Some individual thought will promote a more general understanding.
What Bart said.
As others have observed above, Willis’s argument has a then-a-miracle-occurs step: he postulates a heat engine that is free of the gravity to which the gas column is subjected. He skipped the the step where Harry Potter removes the gravity from the heat engine’s location.
The mean single-molecule kinetic energy in a system of the type Willis describes is (3E/(5N-2))(1-mgz/E), where N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude. If you put numbers to it, you see that the lapse rate is exceedingly small but non-zero.
Imagine Willis’ container to be finite, and filled with air with a pressure of 1 bar, and being in his isothermal condition, with constant pressure and temperature throughout, as required by the Ideal Gas law. Now consider a second container on topo of it, with infinitesimal height, with vacuum inside. Now we remove the lid that separates these two containers. Brownian motion will make gas molecules dissipate into the vauum but the gravitational field pulls them back; and a pressure gradient develops, and with it, according to the Ideal Gas law, a temperature gradient.
And that is the stable configuration, not the isothermal one. What we end up with is the dry lapse rate, assuming that we have no radiative energy redistribution.
The notion that in equilibrium, there must be the same density of kinetic and potential energy in every partial volume must be false, because it would require infinite temperatures at the top end of the gas column.
I failed to mention that the formula I gave above applies to monatomic gases.The factor on the right is different for gases with higher degrees of freedom, but the overall expression remains dependent on altitude.
Another never ending debate is seems. How about escape velocity?
The atmosphere of a planet cannot be completely isothermal since the velocity i.e. energy of the uppermost molecules at some point would exceed escape velocity. Venus and Earth both have roughly the same mass so they would have the same escape velocity. If Venus is a perfect greenhouse effect, since its black body temperature is 184K (65Wm-2) or -89C, the Earth would have the same constraints, -89C, Anyone noticed the minimum temperature of the Antarctic lately?
Gravity places limits on atmospheric energy, if gravity fluctuates it would change those limits. Earth’s gravity doesn’t fluctuate enough to have a significant impact on temperature at the surface. If someone would like to show that Earth’s escape velocity changes enough to have an impact on surface temperature, I would love to see it.
Since mass is energy, Gravity and the Geomagnetic field both help retain atmospheric mass, perhaps a mass balance would be an interesting exercise?
This argument should be possible to resolve in 5 min by any reputable physics professor.
So why not invite one (or several) to comment/clarify?
In 2010 I had a discussion about Hans Jelbrings theory with John Wallace, atmospheric scientist at the university of Washington. He wrote me:
“To understand how radiative transfer influences surface temperature, one needs to go beyond the concept of the adiabatic lapse rate and consider an atmosphere in “radiative-convective equilibrium, as discussed on p. 421-422 of the 2nd edition of our textbook. In such an idealized 2-layer atmosphere, the lower layer, which is comparable in depth to the troposphere, has a lapse rate equal to the adiabatic lapse rate . Two points emerge from this simple analysis:
(1) Were there is no greenhouse effect, the lapse rate in a planetary atmosphere would be isothermal (i.e., temperature would not change with height. In this case, the dry adiabatic lapse rate would be unchanged from its present value, but it would be completely irrelevant to the interpretation of the observed lapse rate.
(2) Greenhouse gas concentrations have no effect on the adiabatic lapse rate in the lower “convective” layer, but they determine the depth of that layer: increasing greenhouse gases increases the surface temperature of the planet not by changing the lapse rate, but by deepening the convective layer. “.
I think that’s it.
There is a difference between temperature and heat. whilst temperature is a measurement of heat we need to know the specific heat of a substance to know how much heat is present. So this imaginary Km high column could have identical temperatures top and bottom but heat content will be different due to the density difference.
I think I have said this before. Sorry for a boring repeat.
Hi Willis,
I am pleased that you have retruned to this.
First, something I said on the previous thread was both hasty and in error. In terms of Potential Energy the change from isothermal and the DALR modes does not lower the centre of mass of the atmosphere as I speculated, it is not energetically prefered as I suggested, in fact it makes no difference as far as I can tell. Now on to more interesting things.
As I see it, you now expose their model as being inviolation of both the 1st and 2nd Laws. That is the way I have seen things.
Having derived the isothermal as being the prefered profile, a way is open to you to do something rather useful, in my opinion. That is to determine the GHE without appeal to the notion of “heat trapping”, which I consider to be bogus and generally unhelpful.
The naming of the GHE, is I suggest, due to our inhabitting the surface. To illustrate this and my issue with “heat trapping”, I will pose the following question in an idealised atmosphere (see below for part of the idealisation).
In terms of just the atmospheric part, does the addition of GHGs move an atmospheric system as a whole towards a warmer or cooler state, or make no difference?
My preference, or prejudice, would be to say that it tends to cool an atmosphere as a whole but also redistribute energy towards the DALR profile. The notion of a GHE being solely a POV issue. For a denizen of the upper atmosphere it might be termed the IBE (Ice Box Effect). I like to put it this way for I find that everybody hates the notion. It seems clear to me that the effect of adding GHGs when viewed from an atmospheric system perspective is to produce a strong cooling tendency that is acted against by a strong response in the form of sensible and latent fluxes.
There is a level of idealisation going on here, Notably an absence of SW absorption and the production of warming in the stratosphere of our Earth. For those who can put that to one side, I think that there is an opportunity for insight into an origin for of the surface GHE in terms of an overall cooling tendency. The complication due the real stratospheric warming and the way that may contribute to the production of the tropopause and in the determination of its height puts the question of whether the real atmosphere as a whole would actually warm, cool, or stay the same into some doubt. That is by the bye in terms of the idealisation I have assumed.
I view the GHE and the IBE as two inseparable sides to the same coin. Compared to the non-GHG isothermal case where the whole system assumes the SB equillibrium temperature, the addition of GHGs causes the surface to be warmer (GHE) but crucially the upper troposhere (on the real Earth) to be cooler (IBE). If people must have “heat trapping” I say they must also have “heat releasing” but I would rather they simply dropped that metaphor altogether. GHGs couple the atmoshpere to the radiative field and whether a cooling or a warming takes place locally, is determined by the spectra of the GHGs, their local density, the local temperature and the local strength of the radiative field at each frequency.
I don’t expect anyone to agree with this, amongst skeptics it may be viewed as a AGW trojan horse, amongst the staunch AGW dogmatists it seems to be viewed as deeply unhelpful perhaps largely due to the loss of the “heat trapping” metaphor, or simply wrong. Hopefully it is a POV that people can have fun with and might just break down some barriers to thinking about GHGEs (Greenhouse gas effects).
It is not all about warming.
Alex
Willis,
please get back to me via email, not here – I just don’t have the time.
Put simply, ignoring the electrical input into the earth system will lead you down many, intellectually attractive, paths.
Have a look at plasma physics and its applicability to your topics – you might be more than surprised,
Louis
David says
The air moving up and down exchanges potential energy (PE) for kinetic energy (KE). The air moving down loses PE but gains KE, and vice versa for the air moving up.
——-
I have not entirely digested the article but an important consideration may be that Willis is referring to an atmosphere that is in equilibrium, without external energy sources. This means that there is NO motion of the air.
David you are referring to air in motion. Which means there must be an external energy source to keep the air in motion.
So David and Willis are describing different situations.
An inverted cone 1klm in DIA, 5klm high, 10m DIA at top, on stilts above water, so atmosphere can enter down low.
Intoduce Co2 @ 5,000ppm with a Co2 drip feed relative to the life of atmospheric Co2. Headed by a turbine.
Just like GH theory, the Co2 enriched atmosphere inside must warm more than the atmosphere coming in below, and because of thermodynamics the heat at the top will turn that turbine.
I’m looking at the problem from a different direction. That of stability. Think of a column of air running from the ground to space. The temperature of this column of air at the very top is fixed. It is the temperature of space itself. As you descend through this column of air, the rate at which the air warms is determined by the lapse rate. Anytime a particular patch of air gets above the temperature determined by the lapse rate for that depth, it immediately becomes more boyant than the surrounding air and starts rising. It cools, but it still remains above the temperature of the air in it’s new surroundings. It rises until it reaches the top of the column, where it radiates it’s excess heat out into space.
This is not a perpetual motion machine, because it isn’t gravity that is causing the heating. It’s the sun. Take away the sun and quickly the whole column of air freezes. Increasing or decreasing the amount of energy coming from the sun will increase or decrease the total energy in the column of air, but the result of this is and expansion or contraction of the air, causing the column to grow or shrink. When the column expands, the average density of the air changes, which decreases the lapse rate. However the temperature at the top of the column and the bottom of the column remain the same. Decrease the energy from the sun, and the density of the air increases, and the height of the column decreases, but the temperature at the top and bottom still remain the same.
This relationship stays true until air cools to the point that it becomes a liquid.
The basic assumption of the radiation theory is that as mass gains energy when it enters a gravity field, all that energy will eventually be radiated away to space as EM radiation. This could take some time but for a gas it wouldn’t be that long.
All the thermal energy gained from gravitational potential energy is then eventually radiated away to space as EM radiation (give or take some cosmic background radiation re-entering the system).
Therefore: Gravitational Potential Energy = EM Energy
Therefore: the gravitational energy will eventually be radiated away to space – for gasses this should be relatively fast.
Well, that now gives us the theory of everything.
Except gravitational energy is not observed to decrease/radiate away over time unless the mass declines so the basic assumption is not complete. Thermal energy radiates away, but gravity does not.
That then implies:
– when an object which has gained energy in a gravity field loses EM energy through EM radiation, it must gain that back from the gravity field; or,
– gravitational energy is only interchangeble with EM/thermal energy at a limited level. It must be so small that we cannot detect a decline in gravity. Higgs bosons do not directly turn into EM photons, or only rarely; or,
– whatever gravitational energy that is not converted into EM energy remains in the system and is what we perceive as gravity.
There are all kinds of issues with this picture that are not understood at all.
Birdieshooter says:
January 20, 2012 at 2:30 am
My reply to all the questions raised above on both sides is WWES…….What Would Einstein Say? I would genuinely like to know. I wonder if he would have enjoyed these blogs. Anyone channel him lately?
He would say:
“If Jeremy and Willis are right, there goes relativity!”
Willis (or Dr Brown) the case for the isothermal cylinder seems to be fairly well established.
But on Earth, any given area on the surface has a conical section of atmosphere above it, and by my sums, I’d expect that if you have 15C at the surface, you’d expect to see about 13C at 15km simply due to the larger volume.
Not sure this has any implications for the overall argument or not.
Rob said, “(1) Were there is no greenhouse effect, the lapse rate in a planetary atmosphere would be isothermal (i.e., temperature would not change with height. In this case, the dry adiabatic lapse rate would be unchanged from its present value, but it would be completely irrelevant to the interpretation of the observed lapse rate.”
That is not entirely true. The lapse rate would attempt to approach isothermal. Even without “greenhouse effects” there is conductive heat transfer. The viscosity of the lower atmosphere would increase to improve conductive heat transfer reducing convection. Gravity would limit the amount of energy that could be contained in the atmosphere. The energy of the upper most molecules in steady state would be less than gravity imposes for the escape velocity of the planet. This is another, “all things being equal” example.
“(2) Greenhouse gas concentrations have no effect on the adiabatic lapse rate in the lower “convective” layer, but they determine the depth of that layer: increasing greenhouse gases increases the surface temperature of the planet not by changing the lapse rate, but by deepening the convective layer. “.
Another all things being equal example. “Greenhouse Gases” have differing thermal conductive properties, molecular weights and heat capacities. The “mix” in a mixed gases environment has more than just radiant effects.
Conductivity is generally assumed negligible in radiant atmospheric physics. The possibility that it is not negligible, reflects poorly on the estimates of the “radiant” portion of the ATMOSPHERIC EFFECT.
Willis,
I think Dr. Brown’s thought experiment makes a mistake by linking the temperature to the vertical component of the molecule speed, only, when he says: “On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down, with zero energy transport, zero mass transport, and zero alteration of the MB profiles above and below, only when the two slices have the same temperature”.
If there is a difference in horizontal speed (i.e. wind) between the layers, the temperatures can be different while the layers do adhere to all of the conditions mentioned in the quote.
On another note, Willis, when you tell Jeremy: “You are a hundred percent correct, gravity can’t do ongoing work to change the temperature”, I hope you’re not forgetting that the Earth does negative work on the molecules going up, which in equilibrium cancels out the positive work being done on the molecules going down, the net energy effect being zero.
Bart: “Without radiating gasses to draw the heat energy away, the atmosphere of a planet is like an ideal electrical capacitor hooked up to a constant voltage source”
I know you changed this to “constant-current source,” but, actually, you had it right the first time If voltage is temperature and the earth’s surface is the voltage source. An atmosphere of the type you describe would acquire energy from its source only when its temperature is less than the surface’s. Of course, the earth’s “voltage” isn’t constant, and its outgoing radiation provides a (highly nonlinear) parallel conductance, but that doesn’t detract from the point: the capacitor won’t pop.
AusieDan says:
January 19, 2012 at 8:14 pm
We need a theory to explain why the surface temperature of the various solar bodies can be derived as a function of distance from the sun (solar radiance) plus near surface atmospheric pressure.
I’ve given it several times now. It’s called the standard GHE at its maximum. That maximum effect is determined by atmospheric pressure. Nothing else is needed. This conjecture explains all the data.
Now, why is there a maximum? I’m not sure, but it could be where the “cooling effect” of GHGs exactly matches the “warming effect”. The latter one is the only one climate scientists look at. The former one is what you get when a GHG extracts energy from the atmosphere and radiates that energy to space.
All right, here is my attempt at the elevator speech for what I understand the Jelbrings paper to be, based on the discussions here. I’m sure someone will find something wrong with it. It’s a little bit longer than Willis’ elevator speech, but I believe that his misses some crucial pieces.
* If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest temperature at the top and the warmest temperature at the bottom.
* In an undisturbed environment that column of air will eventually reach energetic equilibrium, meaning that any given cube of air will contain exactly the same amount of energy, whether at the top or the bottom.
* Because an air molecule at the top of the column (gravitational body) is further away from the earth (gravitational body) than a molecule at the bottom of the column, by the universal law of gravitation less force is being exerted on the molecule at the top of the column.
* Therefore the pressure at the top of the column will be higher than the pressure at the bottom of the column.
* It also then follows, according to the ideal gas law, that P1 * V1 / T1 = P2 * V2 / T2 where 1 is the top of the column and 2 is the bottom of the column.
* Therefore, when examining two identical volumes of air, one at the top of the column and one at the bottom of the column, the one with the greater pressure (bottom of the column) must also have a greater temperature for energetic equilibrium to have been reached.
* This property of an undisturbed column of air in a gravity field, is the cause of what we erroneously refer to as the “greenhouse effect”.
Willis, such a column would never be isothermic throughout the whole column of air. Instead, it would be isothermic at any given layer of the column. In other words any given horizontal slice parallel to the earth would be isothermic, but they would not be isothermic from one layer to the next. The faulty assumption necessary to yield an isothermic column is that the strength of gravity field acting upon any given molecule in the column is uniform throughout the column. It clearly is not, both theoretically and empirically.
Bomber Cat:
Real oceans and lakes are not a good facsimile for this thought experiment. The reason, their primary thermal input comes from the top of the ocean (i.e. the warming rays of the sun) because the depth of water does not allow the sun’s radiation to reach the bottom of the ocean floor to be absorbed and redistributed back upwards.
If you look at how a solar pond works (when the sun can penetrate to the black bottom) you see the warmer water at the bottom and the colder water at the top. The problem in your example is the lack of transparency of the ocean and an energy source at the top.
Why is the N Z paper called a gravity model? I always interpreted their model as a delay model. Sun heats the surface through a largely transparent medium. The surface heats this transparent medium though particle contact. Eventually this distributes through the transparent medium.
The delay in the heated surface to lose it’s temp because of the transparent medium allows the surface to heat more than it would without the transparent medium.
I read what was posted in this article that ignores any of the above as completely missing the point.
Consider a 10km long box 1m x 1m cross section of perfect insulating material (aerogel?). Coat the outside with 100% reflective material (to stop the aeogel radiating it will not conduct)
Fill it with argon.
Lay it horizontally 5km above ground level. It will of course be perfectly balanced
Leave it for a few days.
The temperature must stabilise so that all the gas is at the same temperature – there are no external influences on the gas and it cannot radiate and cannot conduct out of the box
Rotate it 90 degrees. (it will do some of the work itself as the argon compresses in the lower end and decompresses in the higher unbalancing the system.
When vertical the argon will be heavily compressed in the lower end and therefore much hotter and very decompressed and therefore cooler in the high end. In fact it will have the profie dictated by the lapse rate.
Leave this for a few months. Those warm and cool atoms of argon will be wizzing about in the box but they will be loosing no energy. All that wizzing eventually equalizes the temperature throughout the gas – the low pressure gas will have the same temperature as the high pressure gas.
There is no energy in or out to disturb this. Maxwells deamon is not in the box to sort the hot from the cold! the gas is isothermal.
It will stay in this state forever.
Now rotate the box through 180 degrees and magically the low pressure end becomes the high pressure end and therefore the lapse rate is re-established.
HOWEVER
You have just turned a box that is heavy at the lower end. raising it 10km up in the air, working against gravity. There has been much work done.
Leave the box for a few more months and the gas will again become isothermal.
BUT
The temperature will now be hotter than before because you put all that work in rotating the box.
You can continue this rotation untill the argon is white hot of course!
BUT
it only gets hotter because you put in work
Leave the box vertical for as long as you like and no work will be done. Gravity is static it does no work. If gravity maintained the adiabatic lapse rate temperatures the it would be magically operatinhg as a maxwell deamon.
By the way look up “vortex tube” for a way of separating hot and cold air! A maxwell deamon powered by compressed air.
As a follow-on to the solar pond comment I just left, I do not consider it a reasonable facsimile of the experiment either. I was only pointing out that there are bodies of water that violate the example of oceans and lakes. In order for it to be a reasonable facsimile, there would have to be no light source, i.e no sun.
Tallbloke says: “No, as we’ve been saying all along, as have other people on this thread, at the lowest energy state, molecules at the top of the atmosphere have the same total energy as those at the bottom, but less of the total is available as kinetic energy …”
As has been stated many times in many way, science is not decided by consensus, and science is certainly not by consensus of non-experts. Of the people on this list who seem to have formal training in physics, the agreement seems quite strong that isothermal is the the equilibrium condition for the given thought experiment.
Among OTHER trained physicists I have consulted, that is ALSO the agreement for isothermal rather than a lapse rate.
For instance, they say:
“all of the following are true:
— Boltzmann distribution of kinetic energy
— Boltzmann distribution of potential energy
— Boltzmann distribution of total energy.”
This is in contrast to Tallbloke’s claim that ONLY total energy follows the Boltzmann distribution.
EACH of these separately will have the same distribution and the same temperature. I could also go into the rather involved discussion of the partial derivatives involved, like
∂S β = -------- ∂E | Nbut I don’t think it would help much in this discussion.
[COMMENT: I fixed the formatting, Tim, using the “pre” tags (for “preformatted”). WordPress ignores leading blank spaces. —w.]
Oops… I reversed on point in my elevator speech…
* Therefore the pressure at the bottom of the column will be higher than the pressure at the top of the column.
The whole thing should read:
* If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest temperature at the top and the warmest temperature at the bottom.
* In an undisturbed environment that column of air will eventually reach energetic equilibrium, meaning that any given cube of air will contain exactly the same amount of energy, whether at the top or the bottom.
* Because an air molecule at the top of the column (gravitational body) is further away from the earth (gravitational body) than a molecule at the bottom of the column, by the universal law of gravitation less force is being exerted on the molecule at the top of the column.
* Therefore the pressure at the bottom of the column will be higher than the pressure at the top of the column.
* It also then follows, according to the ideal gas law, that P1 * V1 / T1 = P2 * V2 / T2 where 1 is the top of the column and 2 is the bottom of the column.
* Therefore, when examining two identical volumes of air, one at the top of the column and one at the bottom of the column, the one with the greater pressure (bottom of the column) must also have a greater temperature for energetic equilibrium to have been reached.
* This property of an undisturbed column of air in a gravity field, is the cause of what we erroneously refer to as the “greenhouse effect”.
There must be some Peter Principle corollary or perhaps Parkinson’s Law of Triviality corollary with this discussion.
Luke says:
January 20, 2012 at 5:41 am
“* This property of an undisturbed column of air in a gravity field, is the cause of what we erroneously refer to as the “greenhouse effect”.”
This conclusion is not supported. The temperature at the bottom of the column would be the same with or without gravity! Gravity merely trades off temperature for gravitational potential energy with increasing column height. That’s all it does. It can’t add or substract energy from the column, it can’t change the distribution of energy in the column, but it CAN change the distribution of sensible/insensible energy and that distinction is important because potential energy won’t save any brass monkeys from disfigurement.
Here is a Car-Talk Puzzler-type question that (hopefully) will illuminate why Luke and other posters are mistaken.
A WUWT Puzzler
Alice has a cannon that shoots vertically, with a random initial vertical velocity, whose root-mean-square initial value is 100 meters/second, whose mean value is zero, and which is normally distributed (a Bell-shape curve).
Puzzler Remarks:
(1) on half of the firings, Alice’s shell shoots down into the dirt
(2) the other half of the firings, Alice’s shell shoots up-in-the-air
(3) a typical maximal height is (100^2)/(2*10) = 500 meters
(4) some shells fly higher, others lower
Puzzler Question: Of all the shells that pass the “X” meter height, what is their mean square velocity, as measured at “X” meters altitude?
Asserted Puzzler Answer: No matter what the value of height “X”, the shells that pass through height “X” have a root-mean-square velocity of 100 meters/second.
So amazingly, our “Puzzler Shells” do not “cool off” as they fly to higher altitudes. Rather, there are simply fewer-and-fewer of them.
As with “Puzzler Shells”, so with gas molecules: their temperature is independent of elevation, but their density decreases.
And that is my “elevator explanation” of Willis’ problem.
John Mason says: January 20, 2012 at 5:56 am
… The faulty assumption necessary to yield an isothermic column is that the strength of gravity field acting upon any given molecule in the column is uniform throughout the column. It clearly is not, both theoretically and empirically.
============
The force of gravity changes little over a 10km distance above the earth:
http://en.wikipedia.org/wiki/File:Erdgvarp.png
Surface g=9.81
10km g=9.78
less than 0.5% change
Luke says: January 20, 2012 at 5:41 am
All right, here is my attempt at the elevator speech…
* If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest temperature at the top and the warmest temperature at the bottom.
You are starting from a wrong hypothesis. My first thought was also that this might be the equilibrium condition, but a bit of actual study of the issue made it clear this is wrong
“* This property of an undisturbed column of air in a gravity field, is the cause of what we erroneously refer to as the “greenhouse effect”.
If you start with a wrong postulate, then is is easy to come to all sorts of incorrect conclusions. For example, this property of an undisturbed column is ALSO what allows you to run a perpetual motion machine from this air column.
I recently made a very relevant comment to Dr. Robert Brown on one of his past threads. Maybe some here might also read and consider the topic on multiple horizontal Boltzmann distributions at each and every level upward a tall gravitationally held column under an actual DALR. See: http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/#comment-870527
John Marshall says:
Because Jupiter is a gas giant undergoing slow gravitational collapse. I.e., the thermal energy is generated by the conversion of gravitational potential energy. See for example the discussion here http://nineplanets.org/jupiter.html :
This is not happening for the Earth.
“Gravity has NO AFFECT ON TEMPERATURE.”
As far as planet Earth is concerned, the correct statement would be:
Gravity has NO EFFECT ON POTENTIAL TEMPERATURE!
In a gravity field, with a costant source of energy (the Sun), in presence of two sinks (the Poles), the vertical profile of temperature would be that of the adiabatic lapse rate.
The presence of water, other GHGs and ozone makes the thinks more complicated, but vertical mixing can’t be ignored.
Of course gravity doesn’t make the difference between two planet, one with GHGs and the other without. It just rearranges the vertical distribution of absolute temperature (thanks to density), but not that of POTENTIAL temperature that would be costant. Actualy, potential temperature in our troposphere is higher aloft due to the release of latent heat after (mainly) wet convection has occurred
Willis,
N & Z have indeed produced a game-changer here and no thought experiment is needed to understand their ‘Unified Theory of Climate’. Simply stated, their hypothesis consists of their two key equations, (7) and (8).
The existence of a dimensionless ATE ‘factor’ does not imply that “gravity causes heating of the lower atmosphere”, in defiance of the 2nd Law of Thermodynamics. I believe the relevant science is nicely encapsulated in their (non-linear) equation (8),
Ts = 25.3966 (So + 0.0001325)^0.25 NTE(Ps).
That equation enables them to calculate the surface temperature (Ts) of ‘any planet with an atmosphere’, knowing only the TOA TSI (So), the surface pressure (Ps) on the planet and the dimensionless Atmospheric Temperature Enhancement factor (NTE).
The NTE factor for any planet is derived from observations already made (see N & Z, Table 1) and a thorough reassessment of the physics of what they term ‘grey body’ temperature (Tgb) of a real, airless planetary object (our own moon) – a conceptual physical model which is used in equation (7) to derive their NTE values.
The experiments have been done, the data have been analyzed; their ‘grey body’ temperature model and their simple regression are now out there to be pondered for veracity and significance.
The fact that nobody can explain “how it works” (to your satisfaction) is not a valid reason to try to demolish their hypothesis by thought experiments (which include unproven assumptions).
Try to understand it Willis, please. All your thought experiments are irrelevant unless you can demonstrate where their math, or their data, or their ‘grey body’ model, or their regression is wrong. To do that you need to understand what they have done and, by your own admission, you do not yet understand what they have done.
Tim Folkerts says:
January 20, 2012 at 6:06 am
Of the people on this list who seem to have formal training in physics, the agreement seems quite strong that isothermal is the the equilibrium condition for the given thought experiment.
Whereas the engineers and meteorologists tend the other way. And this is interesting, because it shows who has more common sense. 🙂
Tim, it’s no good appealing to Boltzmann when he is one of the protagonists in the unresolved dispute.
Loschmidt, Lagrange, Laplace, Jelbring and me vs Willis, you, Jeremy, Boltzmann and Maxwell
Outside, now!
lol.
I’m reading up on what happened between Maxwell’s formative thinking about classical molecular mechanics (engineers and meteorologists) and his part in the development of statistical mechanics (physicists and mathematicians) to see if I can find the key to this fascinating puzzle.
Last year this was discussed in a Climate Etc. thread. Perhaps a conceptually simpler Perpetuum Mobile machine can be constructed from two identical vertical columns of gases of different heat capacity, e.g. helium and argon thermally connected at their bases and totally insulated otherwise. If the lapse rate is an equilibrium property, there will be a constant temperature difference at the tops of these two columns. Making identical thermocouple connections between these points at the same gravitational potential, one has an electric potential difference and can then use this to dissipate energy. Energy is being extracted from both columns, cooling both while a constant temperature difference persists at the thermocouples.
If one is mathematically inclined, it is not a difficult exercise to start with a fixed volume of an isothermal, ideal gas and show that any perturbation of its density and thermal profiles leads to reduced configurational entropy, total energy held constant.
When one sees the expression “radiative-convective equilibrium” employed, it’s a giveaway that the writer is unconditionally misinformed. There can be radiative-convective steady-states, but any steady-state requires a constant dissipation of either mass or energy for maintenance. That the troposphere might be described as a steady-state implies such an energy of dissipation and, perhaps not too surprisingly, convective stirring takes a steady input ca. 240W/m^2 to be dissipated, but that’s reverting to pre-postnormal science.
DirkH says:
January 20, 2012 at 3:54 am
“And that is the stable configuration, not the isothermal one.”
Looks like I misunderstood the definition of “isothermal” – and my explanation is in no contradiction to Willis’ isothermal configuration. So it seems Willis and I agree that we will observe the lapse rate as defined by the Ideal Gas Law. Sorry for any confusion.
After 24 years of being inundated with “greenhouse effect” theory lectures, first from James Hansen in 1988 which those of us old enough to remember was quite a display of fear mongering (very convincingly so I might add), multiple documentaries on television and never ending use of the term since then in the media and science publications, the whole damn thing has evolved into an unfalsifiable hypothesis . Anyone can use the internet to find these “high school physics” descriptions of how the GHE works.
All I would like an answer to is why the very basic tenet of the GHE, that being the tropical troposphere should be warming at a faster rate than the surface, is unsupported by observational evidence. This was culminated by the debunking of Santer 08, perhaps the most dubious peer reviewed material released for public and scientific consumption since MBH 98. To keep the pseudoscience alive, scientists promoting the failed experiment tell us the observations must be wrong, not the “theory”.
I was fully convinced in 1988 of the “greenhouse effect” and it’s deleterious effects it would have on the earth, yet here it is 24 years later and it isn’t working as advertised. If anything, it is upside down.
There is something wrong with the GHE hypothesis as it has been promoted lo these many years.. I’m not qualified to enunciate what it is in technical jargon, but I don’t need to be an atmospheric scientist or physicist to know something is wrong. Does anyone else feel like they’ve been sold a lemon? Even here on WUWT,with the arguing going back and forth, the “theory” is no better explained or proven than it was 25 years ago.
“If it disagrees with experiment, it’s wrong”.
Regarding N and Z, my understanding of it is this (could be totally wrong, IANAPhysicist):
When a star is accreting, gravitational contraction converts gravitational potential energy into thermal energy. It is this thermal energy which acts as an opposing force to gravity, preventing the star from collapsing completely. As the star gains mass, its gravitational attraction increases as does its thermal energy. At some point thermal energy becomes large enough that fusion occurs and the star becomes stable. What does this have to do with N and Z? As I understand it, they claim that they can estimate atmospheric temperature from the density of the atmosphere. A volume of air that is more dense has more mass than the same volume of air that is less dense. Therefore, shouldn’t the dense volume of air have more gravitational potential energy that can be converted to thermal energy? And, can we assume that any volume of air that is being contracted by gravity will have some thermal energy above the black/gray body energy?
richard verney
The affects of the tides and other orbital parameters will indeed affect the atmosphere. Work is indeed done in the thermodynamic sense of external force acting on a closed system. In fact the Moon and Earth system will experience a loss in Kinetic energy from the “gravitational drag” of tides. These are third or fourth or even higher order effects. Like variations of the trace gas CO2, the affect of these things on “atmospheric temperatures” are extremely small and inconsequential compared to radiative forcing from our Sun.
JJThoms says:
January 20, 2012 at 6:03 am
“Leave this for a few months. Those warm and cool atoms of argon will be wizzing about in the box but they will be loosing no energy. All that wizzing eventually equalizes the temperature throughout the gas – the low pressure gas will have the same temperature as the high pressure gas.”
But according to the Ideal Gas Law, this is not possible. The temperature distribution must follow the pressure distribution so that PV = NkT
where P is the absolute pressure of the gas measured in atmospheres; V is the volume (in this equation the volume is expressed in liters); N is the number of particles in the gas; k is Boltzmann’s constant relating temperature and energy; and T is the absolute temperature.
http://en.wikipedia.org/wiki/Ideal_gas_law
They very fact that there is a vertical pressure gradient results in a temperature gradient.
So in this mythical elevator….
First it has no sides. It has no top. The bottom can vary widely…. Density can be 50% different.
With temperature changes the hight as well as the width of the elevator can grow or shrink.
What is so hard to figure out? As the atmosphere has heat added it grows. This growing increases it’s surface area. This increased surface area. Increases the thermal exposure to the cold of space. It cools.
The atmosphere as well as earths temperature is self regulating.
Jeremy says:
January 19, 2012 at 7:52 pm
Gravity has NO AFFECT ON TEMPERATURE.
How many times must it be said.
You have to do WORK to create a change in temperature – this is basic thermodynamics!!!!
Jeremy please use the standard equation to state your case. simply (Joel I said simply) Q= U+W.
If W=FD and F = mg then W can be done by heated air rising IF volume changes. We have been told there is a diurnal bulge. As W is a path function and W1-2 (air going up) does not have to equal W2-1 (air coming down) then W could be being done via gravity.
KE and PE in U do not change IF the center of gravity does not change.
Bomber_the_Cat :
January 20, 2012 at 3:40 am
In my books an ocean is not a closed system, the way this column of gas is proposed. It is not a good example.
Cosmologists believe in perpetual motion. They believe light travels through the universe ad infinitum without using any energy!
Good luck in this quest to make those who do not understand begin to understand….don Quixote….. er ….. Willis.
For me this is a very simple issue:
1. The Earth is only receiving enough power (or energy flow) from the sun to emit an average of about 240 W/m² from the surface of the Earth. That is derived first, from the solar constant divided by four, the ratio of flat spherical surface area of the Earth to the area of the disk of solar radiation absorbed by the Earth and second, by applying an assumed 30 percent optical reflection factor. Reflected solar energy is treated as never having arrived.
2. According to the Trenberth diagram and the Stefan-Boltzmann equation, the Earth is actually warm enough to be radiating something on the order of 396 W/m² average power from the surface.
3. A transparent, non-greenhouse gas containing atmosphere cannot, by definition, prevent or stop the surface from radiating all this power to outer space. Thus, the surface would be continually losing a net average of 156 W/m² as long as it remained that warm. Fixing this would require a close relative of the perpetual motion machine–the perpetual power creation machine.
4. An average surface power radiation of 396 W/m² can only continue with an atmosphere that can absorb a net 156 W/m² from the outgoing radiated power and return it to the surface.
All technical details of exactly how that happens are irrelevant to this power balance requirement.
Kevin Vaughan,
No they don’t, light is red shifted if it moves opposite a gravitational field, and blue shifted if moving with the field. The frequency shift shows a change in energy due to the gravitational field. Gravity influences everything, even electromagnetic waves.
Willis, or anyone else.
First, the gravity theory is a non-starter. Willis is correct and people seem to keep missing some very fundamental properties here. Heat is not temperature. Heat is temperature at volume. Keep molecular temperature the same, but increase density and you have more heat. Willis is exactly correct in saying that the upper less dense molecules can have the same temp as the lower more dense molecules and it all be at equilibrium.The PE and KE will vary over height (as related to the gravity point source) so total energy (TE) of any specific given volume of gas can and will come to complete equilibrium over time as the temperature of each given molecule reaches the same temperature. It’s a volume issue. His proof is also correct. If this didn’t happen, you would and could have a perpetual motion issue to deal with. Case close, can we move on to something new please? This doesn’t prove GHG theory correct, it just says the alternative theory is incorrect and people are looking like buffoons. (Engineer by schooling)
I do have a question though. As has been said over and over, our atmosphere volume has to increase if it warms (Total heat). Why are we trying to measure average temperatures, which is meaningless and damn near impossible, instead of not just measuring the height of the atmosphere instead? You would think that an average height of the atmosphere would be much easier to do as a function of time and any trends would then be easy to spot.
AJ
Tim Folkerts says @ January 20, 2012 at 6:06 am
Tim, please refer to this page and specifically to the sentence that reads “Ignoring tiny corrections for gravity, the gas will be distributed uniformly in the container, so the only unknown is the velocity distribution function”.
OzWizard says:
January 20, 2012 at 6:43 am
Try to understand it Willis, please. All your thought experiments are irrelevant unless you can demonstrate where their math, or their data, or their ‘grey body’ model, or their regression is wrong. To do that you need to understand what they have done and, by your own admission, you do not yet understand what they have done.
Their math, etc can be completely correct AND Willis can also be correct. All it takes is understanding that the GHE has a maximum value determined by exactly the parameters K&Z use. So, you have a GHE, just as Willis’ thought experiment requires, to warm the surface. You also have many planets that all have reached their own maximum GHE based on atmospheric mass, gravity and irradiance.
“For such machines to work, they’d have to create energy”
I certainly don’t believe in perpetual motion machines, but the above statement is dead wrong. For such machines to work they would have to reduce entropy, not create energy.
One of my imaginary friend lives on a spherical planet that is not rotating, has an atmosphere consisting of one type of gas and a spherically uniform gravitational field.
There was a time when the planet was uniformly lit from a light source in the surface fed by geothermal energy and some of that geothermal energy uniformly warmed the atmosphere from the surface.
Unfortunately the geothermal fuel ran out and now the planet has no light or heat since.
Before this event my imaginary friend built a tall tower of regularly spaced temperature guages.
They were fortunate that the heat source was not strong enough to cause atmospheric molecules to achieve escape velocity. They found a magic point where the atmosphere reached a maximum height at which the molecules had zero kinetic energy and the temperature read 0 deg K. But the temperature below was none zero because of the geothermal heat source and heat was lost to space through radiation from the atmosphere to maintain equilibrium.
Since the heat source died it has been found that the height at which the molecules of atmosphere have zero velocity is decreasing and the molecules of atmosphere below are cooling.
They have a theory of global cooling that predicts that unless a heat source is found, one day in the future the whole atmosphere will have a uniform temperature of 0 deg K but until then the surface will be warmer than the top of the atmosphere because it is always 0 deg K at the top.
anna v says:
January 20, 2012 at 7:24 am
Bomber_the_Cat :
January 20, 2012 at 3:40 am
In my books an ocean is not a closed system, the way this column of gas is proposed. It is not a good example.
Not only that but water is incompressible.
From the “lay” sidelines:
I know of a perpetual motion machine: a priori scientific debate. Generates its own energy. Of course that’s debatable.
AJB says: January 20, 2012 at 7:51 am
“Ignoring tiny corrections for gravity, the gas will be distributed uniformly in the container, so the only unknown is the velocity distribution function”.
I agree completely. If you can ignore gravity, then gases are distributed uniformly in a container. This is a very good approximation for car engines or steam turbines, and so engineers can reasonably ignore the effect of gravity that makes the pressure at the top of a steam vessel 0.0001 atm less than the pressure at the bottom.
However, the thought experiment here specifically removes ALL OTHER EFFECTS BESIDES GRAVITY. Now we are ONLY dealing with those “tiny corrections”. And for a column 10 km high (sort of like the atmosphere), then those gravitational effects become quite important. Density and pressure are known to drop with altitude due to those effect. The question remaining is “does temperature also drop with elevation in a perfectly insulated container?” I say “no”.
Tim,
Have a think about why you say no. Then let the rest of us know. And please don’t just throw names around. Describe the physical phenomenon that leads you to the conclusion.
Thanks
“Otherwise heat will flow from the hotter (right-shifted MB distribution) to the colder (left-shifted MB distribution) slice until the temperatures are equal.”
Robert, your mistake seems right there in the last statement. The molecule members of the upper hotter right-shifted MB distribution cannot equalize (they are actually already in equilibrium) with the lower cooler left-shifted MB for the acceleration that changes the molecules vertical velocity. I think I am right on that, would you reconsider?
You keep want to say the molecules at various levels must be at the same mean velocity but by your own same example, they can’t in the gravity well case. Maybe the MB derivation doesn’t have a gravitation terms when applied vertically. I’ll stop and check it now.
Now, I see where I went wrong. Following the logic of my question to Dr. Brown, I incorrectly thought the final equilibrium arrangement would be where the average energy per molecule was evenly spread out from top to bottom, with the molecules having the same average total energy everywhere. This leads to warmer temperature at the bottom and colder temperature at elevation. Instead, at thermal equilibrium, the average energy per volume is the same from top to bottom, with every cubic metre having the same total energy. To do that, the gas needs to be isothermal, with the same temperature in every part.
Hi Willis,
Permit me to make a small change to this (most of the rest of what I’ve read so far of your post is fine). The final equilibrium arrangement is one where the average kinetic energy per molecule is the same everywhere in the gas column. There are a lot more molecules in the denser gas at the bottom, so there is a lot more internal kinetic energy per unit volume down there, even though the kinetic energy per molecule does not change.
One has to be very careful about how one combines internal kinetic energy and work-based energy. The usual approach is to turn them into the enthalpy of the gas, especially when the gas is in thermal equilibrium with a non-uniform density profile. The usual formula for enthalpy of an ideal gas, E = U + PV, assumes a closed volume, thermal equilibrium, and more or less uniform pressure. Thus one can look at how the enthalpy changes as one e.g. compresses the gas or adds heat. This again becomes complicated when one takes into account variation in density and is the sort of thing that goes into estimating the DALR.
The simplest way to view this is in terms of the heat capacity. The heat capacity of the gas in any volume large enough to hold “many” molecules and be at thermal equilibrium is proportional to the number of molecules in the volume. The gradient in the density that gravity generates in the compressible fluid is accompanied by a gradient in the heat capacity, so that in equilibrium (at the same temperature) the gas at the bottom does have more energy per unit volume than the gas at the top.
The air temperature in the thermosphere can be well over 1000C — in principle hot enough to melt metals. But the atmosphere out there is very thin and the heat capacity is miniscule. The stratosphere — a layer named because of the lack of vertical transport and turbulence — gets warmer from the bottom to the top. That doesn’t necessarily mean that the energy per unit volume increases though, as the density still decreases. It just means that air molecules at the top of the stratosphere are moving, on average, faster than the molecules of air at the top of the troposphere, exactly the opposite of what one would expect from a naive “greater density equals greater temperature” model.
We now return to the regularly scheduled discussion (I haven’t finished reading your post:-).
rgb
Tim Folkerts says:
January 20, 2012 at 6:06 am
Tallbloke says: “No, as we’ve been saying all along, as have other people on this thread, at the lowest energy state, molecules at the top of the atmosphere have the same total energy as those at the bottom, but less of the total is available as kinetic energy …”
–As has been stated many times in many way, science is not decided by consensus, and science is certainly not by consensus of non-experts. Of the people on this list who seem to have formal training in physics, the agreement seems quite strong that isothermal is the the equilibrium condition for the given thought experiment.
Among OTHER trained physicists I have consulted, that is ALSO the agreement for isothermal rather than a lapse rate.
For instance, they say:
“all of the following are true:
— Boltzmann distribution of kinetic energy
— Boltzmann distribution of potential energy
— Boltzmann distribution of total energy.”
This is in contrast to Tallbloke’s claim that ONLY total energy follows the Boltzmann distribution. —
Hello Tim,
I am just intrigued by your logic. You state about the Tallbloke´s statement: “science is not decided by consensus, and science is certainly not by consensus.” This is absolutely true regarding scientists that are not corrupt or ingorant.
Then you state:
“Among OTHER trained physicists I have consulted, that is ALSO the agreement for isothermal rather than a lapse rate.”
Then you use the consensus argument to “disprove” what Tallbloke claimed.
Do you recognize your fallacy? I am making the comment since you pretend to be scientific.
I can ssure you that I didn´t bother about consensus when I wrote my E&E article. If I had done it would never had been written.
Best Hans Jelbring
Why not place a 10metre double insulated tube on a long arm centrifuge, fill tube with dry gas at surface pressure, allow temperatures to settle, the temp should be the same at both ends of the tube.
Rotate centrifuge to give 20g.
We should have multiplied the gravity effect by 20, the short length of tube 10m not 1km, would reduce the effect by 1/1000 so the temperature difference measured should be in the order of 180 millidegrees C assuming 9degrees/Km.
Any temperature difference other than zero would be worth investigating…..
Tallbloke says:
“Loschmidt, Lagrange, Laplace, Jelbring and me vs Willis, you, Jeremy, Boltzmann and Maxwell
Outside, now!”
Thanks for the laugh. ☺
Let me make one purely semantic argument. The adiabatic lapse rate is ~ 10 K/km. “Adiabatic” means “no energy flow”. So if you can make the approximation that there is little energy flow compared to other energies involved, then the adiabatic lapse rate would be a good estimate of the situation ion the atmosphere. Conduction of heat from one part of the column to another is by definition non-adiabatic. Therefore if conduction is the primary means of energy transport, the situation would not be expected to follow the adiabatic lapse rate.
And that lead me to one slight variation of the though experiment. Consider our perfectly insulated column. I will modify this SLIGHTLY by putting a heat reservoir at the bottom that holds at exactly 300 K (this would be similar Willis’s uniformly lit planet where the SB temperature is 300K at the surface). Now I force the air in the column to be well beyond the adiabatic lapse rate (perhaps I add temporary barriers every 100 m and cool each section by 2 K from the one below ie twice the adiabatic lapse rate). If the barriers are removed, two things will happen.
1) the column of air will be unstable to convection, and the air will start mixing like crazy.
2) there will be heat conduction P/A = k * (Delta_t).
Process 1 will continue until the lapse drops to ~ 10 K/km, at which point it will stop. Process 2 does not need to stop when Process 1 stops. In fact, Process 2 will continue as long as there is any temperature gradient.
NOTE 1: Process 1 will be MUCH quicker than Process 2. I guesstimate Process 1 would be mostly completed within a few hours. But even after convection stops, Process 2 would drive the whole column toward a uniform temperature of 300 K. I guesstimate this would take several months due to the low thermal conductivity of air and the great distances involved.
NOTE 2: If there the air is not perfectly isolated from the rest of the universe causing even a TINY heat sink at the top of the column (even a few mW/m^2), then the rate of conduction will not be enough to erase the lapse rate. GHGs (and even tiny bits of dust) in the upper stratosphere are enough to radiate thousands of mW/m^2 (ie several W/m^2), which means any real atmosphere will have a lapse rate. This tiny leak from the top is enough to maintain the lapse rate, and hence explain why the thickness of the atmosphere plays a major role in surface temperature.
A physicist says:
January 20, 2012 at 6:18 am
“Here is a Car-Talk Puzzler-type question that (hopefully) will illuminate why Luke and other posters are mistaken.”
“Alice has a cannon that shoots vertically, with a random initial vertical velocity, whose root-mean-square initial value is 100 meters/second, whose mean value is zero, and which is normally distributed (a Bell-shape curve).”
Is there any relevance about your shooting of she is all the time shooting fron the same altititude?
Will she shoot with the same initial root-mean square 100M/s from any altitude?
I am not a physicist so excuse me if misunderstanding you.
Yet another elevator argument for isothermal (same temperature) atmospheric equilibrium is as follows: we imagine a very tall (10 km tall) thermopile column (a device that converts temperature differences to electricity), and we insulate the body thermopile column so that only its top and bottom exchange heat with the atmosphere.
Now supposing that the upper air is colder than the lower air, our thermopile generates electric power continuously and forever, with no external source of power. Which is impossible. And so we conclude that, at equilibrium, the entire atmosphere must be at one temperature.
Of course, in the real world, such a thermopile column would generate electricity. And this electricity would constitute (ultimately) a form of solar power, deriving from sunlight acting to warm the earth, thus creating rising thermals that stir the atmosphere, creating a temperature gradient that the thermopile can exploit.
This is one more line of reasoning showing that the isothermal folks have got the thermodynamics right.
Tim Folkerts says:
January 20, 2012 at 6:20 am
“You are starting from a wrong hypothesis. My first thought was also that this might be the equilibrium condition, but a bit of actual study of the issue made it clear this is wrong”
If you read my E&E, 2003 carefully you will realize that it is based on two major assumptions and these are:
1) The first law of thermodynamics and
2) Second law of thermodynamics.
The application of these laws in the thought experiments leads to a constant energy content in any two equal submasses of the inclosed insulated atmospherea after relaxation time has passed (approximately 2 weeks).
This was not declared explicitly in the text since it was a topic I wanted to debate but few scientists wanted such a debate. It took 8 eyars for the debate to flourish. Do notice that these laws apply to energy and not temperature.
In all the posts I have scanned, with the possible exception of Joe Born, no one seems to be looking at the extreme case of the very top of the air column. Pick an arbitrary altitude where one N2 diatom is occupying one cubic meter of near vacuum (1 km^3?). What is its PE, KE, freq, temperature? Does it just pop off to space? I think that coming from that direction the argument quickly gets to ‘turtles all the way down.’
” For such machines to work, they’d have to create energy, and energy cannot be either created or destroyed, only transformed.”
This is another conventional wisdom based upon the incomplete information available at this time. Where did all the matter and energy that exists today come from? I guess if you buy one of the oscillating universe theories, it has always existed. Or in the multiple universe theories it may have been “transferred” through collision with another universe or leaked into our universe. But it could have been created as well. The how or from where or Whom of the potential “big bang” is as yet not explained by science.
Tim Folkerts says:
January 20, 2012 at 8:43 am
“Process 2 [thermal conduction] will continue as long as there is any temperature gradient.”
Well said, Tim
Except that you meant to say not. Not locked up. How can I put this gently, firmly, and understandably.
How about temperature has nothing to do with gravitational potential energy.
Look, if you want to comment on thermodynamics and temperature, learn what temperature is and what it isn’t.
As far as an ideal gas — the kind considered throughout this discussion, including by Jelbring and N&Z — is concerned, temperature is a direct measure of the average kinetic energy of a molecule of the gas. Note well, I did not say average potential energy and I certainly did not say average total energy or a jar of matter would get hotter or colder every time we lift it or lower it. Here, let me chill my beer by picking it up off of the table and lifting it to my mouth. No.
Please, please, please. Buy an introductory physics textbook that has a halfway decent thermodynamics section. Pretty please with sugar on top. I beg you. Read it.
Look, you have a choice. Either you can pretend that the Laws of Thermodynamics don’t exist and reinvent them at will, making up a brand new definition of the word “temperature” and pretending that your definition will still work to describe things like equilibrium, the flow of heat and entropy, ideal gases, and so on, or you can learn the ones that we already have. Personally, I think your contributions to the discussion would be better if you did the latter, but suit yourself.
Note well, however, that you will not, and should not, be taken seriously if you assert that temperature of a monatomic ideal gas is related to anything but:
U = 3/2 NkT = 1/2 Nm v^2_avg.
or, “the total internal energy of the gas is equal to the number of degrees of freedom times kT per molecule”, for three — note well, three degrees of freedom. Gravitational potential energy is not a degree of freedom, and if it were it still wouldn’t affect equipartition of energy so kinetic energy is a different one.
rgb
Paolo M. says:
January 20, 2012 at 6:40 am
I am not sure you understand what “potential temperature” actually means. It is a misnomer, at least within the science of meteorology. The meaning is actually constant total energy per mass unit. This state will be be found almost every sunny day above land from the surface up to 1000-4000 m or more. It is best devoloped about 1 hour before sunset. The observational evidence for its existence is just overwhelming.
Another way to put is that at such occations the measured dry adiabatic temperature lapse rate will be very close to -g/Cp or -9.8 K/km (no clouds allowed).
I’ve been trying to wrap my head around the thermodynamic arguments for the last few days. The isothermal column of air argument seems to have some problems:
1) The atmosphere has a measurable temperature gradient. An argument that suggests it should be isothermal seems to immediately be falsified by contradiction.
2) Although it’s an equilibrium system overall it is not in thermal equilibrium. We are discussing a system that has a constant energy input (insolation through a transparent atmosphere) at the bottom where the planet’s surface forces a boundary condition on the temperature of the air at sea level. If the planet/atmosphere system is not to heat up then the energy output at the top of atmosphere must match the input. Therefore there must be a flow of energy from the planet’s surface to the TOA and this can only occur if the atmosphere has a temperature gradient and thus is not in thermal equilibrium.
Or have I missed something basic?
Hans asks me “Do you recognize your fallacy? ”
I recognize your point. But I think it is not quite the fallacy you think it is. I am saying that even a single person who knows what they are doing has applied the principles of known science. He/she has a “proof” (actually several proofs) that the temperature profile must be uniform for the conditions given (and assuming that “textbook thermodynamics” is correct) . That proof has been checked by others to make sure there is no error. This is more akin to “spell-checking” than “consensus”. We now have a new addition to “textbook physics”. (Actually, this is very OLD textbook physics.)
To counter this proof, you need to show a specific error. Maybe there was a sign error. Maybe they took a partial derivative incorrectly. Maybe you can show that a perpetual motion machine IS possible and the 2nd law of thermodynamics IS NOT correct. But if we simply throw back and forth intuition or soundbites, this will not cut it (from either side).
PS This is precisely why “consensus” in climate science is NOT so useful. In the case of the column of air, the situation is very well-defined, so it is easy to apply basic physics and come to a clear solution. For climate science, the situation is very poorly defined. There are sources and sinks of energy all over the place; there are feedbacks; there are continuing subtle changes in orbits, the initial conditions are not well known, etc. All of these mean that you have to include MANY factors in the calculations. This means a computer to determine the predicted affect.
And now there are MANY places for problems. Basically, each person studying climate can only say “I took into account as many of the affects as I could, and here is what I found”. There are (nearly) endless “what if” questions. There are (nearly) endless lines of code to check. This means that there will be considerable uncertainly in the results.
So it is “settled science” that “CO2 radiates IR well and will warm the surface”, because short, easily verified theories (and repeatable experiments) lead inexorably to that conclusion.
It is “settled science” that the air column will be isothermal, because short, easily verified theories (and repeatable experiments) lead inexorably to that conclusion.
It is not “settled science” that “doubling the CO2 levels will cause a 3.7 K increase in temperature” because there are so many other factors and feedbacks that nailing this number down precisely is a damn difficult problem.
PPS Of course, no science is ever 100% “settled”. Relativity showed that newtonian mechanics was not quite right. But overturning “settled science” requires extraordinary evidence. So far I have seen no “extraordinary evidence” that a perpetual motion machine is actually possible and that there could be a continued temperature gradient in a perfectly insulated air column.
DR says:
January 20, 2012 at 6:58 am
” Does anyone else feel like they’ve been sold a lemon? Even here on WUWT,with the arguing going back and forth, the “theory” is no better explained or proven than it was 25 years ago. ”
No, I don´t feel like that since I have been fighting IPCC and its unscientifc statements since it was created. However, Willis is good at keeping the confusion alive which favours the IPCC organization.
A physicist says:
“Yet another elevator argument for isothermal (same temperature) atmospheric equilibrium is as follows: we imagine a very tall (10 km tall) thermopile column (a device that converts temperature differences to electricity), and we insulate the body thermopile column so that only its top and bottom exchange heat with the atmosphere.”
Have you factored in the resistance of the 10km copper(lets say) leads to your thermopiles?
Once you use real thermopiles and realistic conductors you will realise why this experiment will not work.
Final overall comment and off to work.
First, there are actually two kinds of perpetual motion machines that people propose. They are called perpetual motion machines of the first and second kind.
Perpetual motion machines of the first kind violate the first law of thermodynamics. The perform work with no (net) input of energy at all, and thereby increase the mass-energy content of the Universe as they function. They thus violate a very, very basic physical principle as well as a law of thermodynamics.
Perpetual motion machines of second kind violate the second law of thermodynamics. No energy is created or destroyed, it is just moved around so it can be reused again and again.
Both are magic, and actual mythological magic can be classified identically — magic of the first kind is responsible for creating Universes out of nothing, turning lead into gold, and so on. Mass-energy violating magic. Magic of the second kind is more subtle — rising from the dead, healing the sick, walking on water. No energy is created or destroyed, these things are all technically possible, they are just enormously improbable.
Jelbring’s hypothesis enables one to create a perpetual motion machine of the second kind to light stygia. The work done by their Carnot cycle engine and turned into light eventually turns back into heat, so the total energy of Stygia remains unchanged. It is just re-sorted by gravity acting as a Maxwell’s Demon into separated hot and cold reservoirs so that it can be used once again to drive the generator to make more light. The same energy is made available over and over again.
It is this that should make your “horseshit” detectors give a ring. You would have to have been born yesterday to think that Nature gives you any sort of free lunch like that. That hasn’t stopped optimists from seeking PMMs of type 1 or 2, or physicists from proposing theories that violate the laws of thermodynamics, imagining that they are more like “suggestions” than actual laws. But they aren’t suggestions. They are common sense.
* Fact 1: One can run a heat engine between any two reservoirs of energy maintained at different temperatures. Proof: Every heat engine in the world, all of thermodynamic theory, massive engineering…
* Fact 2: Heat engines cannot run indefinitely. In a closed system, they cannot just take random energy in a complex environment and continuously turn it into work. Proof: It’s the second law of thermodynamics Kelvin statement, supported by enormous amounts of evidence and common sense. So much so that if anyone doubts it, I have a bridge that I’d like to sell them in Brooklyn, it should be worth a lot.
* Assertion Gravity sorts air in an adiabatically isolated environment out into hot air at the bottom and cold air at the top. This arrangement is thermodynamically stable and will spontaneously occur and be sustained.
* Argument If the assertion were true, then due to Fact 1, a heat engine placed in the container and run between the top and the bottom would run forever. As fast as it made the air at the top warmer, the heat would somehow “fall” back to the bottom, re-creating the thermal gradient that we know can drive all sorts of heat engines. This violates Fact 2.
* Conclusion The assertion is therefore false. It contradicts two everyday, well-known facts. Anybody who believes Jelbring’s conclusions is invited to make themselves infinitely wealthy, as they have just solved the energy crisis. Just don’t ask me to invest.
“Then the people living in the stygian darkness inside that impervious shell could use that temperature difference to drive a heat engine.”
Wrong. If this argument were correct we wouldn’t need to care about temperature differences. The perpetual motion machine could be driven be the pressure difference alone.
The people would create a massless container to send up and grab some of the low pressure air and bring it down to where the air is is at higher pressure. Then, they could use the pressure difference to drive an engine and do work. Right?
No, of course not. As they bring the massless container of lower pressure (and lower density) air from above it would become a lighter than air balloon. So, it would take work to force it down to the level of the high pressure air. That would negate work done by the pressure difference. No perpetual motion machine.
Couching the argument in terms of temperature doesn’t change the result.
IMhO, there is no way to answer this question about gravity and lapse rates without discussing entropy. Entropy is king. Every other result in thermodynamics – including Boltzman distributions and all the rest, derive from the principle of Equal Apriori Probabilities and the consequence that a system will, in time, inevitably migrate to the macrostate with maximum entropy.
If the isothermal macrostate has higher entropy, that’s where it will go. If a macrostate with a lapse rate has higher entropy, that is where the system will go.
Spector says:
January 20, 2012 at 7:34 am
“For me this is a very simple issue:
4. An average surface power radiation of 396 W/m² can only continue with an atmosphere that can absorb a net 156 W/m² from the outgoing radiated power and return it to the surface.”
Any surface radiation power exceeding 100 W/m^2 is bull regardless if it is from equatorial, midlatitude or polar regions during days or night. Just show how this fantasy power radiation changes between day and night in polar regions as an exsample.
Bart,
Your argument begs the question. Your postulate that the atmosphere must decrease in temperature with altitude assumes your conclusion. But it doesn’t have to decrease in temperature. The pressure and the density must decrease with altitude. But a transparent atmosphere is perfectly thermally insulated at the top. It can’t lose energy to space. If the surface is at constant temperature, then eventually, so will be the entire volume.
In the other thread you asked how to define the top of the atmosphere. Here’s a definition: The top of the atmosphere is the altitude which includes 99.9998% of the total mass of the atmosphere. That’s ~100km.
As pointed out above, your first crack at the capacitor example was correct. It’s a constant voltage source. Even a constant current source reverts to a constant voltage source at some voltage.
OzWizard says:
All they have done is fit some data using a form with many free parameters: There are 4 free parameters in Equation (7) and that is not even including any freedom they may have exercised in choosing the fitting form, choosing how to define T_gb, or even which estimates of the average surface pressure and temperature of various bodies to use.
Hence, it is not surprising that they have fit the data. I got almost as good a fit to the data restricting myself to their particular fitting form when I change 3 of the data points (basically by changing the average temperature of the 3 planets that have a substantial radiative greenhouse effect so that their average temperature is taken to be the conventionally-determined blackbody temperature instead of the observed temperature).
And, that’s another point: Only 3 of the 8 celestial bodies they fit to have a significant radiative greenhouse effect and only for one of them, Venus, is it large enough to be the majority of their calculated surface temperature enhancement. Hence, they are not even fitting data for the greenhouse effect…They are mainly fitting to the effect that a planet can have a number of different average temperatures that are compatible with radiative balance…with airless planets having low average temperatures because of a wide temperature distribution and planets with more atmosphere having higher average temperatures.
One of the strange things about N&Z is how few people have investigated it well enough to even have the most primitive understanding of what they have done!
George Turner and WIllis: I’m interested in the idea of turning the tall cylinder of gas, but let’s start with a horizontal cylinder of gas, which should have the same temperature and pressure throughout its length. Let’s say the cylinder is 1 m2 in diameter, contains 10^4 kg of ideal gas (the same weight of gas as above ever m2 of the earth’s surface), and is 20 km/10 mb tall (99% of the atmosphere). Let’s imagine that there are barriers every meter that are closed during rotation and later opened, so we don’t have to worry about what happens during rotation. Alternatively. we can imagine piston barriers that will allow changes to be made reversibly or irreversibly before opening the barriers. What happens when we rotate the cylinder to vertical and open the barriers reversibly or irreversibly?
Based on what we know about our atmosphere, we can be confident that most of the gas will “fall” to the bottom of the cylinder, increasing the kinetic energy/temperature of the gas at the bottom of the cylinder and therefore it’s pressure (ideal gas law). It certainly seems to me that the cylinder MUST be cold on top and hot at the bottom after rotating. If we didn’t just violate the 2LoT by transferring heat by spontaneously creating a temperature gradient where one didn’t exist before (and I assume we didn’t), then we need to be careful about how we describe entropy in this system.
Alternatively, we could say that the gas at the high end expands under reduced pressure and the gas at the lower end is compressed under higher pressure. However, first we need to explain why the pressure on the gas has changed in these regions. We say the weight of the gas above contributed to the pressure on the gas below, but the gas in this cylinder HAD a pressure before it was rotated. Why is pressure in the vertical position defined by the weight of the gas above while pressure in the horizontal position was not? The answer is that pressure is not really created by the weight of the gas above, it arises (according to the kinetic theory of gases) from momentum transferred by collisions (to the walls of a container or whatever is measuring the pressure). We usually assume that motion in all three directions is ISOTROPIC, but in a gravitational field the speed of the molecules moving upward is slightly less that the speed of the molecules moving downward. The “weight of the gas above” appears to transferred downward by non-isotropic motion of the gas molecules in a gravitational field. (See Section 2.3 of your Caballero reference.)
In Brown’s explanation of molecules crossing a plane, he says that the molecules moving up and down :
“have to have exactly the same velocity distribution moving in either direction”
This statement appears to be incorrect. If there weren’t a velocity difference, the pressure at the top and bottom of the atmosphere would be identical. The molecules moving upward have very slightly less energy that the average for their altitude (given the local temperature) and those moving downward have slightly more energy. As they move past each other, they will create a temperature gradient.
Can we lose the argument that the collision rate, i.e. pressure, has an effect on measured temperature. It doesn’t. A thermometer in contact with a gas at temperature T at low pressure will simply take longer to equilibrate than a thermometer in contact with a gas at the same temperature but higher pressure.
For a gravitationally bound atmosphere at constant temperature, the total energy content per cubic meter decreases exponentially with altitude. The density drops much faster with altitude than the gravitational potential energy increases. See graph here.
mkelly said, “Gravity has NO AFFECT ON TEMPERATURE.”
Everything is relative 🙂
http://redneckphysics.blogspot.com/2012/01/that-dang-cartoon-again.html
Hans Jellbring,
Here’s a plot of upwelling IR radiation measured over 24 hours at Desert Rock, NV by a SURFRAD station there. It looks to be more than 100W/m² to me. Note that the time axis is UTC. Desert Rock is -8 hours from UTC so local noon would be 2000 on the time axis.
There are seven SURFRAD stations in the US. You can access the data here.
Brent Hargreaves says:
January 20, 2012 at 1:51 am
I have considered that. The difficulty is the same as with CO2. In a noisy system like the climate, it’s hard to establish any effect of something which changes extremely slowly.
Regards,
w.
Geoff Sherrington says:
January 20, 2012 at 2:05 am
Thanks, Geoff. If you want a comment, please provide a link.
w.
John Marshall says:
January 20, 2012 at 2:13 am
John, here’s the problem. I can’t make sense of this. Who is “you”? Where is their argument about a “heat engine not working”?
Folks, quote or link to what you are talking about. I’m sure it’s crystal clear in John’s head what he is discussing. Out here, not so much. Quote it or link it or forget it.
w.
Joe Born says:
January 20, 2012 at 3:46 am
Not sure what that means, but let’s take a thermocouple … is it subject to gravity? Is it free of gravity? Where did I assume that there was a heat engine free of gravity? What are you talking about? Quote it or link it or it will be ignored.
If that were true, then we could pull an exceedingly small but non-zero amount of work from it … which would make it a perpetual motion machine.
w.
Johan i Kanada says:
January 20, 2012 at 4:09 am
You’d think so, but we have a couple of reputable physics professors posting in the thread and they are roundly ignored by the perpetual motion crowd. It’s a problem with “gravito-thermal” theorists. To believe that gravity can affect temperature, you have to have a weak grasp of physics, an unshakable belief in your correctness, and a willingness to ignore a bunch of folks who actually understand physics. That’s a bad combo.
w.
Alexander Harvey says:
January 20, 2012 at 4:36 am
Alex, I’m sorry but I don’t have time to wander through all of that. My suggestion is to boil it down, then boil it down again, and re-post it.
w.
Willis says
” To believe that gravity can affect temperature, you have to have a weak grasp of physics, an unshakable belief in your correctness, and a willingness to ignore a bunch of folks who actually understand physics. That’s a bad combo.”
Yet only last week Willis believed in an adiabatic distribution for a thermally isolated column of gas in a gravitational field.
This week he has ‘seen the light’.
Don’t be so hard on yourself Willis.
Willis,
You contend that no one can explain the Jelbring gravito-thermal hypothesis clearly, which means that no one understand the Jelbring gravito-thermal hypothesis, which is prime facie evidence that it is incorrect. Moreover, you indicate in this post that if the Jelbring gravito-thermal hypothesis were correct, it would contradict the second law of thermodynamics, another good reason for assuming it to be false.
But then you chide Qark for ridiculing the discussion, of something no one can explain.
I can only conclude that you are merely trying to keep the pot boiling for no reason but the fun of it. In which case, I think Qark’s contribution was as worthwhile as most others.
But if that is not the case, I should be glad of your response to this seemingly simple plece of logic, which seems to settle the matter:
At equilibrium, planet-wide outgoing radiation at the top of the atmosphere matches planet-wide incoming radiation at the top of the atmosphere. But if the atmosphere is transparent (i.e., without significant GHGs), outgoing radiation at the top of the atmosphere must be the same as outgoing radiation at the surface of the planet, which means that the mean surface temperature must be the same with or without an atmosphere.
If that is correct, it means that gravity does not account for the greenhouse effect.
If it is incorrect, for the reason Jelbring contends, i.e., that gravity causes an atmospheric temperature gradient that accounts for the greenhouse effect, then planet-wide radiation at the top of the atmosphere must exceed planet-wide incoming radiation, meaning that the planet is luminous. However, we know that the planet is not luminous.
QED
I’m probably joining this discussion too late to add anything useful. To my simple mind there are two things to consider which I have difficulty applying to the atmosphere
First, the diurnal bulge/atmospheric bulge is a bit like a pump which compresses, and as a result heats, the atmosphere and then expands and cools the atmosphere? I assume the heating, however small is greatest at the earth’s surface.
Secondly (as a result of all the references to one molecule at time) I remember when we did a short course on Information Theory http://en.wikipedia.org/wiki/Information_theory we had to work out the theoretical maximum temperature we could heat a teapot to with a fixed amount of boiling water, in order to make a decent cup of tea. http://englishtea.org.uk/how_to_make_tea.html. The maximum temperature is achieved when water is added and removed from the teapot 1 molecule of H2O at a time. As we are dealing thermodynamics, presumably Information Theory is relevant but rarely if ever mentioned
Despite being described as a “gravito-thermal theorist”, Nikolov & Zeller seem to have gone to extraordinary lengths to actually mention the word “gravity”. It appears in the text of their Unified Theory of Climate precisely twice.
Gravity, by itself, does not produce heat – surface temperatures on the Moon for example would be identical regardless of whether it’s gravity is one-sixth or six times that experienced on Earth. So to that extent at least Willis, you are entirely correct.
Now consider this thought-exercise :
Introduce an atmosphere equivalent in mass to that of the Earth to the Moon. The vertical structures of those atmospheres would surely be profoundly different according to whether the Moon’s gravitational attraction was one-sixth or six times that of the Earth.
Now consider two “Moons” with identical incoming radiation, identical atmospheric mass and composition, but atmospheres with profoundly different structures due to their differing gravitational attraction.
Identical surface temperatures or not?
Hope that helps……..
I skipped most of the comments in this thread, my apologies if this has been said already.
Willis,
You are not going to get the elevator pitch explanation of N&Z that you demand because N&Z relies not on a single paradigm shift, but on several. No elevator pitch can encompass all of them while still articulating a cohesive whole. If you want to understand N&Z, your going to have to allow it to be broken up into pieces, the paradigm shift of each piece understood, and then put the whole thing back together.
I suggest you start with the constant and continuous misaplication of SB Law as it applies to this discussion as I think that is the single major hurdle to get passed. Every thread I see this topic being discussed, I see the same claims. 240 w/m2 = 253K and 288K = 390 w/m2. These numbers are totaly and completely wrong.
SB Law is valid for a very specific case, and that case only. It is valid for a body at equilibrium exposed to uniform radiance and which is at uniform temperature. The earth is NOT exposed to uniform radiance, it is exposed to radiance which ranges from 0 w/m2 to over 1,000 w/m2 and which fluctuates wildly in both space and time. The temperature of the earth as measured is similarly not uniform, it varies in space (latitude, longitude, altitude, time of day, season of year, and orbital variance).
Until we drop the notion that we can arrive at meaningful numbers for earth temperature based upon averages of insolation and temperature compared via SB Law, there is no value whatsoever in discussing the balance of the N&Z hypothesis.
I note in closing Willis that your approach to questioning N&Z has been much more “attack mode” than anything else, and I think that is becoming a stumbling block in terms of having a productive discussion.
dmh
Robert Brown says:
January 20, 2012 at 9:34 am
Final overall comment and off to work.
“Jelbring’s hypothesis enables one to create a perpetual motion machine of the second kind to light stygia. The work done by their Carnot cycle engine and turned into light eventually turns back into heat, so the total energy of Stygia remains unchanged. It is just re-sorted by gravity acting as a Maxwell’s Demon into separated hot and cold reservoirs so that it can be used once again to drive the generator to make more light. The same energy is made available over and over again.”
Well, if you have read my paper I am not talking about an PM. Willis does so it his problem. I just want to know what is wrong with my paper. For your information an energy generating machin has been constructed to use the energy difference in the oceans. It will of course diminish the temperature difference as time is passing. I don´t see theoretical problem to do the same between the surface temperature and the temperatur at the top of Mount Everest. It is a practical problem of course.
In the case of my model atmosphere there is a temperature difference that can be used for energy extraction. That can only be done by moving energy outside the closed insulated atmosphere. In such a case energy will be removed from the inclosed atmospherea and its average temperature would sink. However such a machine is not allowed since the atmosphere was inclosed and no energy at all was allowed to enter or leave the system. You and Willis seem unable to recognise the assumtions that has been made in the text.
The theory tells that no energy can be extracted within a system that is at maximum entropy and that is also my opinion as long as we stay within the closed system. Willis is favouring a PM and I am not, he has been doing so for 8 years. To avoid more irrelevant comments I want to make clear that ALL content within the closed system consists of ideal gases as is told in the assumptions.
An Elevator Speech to prove isothermal result using two tubes of different gases.
The situation we analyze is a uniformly insolated ground in a spherically symetric system with some gravitational gradient. The ground is at a constant temperature T(r=ground). We erect TWO vertical tubes, insulated from the environment and each other, except at two points r=ground, and r=B. We put a different gas (ideal or non-ideal) in each tube. The only difference we require is that they have different specific heat values (Cpx and Cpy). The system must be at equilibrium by definition.
Let us assume that at equilibrium the atmosphere is NOT isothermic, but is a function of r. If so, there must be a real lapse rate in each tube. The Lapse rate (dT/dr) is a function of the Specific Heat of the gas(Cp(i)) and the gravitational accel (g). Both tubes experience the same g, but they have different Cp gasses. Therefore, there must be Difference Equilibrium Lapse rates in the two tubes. At point A, they can have the same temperature, but at point B they must have different temperatures. If they have different temperatures at B, then you can have heat flow at B, which means the system is NOT in Equilibrium. If not isothermal, then not in equillibrium if Cp’s are different.
Only if the gases are isothermal at all z, can the system stay in Equilibrium.
capt. dallas says:
January 20, 2012 at 9:51 am
mkelly said, “Gravity has NO AFFECT ON TEMPERATURE.”
It was actually Jeremy that said “Gravity has no affect (effect) on temperature.” I was asking him to put it in the form of the first law Q=U+W.
Tallbloke said:
I don’t think so Tallbloke. I have spent the latter half of my 60+ years on Earth discovering that what I “knew” is often enough false. That is, my ignorance has increased. Whenever someone says to me “you are ignorant” I take it as a compliment 🙂
That said, a very great deal of what I know has been well-tested and therefore I currently take to be true. In order to learn, we must continually test what we believe to be true.
Additionally, in order for any significant amount of comments on this thread to be not-ignorant we would have to discard the Law of Contradiction.
Embellishment to Rasey 11:09 Two tube example.
Modification 1: Both tubes are optically transparent, but thermally insulated. Fill them with different NON-greenhouse gases, different Specific Heats (Cp). Result must be isothermic.
Modification 2: Now fill one of the tubes with a strong Greenhouse gas. It is optically transparent so it should be absorbing and emitting lots of IR. By the same argument as 11:09, if that tube is not isothermal, then there will be different temperatures at B and non-equilibrium heat flow. So this GHG tube must also be isothermal. But how can it be so if the pressure of the GHG varies with r, and therefore its alleged Greenhouse capacity varies by r. How can it remain isothermal?
(shhh… unless GHG doesn’t matter?)
“His thought experiment is a planet with an atmosphere. The planet is isolated from the universe by an impervious thermally insulating shell that completely surrounds it, and which prevents any energy exchange with the universe outside. Inside the shell, Jelbring says that gravity makes the upper atmosphere colder and the lower atmosphere warmer. ”
This is true.
But two things.
One it is gravity which causes more density near surface and less density higher- that would be a difference of temperature.
Two: the atmosphere isn’t creating heat. But gravational body can create heat- it will have radioactive elements which create heat and with “an impervious thermally insulating shell”
there isn’t temperature limit, heat keeps building.
I suppose if got rid of all radioactive elements [or only used elements with extremely long half-lives] one would lower the amount of heat generated. I think a pure sample with no radioactive elements wouldn’t be easy to make or find- not sure if or what this could done- or if it would remain so.
Now if you have atmosphere so large it enter the realm of being a planetary type mass, then the atmosphere itself could generate heat. But Earth type atmospheres or Venus type atmospheres are insignificant in regard to planetary masses. Whereas neptune’s mass is mostly “ocean” and “air”- Neptune and other gas giants do have enough atmosphere to have planetary scale atmospheres.
So with this “impervious thermally insulating shell” one could get endless energy from a planet with an atmosphere.
Willis,
With some trepidation I offer this change of your thought experiment…
Assume a long, perfectly insulated tube. One end is anchored on a planet’s surface the other end extends to vacuum ( where it is also perfectly insulated). Assume the tube contains an ideal gas. Assume also that the planet’s surface gravity equal s earth’s (1 g ) and the planet’s radius is also equal to earth’s. No energy is added to the tube, and no energy escapes the confines of the tube. assume the density of the gas is equal to the the density of earth’s air at sea level. Last assumption: Assume the ideal gas is already at vertical (and horizontal) thermal equilibrium.
So the initial conditions are the tube is at the same temp for all altitudes and the density profile is similar to earth’s atmosphere.
Rhetorical question: What happens to the temp as a function of time, and altitude?
Here’s my answer: The temp says constant throughout the length of the tube forever.
here’s how I look at the situation…Consider a plane P at, say, 1Km above the surface (and parallel to the surface) that divides the tube into 2 sections, A and B. The plane does not hinder (or aid) the gas molecules.
In order for the temp to change so that there is higher temp in section A, the average kinetic energy of the molecules in A must somehow increase (that’s the definition of increased temp). However this would be equivalent to Maxwell’s Demon setting up shop at plane P, only allowing fast molecules into A, while only allowing slow molecules into B. Since that cannot happen, the temp in A can not change (also true for B).
I look at this Thought Experiment as your T.E. running backwards in time.
Thanx for all your postings. You are an inspiration.
Tom M.
quondam said @ January 20, 2012 at 6:57 am
Thanks for that insight, Quondam.
Johan i Kanada says:
January 20, 2012 at 4:09 am T
“This argument should be possible to resolve in 5 min by any reputable physics professor.”
Unfortunately, whatever this physics professor said, he would be immediately overwhelmed by lots of commentators here saying that it wasn’t true.
You have now entered the twilight zone, where idiotic comments far outweigh the occasional science that you may come across. The denizens of this zone do not believe in the greenhouse effect, because they don’t want to. They do not understand science (but that doesn’t matter) and so latch on to any passing pseudo- science – as promulgated by various anti-science blogs. As Willis Eschenbach said ” you are fighting basic ignorance of science, you will be deluged with ignorant people”. And in this twilight zone, ignorance swamps knowledge,
Having said that, the problem here is that not easy to solve. There is no empirical test which can resolve it. I think that different physicists could easily arrive at different conclusions based on mind experiments.
From an earlier post:
“The water at the bottom of the ocean is under great pressure. Does this make it warmer? No, we all know that warm water rises. The hotter water is at the top. You can verify that in your bathtub. So pressure does not cause warming. QED.”
Luke says this is because the oceans are heated from the top. But this wouldn’t explain why the water in a domestic hot water tank is hotter at the top. So, good guess Luke, at the obvious wrong answer, but null points.
Tallbloke says that “Not only that but water is incompressible”. The right answer! – have a cigar.
!
Robert Brown says:
January 20, 2012 at 9:11 am
Robert, I don’t provide long answers to condescending replies so I’ll just repeat my earlier request:
tallbloke says:
January 19, 2012 at 4:34 pm
Hi Robert,
I think the laws of thermodynamics talk about energy, rather than temperature or heat, but there are several formulations of them, so maybe we’d better discover who is using which definitions. We’d better do this, because in the application of classical mechanics to energy distribution in the model atmosphere, as defined by Hans Jelbring, there will indeed be a thermal gradient, as confirmed by Graeff’s empirical experimental data (Which should be replicated by an accredited laboratory).
And while your checking those out, would you be so kind as to consider my refutation of your ‘slices’ argument:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
Thank you.
I read the contribution by George Turner:
“I take your kilometer’s tall cylinder of atmosphere in thermal equilibrium and flip it over, like flipping an hour glass. It involved no input of work as its height did not change, so the column’s energy remains constant. But now the gas at temp T that was at the top has been wildly compressed, making it much, much hotter, while the gas at temp T that was at the bottom has been expanded, making it much, much colder.”
And Willis’ reply that there is more than half the (mass) of the atmosphere near the bottom. I was thinking as I read George that I would have flipped it at the centre point of mass in which case the objection does not apply. Flip a cylinder of atmosphere that is well above the surface about its centre of mass and allow it to float vertically (as it wll tend to do). The mass remains the same, the centre of mass will remain the same distance above the surface at all times even if the (‘mechanical’) flipping point rises or falls, the flipping requires no energy. The same compression and decompression mentioned by George applies and there will not longer be an isothermal state. Having now perturbed the initial isothermic condition, it will not re-establish itself even if perfectly insulated.
Why? As the atmosphere is not a solid, gas molecules will move up and down randomly. As a parcel of molecules rises the temperature will drop due to expansion. Yes, the potential energy with respect to the surface rises, but the temperature drops and this is about temperature which is a measure of only one type of energy. The thesis states that there will be a difference in temperature, top and bottom, it does not make claims about a difference in total energy with respect to the surface of a gravitational object. Work accomplished heating a falling molecule has to be accounted for by cooling taking place elsewhere in the system. It cannot come from the bottom of the system.
One can make an argument that the total energy in any molecule is constant at all heights but I doubt that applies to any molecule in a spherical atmosphere. If it applies to a particular sphere (because of a perfect balance between sperical expansion and elevation) it cannot also apply to a ‘flat planet’ with a cylinderical sample of atmosphere because the volumeetric expansion rate is different (trumpet v.s. clyinder). Because it cannot be correct for both, I suspect it is generally correct for neither save in special cases.
From an initial isothemal state all molecules falling will warm and all molecules rising will cool. The heat will accumulate at the bottom. There is no work done. Gravity does not do any work. The heat will shift towards the bottom because gases move without input of energy and cool or heat adibatically.
The idea of using the temperature difference between the hot and cold zones to generate power is a good one. It will work, but will cool the whole atmosphere in the process. It was written above as if all the heat from below would be vented near the top of the atmosphere, therefore that would allow the heat to migrate down again and this be recycled, but this description is incomplete. Energy extracted in the form of energy would be subtracted from the heat processed, providing net cooling by exactly the amount of the energy extracted (including friction). So it would work for a while, converting the atmosphere’s heat into electrical or mechanical energy. It is exactly the same as using the temperature differential between the surface and the ocean deeps to generate electricity. It works (quite well) but it cools the ocean in the process and it is not perpetual.
In the Elevator:
1. Gas molecules move about with no external input of energy. Therefore they may rise or fall in the atmosphere over time.
2. It is well established that gases cool when decompressed and heat when they are compressed. In a free atmosphere above a spherical, insulated planet, molecules are free to compress and decompress as they move vertically. There is no work performed during this process because they are drifting in free space. The total heat content of the system remains constant.
3. If an atmosphere was initially isothermal, the higher elevation gases would have to have been artficially heated to be at the same temperature while also being in a decompressed state. The lower gases would have to have been artificially cooled to be as the same temperature and simultaneously in a compressed state.
4. As the gases move randomly about the vertical column, the hot molecules dropping from above will carry heat downwards, increasing in temperature above the initial average. Lower gases rising will cool by adiabatic expansion below the initial average. There is no net heating or cooling in the atmosphere.
5. After a time, when equlibrium is established (no net change in temperature of any horizontal slice of the atmosphere) it will be warmer at the bottom and colder at the top. The direct cause for this temperature difference is gravity and the effect is observed because of the size of the system (large) and the lack of any external, gravitational perturbations.
6. If gravity is increased, or if the total volume of atmosphere is increased, the equilibrium temperature at the bottom will increase. An average temperature can be calculated, as can an elevation at which this temperature will be observed.
Very interesting discussion Willis, and nicely done.
The only thing that gets me is the idea that the gas would be isothermal (temperature being kinetic energy) violates the conservation of energy. As a gas molecule moves away from the gravity source, it MUST lose kinetic energy to potential energy. As it falls back down, it’ll convert that potential energy back to kinetic energy. Ergo, energy is conserved. But -temperature- is the observation of kinetic energy, not potential energy.
In that regard, a higher elevation must be cooler in -temperature-, even if the heat content (total energy) is the same. I think that’s the problem. We conflate temperature with heat, when they are not the same thing.
You can have two objects at different temperature, but if they are equal in heat, so that energy cannot transfer one way or the other, this temperature difference must be maintained. And we can do this by turning kinetic energy (temperature) into potential energy. No heat is lost.
I’m not defending the gravity thermal theories, but I am expressing how I cannot understand this thought experiment in terms of energy conservation. And I wonder if the ideas about heat and temperature have gotten crossed. Or maybe I am fundamentally misunderstanding?
But still, raising up means losing kinetic energy to potential energy, and temperature is kinetic energy not potential energy. Heat transfer can only happen through kinetic energy, not potential energy… So to conserve energy, temperature must decline while raising above a gravity source… Likewise, if you stand in a gas close to the gravity source you’ll get hit by more collisions than if you sit above where more energy is locked in potential energy. And, if you used your body’s temperature (kinetic energy) to push you upwards against gravity, you’d cool your body down as more energy was put into potential, right? You’d still have the same amount of energy in you, but you’ve transformed it. That’s the only logical conclusion I can come to. Isothermal (temperature wise) while moving in a gravity well doesn’t make sense to me at all, and nor do we ever see that in nature in atmospheres.. to my knowledge.
Again, maybe I am missing something.
I have to disagree here. It is possible to make energy between the ground and the top of the atmosphere because the ground is warmer than space from the point of view of the atmosphere. The atmosphere does not see the Sun because it does not absorb in the UVs. So space is very cold for it. If there was a blob of nitrogen in orbit around the Earth, it would be cold. So it would be possible to make energy between this blob of gas and the ground. Don’t forget, the ground is warmer than space in the IR.
Think about a one molecule atmosphere. It receives or gives energy to the ground every time it hits it. The vertical kinetic energy of the molecule near of the ground can be different every time. But each time, it slows down as it goes up and then it speeds up when it comes back down. If you make an average over multiple hits to the ground, you will find these 2 facts:
1- It is more often near of the ground(higher pressure at the ground level).
2- It has more kinetic energy near of the ground(Higher temperature at the ground level).
If you increase the number of molecules, each of them will still lose kinetic energy as it goes up.
Now, let’s make a little thought experiment. Let’s imagine an invisible layer around the Earth at a certain altitude. The number of molecules passing from under to over the layer has to be the same as the number of molecules going the other way at equilibrium. The amount of kinetic energy also has to be the same at equilibrium. Now let’s put a second layer very near of the first to look at the derivative. What’s been said in the first part of this paragraph still holds true. But sometimes, a molecule will pass through the lower layer and come to vertical halt between the layers and it will go back down through the lower layer without passing the higher layer. This is where the differential of pressure appears. Every time a molecule passes through both layers, it either loses kinetic energy on its way up or gain kinetic energy on its way down. If 2 molecules hit each other, you have to look at the pair of molecules. The center of mass can turn around between the two layers, pass the two layers in an upward or downward motion. And the pair of molecules will act like a single molecules if you calculate the differential of pressure or the differential of kinetic energy between those two layers.
John Marshall says: “I also ask my Jupiter question again. Why does this gas giant radiate more heat than it receives from the sun. your argument above makes this impossible.
My astrophysics instructor said he believed there were decaying fissionables at the core.
Stephen Rasey says:
January 20, 2012 at 11:09 am
“An Elevator Speech to prove isothermal result using two tubes of different gases.
Let us assume that at equilibrium the atmosphere is NOT isothermic, but is a function of r. If so, there must be a real lapse rate in each tube. The Lapse rate (dT/dr) is a function of the Specific Heat of the gas(Cp(i)) and the gravitational accel (g). . If they have different temperatures at B, then you can have heat flow at B, which means the system is NOT in Equilibrium. If not isothermal, then not in equillibrium if Cp’s are different.
Only if the gases are isothermal at all z, can the system stay in Equilibrium.”
I certainly don´t believe that an elevator speach is the way to perform scientific work.
You are free to express your opinion whatever it is but in this case it is not based in scientific reasoning and definitly not based on first principle physics.
The adiabatic temperature lapse rates that will develope (if long time enought will pass) in your examples are given by dT/dZ = -g/Cp1 and the other will be given by dT/dz= -g/Cp2. A remarkable fact is that both these laps rates are independant of the absolute amount of energy that was inclosed in each tube. In both cases the energy in two arbitrarily chosen equal mass units will carry the same amount of energy. Remember that the 1:st a and 2:nd law of thermodynamics apply to energy, not to temperature. When gravity is not included the second law can be formulated in terms of temperature even if it is not recommended (but very common in anglo litterature).
davidmhoffer says:
No…They are not “totally and completely wrong”. The 288 K = 390 W/m^2 is in fact quite accurate for the Earth’s present temperature distribution to a good approximation (introducing about 2K of error with a generous estimate).
The 240 W/m^2 = 255 K is a bit dicier since it is for a hypothetical world that is a little hard to specify and could have a somewhat broader temperature distribution. However, what can be said rigorously is that a blackbody radiating an average of 240 W/m^2 could have an average temperature no higher than 255 K. It could have an average temperature lower than that (which is what N&Z show with their T_sb calculation that assumes an extremely broad temperature distribution since the local instantaneous temperature is determined by radiative balance with the local instantaneous solar insolation). [To the extent that the Earth is not exactly a blackbody in the mid- and far-infrared, the average temperature could be higher, but for the terrestrial surface, we are talking about 3 K at about the outside limit. And, the fact that the “non-uniform effect” and the “not exactly a blackbody effect” go in opposite directions means the two effects tend to offset each other to some degree.]
By the way, for those who want a more precise quantification, I have derived a formula for the difference between the fourth root of the average of T^4 and the direct average of T. The formula is that this difference is given by
difference = (3/2)*(sigma^2)/T_ave
where sigma is the standard deviation of the distribution of T and T_ave is the average value for the distribution (in absolute temperature units, like Kelvin). [Although this formula is an expansion good for small sigma/T_ave, I have verified that it gives answers within 2% of the actual difference for two distributions (Gaussian and flat “box” distributions) for sigma/T_ave < 0.10 (which is larger than any reasonable estimate I can make of the relevant value for Earth).]
Dave, we all know that you have become completely obsessed with this particular issue…but you have failed to refute both my and Willis's rough quantifications of its magnitude and hence you are continuing to "make a mountain out of a molehill" on this subject.
ShrNfr says onJanuary 19, 2012 at 4:03 pm :
“However, we do have a real heat source in the earth’s core with the fission of heavy nuclei. Not perhaps a lot, but some. I never have gotten any really good estimates of how large the effect is, and I am not enough of a geologist to derive it. Anyone around have an idea??”
===
If it can help – my advice to you – is to go back in history and look up where it all originates and how “Global Warmists” or “CAGW” alarmists get it wrong, even right from the start, with written history.
The 18th and 19th Centuries saw many famous Philosophers/researchers. (They did not all live long enough to be called Scientists – as that title is fairly new)
Let me quote, if I may, Timothy Casey B.Sc. (Hons): Consulting Geologist who has put the papers of some of them online: “According to Weart (2003, Flannery (2005) and Archer (2009) the “Greenhouse Effect” originates with Fourier, ——.” – And further on: “Arrhenius claimed:
Fourier maintained that the atmosphere acts like the glass of a hothouse, because it lets through the light rays of the sun but retains the dark rays from the ground.”
Nothing could be more wrong – or further from the truth – Put “Fourier (1824) as translated by Burgess (1837)” into your “Computer search engine” or http://geologist-1011.net and see what comes up. You will learn a lot – including, maybe, why davidmhoffer says what he says on January 20, 2012 at 10:49 am.
Reading Fourier’s paper from 1824 really is worth the effort and Timothy Casey B.Sc is a pretty good teacher too.
O H D
Oh, I think I should also point out that I was ignoring radiative transfer for heat, since that isn’t part of the thought experiment. But that necessarily energy is lost to space as radiation which cools everything over time, and without energy input, there will be no energy to propel gas molecules up against gravity. So then, after a certain amount of time, the entire atmosphere will settle down upon the surface with no more “bouncing” molecules jumping around.
But since the sun warms our surface, it acts to “bounce” molecules upwards, where they bounce against each other, etc, and act as a kinetic battery for storing energy. But again, as a ball, or molecule, bounces upwards, it loses energy and slows down (temperature decrease if temperature is the only thing driving this as it is in the atmosphere of our thought experiment), then plummets and warms back up, bounces again, and so forth. It’s just like watching balls bounce around a vibrator machine.
So… it seems to be you will have a thermal gradient, no matter what! It’s just will this gradient make apparent temperatures in the atmosphere at the surface higher than the S-B surface temperature?
Again, not defending these gravity-thermal theorems, but I do fully believe that gravity creates and maintains a thermal gradient for altitude above the surface (this also ignores the increasing volume of the sphere as we raise above the surface which decreases temperature and/or pressure).
I dunno, those are my thoughts, Willis. I actually think Dr. Brown was wrong then.
And could you generate energy from the atmosphere using a thermocoupler? Of course! But it would be ENORMOUS to have to reach high enough from the surface up to altitudes where the temperature drops (30,000 feet is pretty dang cold 😉 ).
Stephen Rasey good post.
For your first proposed experiment if you use argon and hydrogen you should get about 2K difference in 100m tubes which should be reasonably practical and measurable.
I like even better your second proposed experiment.
….” But how can it be so if the pressure of the GHG varies with r, and therefore its alleged Greenhouse capacity varies by r. How can it remain isothermal?
(shhh… unless GHG doesn’t matter?)”…..
Just goes to show that for isothermal/adiabatic distribution nothing can save the greenhouse effect.
I’ve always loved the ingenuity of those who tried and in some cases succeeded”
http://en.wikipedia.org/wiki/Beverly_Clock
A six-degree Celsius temperature variation over the course of each day creates enough pressure to raise a one-pound weight by one inch (energy extracted = .11 joules), which drives the clock mechanism.
Just ignore the next line in the description that says “It is not therefore an example of perpetual motion.”. Okay it is “naturally assisted” perpetual motion but still a darned impressive piece of work providing real output.
There were earlier clocks as well that used this principle.
Willis Eschenbach: “Where did I assume that there was a heat engine free of gravity?”
I believe you did so without realizing it. But perhaps a better way to express my meaning would have been to say that you have begged the question by assuming a heat engine could be designed that could perform work operating between the differences in altitude and temperature that Velasco et al.’s lapse rate specifies.
My contention, which I at least for the moment believe Velasco and his colleagues have established through a statistical-mechanics-based proof, is that the maximum-entropy configuration of the system you describe exhibits a small but non-zero temperature lapse rate. If that is true, then isn’t it true that a heat engine operating between the same differences in altitude and temperature would have to reduce entropy to operate? If not, then perhaps you could reveal the heat-engine design you have in mind. Otherwise, you’ve just allowed invocation of the term “heat engine” to short-circuit further analysis.
I have to confess that two weeks ago I was arguing over at tallbloke’s place for the same position you’re taking now: http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/#comment-12926. Currently I think I was wrong then and that you’re wrong now. Maybe by tomorrow you and Dr. Brown will have convinced me otherwise.
Bart,
Your solution of the heat equation in spherical coordinates ( ∂T/∂t = αΔT ) is wrong. As you point out above, α is not a constant with r. α is the thermal diffusivity = thermal conductivity, k (which is a constant), divided by the product of the density, ρ, and the heat capacity at constant pressure, Cp = k/(ρCp). But density decreases exponentially with altitude so the diffusivity increases exponentially with altitude. That means the heat equation must be derived again with this condition and solved. According to the Wikipedia page on the heat equation
Then you say:
You can’t have it both ways. You have a logical contradiction. You say the temperature must always decrease with altitude but then you say the temperature increases at altitude basically without limit. You also can’t wave your hands and say the surface temperature also increases without limit because it’s warmed by the atmosphere. It can’t be by your condition that the temperature gradient is always negative (positive lapse rate). So what you’re actually saying is that the surface temperature increases without limit. But that can only happen if the emissivity of the surface is zero, or at least very small. But of course, it isn’t. In Willis’ example, it’s defined as equal to 1. You can also assert all you want about the SB equation not applying, but you have given no evidence or citation that it doesn’t. Even if the emissivity does vary with wavelength and angle, one can still integrate the Planck equation, B(λ,T) and get the total emission. The SB equation will still be a very good approximation for most real materials.
Oh, and molecules at high altitude are not in orbit even if they are moving at the same radial velocity as the surface below them. The correct reference frame for determining the RMS velocity is a rotating spherical coordinate system with the rotation rate equal to the rotation rate of the planet. A molecule that was actually in orbit would indeed have very high kinetic energy relative to the molecules around it. If that were not the case, then satellites in low orbit would not be subject to atmospheric drag.
Bryan says:
January 20, 2012 at 10:24 am
Since I obviously don’t have an unshakeable belief in my correctness (I publicly admitted I was wrong), and since I obviously listened to the ideas of Robert Brown, who actually understands physics, I fear your reading skills appear to be in desperate need of improvement …
w.
PS—This week, I learned something. I expanded my understanding of physics. You can disparage that all you want, by calling it “seeing the light”, but to me, learning is invaluable gold, and your childish sneers can’t tarnish it. You ought to try it yourself, Bryan, you might like it.
CanSpeccy says:
January 20, 2012 at 10:27 am
Quote my words, you are off in fantasy. I gave a “elevator speech” laying out the Jelbring hypothesis, which demonstrates that I understand it even if you don’t.
In addition I have never argued that it must be incorrect if no one understands it.
QUOTE MY WORDS that you disagree with, CanSpeccy, I haven’t a clue what your are talking about.
w.
jorgekafkazar says:
January 20, 2012 at 12:02 pm
John Marshall says: “I also ask my Jupiter question again. Why does this gas giant radiate more heat than it receives from the sun. your argument above makes this impossible.
My astrophysics instructor said he believed there were decaying fissionables at the core.
Was he waving his arms in a wiggle matching sort of way when he said it? 😉
davidmhoffer says:
January 20, 2012 at 10:49 am
You can believe that if it helps you to sleep. Me, I’m still waiting for someone to give a clear explanation of N&Z. Your claim, that it cannot be explained simply because its all too complex, just tells me that you don’t understand it.
But heck, let’s play it your way. You say you can do it if we break it up into pieces. OK, give me the elevator speech for the first piece of the lot, we’ll start with that.
w.
Willis says: “Thanks, Jeremy. You are a hundred percent correct, gravity can’t do ongoing work to change the temperature.”
Gravity does not do work it causes work to be done.
Q= U + W (first law)
W=FD
F=ma
a=g (g is gravity)
W=mgD
W=PdV
This is stated simply on purpose.
It is the gas that is doing the work against the force of gravity else why would a heated column of air stop rising. What the magnitude or if it is? Or is it offset somewhere else is a different story.
Bart says:
January 19, 2012 at 11:06 pm
Look at the earlier thread. I’ve got it nailed. There is no doubt about it.
Such hubris, actually you’ve got it wrong, but refuse to acknowledge it.
Your explanation contains a basic thermodynamic error as shown below (again).
F) the transferred heat accumulates in the atmosphere until:
1) highly energetic emissions are stimulated, which balances the energy fluxes all around in the same way GHGs would in the standard “greenhouse” theory
OR
2) the heat accumulates until the atmosphere achieves escape velocity and vanishes.
This is where you introduce the fundamental error in your analysis.
What it should say is that ‘the heat accumulates in the atmosphere until the layer of gas nearest the surface reaches the surface temperature at which point heat transfer ceases due to ΔT=0.’ There can be no more heat transfer to the atmosphere unless it cools down and ΔT again exceeds 0. This is the fundamental flaw in all the ‘hotter and hotter’ arguments (it’s independent of the profile with altitude).
Hans Jelbring says:
January 20, 2012 at 12:15 pm
Hans, like you I “don´t believe that an elevator speech is the way to perform scientific work.” So we are in agreement on that.
What it is useful for, as I have said before, is three things:
1. It helps me to clarify my own thoughts. When I only have a few sentences I am forced to be clear.
2. It helps me to explain ideas to other people.
3. It helps me to determine if someone else understands something, under the principle that if you can’t explain it clearly, you don’t understand it clearly.
You have claimed in the past that no “elevator speech” for your theory could be any shorter than your original paper. And indeed your paper is concise … but a much simpler explanation could be given.
I have given my simple explanation, my elevator speech for your hypothesis, in my head post. If you think any of the statements is wrong, please let me know a) which one is wrong, and b) what’s wrong with it. If you think a logical step is missing, please provide it. Blanket statements and denunciations will not serve in this case, please avoid them.
My best regards, Hans,
w.
@Hans Willis is NOT indicating the heat engine will REMOVE energy from the system! He is indicating that the temperature difference used to run the heat engine can do work on the fluid in the system, and infinitely do so per the theory. Hence the theory fails.
Bart,
I’ve calculated the thermal diffusivity, α, of air as a function of altitude at a temperature of 255 K.
km α(m²/s)
0 1.65E-05
10 6.30E-05
20 2.39E-04
50 1.28E-02
100 8.91
For reference, the thermal diffusivity of pure silver is 1.66E-04m²/s. You can see why the IPCC allows the stratosphere to equilibrate before calculating forcing. The equilibration time is going to be short.
If I’m really bored sometime, I’ll try to calculate numerically the temperature profiles for a gravitationally bound vertical column of atmosphere 100 km high with an initial temperature of 2.7 K for a monatomic, non-condensable ideal gas with a heat flux into the surface of 240 W/m² and a surface emissivity of 1.
Werner Brozek says:
January 19, 2012 at 7:30 pm
That’s the problem – outer space has no temperature. Temperature has meaning for substances (stuff), not for a vacuum. The only way to transfer heat to space is via radiation, and the only way for the atmosphere to loose heat to space is to have greenhouse gases that radiate energy.
Hans Jelbring says:
January 20, 2012 at 11:03 am
And therein lies the problem, that you’re talking about perpetual motion and you don’t know it. Hans, you are indeed proposing perpetual motion. You are saying that gravity separates the warm and cold molecules.
If that were true, we could use the temperature difference to perform work forever in your thought experiment world which is sealed off so no energy comes in or out. How is that possible?
Either you need to:
a) show that there is no problem with a sealed system performing continuous work with no energy inputs or exchange with the outside world, or
b) show that for unknown reasons, we could not use the temperature difference to perform work.
I see no reason that we could not extract work from your claimed temperature difference, we’ve been doing that for hundreds of years. So if gravity creates a temperature difference, the default assumption has to be that we can extract work from that temperature difference.
But if so, you could get unceasing work out of gravity forever … and while that would be a nice trick, no one has ever done it, there is no physical theory saying it is possible, and the laws of thermodynamics say it won’t work.
That’s why we call it a “perpetual motion machine”. If you think not, then you will have to do either a) or b) above. I see no way to do either.
In fact, your thought experiment is an excellent one to show that gravity can’t possibly affect the temperature of an atmosphere such as you postulate. If it could, you could use the gravity-created temperature difference to light Stygia forever, and you’d be a hero for solving the energy crisis, we could just run the world off of gravity …
Thanks,
w.
Willis Eschenbach says (in response to davidmhoffer):
Oh…This sounds like fun! I volunteer to give the elevator speech for Section 2.1)B:
“In Section 2.1)B, N&Z demonstrate that when you add convection to a simple radiative model of the greenhouse effect in such a way that it drives the atmosphere to an isothermal temperature distribution with height, then you no longer have a greenhouse effect. Unfortunately, however, in the real atmosphere, convection only drives the atmospheric distribution as far as the (appropriate) adiabatic lapse rate and the greenhouse effect does not disappear. That the temperature at the ‘effective radiating level’ must be colder than the temperature at the surface in order to have a greenhouse effect was already well-known.”
O H Dahlsveen says:
January 20, 2012 at 12:16 pm
Let me quote, if I may, Timothy Casey B.Sc. (Hons): Consulting Geologist who has put the papers of some of them online: “According to Weart (2003, Flannery (2005) and Archer (2009) the “Greenhouse Effect” originates with Fourier, ——.” – And further on: “Arrhenius claimed:
Fourier maintained that the atmosphere acts like the glass of a hothouse, because it lets through the light rays of the sun but retains the dark rays from the ground.”
Nothing could be more wrong – or further from the truth – Put “Fourier (1824) as translated by Burgess (1837)” into your “Computer search engine” or http://geologist-1011.net and see what comes up.
Class. I’ve reposted the Fourier translation, thanks muchly.
“It is equally probable, that in respect to most of the planets, the temperature of the poles is little above that of the surrounding space, with respect to the temperature which each of these bodies owes to the sun, it is not known; because it may depend on the pressure of an atmosphere and the condition of the surface.”
Geez, folks here move around more than ideal gas molecules > 0K. Including me.
Willis says 1/20 10:49am: “Me, I’m still waiting for someone to give a clear explanation of N&Z.”
Robert Brown says 1/20 9:34am:
*” Fact 1: One can run a heat engine between any two reservoirs of energy maintained at different temperatures. Proof: Every heat engine in the world, all of thermodynamic theory, massive engineering…
Fact 2: Heat engines cannot run indefinitely. In a closed system they cannot just take random energy in a complex environment and continuously turn it into work….
* Assertion Gravity sorts air in an adiabatically isolated environment out into hot air at the bottom and cold air at the top. This arrangement is thermodynamically stable and will spontaneously occur and be sustained.
* Argument If the assertion were true, then due to Fact 1, a heat engine placed in the container and run between the top and the bottom would run forever. As fast as it made the air at the top warmer, the heat would somehow “fall” back to the bottom, re-creating the thermal gradient that we know can drive all sorts of heat engines. This violates Fact 2.”
Trick says “No, this heat engine won’t run forever”. My view: the N&Z assertion IS true since it can be shown to comply with natural laws. Can Willis’ find a fail? Here’s my view:
As Robert Brown writes, the air container in question is an isolated closed system so by Robert’s Fact 2, that heat engine cannot run indefinitely & thus it will run to control volume (cv) equilibrium and stop. The heat can’t “fall” once in equilibrium. There is no way to perpetually “maintain” the non-equilibrium unless the heat reservoirs are infinite & they are not. Eventually, in the same control volume, every real hot & cold irreversible heat engine equilibrates the two non-infinite hot & cold “fuel” reservoirs and stops making work – up until the reservoirs are “maintained” i.e. replenished across the cv. Like my refrig. with the electricity off. Gotta’ plug it in.
In the N&Z thought experiment case at hand viz. a tall adiabatic (occurring without gain or loss of heat across cv) control volume of GHG-free air i.e. gas in the presence of gravity will reach equilibrium and the heat engine will stop. No way can it be made to run except by being an outlaw. This occurs in the non-outlaw earth & near earth planets.
There are at least three important laws applicable to understand the tall air column cv of height h in the presence of gravity, there may be more, but each of these must ideally operate at same time & are sufficient to arrive at equilibrium in the cv of interest & some conclusions:
1st law: conduction always operates from high T to low T objects made of normal matter & they touch, to equilibrium.
2nd law: energy is always conserved & usually written PE + KE = constant (here P*V = KE, ngh = PE).
3rd: ideal gas law: PV = nRT (where in the cv of interest we hold V, n, and of course R constant)
So here’s what the 3 natural laws together mean will happen at equilibrium in the cv: the inexplicable gravity field will drive the tall column of gas to stratify itself with higher pressure at the bottom (2nd law: P*V + 0 = const. at h=0) due to the “weight” of the gas above (ngh). This is sustainable in adiabatic cv.
At column top, there must be lower pressure written by 2nd law also: P*V + ngh = same const. P*V is thus lower and with V constant, it is pressure P that has to be lower. No way around it, fight the laws all you want (I know y’all will try).
That means if you put a thermometer at h>0, it will show temp. T at h>0 . Put the same thermometer at bottom (h=0) read: T + delta T. The delta T from increased KE which has to be larger since V is constant to get (P*V + 0) = same const. = (P (at h>0) * V + ngh). Try that with P0 and Ph subscripts for practice. You can do it Willis.
The temperature (KE) of the atmosphere reduces as h (really ngh PE) increases above 0 for earth & nearby planets. N&Z stands. AND with all three laws operating from top to bottom, we’ve achieved closed system equilibrium so no heat engine can operate Robert: Fact 2 n’est ce pas? Even with T at top and T+delta T at bottom, there is equilibrium with all 3 laws. Go figure.
PS: Somewhere in there is an elevator speech to understand N&Z, I have one already posted but will leave it up to the reader to check their N&Z understanding.
Note: the density anywhere in the air column control volume is given by ideal gas law rearranged or density = n/V = P/RT so at any h in the cv, density can be computed from measuring P and T at that h, same time.
Got that? Move on….ha. Never happen, let entropy calmly increase. I dislike THAT entropy law: more reading.
Lets be clear a closed system with an atmospheric temperature gradiant present could exist indefinitely without breaching thermodynamic laws.
However, that is only possible if absolutely no work is being done at all. If work is being done, and I cannot imagine a system, with a temperature gradiant, that could prevent work being done, then entropy must increase.
If work is being done then in the long run you are creating energy somewhere or you are preventing the entropy of the system from increasing. That is impossible and surely obvious to any reasonably intelligent person.
However, in an open system we have the question, can gravity affect the temperature of Earth on a continual basis. Well we know gravity can certainy affect temperatures in a system at least once.
Take a difuse cloud of particles and wait for gravity to compress them and you will certainly see a localised increase in temperature of the system. Not only that the gravitional effect is the ultimate cause of something like the Sun continuing to increase its temperature for many billions of years. So you would need to define the statement ‘once’. If ‘once’ is taken as the period between the life and death of the Universe then that would be continuously for ever.
We know that Jupiter is radiating more energy than it receives and this is ultimately due to the gravitational effect. This process has been going on and will continue to go on, for billions of years.
Is this just ‘once’ or continuous? It depends on your definition of once.
In addition is any of this gravity induced temperature difference helped along by the addition of energy from the Sun moving particles to a higher potential energy within the system making more energy available to the gravitational effect, which we know for sure increases the temperature of the planet. It must do, the exchange of energy must be producing work in the atmosphere.
So the question is, given that the Earth also receives energy continuously from the Sun, does this mean that there is a continuous temperature increase in the Earth’s system caused by the effect of gravity.
Why wouldn’t it? It is just expanding the period of ‘once’, which we all agree gravity can iat least increase the temperature of a system by.
Alan Millar
Hans Jelbring says: January 20, 2012 at 11:03 am
In the case of my model atmosphere there is a temperature difference that can be used for energy extraction. That can only be done by moving energy outside the closed insulated atmosphere. In such a case energy will be removed from the inclosed atmospherea and its average temperature would sink. However such a machine is not allowed since the atmosphere was inclosed and no energy at all was allowed to enter or leave the system.
======================
The average temperature may fall but the thermoelectric effect relies on temperature difference NOT absolute temperature.
Therefore, providing your hypothesis that the adiabatic lapse rate is fixed (10K/km), then your theory predicts that the temperature difference between top of column and bottom of column will be maintained. Therefore it is possible to extract the same power from the column for ever (or at least until the lower electrode reaches absolute zero!
Louis Hissink says:
January 20, 2012 at 4:46 am
Thanks, Louis. Very little is known about the effect of electricity on the climate, despite its obvious involvement (lightning, charge separation in clouds, auroras, the ionosphere, sprites and jets, cloud nucleation, the list goes on).
For this topic, however, we’re away from electricity. I do have a keen interest in the question, and I think it is an important one we’re just starting to investigate but this thread is about the Jelbring hypothesis, a very different question.
Thanks,
w.
MarkW says:
January 20, 2012 at 5:11 am
Let me stop you there and say that a) space has no temperature and therefore b) the top of the atmosphere cannot be “the temperature of space itself”. Since your exposition clearly depends on those errors, you’ll have to start again.
Sorry,
w.
steveta_uk says:
January 20, 2012 at 5:13 am
Since gravity has no effect on temperature, why would the shape of the container have any effect at all? The shape of the container, whether a cone or a cylinder or a pentagram, doesn’t have some magic ability to mystically give gravity the power affect temperature. Well, maybe the pentagram …
I’d say you have a problem in your assumptions, steve.
w.
Willis,
This explains why no one can give a clear explanation of quantum mechanics. According to Feynmann, nobody understands quantum mechanics.
capt. dallas says:
January 20, 2012 at 5:20 am
I’ve not usually seen that assumption made. In general, conduction of both sensible and latent heat is not “assumed negligible” unless it is in fact negligible for the particular phenomenon under study.
w.
Joel Shore;
Dave, we all know that you have become completely obsessed with this particular issue…but you have failed to refute both my and Willis’s rough quantifications of its magnitude and hence you are continuing to “make a mountain out of a molehill” on this subject.>>>
Get off of it Joel. I keep pointing out that the upper bound is not the important one to understand, it is the lower bound, and you keep coming back with noise about the upper bound. I’ve even provided sample calculations showing what a realistic estimate of the EFFECTIVE black body temperature of earth should be based on an “average” of 240 w/m2 which you have studiously avoided.
If you calculate a realistic surface temperature for earth based on actual variance of insolation, you arrive at a value of about 140K. That is the EFFECTIVE black body temperature of earth. 253K can only happen if insolation is 100% uniform and with half the earth being 0 w/m2 at all times, that notion is totaly ludicrous.
140K is far more realistic. If I accept for the moment that the earth average is anywhere near 288K, that leaves one looking for an explanation of a temperature increase of nearly 150K, not the 33K we keep on seeing quoted.
As I have asked you previously, stop modeling the error in the 288K number and start modeling the effective black body temperature of earth based on insolation varying from 0 to 1000 w/m2 on a daily day/night basis and on a tropics to poles basis. Show me what a realistic EFFECTIVE black body calc for 240 w/m2 “average” should be. Why are you afraid of coming to grips with this number? How can we possibly have a discussion of the physics required to raise surface temperatures by some amount due to the GHE when we start the discussion with a GHE number that it totaly, completely, and ridiculously, WRONG?
Willis Eschenbach;
But heck, let’s play it your way. You say you can do it if we break it up into pieces. OK, give me the elevator speech for the first piece of the lot, we’ll start with that.>>>
Already did. Several times.
John Mason says:
January 20, 2012 at 5:56 am
Despite repeated calls, no one has yet given the “elevator speech” about how N&Z actually works. Someone above said it’s all far too complex to be explained simply … riiight.
My point is that I don’t know how the N&Z model works, and if you do, please give us the elevator speech so we can all understand how it’s supposed to go.
w.
PS—I also don’t know what you are calling a “delay model”.
RE: Jelbring at 12:15 pm
You criticize my use of elevator speech (or verbal proof, take your pick) without addressing its conclusion. You say that there should be two different lapse rates, then go on to say something about equal masses. I never specified equal masses, in fact with two different gasses, with different Cp and probably different densities, there is no reason to expect equal masses or densities at intervals r from A to B.
You agree with me that there will be different lapse rates, so therefore, there can be only one point C along the tubes where they have the same temperature. At all other points (not C), the temperatures will be different between the tubes. Different temperatures means heat flow at the window at point B. Heat flow is not equilibrium, therefore there is a contradiction. Modus tollens.
The tubes must remain isothermal, same temperature at all r.
It problem is the uniform insolation, dead planet, without day-night, assumption. There isn’t a planet in the universe above 3 deg K that fits that assumption. Because planets are not uniformly heated, have a night and day, some deg of radioactive internal heating, they have non-zero real lapse rates, so we feel that it must also be so in the dead planet case. That is was the bugaboo that haunted me. Until I realized that uniform insolation requires a uniform radiator at distance at the same temperature.
My take away from these long threads this week is that the uniform insolation case requires an isothermal atmosphere (zero real lapse rate ) regardless of composition. Isothermal atmosphere is worthless to study in climate science. Therefore uniform insolation cases are worthless to study. [Watch the fur fly now!]
JJThoms says:
January 20, 2012 at 6:03 am
Quite true, many thanks.
I first read about the Hilsch Vortex Tube in the “Amateur Scientist” column back in the sixties when “Scientific American” used to actually be about science. I machined one up a decade or so ago out of a block of aluminum, they are amazing. I could get about 100°F difference between the hot and cold ends.
w.
Wrong. The mechanism of heat conduction cares nothing about gravity. It only cares about a temperature difference (gradient). Is the temperature at the opposite ends of an insulated (say enclosed in a high vacuum chamber) bar of copper different if it’s vertical rather than horizontal? The gravitational potential energy of the copper atoms at the top of the bar is higher than for the atoms at the bottom of the bar.
Exactly the opposite. It is only possible if work is being done, i.e. something is moving the air up and down in the column to establish the adiabatic profile. At the adiabatic profile, the work done would be very small, but even with neutral buoyancy at all altitudes, you can’t move something without doing some work. Since work is being done, the average temperature in the volume will also increase over time Conduction does no work. It simply increases the entropy of the system.
OzWizard says:
January 20, 2012 at 6:43 am
Great! You understand it! So give us the clear, concise version of their theory called an “elevator speech”, that not only says what it does, but how it does it.
I’ve asked a lot of people but you’re among the first to claim to understand it, and the other claimants haven’t come through with their elevator speeches, so you could be the first off the blocks.
w.
PS—please don’t mention atmospheric radiation in the elevator speech, as N&Z say the following:
In other words, their theory doesn’t require GHGs to work … and so neither can your explanation.
All the best,
w.
DirkH says:
January 20, 2012 at 6:57 am
Say what? QUOTE MY WORDS. I have no recollection of anything even remotely resembling that. I say the column will be isothermal, meaning all at the same temperature top to bottom. If not, you could pull work out of it, and that would be perpetual motion.
w.
JJThoms and Willis,
You can often see Hilsch Vortex tubes in use in Formula 1 pits between sessions. They use them to keep the drivers cool while they’re waiting in the cockpit to go back out on the course. The air hose connected to the middle of a metal tube with a small diameter tube on one end and a large diameter tube on the other end is a dead giveaway.
davidmhoffer says:
Dave, repeat after me, 100 times if you have to: “The local value of the insolation does not determine the local temperature of the Earth. It does not even come close…It does not even come sorta, a little bit close! It is freakin’ ridiculous!” I don’t know about where you live, but where I live I’ve never seen the temperature drop down even close to the 3deg background temperature at space at night. In fact, the coldest temperature recorded on Earth isn’t even close to that!
There is absolutely no justification for using the variation in local insolation to determine an effective blackbody temperature of the Earth…None, whatsoever. Nada! It is an extremely silly thing to do. What could possibly be your justification for thinking this number is at all useful? Even on the airless moon, where the diurnal cycle is much longer and the heat capacity is much lower and convective transport of heat is practically non-existent, this is not all that good approximation for what the local temperature is going to be…at least once you get just a few cm down below the surface.
I think the problem with this experiment, the more I read it, is that it’s being forgotten for a gas to be a gas it has to be receiving a source of energy. Otherwise, it would cease being a gas and settle on the bottom of its containing (i.e. Pluto’s atmosphere); as it would lose all its energy to its surroundings (which ultimately lose their energy to space, which ultimately spreads out across the universe as the CMB, which will ultimately lead to the heat death of the universe according to Entropy and how we currently understand physics). The atmosphere’s temperature difference could do work, as that difference is captured energy from the Sun, and thus can be recaptured if one figured out a way to do so.
Basically, we cannot create a rational situation where energy input, and thus energy gradients, are separated out from the system. We can’t just say “here we have an isothermal gas doing nothing”. Shoot it off into space and it’ll soon turn solid! The container its in itself would act as a heat exchanger, as there is no perfect insulator.
In short, are we saying anything logical or reasonable about reality with these thought experiments I see people running?
Robert Clemenzi says: “The only way to transfer heat to space is via radiation, and the only way for the atmosphere to loose heat to space is to have greenhouse gases that radiate energy.”
Robert – not so. All gases radiate. Search for “emissions spectrum”. wikki will give you a good introduction.
“DeWitt Payne says:
January 20, 2012 at 2:14 pm
Alan Millar says:
January 20, 2012 at 1:39 pm
Lets be clear a closed system with an atmospheric temperature gradient present could exist indefinitely without breaching thermodynamic laws.
Wrong. The mechanism of heat conduction cares nothing about gravity. It only cares about a temperature difference (gradient). Is the temperature at the opposite ends of an insulated (say enclosed in a high vacuum chamber) bar of copper different if it’s vertical rather than horizontal? The gravitational potential energy of the copper atoms at the top of the bar is higher than for the atoms at the bottom of the bar.
However, that is only possible if absolutely no work is being done at all. If work is being done, and I cannot imagine a system, with a temperature gradiant, that could prevent work being done, then entropy must increase.
Exactly the opposite. It is only possible if work is being done, i.e. something is moving the air up and down in the column to establish the adiabatic profile. At the adiabatic profile, the work done would be very small, but even with neutral buoyancy at all altitudes, you can’t move something without doing some work. Since work is being done, the average temperature in the volume will also increase over time Conduction does no work. It simply increases the entropy of the system.”
Are you bereft of understanding?
Did I not say that I could not imagine any system with a temperature gradiant that could stop work taking place?
You have just said the same thing using different words!
Sheesh!
Alan
davidmhoffer says:
We know the explanation of the other ~100K: The explanation is simply this – Different temperature distributions will have different average temperatures. Any temperature distribution that emits 240 W/m^2 into space is compatible with the energy balance of the Earth. Hence, you can have all sorts of temperature distributions that will have average temperatures less than 255 K. Which one is chosen for the Earth or for any other celestial body depends on things like the variation (in space and time) of solar insolation on the body, the heat capacity of the solid and liquids at the surface, the heat capacity of the atmosphere, the heat transport in the atmosphere and in the solids and (especially) liquids at the surface.
You think there is some great mystery to explain…but there is not.
The more and more I think about this…
“If not, you could pull work out of it, and that would be perpetual motion.”
But Willis, all gases can do work. To be a gas means it has to have stored energy in kinetic motion, making the molecules move too much to stick together as a liquid and then a solid.
You can get energy back out of any gas when it condenses, and from any liquid when it solidifies. The heat of condensation and heat of fusion. It would not be perpetual motion, because any energy we take out of a non isothermic gas will simply drive the entire column towards condensation, and eventually the “atmosphere” will simply settle as a solid. And, if we drag out every ounce of kinetic motion, we can get all the energy from it till it hits 0 K.
I think this is where the logic is going awry. There is always a kinetic energy difference between each molecule in a gas (a distribution as you point out), and that difference turns to potential energy as one moves against a gravity field. But if you try to use this difference to do work, you lower the average temperature of the entire column, even if there is a temperature gradient in the column (the top colder than the bottom). The top will cool fastest, and the gas will begin to settle, forming condensate on the sides of the container. The bottom would stay warmer and condense last, allowing some molecules to continue zipping around (the highest energy part of the distribution); but the pressure of the gas would steadily decrease.
Isothermal has nothing to do with anything from what I see the more I think of it. Work is done by the entire column or not; any gradient itself does and cannot do work, as affecting the gradient changes the entire average distribution and removes energy from the entire column of gas.
Again, perpetual motion comes in nowhere; because gas is a high energy phase of matter, and matter can change states.
Alan Millar,
That’s not at all what your words said. You said:
And that bolded part of your statement is still wrong. Unless you meant “could not exist indefinitely” instead of could. I’m commenting on what you actually posted, not what you may have meant.
Surely centripetal force has also to be taken into account?
Joel Shore said:
“Which one is chosen for the Earth or for any other celestial body depends on things like the variation (in space and time) of solar insolation on the body, the heat capacity of the solid and liquids at the surface, the heat capacity of the atmosphere, the heat transport in the atmosphere and in the solids and (especially) liquids at the surface.”
Hey Joel, I think you are nearly there but not quite.
This is how it works:
The oceans control air temperatures as per my Hot Water Bottle Effect.
I’ve also said that the oceans should be considered as part of the atmosphere and I have explained in detail why the ocean heat content and the rate of energy flow from ocean to air is also pressure dependent just as is the ATE of Nikolov and Zeller.Therefore both being pressure dependent the oceans cannot alter the ATE. In fact they are an important part of setting it on our particular planet.
The ATE effect is dominant in the atmosphere and governs the temperature gradient from surface to space. We could adopt the term Ocean Temperature Effect (OTE) for the region below the ocean surface but I prefer the term ATE to cover both, with the oceans just being considered as a part of the atmosphere. Obviously there is a discontinuity at the water/air interface but the evaporative process deals with that to leave ATE in the air undisturbed.
ATE is far more powerful than the misleadingly named and probably non existent GHE from GHGs because the former involves the entire atmospheric mass whilst the latter involves only GHGs which are a not a sizeable proportion of total mass.
What I think happens is that ATE is in complete control and all the other features of the system from the bottom of the oceans to the top of the atmosphere organise themselves around ATE to maximise system entropy (the tendency of any system to become less organised over time). In the case of an irradiated planet the concept of it becoming less organised over time involves the removal of energy to space as fast as possible given the constraints of basic physics.
The system always responds negatively to any factor that tries to alter the surface temperature fixed by the ATE because any deviation from the ATE represents a reduction in efficiency as regards the rate of energy loss to space. As far as we can tell it always succeeds leaving ATE unaltered.
The way that the system organises itself depends on the composition of the component elements but due to the dominance of the ATE the only effect from composition differences is to change the rate of energy throughput within each section of the system so that the surface temperature does not change.
However a faster throughput of energy within a particular section of the atmosphere will result in higher temperatures wherever more warm air passes more often across sensors situated within that section but that is a result of energy redistribution and not a sign that the equilibrium temperature of the system has changed.
The ability to redistribute energy in that way (differentially in different sections of the system) is in fact what makes the system flexible enough to maintain the ATE.
Thus pressure and energy input give us the ATE but everything else including GHGs only affects the rate of energy throughput and not the ATE itself.
If GHGs increase then the energy throughput increases to leave the ATE stable and if GHGs decrease then the energy throughput decreases to leave the ATE stable.
N & Z are carefully collating data to verify that proposition. If they do indeed get accurate enough planetary surface temperaures to prove the dominance of the ATE from planet to planet then all that other factors can achieve is to work around the ATE just as I suggest.
So far they have firmed things up by getting a more accurate lunar surface temperature and corrected an apparent error in the application of the S -B equations.
The resulting figures are in supprt of the proposition that ATE governs the surface temperaure regardless of all other potentially confounding factors.
DeWitt Payne says: “The mechanism of heat conduction cares nothing about gravity. It only cares about a temperature difference (gradient)”
I know what you mean – but are you missing something?
Imagine a heat engine with a low altitude heat source (extract heat here) and high altitude sink (dump heat at the assumed cold upper atmosphere). The plan is to exploit a temperature difference. The link between the source and sink is solid (e.g. electrical conduction through copper).
To simplify things, consider operation in terms of a single molecule of the atmosphere. The heat sink adds kinetic energy to the molecule at altlitude. The molecule (in the fullness of time) falls towards the heat source, and gains kinetic energy. At the source, it has the capacity to pass an additional chunk of kinetic energy to the machine.
As you’ll know, this doesn’t work. We appear to have passed energy from the sink to the source to create a boost for the engine.
What went wrong?
And therein lies the problem, that you’re talking about perpetual motion and you don’t know it. Hans, you are indeed proposing perpetual motion. You are saying that gravity separates the warm and cold molecules.
Don’t expect maths terms used correctly, but, isn’t weight a ‘function’ of gravity? Molecules only have weight because of gravity, which is why the ideal gas law doesn’t have gravity, the ideal gas doesn’t have weight or volume, so, a hotter real molecule has weight and will become less dense and therefore lighter as it heats up and so will spontaneously rise unless work is done on it to stop this, and as it rises into the colder regions it cools, becoming heavier it sinks where it will again gain some heat and so the cycle continues, doesn’t it?
Isn’t the Earth as is a fairly good approximation to this if one takes out the spin? All the spin does is put in a some interesting routes for volumes of air, which with or without the spin naturally rise as heated and gravitate to the colder poles (heat flows spontaneously from hotter to colder). Did you like the use of gravitate… 🙂 I only noticed after I’d typed it. Anyway, this is what I read when looking up how our winds work.
Our weather is very energetic because we have the Sun’s imput, but that only exaggerates what gets it all moving, which is the difference in temps between land and seas and so cold air moving in to under hot air rising such as coastal regions, morning and evening differences, and cold air compressing the molecules beneath as it descends again after being for a while hot air rising, both these create winds. Without the ocean one of the wind systems would be out, but, I suppose, as long as there is a temp difference among the rocks or whatever is on the surface, there should be winds created from the temp differences of the volumes of air above them.
If all the surface was of the same material, the same kind of rock, I suppose it would just, somehow, rise and fall simply from the effects of gravity, that is, weight of the molecules, as they changed densities on becoming hotter and colder.
Perhaps one could design an optimum mix of gases to exaggerate these differences in weight of gravity..?
Anyway, gravity separates molecules by weight, hot or cold doesn’t make any difference to gravity, that just makes a difference to the molecule – expanding, becoming less dense with heat, contracting, becoming more dense with cold. the molecules have a different weight relative to each other at the same temperature anyway.
I think I have an answer that I actually believe.
1) Temperature, BY DEFINITION, is the inverse of the rate of change of entropy with respect to energy (with a proportionality constant). That means, when a unit of energy is transferred from a relatively warm object to a colder one, the entropy of the warmer object decreases, but by a relatively small amount, because high temp means low rate of change of entropy, and the entropy of the colder object increases by a larger amount (low temp means high rate of entropy change). So, the total entropy increases.
2. The total entropy of an isolated system must stay the same or increase (2nd law of thermodynamics).
3. If we make the Earth an isolated system by putting a perfect, reflective bubble around it, then it must obey the 2nd law: its total entropy can remain the same or increase; it can never decrease.
4. We neglect the interior of the Earth, and the oceans, because the effect we’re discussing should apply on a planet with a cold interior and no water.
5. If we initialize the atmosphere of this world with regions of unequal temperature, then occasionally, through simple conduction or via radiation, net energy will get transferred from a warmer region to a colder region and when it does, the total entropy of the system will increase IRREVERSIABLY – the system as a whole will never be able to return to the lower entropy state.
6. Therefore, there is an inevitable process by which temperatures will equalize, pushing entropy up, or there is a state with regions of unequal temperature but no possible means for net energy transfer between them – a highly improbable scenario and certainly not the general case.
7. Willis is right. Gravity won’t produce any temperature gradient. It does, of course, produce gradients in pressure and total energy content (as a function of temperature), but not temperature.
I think this thread has jumped the shark.
A physicist says:
January 20, 2012 at 8:53 am
This to me is the fundamental argument.
This is why any and all arguments about the means of how gravity supposedly separates by temperature, or the strong arguments from various people about why it clearly must separate by temperature, are meaningless. Not because the means are wrong or falsifiable, but because the outcome is impossible. Robert Brown has capably and clearly given (and continues to give) the “why it is isothermal” speech. He has done the best job to date of explaining the means and the whys.
But set all of that aside:
IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!
If sealed thermally isolated containers of air were to gravitationally separate, with warm air at the bottom and cool air at the top, then we could build tall insulated cylinders and pull power out of them forever.
Does anyone really, in their heart of hearts, regardless of the logic of any and all arguments, believe that we can constantly and unceasingly pull energy out of a thermally insulated column of air?
Because make no mistake about it. That is what Hans Jelbring is claiming.
w.
Gravity generates the adiabatic lapse rate by converting kinetic energy to potential energy. If the atmosphere was full with greenhouse gases, it would equilibrate with radiative transfer. The only reason the atmosphere would have an adiabatic lapse rate with greenhouse gases is if the ground was warmer than space. Because with radiative transfer, different bodies will equilibrate to an identical temperature.
Now, I think it is a question of optical thickness to IR. Space does not emit a lot of IR, so there is a differential of temperature at the radiative level between space and the ground. Now, here’s a small piece of thinking that is probably wrong, but I don’t know why. The mass of the atmosphere is equivalent to 5m of ground. The interior of the Earth is very warm and, I guess, optical thickness is probably the reason why the ground is not has warm. The highest variation of temperature you can find in the ground is about 30K/km. This is equivalent to 0.15K for 5 m. So, if the atmosphere is able to generate 33K for the whole atmosphere, it would mean that a molecule of a greenhouse gas is around 220 times more optically thick than a molecule of the ground.
From that I have 2 questions:
1- Are greenhouse gases 220 times more optically thick than the ground?
2- Why?
Gravity does not separate by temperature. It separates by mass.
Jelbring, Nikolov and Zeller (and me) accept that in the absence of an energy source the column will become isothermal.
It is only when energy is added that the temperature gradient forms as a result of more densely packed molecules converting a greater proportion of the incoming radiation to kinetic energy.
davidmhoffer says:
January 20, 2012 at 1:55 pm
If you calculate a realistic surface temperature for earth based on actual variance of insolation, you arrive at a value of about 140K.
That is a totally unrealistic surface temperature as it assumes the dark-side of the planet is at absolute zero, which is a totally unrealistic case and far from what is observed, it isn’t even realistic for the moon which has a 14-day ‘night’ and no atmosphere!
Robany says:
January 20, 2012 at 9:20 am
Robany, you are right about the real atmosphere. We are talking about the atmosphere in Stygia, the planet described in Hans Jelbring’s thought experiment. Our real atmosphere is not at equilibrium so (as you note) it has a temperature gradient. Take away all external sources of energy, as in Jelbring’s thought experiment, and at equilibrium it will be isothermal.
w.
Willis says at 1/20 2:19pm:
“I say the column will be isothermal, meaning all at the same temperature top to bottom.”
This violates the 2nd law, KE + PE = constant at each h in the presence of an inexplicable gravity field in the gaseous cv of interest, namely an adiabatic (no gain or loss of heat from CV) GHG-free air column.
Willis’ statement, I believe, derives from the top post which says in part:
“On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down…”
This too is a violation of 2nd law conservation of energy PE + KE = constant since in my view: each molecule moving up in the presence of an inexplicable gravity field cannot have exactly the same velocity/kinetic energy as those moving down in presence of gravity field PE (= ngh) since KE changes with h for energy constant to stay the same, therefore KE must change. Only outlaws need run from this sheriff.
The kinetic energy will vary with h thus P*V will vary and since V is constant in the cv, P must vary. Thus ideal gas thru PV=nRT must vary the temperature so the cv of interest is a non-isothermal air column. No work can come out of it b/c air column’s one macro object is in equilibrium given all 3 laws so we do not have free energy forever (Willis says at 1/20 3:18pm).
Bart says:
January 19, 2012 at 11:28 pm
Unfortunately, I find that my argument regarding non-constancy of the solution is lacking. The modified and adjusted zeroth order Bessel function of the second kind is an eigenfunction, but so is the modified and adjusted zeroth order Bessel function of the first kind, and these can be independently weighted with a time dependent amplitude function. I believe, offhand, that these may be added in such a way as to produce a constant. Therefore, a result which thus converges to a constant is possible. I think it is not physically realizable, but I haven’t <i<proven it yet.
So, I must come up with another argument justifying the approach to 0K at infinity. When I resolve this, one way or the other, you all will be the first to know.
It seems self-evident, though. Using the Earth as an example, with the standard greenhouse hypothesis, it otherwise never would get warm enough to produce water vapor to initiate the greenhouse warming. This is an unstable situation, because the Earth always has the capability to nosedive back to -18C, and become a bleak and barren ice planet. In Nature, whatever can eventually does. That which is not forbidden is compulsory.
Anyway, it’s back to the books for proof. Will keep you all posted.
Willis Eschenbach says:
January 19, 2012 at 11:48 pm
<So in the Jelbring thought experiment you are saying the atmosphere never, ever achieves thermal equilibrium? If not … why can’t we pull work out of it with a thermocouple?"
You can. Remember, there is an outside energy source which is constantly adding energy to the system. Is the Earth at equilibrium? How do windmills work?
Joe Born says:
January 20, 2012 at 5:30 am
“An atmosphere of the type you describe would acquire energy from its source only when its temperature is less than the surface’s.”
This is not a constraint. The surface can get hotter at the same time. Stefan-Boltzmann does not limit it as A) the system is not in steady state and B) even in steady state, SB only puts an upper bound on outgoing radiative energy, but not an upper bound on temperature.
Besides, constant flux from the external heating radiator is more like current than voltage. See section 2.3 here.
DeWitt Payne says:
January 20, 2012 at 12:38 pm
“You can also assert all you want about the SB equation not applying, but you have given no evidence or citation that it doesn’t.”
But, you have no evidence or citation that it does. SB is always couched in terms of steady state, or at least a quasi-steady state.
SB radiation is an outlet for energy flow, so that incoming and outflowing are equal in the steady state. When you have energy conducting outward in large volume, then the individual particles of the surface do not have to be radiating to maintain balance.
“Even if the emissivity does vary with wavelength and angle, one can still integrate the Planck equation, B(λ,T) and get the total emission.”
The Planck distribution describes the energy distribution of particles making up the surface when in thermodynamic equilibrium and unable to transition to a lower energy state through any means other than radiation.
“A molecule that was actually in orbit would indeed have very high kinetic energy relative to the molecules around it.”
We don’t consider that in calculating temperature, though. As I pointed out, satellites moving in LEO at 7500 meters per second do not radiate as though they were at 30,000K. A full treatment would need to be couched in terms of relativistic four-momentum, which is always conserved, and for which the time-like element is the energy. I don’t think we really want to get into that, here.
DeWitt Payne says:
January 20, 2012 at 9:39 am
“As pointed out above, your first crack at the capacitor example was correct.”
As I pointed out in the reply to Joe, no it isn’t.
“Even a constant current source reverts to a constant voltage source at some voltage.”
Not an ideal current source. This is like the assumption of an ideal non-radiating atmosphere. In fact, at some point, every real current source breaks down, and every real atmosphere radiates.
“Bart,Your argument begs the question.”
You are right. It does. See the top response to myself in this post. I did not recognize it because I had dismissed the other eigenfunction and thought no more of it.
Phil. says:
January 20, 2012 at 1:02 pm
“What it should say is that ‘the heat accumulates in the atmosphere until the layer of gas nearest the surface reaches the surface temperature at which point heat transfer ceases due to ΔT=0.”
The surface temperature is increasing, too. There is no limit on the temperature of the surface, only on radiated energy. If you are not at equlibirum, then you are not generally radiating at the SB limit. That which is not forbidden is compulsory.
I am still puzzling over the idea that a column of air will reach an isothermal state or not. I took Dr. Brown’s explanation as a starting point and made changes to it that reflect my view. I hope he doesn’t mind. Who would think that such a simple thought experiment would be so complex.
Imagine a plane surface in the gas. In a thin slice of the gas right above the surface, the molecules have some temperature. Right below it, they have some other temperature. Let’s imagine the gas to be monoatomic (no loss of generality) and ideal (ditto). In each layer, the gravitational potential energy is constant but the upper layer has a higher potential energy. Bear in mind that only changes in potential energy are associated with changes in kinetic energy (work energy theorem), and that temperature only describes the average internal kinetic energy in the gas.
In equilibrium, the density of the upper and lower layers, while not equal, cannot vary. Which means that however many molecules move from the lower slice to the upper slice, exactly the same number of molecules must move from the upper slice to the lower slice. They have to have exactly the same velocity distribution moving in either direction. However, the molecules moving downward are accelerating, therefore increasing in kinetic energy while the ones moving upward are losing kinetic energy for potential energy. The molecules below had a higher temperature, and have a different MB [Maxwell-Boltzmann] distribution, with more molecules moving faster. Some of those faster moving molecules would have the right trajectory to rise to the interface (slowing and cooling, sure) and carry energy from the lower slice to the upper. The upper slice (lower temperature) has fewer molecules moving faster —However, at the interface, the MB distribution is equal owing to the average energy of the upward and downward flows being equal There are therefore equal molecules that move the other way at the same speeds that the molecules from the lower slice deliver (allowing for gravity). Thus, average kinetic energy is unchanged in both layers. On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down, with zero energy transport, zero mass transport, and zero alteration of the MB profiles above and below, only when the upper slice is cooler than the lower, due to the conversion of potential energy to kinetic energy as they fall. Therefore, a vertical column of air at equilibrium will be cooler at the top than the bottom.
Willis said, “I’ve not usually seen that assumption made. In general, conduction of both sensible and latent heat is not “assumed negligible” unless it is in fact negligible for the particular phenomenon under study.”
Convection and Latent are considered. I have not see conduction properly considered. CO2 impacts the conductivity of mixed gases as well as the radiant impact. That impact is considered negligible when it appears not to be. Trenberth’s latest cartoon reduced the conductive impact, (thermnals he uses for combined conductive/convective) even while increasing the surface temperature. If you have a link that addresses conductive change, I would like to see it.
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!”
This assumes a heat engine can transfer energy from its source as efficiently as transfer of energy into its source. I do not believe this assumption has been justified where conditions at the source are not the same as conditions at the sink. .
An ideal sink assumes the same temperature as its surroundings because there is perfect heat transfer from the sink. Same for an ideal source. If the source is not ideal, the “effective” source temperature will be at lower temperature compared to its surroundings. Vice versa for a heat sink.
Consider the effect of gravity on the density of gases and the impact this can have on the ability to conduct thermal energy by conduction.
Until it is demonstrated that the physical conditions for transfer of thermal energy around the source and sink are favourable (i.e. the do not fully negate the temperature difference we are trying to exploit), the arguments in this post cannot be supported using notional heat engines.
Willis,
You said
” I offered up a proof that no possible mechanism involving a transparent, GHG-free atmosphere could raise the temperature of a planet above its theoretical Stefan-Boltzmann temperature.”
Sounded like a refutation to me, since there is nothing in the theory that requires greenhouse gases.
And you said:
“if you can’t explain it in an elevator speech, it [sic] you don’t understand it”
which sounds like an assertion that no one understands it, or did someone win the prize for a coherent elevator speech demonstrating clear understanding? If so I missed it. But in any case understanding a hypothesis that is wrong seems almost a contradiction in terms.
I have here some background on the journal in which the Jelbring paper appears. It provides some insight into the way in which peer reviewed journals validate or fail to validate what they publish.
“This assumes a heat engine can transfer energy from its source as efficiently as transfer of energy into its source”
Oops – “transfer into its sink”
PeterGeorge said, “7. Willis is right. Gravity won’t produce any temperature gradient. It does, of course, produce gradients in pressure and total energy content (as a function of temperature), but not temperature.”
Completely isolated as in your reply, true. Gravity appears to impose a low limit on the atmospheric sink temperature to the energy required for obtaining escape velocity. Gravity interactions with the moon and sun do add some tidal energy to the system. I think the escape velocity limit is what is causing the confusion. I haven’t seen it explained very well anywhere. There are a number of things about the tropopause I have not seen explained very well.
Willis wrote;
“If you are fighting basic ignorance of science, you will be deluged with ignorant people. Not much I can do but just keep putting the facts out there.”
With all due respect, I am most certainly NOT ignorant of science. With multiple Master’s Degrees in Optics and Electronics and three decades of reconciling computer predictions with actual observations I am quite familiar with all of the relevant disciplines necessary to understand the “Greenhouse Effect”.
Ironically enough, the FLAW in the GHE HYPOTHESIS is actually more of an accounting error than a science error. When the energy returns to the surface from the “GHG” you cannot ADD it to the energy arriving from the Sun to produce an alleged “energy budget”. The energy returning from the “GHG” has already travelled once through the system leaving cooling (at the previous location it departed from) in its wake. So we have sequential warming/cooling/warming/cooling…. events occurring from ONE bundle of energy that came from the Sun. So one bundle of energy from the Sun simply warms the surface multiple sequential times, since these warming events are followed by equivalent cooling events and are separated by finite time delays YOU CANNOT ADD THEM TOGETHER and get a correct result. Doing so is the equivalent of creating energy, which you must admit violates the First Law.
One of the clues to this accounting error is the use of the terms; “Net Energy Gain” and “Extra Energy” as used in the climate science community.
In the engineering community we use the existence of “extra energy” as a RED FLAG to tell us our analysis is wrong.
Engineers that routinely calculate “Net Energy Gains” either get fired or promoted into management.
Regarding why the Earth is at the average temperature it is, I have yet to hear any one explanation that makes common sense. But knowledge always expands and we will likely know why sometime. Although from a practical sense it seems to make very little difference.
But it most definitely is not the result of the GHE.
The GHE appears to cause some energy to travel through the system (bouncing as it where) between the gases and the surface all the while dissipating energy to Space via radiation. The end result is a slight delay to the energy as it travels from the Sun to the Earth through the Atmosphere to the Universe.
The “missing heat” is currently travelling as a spherical IR wavefront that is “X+d” light years away from the surface. In this equation X represents the elapsed time since the sunlight arrived (i.e. 100 years for sunlight from 1912) and d represents the slight delay from the GHE and is likely about 5 milliseconds. “d” is actually a statistical distribution which will of course have a different specific value for each photon travelling through the system. Some will bounce many times and take longer to exit, while others may not bounce at all and exit directly to space.
Just because people disagree with you does not necessarily mean they are ignorant of science, another possibility is that you are mistaken in some way. I know the GHE theory has been written down in papers and textbooks, but so was the theory that stress and spicy food
caused stomach ulcers.
Cheers, Kevin.
I cross posted this at an aerospace blog I frequent:
There are two ways of looking at the problem:
1) You can claim that the atmosphere will continually conduct heat away only until everything is at equilibrium, so that the surface is at the temperature set by the Stefan Boltzmann formula such that incoming and outgoing radiation balance. At this point, the atmosphere is at constant temperature throughout, so there are no more heat flows to disrupt the radiative equilibrium.
The problem with this point of view is that it is a circular argument. It iassumes a priori that equilibrium will be established, therefore the Stefan Boltzmann relationship will hold.
2) Another, equally plausible, scenario is that the system is unstable, such that equilibrium is never attained, even asymptotically. Then, the Stefan Boltzmann relationship does not hold, and there is no limit to surface temperature.
The temperature profile of the atmosphere due to the surface/atmosphere conductance has to satisfy the heat equation. Solutions of the heat equation exist for both of these alternatives, as I have had to acknowledge in my lastest posting here. But, the runaway scenario #2 is more satisfying from a stability viewpoint. If scenario #1 is the case then, since water freezes at the non-atmospheric equilibirum temperature of the Earth, a question arises as to how the effect ever got started, and there remains a possibility of slipping back to that lower energy equilibrium state.
And, if you want to know the perils of dismissing the possibility of transition to a lower energy state from a quasi-stable higher energy state… Well, the denizens of this site know where to look.
I am sifting through the problem to try to find some requirement which will decide the question one way or the other. When I find it, I will let you know. Other inputs which would contribute to that search are welcome, but baseless assertions, based on the standard way most people think things should be, are not. As of right now, mostly what I have gotten are baseless assertions that SB must hold regardless of anything else going on, and that argument has no foundation.
“If W=FD and F = mg then W can be done by heated air rising IF volume changes. ”
Agreed. There is no denying that hot air heated by blackbody radiation near the earth’s surface gains energy, the gas expands as it heats up and rises. As it rises it cools loses energy and the molecules have higher potential energy because they are higher up the gravity field. In essence, work done by the Sun (radiative energy) has heated the Earth and in turn heated the air. The Work is added by the SUN not gravity. Another transformation happens when the Suns radiative energy causes warmth and water to evaporate however this energy gets stored in another form of potential energy or properly described as “latent heat” – later on when water condenses this latent heat is released.
The point is that it is not “gravity” which is actually doing any of the Work or creating any energy or any heat. It is just a passive bystander. For sure as molecules move up and down in a gravitational field they gain and lose kinetic energy in exchange for potential energy but there is no Work or net energy being added by gravity itself to the system as a whole. Gravity plays a strong influencing role on atmospheric properties and how the atmosphere behaves but it does not create or add net heat to the system. External sources of Heat and Energy come from things like meteor impacts, cosmic rays, solar wind, radioactive decay, the Sun and chemical reactions.
When you drive your car or turn up the heat in your home or have a meal or pretty much anything (like breathing) then you are using up stored energy from our Sun. The Sun’s energy has been stored by plants and the main storage mechanism is to create Carbon from CO2 (that is how plants grow – by stealing Carbon from the air). Animals consume Carbon by oxidizing it with Oxygen and releasing CO2. When you burn fossil fuels then you are burning carbon energy that was stored by plants using energy from the Sun. When you use electricity then it comes from the Sun also. Even Wind Power and Hydro is all from the Sun while tidal power would be mostly Moon and Sun. Hydro power simply comes from the Sun’s energy heating water that rises and then eventually cools and condenses and precipitates at altitude on some mountain. Even in the case of Hydro power it is not gravity that provides the source of energy.
When Sisyphus pushes his boulder up the hill it would be the energy from the lunch he ate that creates the potential energy of the boulder that is released when it falls back down – gravity has not done anything.
For most of us the Sun is the source for everything including the heat engine that drives our atmosphere. There are some life forms that seem to work from heat of radioactive decay or heat left over from when the Earth formed (Sulphur based?) but the majority of life forms seem to be Carbon based.
Downdraft says:
January 20, 2012 at 3:37 pm
“I am still puzzling over the idea that a column of air will reach an isothermal state or not.”
It will, but if and only if equilibrium is established. So, it becomes a circular argument.
Phil;
That is a totally unrealistic surface temperature as it assumes the dark-side of the planet is at absolute zero, which is a totally unrealistic case and far from what is observed, it isn’t even realistic for the moon which has a 14-day ‘night’ and no atmosphere!>>>
EXACTLY!
The question then becomes:
WHY?
We cannot answer “why?” unless we know what the theoretical starting point is. Beginning with a theoretical starting point of 253K which can only be achieved by a uniform insolation is wrong, wrong, wrong.
How is it that everyone wants to use SB Law to calculate a theoretical equilibrium surface temperature, and when I show that the theoretical surface temperature calculated by averaging P produces a false number, everyone jumps up and says:
“well your theoretical number cannot be correct because when we look at practicaly measured instances like the moon, we don’t get anything like your theoretical number”.
AND THAT IS THE POINT!!!
The moon doesn’t reach that theoretical number because even though it doesn’t have an atmosphere, it still has:
a) heat capacity
b) conduction
so those factors tend to make the temperature more uniform even given a 14 day night. It is the exact same discussion as before. What are the factors that redistribute energy on the moon, and by how much for each, and how does that affect the measured “average” temperature across the entire surface versus the theoretical average arrived at by black body provided that conduction is zero and heat capacity is zero?
The point I’m trying to get at here is that the moon does not in fact meet the theoretical effective black body temperature profile, and hence, we can assume that the difference between the observed temperature distribution and the theoretical distribution is due to factors such as heat capacity and conduction.
The same is true of the earth except that with earth we have to throw in all the atmospheric and oceanic processes that move energy around on top of heat capacity and conduction.
The value of 255K is meaningless as a theoretical value, it applies in no way shape or form to the actual earth. We cannot possibly sort out how much of the observed temperature profile is due to heat capacity, conductance, atmospheric and oceanic processes if we start with a mythical number in the first place. If we’re going to start out with the “theoretical” black body temperature of the earth and then subtract that from the observed temperature of the earth to figure out how much of the difference is due to GHE and other factors, of what value is doing the theoretical math against a theoretical black body temperature that is wrong in the first place? Of what value is pointing out that the moon doesn’t adhere to the theoretical black body temperature in the first place when what we’re trying to measure is WHY it doesn’t match, not if it matches. Unless conduction and heat capacity on the moon are zero, why would anyone possibly expect the measured temperature profile to match the theoretical profile which is calculated in the absence of those factors?
DeWitt Payne said @ January 20, 2012 at 1:53 pm
“Understand: To comprehend; to apprehend the meaning or import of; to grasp the idea of.” [from the OED]
Oddly enough, my understanding of QM came from Feynman’s lectures. Are you saying that he didn’t know what he was talking about? Or are you really Gavin Schmidt putting a different meaning on “understand”?
Willis Eschenbach
For some reason the idea that the atmospheric pressure of a planet plays a dominant role in the temperature make sense to me, after all, the second law of thermodynamics is expressed in terms of entropy, pressure, volume, energy and temperature. While I have only a BSME, I thought I would look at this from an engineering standpoint. In M.E. heat transfer is by conduction, convection or radiation. My text on the subject does have some combined conduction-convection problems but radiation problems are generally restricted to heat transfer between bodies in a vacuum, except for some inside a furnace problems.
So I will looks at this as a free convection problem and while this could be solved directly from the differential equations for, momentum, continuum and, energy or the corresponding integral equation for free convection, we engineers generally use working correlations derived from experimental investigations carried out for a vast number of situations. These investigations resulted in a number of dimensionless properties derived from ratios of state properties. The ones I will use here are:
Re = Reynolds Number = U*l/v
Gr = Grashof Number = (g*l(^3)*beta*Delta-T)/v(^2)
Pr = Prandlt Number = (mu* C(sub p))/ k
Nu = Nusselt Number = (h*l)/k
Statement of the problem: What is the effect of atmospheric pressure on the free convection from a surface? I will calculate the energy needed to be input into a plate 1m X 1m so as to maintain its upper surface (lower surface insulated) temperature of 40C. I will calculate it for the plate being in 0C air at 1 Atm and 25 Atm. We need to gather some data:
L = characteristic length = Area/Perimeter = (1m X 1m)/4m = 0.25m.
v = kinematic viscosity
k = thermal conductivity
beta = coefficient of thermal expansion
Pr = coef. of dynamic viscosity * specific heat / density
From Wolfram Alpha at average air temp of 20C:
For Air at 1 Atm:
v= 15.09 X 10(^-6) m(^2)/s
k= 2.56 X 10(^-2) w/m-C
beta = 3.41 X 10(^-3)/C
Pr = 0.713
For Air at 25m:
V=6.24 X 10(^-7), k=2.68 X 10(^-2), beta=3.41 X 10(^-3) and Pr =0.726
h(what we are looking for) = heat transfer coef = Nu*k/L
Nu = 0.54*Ra(^0.25) for 2.6 X 10(^4) < Ra < 10(^7)
Nu = 0.15*Ra(^.33) for 10(^7)< Ra < 3 X 10(^10)
So at 1 Atm:
Gr = ( 0.25(^3)*9.8*3.4 X10(^-3)*40)/(15.09 X 10(^-6))(^2) = 91.724 X 10(^6)
Ra = Gr*Pr = 65.399 X 10(6)
Nu = 0.54*(65.399 X 10(6))(^2) = 48.56
h = 48.56*2.56 X 10(^-2)/0.25 = 4.98 w/m(^2)-C
Similarly for 25 Atm
h = 54.5 w/m(^2)-C
One would need to input over ten times more heat to maintain the surface temperature at 25 Atm than at 1 Atm. This says to me that I would expect to see a significant difference in the atmospheric temperature profile (with height) for identical planets except atmospheric pressure.
KevinK,
Thankyou for that lucid and well presented argument which I endorse right down to the fate of engineers that get it wrong!
DeWitt Payne says:
January 20, 2012 at 9:59 am
“Hans Jellbring,
Any surface radiation power exceeding 100 W/m^2 is bull regardless if it is from equatorial, midlatitude or polar regions during days or night. Just show how this fantasy power radiation changes between day and night in polar regions as an exsample.
Here’s a plot of upwelling IR radiation measured over 24 hours at Desert Rock, NV by a SURFRAD station there. It looks to be more than 100W/m² to me. Note that the time axis is UTC. Desert Rock is -8 hours from UTC so local noon would be 2000 on the time axis.
There are seven SURFRAD stations in the US. You can access the data here.”
Thank you very much for telling me that there exist data. I don´t know if you ever heard about the Wangara and Koorin expedition i Australia. All types of mesurements were made every 3 hours for 30 days including solar irradiation and IR radiation. “The Koorin expedition, Atmospheric boundary layer data over tropical savannah land. Measurments were taken from 0 to 3000 m.
There were never any IR backradidiation or forward radiation to earth exdceeding 100 W/m^2 during this period. The book is 360 pages mostly containing data. this was during a time before “back radiation was invented”.
If back radiation existed i would be able to go out in the middle of the night and let the radiation hit my face and I would get a nice brown tan in the middle part of winter in northern Sweden when the sun is not shining at all. Even high school kid can figure out that it is impossible. Prefessor Gerlisch has written about this fraud in the G&T falsification of the Greenhouse Effect.
You might know that you quickly get a sun tan in a snowy area since the solar irradiation is coming both directly and is reflected from the snow. Photons works directly from the point they are emitted. If the infamous 390 W/m^2 plus the 330 W/m^2 “downwelling” was real we would get an equivalent to sunshine everwhere regardless if is dark or sunshine. The whole thing is just Alice in the Wonderland Fairtale.
DR wrote;
“There is something wrong with the GHE hypothesis as it has been promoted lo these many years.. I’m not qualified to enunciate what it is in technical jargon, but I don’t need to be an atmospheric scientist or physicist to know something is wrong. Does anyone else feel like they’ve been sold a lemon? Even here on WUWT, with the arguing going back and forth, the “theory” is no better explained or proven than it was 25 years ago.”
I initially started from a position of disbelieving the stated accuracy of the computer models. We use computer models in engineering all the time and have a saying; “If your hardware does not perform as predicted by your model you need to improve your model”. The level of hubris required to predict the Earth’s temperature in 100 years is STAGGERING.
After looking into the GHE hypothesis a little bit, I am now convinced it is fatally flawed. See my previous post (assuming it went through) for details.
So I guess I didn’t actually BUY the lemon, but I did pick it up, squeeze it, look it over and then I took a pass.
Nice YouTube, I actually went to school when we still used chalkboards, is there a chalkboard museum around someplace to preserve them ?
Cheers, Kevin.
Suppose Jelbring is right, and the temperature in the atmosphere inside the shell is warmer at the bottom and cooler at the top. Then the people living in the stygian darkness inside that impervious shell could use that temperature difference to drive a heat engine. Power from the heat engine could light up the dark, and provide electricity for cities and farms. The good news for perpetual motion fans is that as fast as the operation of the heat engine would warm the upper atmosphere and cool the lower atmosphere, gravity would re-arrange the molecules once again so the prior temperature profile would be restored, warm on the bottom and cold on the top, and the machine would produce light for the good citizens of Stygia … forever.
As this is a clear violation of conservation of energy, the proof by contradiction that the Jelbring hypothesis violates the conservation of energy is complete.
The final sentence is incorrect.
What is being described is Perpetual Motion of the Second Kind, in which energy is precisely conserved.
Willis wrote:
a) show that there is no problem with a sealed system performing continuous work with no energy inputs or exchange with the outside world, (my bold)
You stipulate conservation of energy, and then suggest that energy is not being conserved. All that is being done is internal rearrangement of the sealed system.
As for your challenge, it will be done, in an unmistakeable manner.
For now, I offer a single word:
Rain.
In the calculations above I typed Ra(Raleigh Number) for Re by mistake..
Trick,
That doesn’t look like the Second Law to me. Where’s the entropy term?
Adiabatic expansion ignores conduction, which is a reasonable thing to do in a normal atmosphere. But the atmosphere of a spherical planet with an isothermal surface isn’t normal. There would be no circulation once an adiabatic lapse rate was achieved. At that point, conduction dominates even if it isn’t fast. Conduction is proportional to the temperature gradient. Movement of energy from hot to cold raises the temperature of the cold and decreases the temperature gradient. Eventually, Bart notwithstanding, the temperature gradient approaches zero and the atmosphere becomes effectively isothermal. The entropy of an isothermal atmosphere is higher than for an atmosphere with an adiabatic lapse rate. So the Second Law requires that when work isn’t being done on the atmosphere by circulation between areas on the surface at different temperature (ruled out by the definition of the problem), it becomes isothermal.
Bryan says:
January 20, 2012 at 10:24 am
Willis says
” To believe that gravity can affect temperature, you have to have a weak grasp of physics, an unshakable belief in your correctness, and a willingness to ignore a bunch of folks who actually understand physics. That’s a bad combo.”
“Yet only last week Willis believed in an adiabatic distribution for a thermally isolated column of gas in a gravitational field.
This week he has ‘seen the light’.
Don’t be so hard on yourself Willis.”
I would for sure want him to get as much insults from the audience here that he has provided to me during this last week and certainly I don´t pitty this clown that tries to be scientific. He can tell where I am wrong in my paper which is common rutin in science or skip the thing. The article is peer reviewed and it stands as it is. And the clown wants an “elevator speach” from me since he “does not understand” what I have written. What comedy.
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!
If sealed thermally isolated containers of air were to gravitationally separate, with warm air at the bottom and cool air at the top, then we could build tall insulated cylinders and pull power out of them forever.”
Why would the inside of this thermally insulated column of air find a different equilibrium from the equilibrium that existed before you built the containers? Which changes its equilibrium? The air inside the containers or the air outside the containers and why.
Bart says:
January 20, 2012 at 3:35 pm
Bart says:
January 19, 2012 at 11:28 pm
Phil. says:
January 20, 2012 at 1:02 pm
“What it should say is that ‘the heat accumulates in the atmosphere until the layer of gas nearest the surface reaches the surface temperature at which point heat transfer ceases due to ΔT=0.”
The surface temperature is increasing, too. There is no limit on the temperature of the surface, only on radiated energy. If you are not at equlibirum, then you are not generally radiating at the SB limit. That which is not forbidden is compulsory.
You’re dreaming, kindly explain how the temperature of the surface is increasing?
The original post referred to a uniformly heated planet with a transparent atmosphere, the only way that conduction can reduce the radiation loss from the surface is by cooling the surface. If the surface is heating up and still continually losing heat by conduction that’s some bizarre physics you’re proposing!
Willis Eschenbach says:
January 20, 2012 at 3:18 pm
Does anyone really, in their heart of hearts, regardless of the logic of any and all arguments, believe that we can constantly and unceasingly pull energy out of a thermally insulated column of air?
Because make no mistake about it. That is what Hans Jelbring is claiming
Gravity != energy
Einstein! Help!
Houston, we have a problem. A quick trawl around the net reveals at least five conflicting second laws of thermodynamics.
Science != Settled
rwemyss says:
January 20, 2012 at 1:29 pm
“@Hans Willis is NOT indicating the heat engine will REMOVE energy from the system! He is indicating that the temperature difference used to run the heat engine can do work on the fluid in the system, and infinitely do so per the theory. Hence the theory fails.”
My paper deals with the properties of ideal gases inclosed in two sperical shells. I don´t know how Willis is going to construct a perpetuum mobile with the aid of ideal gases only since he has never told me. Ask him that.
I have already pointed out that a “perpetuum mobile” is possible to construct in our real atmosphere since the cold air at the top of Mount Everest for sure is colder than air at the surface.
Elevator speech.
An atmosphere currently in motion driven by solar surface radiance and TOA atmosphere radiance. Lots of convection. Seeking equilibrium
Night comes convection greatly slows but does not stop because TOA atmosphere radiance continues.
You also get an inversion layer at the surface due to the surface being a better radiator and cooling the atmosphere some by conduction without the aid of convection but with the aid of residuals of convection, the wind.
Enter Jelbring and and he waves his magic wand and stops the TOA atmosphere radiance to space. He also stops surface interactions with the bottom of the atmosphere or any input or output to the atmosphere.
He is allowed to do this by virtue of basic math where you are allowed to do this conceptually by subtracting equal amounts from both sides of an equation and the laws of thermodynamics specifies incoming must equal outgoing.
Now with nothing to drive it convection rapidly stops probably within hours. The atmosphere is left with a temperature gradient.
So if this is wrong evidence should be given of a force to continue convection without cooling going up or warming going down, or alternatively if it is acknowledged that no force exists to continue convection; where does the energy go or come from to cool the lower atmosphere and warm the upper atmosphere to an isothermal equilibrium.
I am curious about this force. What is it and where does it come from?
Robert Clemenzi says:
January 20, 2012 at 1:36 pm
” The only way to transfer heat to space is via radiation, and the only way for the atmosphere to loose heat to space is to have greenhouse gases that radiate energy”.
Wrong. Don´t you think that salt particles, droplets (clouds), and dust particles emits IR from the atmosphere. You might also have heard about lightning.
Certainly gravity does not produce thermal energy per se. But it behooves those who contend that a gravity-bound atmosphere devoid of GHGs would be completely isothermal to explain what then happens to temperature at TOA. Sound physics is not just a simple matter of mathematically pursuing a priori concepts to logical extremes. I’m throwing this on the pile for discussion by others, because I’ll not have any free time next week.
small correction to above elevator speech. The sun stops shining and an inversion layer comes to the cusp of forming when Jelbring waves his magic wand. I want to avoid having a layer actually form and then have to wait for the inversion layer to conduct through the atmosphere which would be a lot more than a few hours.
Stephen Rasey says:
January 20, 2012 at 2:01 pm
RE: Jelbring at 12:15 pm
“You criticize my use of elevator speech (or verbal proof, take your pick) without addressing its conclusion. You say that there should be two different lapse rates, then go on to say something about equal masses. I never specified equal masses, in fact with two different gasses, with different Cp and probably ….”
The ponit is that an adiabatic lapse rate is equivalent to the statement “that any two arbitrarily choosen equal massues” wihtion the closed sysem will contain equal energy regardless at which altitude they are situated.
I am sorry if I have been unclear in my wording.
Bart,
You mean besides every textbook in the world about atmospheric radiation? Let’s look at two examples:
Atmospheric Radiation: Theoretical Basis, Goody, R.M. & Yung, Y.L. p31.
A First Course in Atmospheric Radiation, Second Edition, Petty, G.W. , p126
What you have asserted and both references reject is that LTE does not exist for any real object. If LTE did not exist, the the Planck source function would not apply. But you have only asserted this with absolutely no evidence or support. If you were correct, then buying an IR thermometer would be pointless. But you’re wrong and IR thermometers, which use SB, work.
Stephen Wilde says:
January 20, 2012 at 3:25 pm
Gravity does not separate by temperature. It separates by mass.
This is what I meant, but here said more succinctly..
Jelbring, Nikolov and Zeller (and me) accept that in the absence of an energy source the column will become isothermal.
It is only when energy is added that the temperature gradient forms as a result of more densely packed molecules converting a greater proportion of the incoming radiation to kinetic energy.
But, gravity here is the energy source if pressure is an energy creating heat, if pressure is energy and gravity creates it, then as long as there is gravity, which has different pressures at different heights, then there is a temp differential and if there is a temp differential we get movement, right?.
As colder molecules at the top cool down they will sink, not because they are colder, but because they are now more dense and heavier than the molecules beneath which are hotter, and which being hotter are rising anyway. The pull of gravity now on the denser colder causes them to heat up, there are winds which show this principle, they are also increasing pressure on the warmer beneath which will warm them further/ displace them.
It seems to me that it is only a matter of getting the balance right, the colder top has to be cold enough and at the optimum height re the amount of gravity available not to warm up to a degree that will upset the cooling and sinking as molecules give up their heat to become colder and denser.
Gravity, it seems to me, is as near dammit perpetual motion as long as the optimum conditions can be maintained because it is then a constant force as source of energy, is energy since energy does work – just ‘cos it sits around looking as if its not doing anything…
And I’ve just had a thought, much like those sealed terrariums – the water recycles.
I’ve just had a look to see what I could find on pressure and compression and found this:
“You’ll find an equation used to relate temperature and pressure, which is the 3’rd equation under the heading. Temperature is in absolute (ie: Rankin or Kelvin) and pressure is also absolute, not gage pressure. The exponent n should be 1 for a case where there is so much heat transfer that the air stays at constant temperature, and 1.4 for a case where there is no heat transfer at all. The real case of compressing air in a cylinder by applying a piston and doing work on the air will be somewhere in between because there will be some heat transfer, but not isothermal conditions. ” http://www.physicsforums.com/showthread.php?t=212327
Isn’t that what the force of gravity is doing just by being itself?
PeterGeorge says:
January 20, 2012 at 3:06 pm
You seem to be religous pretending to be reasoning by a chain of unvalid logic. Please, read the G&T falsification paper or Willian Gilbert on the Tallbloke blog.
Would Jelbring’s theory predict a measurable difference in temperature for a feasible version of the following setup? Take a tube of gas, place a thermometer at the bottom and at the top, place it in a centrifuge with the bottom facing outwards, and spin it up.
p.s. meant to say Ta, Mods!
Willis:
Your initial statement: ‘Consider a gas in a kilometre-tall sealed container.” Assuming a tall, sealed container that is not free to expand, it makes no difference whether it is laid on its side or vertical. The pressure will be the same throughout the container: top to bottom, regardless of gravity. Assuming no outside input and at equilibrium, then according to the equation: P1V1/T1 = P2V2/T2, then the temperature would be the same throughout as well. It cannot possibly be anything else. The pressure at the bottom of the container will be no different to the pressure at the top. In this situation, pressure differential between the inside and outside of the container will increase with vertical height, but the pressure inside the sealed container will not change. So a tall, sealed container might sound reasonable, but cannot be representative of the atmospheric situation. For imaginative purposes, the container needs to be free to expand over its entire length.
Assume the container weighs nothing, contains a portion of atmosphere and is free to expand and contract thoughout its entire length. In this case, the result will look something like a huge weather balloon prior to launch with the dense gas towards the end closest to the gravity source and the expanded end furthest from the gravity. If position 1 is at the lowest end of the bag and position 2 is at the highest end of the bag, then I would assume that without outside input, the reduction of pressure from P1 to P2 would be negated by an equal increase in volume from V1 to V2. There would be no change in temperature…isothermal. In the situation of the sun heating the earth’s surface…well that is a subject for another discussion.
Myrrh says:
January 20, 2012 at 3:03 pm
And therein lies the problem, that you’re talking about perpetual motion and you don’t know it. Hans, you are indeed proposing perpetual motion. You are saying that gravity separates the warm and cold molecules.
Your lasr sentence is correct. The reason is that maximum entropy or the second law of thermodynamics has to be applied when a gravity field is at work and we are studying a insulated inclosed atmosphere as defined in my E&E paper. Yes, the consequences are tremendous.
Hans Jelbring said @ January 20, 2012 at 4:43 pm
I get my suntan from UV radiation. Where is this fairytale land that enables one to get a suntan from IR radiation? Or are you just saying whatever pops into your head because you know the ignoramuses will believe whatever you say? Sorry if this is rude, but the difference between UV & IR is really basic stuff!
[Mods, you overworked & underpaid paragons of all that is virtuous, this is my third attempt at posting. If the two previous are stuck in the spam filter, please delete them]
Stephen Wilde says:
January 20, 2012 at 3:25 pm
Gravity does not separate by temperature. It separates by mass.
“Jelbring, Nikolov and Zeller (and me) accept that in the absence of an energy source the column will become isothermal.”
I certainly disagree. The temperature lapse rate will be DALR which is equivalent to equal amount of energy in arbrarily equal masses if the system is insulated.
Bart,
A lone molecule that isn’t colliding with other molecules isn’t in LTE and would most likely be in the ground state so of course it wouldn’t emit. A satellite, if charged, would only emit cyclotron radiation based on its velocity and radial acceleration. That wouldn’t be much because the acceleration is low. The molecules in the atmosphere it collides with. though, is a different matter entirely. Satellites re-entering the atmosphere create temperatures high enough to ionize the air and create a plasma.
Trick says:
January 20, 2012 at 3:33 pm
Willis says at 1/20 2:19pm:
“I say the column will be isothermal, meaning all at the same temperature top to bottom.”
“This violates the 2nd law, KE + PE = constant at each h in the presence of an inexplicable gravity field in the gaseous cv of interest, namely an adiabatic (no gain or loss of heat from CV) GHG-free air column.”
Many thanks for your insight.
.
DeWitt says at 1/20 4:51pm –
“That doesn’t look like the Second Law to me.”
My view is you’re right DeWitt, I’m mixing my metaphors – taking a mechanical conservation of energy knife into a thermo gun fight. My view is employing conservation of energy works here in thermo, I think of 2nd law as conservation of energy. Violating energy conservation whatsoever to create energy collapses a thermo or mechanical issue – especially useful to collapse proposals for perpetual motion machines of the 2nd kind.
DeWitt continues –
“So the Second Law requires that when work isn’t being done on the atmosphere by circulation between areas on the surface at different temperature (ruled out by the definition of the problem), it becomes isothermal.”
This might be true! – & you’re right, ruled out here since it expands the cv of interest to include a surface at different temperature. For the adiabatic GHG-gas column in the presence of gravity, there is no cooler surface exchanging heat with it. That’s where this gun fight might need to go once the simple adiabatic GHG-free gas column in the presence of inexplicable gravity gets general agreement about whether or not it is isothermal (if ever, the age of the universe may come into play). If it does, some agreement will no doubt be grudging, esp. me if isothermal! Then progress discussion to add a surface conducting & radiating heat into the CV with a radiating sink to space.
Turn off the inexplicable gravity in the cv of interest, no pressure stratification occurs and get isothermal PV=nRT everywhere in the gas. KE + 0 = constant everywhere. Turn on inexplicable gravity field and the pesky PE (n*g*h) term inexplicably appears & need to show KE + PE = constant everywhere in the cv or machine of interest.
Phil. says:
January 20, 2012 at 5:02 pm
“…the only way that conduction can reduce the radiation loss from the surface is by cooling the surface.”
No, that is not the only way. Stefan-Boltzmann is an equilibrium relationship. You do not know the radiation loss from the surface in non-equilibrium conditions.
You know, it is really annoying when someone proclaims they know everything about a phenomenon which is actively under investigation.
Bill Hunter says:
January 20, 2012 at 5:20 pm
“I am curious about this force. What is it and where does it come from?”
Let´s say that you inclose and isoöate the atmosphere at an arbirary time when there is wind, lightning, rain tornados etc.
Agree with me that there is a certain total energy inclosed that is constant (1:st law)
Wait a month to see what happens. The second law of thremodynamics will make sure that the entropy get maximised. At that moment the adiabatic temperature has to be -g/Cp.
It is simple!
With respect, Fred, your starting statement is just plain mistaken.
Ask yourself, what Newtonian force exerts lift upon a helium-filled balloon? Neglecting the weight of the helium (which is far lighter than the air it displaces), the net force on the balloon simply the (lesser) downward force exerted by the lower atmospheric pressure at the top surface of the balloon, relative to the (greater) upward force exerted by the larger atmospheric pressure at the bottom surface of the balloon. After all, what other force is acting upon the balloon?
More generally, what is dismaying about the WUWT posts on this topic is not the errors, but rather the wholly misplaced confidence of many WUWT posters in their erroneous reasoning; a misplaced confidence that dismayingly tends to persist even when serious mistakes in physical reasoning are pointed out.
WUWT has begun to attract criticism in this regard, see for example the scathing editorial Climate Warming Skeptics Ditch the Second Law of Thermodynamics.
It’s time for WUWT folks to accept the possibility — so great as to amount to an plain fact — that “gravito-thermal” theories have just plain got their thermodynamics wrong.
Bart,
Sorry, it isn’t equally plausible. The surface temperature would be limited by the brightness temperature of the incoming radiation or you violate the Second Law. You don’t get to have high absorptivity and low emissivity at the same wavelength. The only way that a surface couldn’t be in LTE is if the temperature is increasing (or decreasing) extremely rapidly. Not to mention that the original problem in A Matter of Some Gravity stated that the surface of the ‘planet’ was a blackbody.
If it’s plausible, then you should be able to provide examples of solid or liquid materials that don’t obey a Planck source function at 255K and above rather than just postulate their existence.
DeWitt Payne says:
January 20, 2012 at 5:34 pm
“…is obeyed for the levels under consideration.”
“For all common applications in atmospheric radiation…”
“But you’re wrong and IR thermometers, which use SB, work [on the Earth in normal conditions].”
So, all of these things hold in a quasi-equlibrium environment. Who’da thunk it?
Hans,
I almost fell into the same “the adiabatic lapse rate is the equilibrium condition since it implies the same total energy at any altitude”. Read the discussion below — you should find that the isothermal profile is indeed the correct answer. And you will see that physicists who know thermodynamics well struggle with this question, so it is no wonder that you and I could be misled. (I would love to say I was just playing the devil’s advocate, but I ignored some fundamental thermodynamics and let a too-simple argument cloud my judgement.) Since this is a fundamental assumption of your argument, it invalidates all the rest of your conclusions — sorry about that.
https://carnot.physics.buffalo.edu/archives/2012/1_2012/msg00064.html
This is really not saying anything that has not been said here, but it is more technical/mathematical that the discussions here.
DeWitt Payne says:
January 20, 2012 at 5:51 pm
Let’s ditch this line of argument, shall we? Do you deny that there is an upper limit beyond which an atmosphere will boil off, and that your 2nd law objection is therefore inapplicable? If no, then the subject is moot. If yes, then we have reached an impasse.
Willis says: “Any random movement of air from above to below a horizontal slice through the container must be matched by an equal amount going the other way.”
It is worth mentioning that since there are more molecules below the slice than above, individual molecules below the slice must be less likely to move up than individual molecules above the slice are likely to move down. Some attempted counter arguments (such as David’s) fail for not taking this aspect into account.
thepompousgit says:
January 20, 2012 at 4:26 pm
I was in the class of 1965 at Caltech, so I heard the Feynman lectures first hand. I believe Feynman’s comment applies to understanding the mechanism of quantum mechanics. There is no understanding. It just is.
Ultimately, when participating in discussions like this, you have three basic choices.
1) Enjoy it as a rousing debate, where everyone can throw out whatever ideas they want. Listen to the various points — maybe argue for your pet position.
2) Accept the conclusions of those who are the experts. In this case, pretty much everyone with an advanced degree in physics agrees that the isothermal answer is the right answer.
3) Educate yourself so you are yourself an expert. This means actually reading books and/or taking classes and/or sitting down with serious students of the topic. Sometimes the comments here are sufficient to provide that education, but don’t count on random internet posters to provide a free, clear, correct, comprehensive education.
” There is no understanding. It just is.”
So it’s, like, magic?
Er, not I, Willis said that:
Hans Jelbring says:
January 20, 2012 at 5:45 pm
Myrrh says:
January 20, 2012 at 3:03 pm
And therein lies the problem, that you’re talking about perpetual motion and you don’t know it. Hans, you are indeed proposing perpetual motion. You are saying that gravity separates the warm and cold molecules.
Your lasr sentence is correct. The reason is that maximum entropy or the second law of thermodynamics has to be applied when a gravity field is at work and we are studying a insulated inclosed atmosphere as defined in my E&E paper. Yes, the consequences are tremendous.
======
Sorry, didn’t make it clear – I was quoting Willis and responding to him.
Would you take a read of what I actually wrote in response and tell me what you think? Please don’t use maths.. 😉
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871089
My next post on here: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871204
says: “Gravity, it seems to me, is as near dammit perpetual motion as long as the optimum conditions can be maintained because it is then a constant force as source of energy, is energy since energy does work – just ‘cos it sits around looking as if its not doing anything… “
KevinK says:
January 20, 2012 at 4:04 pm
“With all due respect, I am most certainly NOT ignorant of science. With multiple Master’s Degrees in Optics and Electronics and three decades of reconciling computer predictions with actual observations I am quite familiar with all of the relevant disciplines necessary to understand the “Greenhouse Effect”.”
This is really NOT a problem of being smart or dumb. This is an arena in which many misconceptions are deeply embedded and actual data is sparse, leaving smart and able people open to reaching multiple conclusions and interpretations which are not capable of being confirmed or denied.
But, I am afraid you have jumped the gun on this:
“When the energy returns to the surface from the “GHG” you cannot ADD it to the energy arriving from the Sun to produce an alleged “energy budget”.”
Energy is constantly coming in. If some of it is made to hang around longer than instantaneously, before the new batch arrives, then you will accumulate a net offset.
It’s not about increasing energy flow, which is always nearly constant. It’s about impeding that flow so that you keep more close to you.
Hans Jelbring says:
January 20, 2012 at 4:43 pm
Good luck getting a tan with long wavelength IR. One usually gets a tan from direct normal sunlight which reaches 1000 W/m² total flux with a significant fraction of that in the UV at reasonable elevations above the horizon. But you can point an IR thermometer at the sky at night and get a temperature reading, assuming it isn’t so cold as to be below the lower end of the IR thermometer range. Mine goes down to -60 C, so it has to be really cold and dry to not register. If the sky is clear, the IR thermometer reading will always be lower than the ambient near surface air temperature. With cloud covered sky, it will be about the same.
Or, you could build a box like Roy Spencer and see that while the internal temperature goes below ambient, it would go a lot lower if it were looking at a sky with an effective temperature of 2.7 K. The other advantage of building such a box is that you could also repeat the Wood experiment and prove to yourself that he was wrong. But you would need to paint the inside black rather than white. The other problem is that Roy uses foamed polystyrene. It will likely melt at the temperatures that can be achieved exposed to direct normal sunlight even with a polyethylene film cover. I recommend cardboard coated outside with aluminum foil and baked out in a 120C oven before applying the window and sealing. Otherwise you get condensation on the window. You also need to insulate the walls of the box with six inches (at least) of fiberglass insulation.
Smokey says at 1/20 6:32pm:
“So it’s, like, magic?”
LOL, for quantum mechanics, Einstein set the terminology – he didn’t go so far as to say “magic” but was ok with “spooky” action at a distance.
DeWitt Payne says:
January 20, 2012 at 6:13 pm
“The surface temperature would be limited by the brightness temperature of the incoming radiation or you violate the Second Law.”
My capacitor analogy is good here. Incoming radiation is the current. Temperature is the voltage. The capacitor accumulates charge continuously, and the voltage climbs continuously.
Does it violate the 2nd law? Yes, if it is an ideal capacitor. Which is precisely why an ideal capacitor cannot exist
Similarly, a non-radiating atmosphere cannot exist. So, if you’re looking for a violation of the 2nd law, it is embedded in the fundamental premise of the problem. But, that does not mean we cannot draw useful conclusions from the thought experiment.
sky says:
January 20, 2012 at 5:26 pm
Umm, it’s the same as the surface temperature. That’s what isothermal means. Of course it depends on what you mean by the TOA. I take it to mean the maximum altitude where the kinetic energy distribution of the molecules is still accurately described by the Boltzmann distribution. That’s going to be on the order of 100 km and includes 99.9999% or so of the total mass of the atmosphere. There’s still some mass left that is gravitationally bound, but when the mean free path is measured in km, the collision frequency is low and LTE doesn’t apply.
Willis:
“jae, I fear I don’t have a clue which “empirical evidence” you are speaking of. If it has to do with the N&Z hypothesis, I don’t understand the hypothesis so “evidence” means nothing. Let me know which evidence and which theory you’re talking about.
More light and less heat would help here.”
Well, I will assume you are being honest, because you always have been, AFAIK. But…maybe because you get a “headache” when you read Huffman’s stuff (your words, I think), you did not really read his stuff and pay attention to the DATA (aka, EMPIRICAL EVIDENCE!). Same with the other papers. It seems that the atmospheres of other planetoids that have atmospheres have temperatures that are not correlated in any way to the amounts or types of GHGs present in those atmospheres. You continue to ignore all that part of the question, and do not confront it/explain your position. Is it confirmation bias? Old age? Ego? What, Willis?
AND, Willis, we have the little issue of rapidly increasing GHGs (OCO), but steady or maybe decreasing temperatures for the last few years! Although many other factors are involved, this fact does not help the concept the GHG/heating nonsense and must be annoying, no?
DeWitt Payne says:
January 20, 2012 at 6:13 pm
“The only way that a surface couldn’t be in LTE is if the temperature is increasing (or decreasing) extremely rapidly.”
Exponential rise is generally considered “extremely rapid”. What would the exponent be? What would you expect the heat capacity of a non-radiating material to be?
Bill Hunter says:
January 20, 2012 at 5:01 pm
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!
If sealed thermally isolated containers of air were to gravitationally separate, with warm air at the bottom and cool air at the top, then we could build tall insulated cylinders and pull power out of them forever.”
Apparently you’ve never heard of OTEC.
http://en.wikipedia.org/wiki/Ocean_thermal_energy_conversion
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!
Why do you think the oceans are colder on the bottom and hotter on the top? Gravity separates the oceans by temperature.
However, water is incompressible, which explains why the oceans are colder at the bottom. Water can only change density as a result of temperature. Air on the other hand changes density in response to both temperature and altitude.
For Davidmhoffer, if you liked my shorter analysis this longer analysis may be of interest to you;
The CRITICAL flaw in the “Greenhouse Gas” hypothesis….
Lots of attention has been given to the alleged “Greenhouse Effect” over the last few decades. It seems an elegant hypothesis that can explain almost every weather effect that occurs (i.e. droughts, floods, blizzards, warm spells, cold spells, more arctic ice, less arctic ice, shrinking glaciers, etc.)
However this hypothesis has a CRITICAL flaw.
Please allow me to me explain…
In any proper analysis of energy flow through a complex system it is necessary to stop occasionally to perform a “sanity check”, this indispensable tool is applied by engineering professionals to ascertain if our predictions/calculations still make sense in regard to the system we are analyzing/designing.
So let’s do a simple sanity check on the “Greenhouse Effect” (moving forward this will be abbreviated as the “GHE” for simplicity).
To quickly summarize the GHE;
1) Visible light (aka EMR radiation) is absorbed by the surface of the Earth
2) The warmed surface of the Earth emits Long Wave Infrared (LWIR) light
3) This LWIR is absorbed by gases in the atmosphere
4) These “warmed” gases emit energy back towards the surface of the Earth
5) And THEREFORE the “GHGs” cause the surface of the Earth to acquire a higher “equilibrium” temperature
I hope I have captured the essential essence of the GHE with this short summary. Of course, I fully expect some folks to disagree and inform me that I have misunderstood the GHE. So be it, we will deal with that at a future time.
To describe the flaw in the GHE we first need to conceptualize a convenient bundle of energy that we can follow through the Sun/Earth/Atmosphere/Universe system. This bundle of energy needs to be large enough so we can (largely) ignore effects that happen at the atomic level. Also the bundle needs to be small enough so we can track it in a “real time” fashion and determine what happens to the energy.
For the purposes of this discussion I propose we use 1 milliJoule of energy. This is 1 one-thousandth of a Joule. This should be just the right size for my purpose of demonstrating the flaw in the GHE hypothesis.
So let’s get to it, here is the sequence of events involved in the GHE;
1) A bundle (1 mJ) of energy in the form of mostly visible light arrives at a specific location on the Earth’s surface, courtesy of our friendly neighbor the Sun.
2) Some portion of the bundle is immediately reflected and departs at the speed of light in an opposing direction. I won’t bother to get into an interminable discussion of what the Albedo is since this proof does not require this knowledge. In fact it could be any number above zero and less than or equal to 1 without making any difference.
3) The remaining portion of the bundle is absorbed by the surface and converted to heat, this is the FIRST warming event (Warming Event #1) caused by our little bundle of energy. And if we know what material (sea water, soil, etc.) absorbs the bundle we can make a pretty good estimate of the temperature increase, but that calculation is not necessary.
4) Now, several things will happen to this bundle of thermal energy;
a) It can be conducted to locations below (or adjacent to) the surface which are colder.
b) It can be conducted to a stationary gas molecule in contact with the surface.
c) It can be convected away by a moving gas molecule that happens to contact the surface while travelling past.
d) It can be radiated as Infrared Radiation away from the surface.
It is important to note that in most cases all four of these events will happen in parallel (i.e. at the same time), so the actual Portion Radiated can vary from 0% to 100%.
5) No matter what happens in step 4, the surface cools (Cooling Event #1) by an amount which corresponds to the loss of of energy at that location. The amount radiated is somewhere between 0% and 100% (Albedo * Portion Radiated) leaving our radiated bundle at somewhere between 0 and 1 mJ. Again, if we wanted to calculate the temperature drop associated with the cooling we could use the thermal capacity of the material to make an estimate. It is important to note at this time that this absorption/remission process IS NOT instantaneous; it requires a finite amount of time (Time Delay #1) that is greater than zero.
6) The radiated energy (something between 0 and 1 mJ) now travels away from the surface at something close to the speed of light. Note that there are a few “flavors” of the speed of light depending on the material our little bundle is travelling through. The fastest flavor is when the light is travelling through a vacuum. When moving through the lower atmosphere the speed is slightly slower.
7) Our little bundle, speeding happily along and accelerating as the atmosphere gets less dense MAY be absorbed by a GHG. Whoops, that’s inopportune; it was hoping to get to Alpha Centuri before next Tuesday to watch the REAL Miss Universe show (ok a lame joke). The important thing to note is that a finite amount of time (Time Delay #2) has elapsed before the absorption occurs.
8) Once our bundle of energy is absorbed by the GHG it ceases to exist as IR light and is converted to heat (Warming Event #2). This is our little bundle’s second warming event within a few milliseconds, boy am I proud. Note that with any flavor of the speed of light our bundle can make it to the top of the atmosphere (TOA) in a few milliseconds at most.
9) Now, of course the same possibilities shown in step #4 may happen at the GHG molecule. Again all four of the possibilities can occur. To speed up this discussion we will assume that 100% of the energy is emitted as IR radiation. So the best case (or worst case if you still believe in the GHE) is that all of our 1 mJ has arrived at the GHG molecule and is going to be radiated away. A couple of important points need to be made here, first something less than 50% of our bundle can hope to head back towards the surface (as fixed by the geometry of a sphere). And secondly there is some finite time delay involved in the absorption/remission event (Time Delay #3).
10) Once the bundle of energy has been radiated by the GHG molecule the molecule will cool (Cooling Event #2) by an amount commensurate with the energy reradiated. And we could, if we wished, estimate the temperature drop, but it is immaterial.
11) Once again our little bundle is happily speeding along at the speed of light, but slowing down this time as the atmosphere gets denser. Oh well, zipping along at any flavor of the speed of light still beats a Lamborghini. And of course our little bundle is now no more than half of the man (or woman) that it used to be.
12) Damn, we flew our bundle right down into the ground (after Time Delay #4); I hate it when that happens. But the good news is we now have GHE induced warming (Warming Event #3). We could again have the Albedo argument (in the IR portion of the spectrum this time), but it does not matter. The IR radiation has ceased to exist and is now heat. So this takes us full circle and we are back at step #3, EXCEPT, AND THIS IS A REALLY BIG EXCEPT STATEMENT, two subsequent warming and two subsequent cooling events with four time delays interspersed have also taken place in the meantime. TO SUM THE FIRST WARMING EVENT WITH THE THIRD WARMING EVENT (both of which happen at the Earth’s surface) CLEARLY VIOLATES THE FIRST LAW OF THERMODYNAMICS. THE GHE DOES NOT AND CANNOT CAUSE ANY “HIGHER EQUILIBRIUM” TEMPERATURE TO EXIST AT THE SURFACE OF THE EARTH (or anyplace else for that matter).
13) You are free to expand this time series to describe the third, fourth, fifth, etc, etc. sequential warming/cooling events if you wish, but I am extremely proud of my little bundle at this point and do not see the need. It should be noted that as one bundle of energy is absorbed/reemitted multiple times the amount reaching the surface declines as follows, 50%, 25%, 12.5%, 6.25%, 3.125%, 1.5625%, 0.78125%, 0.390625%, 0.1953125%, 0.09765625%. So after as few as ten bounces (taking at maximum perhaps 40 milliseconds if the energy makes it all the way to the TOA) the energy is already less than 1/10 of 1 percent of the initial amount.
I know lots of folks will disparage this discussion. So I will suggest a few FAQs in advance;
1) Q: But, but, but… all the other proud energy bundle parents are sending theirs out to do the same thing, so surely the volume of little bundles will make the GHE occur.
A: NO, all of the little bundles are travelling in parallel and do not sum. If we cannot demonstrate how one bundle of energy can make the GHE real, then we cannot claim that the GHE exists when discussing a “higher equilibrium” temperature
2) Q: But, but, but… the time delays you suggest are so long that energy is left over at the end of each day, and this is really what the GHE is about.
A: NO, at the speed of light each bundle could make 10 round trips to TOA and back to the surface in less than about 40 milliseconds. For reference, each day contains about 86 million milliseconds. So no energy is left over at the end of each day. Besides if energy was left over the Earth would slowly heat up 1 little notch at the end of each day and would have melted a long time ago (ignoring for a moment the fact that as the surface warms the emission spectrum shifts to lower (i.e. more transparent) wavelengths).
3) Q: But, but, but… you assumed the wrong value for the Albedo, Lapse Rate, Amount of
GHGs present, etc. etc. etc.
A: NO, this discussion is about the fact that sequential warming and cooling events occur with finite time delays (more than zero, but otherwise undefined) between each event and the events have finite (greater than zero and less than or equal to one) probabilities of occurrence and therefore CANNOT be summed to yield a “higher equilibrium” temperature.
4) Q: But, but, but… you have not presented any explanation about how the surface of the Earth reaches its average temperature without the GHE.
A: Agreed, I have not yet presented any theories about why the temperature of the Earth is what it is. So here I will;
First, I believe that the application of the Stephan-Boltzmann equation in conjunction with the
Kirchhoff radiative equilibrium equation have been performed incorrectly. Yes, I realize that’s a big deal because it’s written in all the climate science textbooks, so IT MUST BE SO. Well, if we never revised textbooks we would be in a bunch of hurt. Just because it is written down in a 20 year old book does not necessarily make it so. And a computer model that implements the assumptions of a hypothesis IS NOT a proof of the hypothesis, regardless of how fast the computer can collide numbers together.
Second, I suggest that the average temperature of the Earth is determined by the massive thermal capacity of the Oceans and the amount of energy already deposited there (by some undefined prior event(s)). Clearly, the Oceans do not respond to changes in the energy arriving at the surface on a time period of days, or even hundreds of years. A good analogy (as an electrical engineer I must mention this) would be the battery that you (likely) use to start your internal combustion car each day. It has a bunch of energy stored inside (provided by the manufacturer). Each day when you start your car you suck a whole bunch of energy out of it. Then you slowly recharge it (with the alternator) while you drive. So it always has a bunch of energy present, we are missing the fact that the manufacturer filled it up before you bought it. Just like the Ocean was already “charged up” before we invented the GHE.
5) Q: But, but, but… you are ignoring all the “evidence” of climate change, the shrinking glaciers, the floods, the droughts, the heat waves, etc. etc….
A: Two Words; CONFIRMATION BIAS. And unlike our current Vice President, I can in fact perform a simple task like counting the number of words in my statements.
6) Q: But, but, but… you are not a climate scientist that has peer reviewed publications, so we should not listen to you.
A: Yes, I am not a climate scientist with peer reviewed publications. This is in fact a situation I take some pride in. This discussion relies on a simple flow of logic. Read it at your own risk. Find the flaws. Point them out. Be skeptical.
In summary, the critical flaw in the GHE hypothesis is that the warming from GHGs happens sequentially AFTER a previous cooling event WITH an intervening time delay, THUS it cannot be added to the initial warming event. And therefore the GHE does not create “extra energy” or “net energy gains”.
Further, these warming and subsequent cooling events happen so quickly that the GHE has nothing to do with the “equilibrium” temperature present at the surface of the Earth.
Further, the warming from the GHE dissipates so quickly that it cannot be reasonably expected to influence the massive thermal capacity of the Oceans in any way.
Cheers, Kevin.
thepompousgit says (in response to Hans Jelbring):
Yeah…The way I was going to put it is that infrared radiation doesn’t give you a sun tan for the same reason that cell phones don’t give you brain cancer: The energy of the photons are way too small to cause this sort of cell damage to occur.
Einstein figured this out about a century ago.
pochas says:
This may well be the understatement of the century!
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!
Consider ice. Ice is colder than liquid water, yet unlike liquid water, ice floats to the top (because of gravity). Liquid water on the other hand sinks when it gets colder. Again due to gravity.
Ice shows that density work together with gravity to separate the oceans by temperature.
Almost by definition I am the guy for which an elevator speech is supposed to be comprehensible.
I may be asking some bizarre stuff but if somebody doesn’t start answering them I would have to think that there is no hope or desire by anybody to come up with an elevator speech.
So that said here is one more. I have been led to believe 2 things.
1) the adiabatic convention process does not change the kinetic energy of a molecule of gas but does change the temperature. Though kinetic energy might change via radiation to space or to another molecule that lost some of its energy to space.
2) My reading on thermopiles is they do not respond to absolute temperature but instead gradients represented by kinetic energy differences (e.g. frequently also temperature gradients).
Are these substantially correct facts?
DeWitt Payne said @ January 20, 2012 at 6:21 pm
De Witt, I envy you. Nothing to disagree with when you use the modifier. Did you by any chance attend the lectures on the derivation of Newton’s Laws from conic sections? They are my favourite…
Bart wrote;
“Energy is constantly coming in. If some of it is made to hang around longer than instantaneously, before the new batch arrives, then you will accumulate a net offset.”
Bart, have you ever taken any classes in the analysis of AC (alternating current) circuits ? I am very serious in this question.
If you have determined a means to effect a DC (direct current) “offset” by changing the response time of an electrical circuit you are quite a bit ahead of those that do electrical circuit design for a living. We can do this but we use diodes (aka rectifiers) to accomplish this task. the GHE does not behave as a rectifier does (i.e. energy only flows one way).
I suggest you study electrical engineering a bit before you discuss “offsets”, it is quite a BIG topic in our field.
Cheers, Kevin.
(To Moderators: Very Sorry – previous post unfinished and held a contradictory statement – please use this one) [Done -w.]
Taking the most simplistic approach possible:
A uniformly irradiated world, no rotation, no poles, two gas atmosphere (no water/condensation)
Atmosphere is uniformly dense – no pressure gradient:
Assuming there is only 1 GHG gas present in trace amounts and only that GHG can absorb IR energy. Assuming the ‘one other gas’ TG is totally transparent to all incoming and outgoing radiation.
Assuming all energy absorbed by the surface from the sun is then emitted at one IR wavelength which is only (and totally absorbed) by the one GHG gas.
Assuming all IR energy (100%) is absorbed by that GHG in the first 10 meters above the surface.
Assuming that energy is transferred rapidly by collision to TG molecules near the surface.
Assuming it is largely the TG molecules which carry the energy to the upper atmosphere by convection.
Assuming the energy can only be emitted to space by being passed back to a GHG in the upper atmosphere.
Case 1 Double the quantity of GHG, energy is totally absorbed and passed to TG twice as rapidly. After the system equilibrates there is a trace amount of extra energy retained in the atmosphere at any point in time. (related to that held by the extra GHG molecules).
There are now twice as many GHG molecules in the upper atmosphere with the opportunity to emit energy to space, but there is also twice as much opportunity they will intercept each-others’ IR emissions – average emission to space remains the same, and atmospheric temperature remains (almost) the same.
Case 2 Double the quantity of TG, retaining the original level of GHG There is twice the opportunity of collision between GGH and TG at all levels of atmosphere , energy near the surface is transferred more quickly to the TG. The atmosphere can now retain twice the energy of the original world. In the upper atmosphere there is twice the opportunity to pass the energy from the TG to the emitting GHG. Emission rate in the upper atmosphere is dependent on the opportunity of the GHG to collect some energy (now doubled), and the density of GHG at altitude (only another GHG molecules can intercept the transfer of IR energy to space). As GHG density at the upper atmosphere is still the same, emission rate should double (Twice the opportunity to collect energy, but retaining the original opportunity to emit to space.)
The lower atmosphere at equilibrium retains twice the energy of the original world. The upper atmosphere emits it twice as quickly. What effect does this have on atmospheric (and surface) temperature?
Repeat the exercise with graduated layers of from greater to lesser density atmosphere. Do the effects remain the same layer by layer?
KevinK says:
January 20, 2012 at 7:28 pm
“I suggest you study electrical engineering a bit before you discuss “offsets”, it is quite a BIG topic in our field.”
Maybe you need to review laser cavities a little.
KevinK says:
January 20, 2012 at 7:28 pm
“We can do this but we use diodes (aka rectifiers) to accomplish this task. the GHE does not behave as a rectifier does (i.e. energy only flows one way).”
Actually, that is exactly what is happening, so this might help you comprehend. The energy comes in shortwave. It goes out longwave. So-called GHGs are transparent to shortwave, opaque to longwave.
KevinK says:
January 20, 2012 at 7:28 pm
For the other people with whom I have been discussing my hypothesis of “GHG” cooling: Anyway, that’s the standard greenhouse argument. I am not saying GHGs necessarily cause the ground to heat more than it would otherwise. But, what they are doing has the same steady state effect as if they were, and the interpretation in the steady state is merely ambiguous. Kevin has simply gone off the rails, and I am trying to help him understand the GHG argument, and that he has done nothing to refute it.
Ok, I now declare this an OFFICIAL FARCE…….
We clearly have two camps, the believers in the GHE and those that have sound technical reasons to DOUBT it.
I suggest we all shovel the snow from our driveways and reconvene next summer to discuss it again……………..
Cheers, Kevin (see you after the thaw)
KevinK;
Well said. May I offer a simple analogy and then launch into an extension of your explanation that ties back to N&Z? Thanks!
Analogy:
I dam a river and form a lake. At equilibrium, the river flowing in has a flow rate of 240 m3/min. Outflow at the dam = 240 m3/min. I raise the dam one meter. After a short period of time, the flow rate in is… 240 m3/min and the outflow is 240 m3/min. The depth of the water increases by 1 meter, and the potential energy stored as a consequence of that can be calculated, but the flow rate at equilibrium changes not one bit. The same is true of doubling C02. Temporary fluctuation in the system, but one equilibrium is established again, the incoming w/m2 and the outgoing w/m2 are exactly what they were before. There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt. Since everyone wants to average w/m2 and convert that to degrees, I say let ’em. If they cannot show that the incoming watts absorbed has changed, then at equilibrium the outgoing watts have to be the same too. Since the outgoing watts are the same, temperature of the planet is the same. Go ahead folks, use SB Law to contradict that.
Onto N&Z building on what KevinK explained.
Imagine that there are upward bound photons being released from earth surface and from every layer of the atmosphere at every elevation. Some escape, and some don’t. The most photons emerge from the earth surface, BUT… they have the smallest chance of getting out to space because they have the greatest chance of being absorbed on the way up. As we increase in altitude, there are less upward bound photons released from a given layer because a) the temperature is lower and b) the density is lower. BUT…the percentage chance that any upward bound photon will escape to space increases with altitude. Let’s repeat that. The higher the altitude, the greater the percentage chance that any given upward bound photon will escape to space is.
As KevinK explained, once you hit equilibrium, every photon rising from earth surface cools the surface. If it is intercepted somewhere in the atmosphere, then that spot in the atmosphere warms. If the photon is re-emitted, then that spot in the atmosphere cools. If the photon goes downward, it must inevitably be absorbed again, either in the atmosphere or the earth surface, and where ever that happens, that spot warms and re-emitts. At some point though, the photon escapes, taking with it the exact same amount of energy as it brought when it entered the system. The net is zero.
So…as seen from space, let us assume we can “see” an average (ignorant term and wrong but for sake of argument I’ll live with it for this example) we can “see” 240 w/m2 exiting the system. How many watts came from where? Some of the photons would in fact be emitted from earth surface and go straight to space uninterrupted. How many? Not many. Even though the most upward bound photons in the system are being released from the warmest surface (earth surface) not many escape. The frigid TOA though is pouring the photons into space. It might be cold at TOA, but every upward bound photon emitted at TOA has a 100% chance of escaping to space.
Hence….
The greater the mass of the atmosphere, the lower the chance of any energy packet escaping to space. At earth surface, conduction and convection will massively dominate radiance, and photons released via radiance have a near zero chance of escaping directly to space anyway. At high elevations, while the temperature is must lower than earth surface, conduction and convection are vastly reduced. While the number of photons is also vastly reduced due to the lower temperature, the percentage chance that any given one of them will escape is vastly higher.
Hence…
The greater the mass of the atmosphere, the greater the surface pressure and the greater the dominance of conduction and convection and the less impact on temperature of radiance. The greater the mass of the atmosphere, the more skewed the ratio of photons released at high altitudes will be. And if one “averages” the upward bound flux that actually escapes to space from earth surface to TOA, what will one get?
One will get exactly the total flux being absorbed.
If 240 w/m2 is being absorbed, then 240 w/m2 is existing. The greater the mass of the atmosphere, the more likely that any given energy packet will be returned from the surface straight back to the surface, and thus the higher the surface temperature. The greater the atmospheric mass (and hence surface pressure) the more skewed to higher (colder) altitudes the percentage of escaping photons will be. BUT… the average will always equal exactly the mount being absorbed.
Hence, as per N&Z…
The greater the mass of the atmosphere, the higher the surface pressure and so the higher the surface temperature. Calculate insolation absorbed, and you now have two variables, insolation and surface pressure, that define the surface temperature. Double or quintuple CO2 and you know how much the surface temperature changes? It changes by the amount that the mass of the atmopshere (and hence surface pressure) changes. But the incoming and outgoing watts remain (at equilibrium) unchanged.
Just like the water going over the dam.
Dammit
Myrrh says:
January 20, 2012 at 5:38 pm
Stephen Wilde says:
January 20, 2012 at 3:25 pm
Gravity does not separate by temperature. It separates by mass.
This is what I meant, but here said more succinctly..
======
er.., I meant weight…
http://www.exploratorium.edu/ronh/weight/
“jae says:
January 19, 2012 at 7:05 pm
DAMMIT, WILLIS:
PLEASE ADDRESS THE EMPIRICAL EVIDENCE, WHICH WILL ULTIMATELY RESOLVE THIS ARGUMENT! WHY DO YOU REFUSE TO DO THIS?
[Moderator’s suggestion: If you didn’t YELL at him maybe more would be accomplished? Maybe? -REP]”
It seems that some of Anthony’s “moderators” need some moderation. Is it really the moderator’s place to tell me just how to express myself on a blog? I think NOT! Just what the hell do you think you are?
Bart says:
January 20, 2012 at 6:17 pm
Of course there’s an upper limit, but there’s not very much mass between that level, where the atmosphere consists almost entirely of hydrogen and helium, and the level where LTE applies, which would be isothermal at the surface temperature under the conditions specified by Willis in his original Gravity post.
Out of curiosity I calculated the pressure a function of altitude for a purely adiabatic atmosphere. For a surface temperature of 255K, it would reach a pressure of zero and a temperature of 0K at an altitude of about 26.2 km. But there’s no way it would stay that way as the thermal diffusivity gets quite large as the pressure goes to zero.
ferd berple says:
January 20, 2012 at 7:12 pm
“Why do you think the oceans are colder on the bottom and hotter on the top? Gravity separates the oceans by temperature.”
Yes, but only accounts for a small portion of the temperature gradient in the upper ocean.
KevinK says:
Speaking of sanity check, you may want to try one on your argument. For example, can you extend it to show that there’s no advantage to wearing a jacket when you go out on the cold or putting any insulation in the attic of your house? (Basically any case where you have the steady-state temperature of an object determined by the balance of what energy receives and what energy emits.)
It is amazing how difficult basic concepts become for people, even “engineering professionals” when the concepts conflict with what they strongly want to be the case!
Bart says:
January 20, 2012 at 6:50 pm
What useful conclusions can be drawn? A surface that can absorb radiation but not emit it at any wavelength at any temperature while having zero heat capacity goes far beyond postulating a perfectly transparent atmosphere or a surface with an absorptivity identically equal to 1 as limiting cases in thought experiments. It’s so far removed from reality as to be worthy of the term unphysical and the results are completely irrelevant to anything real. It says nothing whatsoever about the atmospheric greenhouse effect on a real planet. Especially it does not say that addition of greenhouse gases cause the surface of a rotating spherical planet with an IR absorbing atmosphere in orbit around a single star to cool compared to an atmosphere that is more transparent in the IR.
“Taking the most simplistic approach possible:
A uniformly irradiated world, no rotation, no poles, two gas atmosphere (no water/condensation)
Atmosphere is uniformly dense – no pressure gradient:
Assuming there is only 1 GHG gas present in trace amounts and only that GHG can absorb IR energy. Assuming the ‘one other gas’ TG is totally transparent to all incoming and outgoing radiation.
Assuming all energy absorbed by the surface from the sun is then emitted at one IR wavelength which is only (and totally absorbed) by the one GHG gas
Assuming all IR energy (100%) is absorbed by that GHG in the first 10 meters above the surface.”
Let’s stop here.
Let’s look at this world. Unless one can see this IR energy [humans without devices can’t], from space this planet would be utterly black.
Next problem is nothing absorbs all the sun’s radiation.
If we assume you mean that a limited amount of the sun’s radiation is absorbed and have the rest reflecting, we fix the problem of having a utterly black world and it’s conceivable that the GHG is absorbing this limited band width. It’s quite simple actually, you have CO2 ocean and a CO2 GHG gas. The CO2 ocean only absorbs and emits what the CO2 gas can absorb and emit.
Venus is almost this- but Venus ocean is not a liquid, but instead it’s a hot dense gas. Though don’t think matters whether it’s gas or liquid. On Venus visible light reaches the surface but we can assume any wavelength the sun emits which CO2 can absorb isn’t shining from the sun to the surface- it’s being absorbed and emitted by the CO2. The CO2 is not a “surface” [liquid or solid] but what difference does it make?
So it seems you are basically talking about Venus. If you don’t mean the planet is utterly dark from space [or on the planet].
Continuing:
“Assuming that energy is transferred rapidly be collision to TG molecules near the surface
Assuming it is largely the TG molecules which carry the energy to the upper atmosphere by convection
Assuming the energy can only be emitted to space by being passed back to a GHG in the upper atmosphere”
Again still Venus. Though difference is CO2 not trace gas, and TG is actually minor constituent
of atmosphere [around 3% but that is about 3 earth atmospheres of N2- so a lot of N2 but dwarf by massive amount of CO2]
“Double the quantity of GHG, energy is totally absorbed and passed to TG twice as rapidly.
After the system equilibrates there is a trace amount of extra energy retained in the atmosphere at any point in time. ”
We could 1/2 the amount of CO2 on Venus and thereby double the amount of N2 relative to the CO2:)
I think Venus would cool if lost half it’s CO2.
But replaced the half of CO2 with same amount of
N2- took out 46 atm of CO2 and replaced with 46 atm of N2, then I would suppose that the Venus surface would become much dimmer- not allowing as much diffused sunlight from reaching the surface. But don’t think that would change the planet’s temperature.
“(related to that held by the extra GHG molecules).
There are now twice as many GHG molecules in the upper atmosphere with the opportunity to emit energy to space, but there is also twice as much opportunity they will intercept each-others IR emission – average emission to space remains the same, and atmospheric temperature remains the same.
“Double the quantity of TG. There is twice the opportunity of collision between GGH and TG at all levels of atmosphere, energy near the surface is transferred more quickly to the TG.”
As said don’t think there would any difference.
Move Venus to earth orbit and Venus would change quite significantly- though require a long time to radiate the energy- so could a million years to change significantly.
“Repeat the exercise with graduated layers of from greater to lesser density atmosphere. Do the effects remain the same layer by layer?”
As saying above, if you half the amount CO2 in the atmosphere, made a 46 atm instead of 92 atm, this would radically change the temperature- lower by say 100 K. Half it again, and I would guess the temperature would lower again significantly and because Venus does not rotate, one could get liquid CO2 on nite side and cascading lowering of atmosphere pressure, and further lowering of temperature. Or have similar effect of moving Venus out to Earth orbit.
Venus would develop very small ocean of liquid CO2- possibly seeping under ground or making lakes in lowest elevations.
thepompousgit says:
January 20, 2012 at 7:28 pm
He did that (those) lecture(s) after I graduated, as I remember. I do have a copy on CD of the lecture on deriving the 1/r² force relationship from Kepler’s Laws, though. It’s too bad there weren’t pictures of the blackboards he used. Newton used conic sections. Feynman used a more readily comprehensible geometric approach. As I remember, Feynman did it that way because he couldn’t follow Newton’s proof. Conic sections were big in Newton’s day.
Cheers, Kevin.
Joel Shore says:
January 20, 2012 at 7:15 pm
thepompousgit says (in response to Hans Jelbring):
I get my suntan from UV radiation. Where is this fairytale land that enables one to get a suntan from IR radiation? Or are you just saying whatever pops into your head because you know the ignoramuses will believe whatever you say? Sorry if this is rude, but the difference between UV & IR is really basic stuff!
Yeah…The way I was going to put it is that infrared radiation doesn’t give you a sun tan for the same reason that cell phones don’t give you brain cancer: The energy of the photons are way too small to cause this sort of cell damage to occur.
Einstein figured this out about a century ago.
=============
Go out in the midday sun in the tropics, photon for photon UV won’t be able to penetrate further than the epidermis, the first layer of skin, to give you a tan while thermal infrared, heat, will penetrate some inches and its energy absorbed by the water in you, you’re mostly that and water is the great absorber of heat, will move the water into vibration, kinetic energy, heat. UV will damage you from the outside in and you won’t know it’s happening until you can feel the effects; thermal infrared, if you’re unable to sweat it out, will boil you from the inside out. You’ll boil quicker than you’ll get skin damaged to death… Put on a shirt, you stop UV in its tracks, no suntan, that’s how strong an energy it is.. it’s tiny, really tiny, compared with thermal infrared.
UV doesn’t give you a tan because it burns you, you can’t feel UV, it doesn’t burn you, it gives you a tan because the melanin in your body is doing its best to neutralise excess of it scrambling your skin on a DNA level. It is essential for the production of vitamin D. Thermal ir is bigger and packs more punch, stand in front of a fire, that’s thermal ir you can feel burning you directly. Compare with sunbed.
There’s a world of difference between UV photo-damage and thermal infrared heat damage.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2671032/
Thermal infrared is the size of a pin head, near ir is microscopic, how big is UV compared with them?
“8) Once our bundle of energy is absorbed by the GHG it ceases to exist as IR light and is converted to heat (Warming Event #2). This is our little bundle’s second warming event within a few milliseconds, boy am I proud. Note that with any flavor of the speed of light our bundle can make it to the top of the atmosphere (TOA) in a few milliseconds at most. ”
What is meant by warming event. Has the GHG gas molecule velocity increased, or is it excited molecule which may soon emit the same wavelength or same different wavelength that it is capable of emitting.
The isentropic state is the state of maximum entropy. If you start with an isothermal atmosphere and thoroughly mix it without adding energy, you end up with an isentropic state which has the adiabatic lapse rate. Once in this state, the air can’t be unmixed (because entropy is maximized), so in this sense it is a final state for a given energy content. This doesn’t mean that an isothermal state or any stable lapse rate cannot last an infinite amount of time. I don’t think the problem is well posed because many states can last infinitely long if only diffusion is acting, but only the isentropic state cannot be transformed to another state by mixing, in my opinion.
gbaikie says:
January 20, 2012 at 8:47 pm
I’ve actually run some numbers on Venus using the line-by-line program at spectralcalc.com . They have a database of high pressure CO2 absorption lines. Because the surface temperature is so much higher, the 15 μm band of CO2 doesn’t contribute much. Even at 92 bar, there are still ranges of relatively low absorption for CO2 alone. However, those gaps are nearly filled by SO2 and H2O, even though the partial pressure of H2O is small. Replacing half the CO2 with nitrogen does cool the surface, but not by much. Another interesting point is that it isn’t completely dark in the visible spectrum at the surface when the sun is above the horizon. One of the Russian probes that made it to the surface managed to transmit some visible light photographs before it failed. You do have to lower the total pressure to get major a major reduction in temperature. We had a big discussion of this over at scienceofdoom.com .
Joel Shore;
Speaking of sanity check, you may want to try one on your argument. For example, can you extend it to show that there’s no advantage to wearing a jacket when you go out on the cold or putting any insulation in the attic of your house? >>>
Good lord man, what brand of cereal did your physics degree come in anyway? The person wearing the jacket and the home with the attic both have INTERNAL heat sources. The increased layers of insulation change the temperature gradient such that is is more pronounced closer to the heat source, hence warmer “inside” the jacket and “inside” the house. Which has what to do with the topic at hand?
While you’re trying to come up with an answer for that:
The earth absorbs 240 w/m2. At equilibrium, the earth emitts 240 w/m2. Temperature (accoding to you and your wrongly applied SB Law) is 255K
CO2 quintuples. The system is perturbed until equilibrium is once again established in which event:
The earth absorbs 240 w/m2. At equilibrium, the earth emitts 240 w/m2. Temperature (according to you and your wrongly applied SB Law) is 255K.
Change from quintupling CO2 = 0.
Bill Hunter says:
January 20, 2012 at 8:35 pm
“Yes, but only accounts for a small portion of the temperature gradient in the upper ocean.”
Any can disagree. Tell us why you disagree. What accounts for the other portion?
Been thinking about an increase over the SB average. It would be higher because of the thermal lag in a purely mechanical atmosphere system without GHG’s. In a GHG free atmosphere the molecules cannot shed the energy gained mechanically from the ground during the day. I do think that GHG’s would raise the temperature further, in line with the equations in Loudon, but not according to Hansen’s empirical relation.
davidmhoffer says:
January 20, 2012 at 8:10 pm
“There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt.
The water retained is the analogy to energy retained. And, if heat capacity is unchanged, that means the temperature must go up in the GHG paradigm.
However, here’s what I believe is wrong about the GHG paradigm: the example is inverted. The dam is already there. Having the dam a particular height means the water behind it is at a particular height. In introducing IR emitting gases, you are, in fact, opening up new floodgates in the dam (more avenues for energy to radiate away), and the water will go down.
DeWitt Payne says:
January 20, 2012 at 8:23 pm
Of course there’s an upper limit,…
I was speaking of a temperature limit. Your objection was that there was a 2nd law violation because the temperature could increase forever. It couldn’t for two reasons: 1) all real atmospheres will radiate somewhere 2) in the worst case, the atmosphere simply escapes and stops heating.
“Out of curiosity I calculated the pressure a function of altitude for a purely adiabatic atmosphere.”
You can’t get adiabatic conditions when there is no heat sink in the upper atmosphere. This is a basic assumption in all the text books when they calculate adiabatic lapse rates, and one of the items which led me to my hypothesis, which is that IR gases in the Earth’s atmosphere are, in fact, heat sinks.
DeWitt Payne says:
January 20, 2012 at 8:46 pm
“…”
I’d appreciate it if you try that again. You’re not talking about anything I’m talking about. I am saying that, on this hypothetical planet, the surface is a blackbody, but its emissions are suppressed while massive amounts of heat are being drawn out of it to feed an ever hungrier atmosphere. At least, that’s my current running hypothesis. It could change to accomodate new facts as I gather them. But, so far, I do not see any evidence that says this scenario is unphysical or impossible, and it is very appealing from the point of view of stability.
ferd berple says:
January 20, 2012 at 7:19 pm
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!”
If Willis says it, then he is pulling your leg. Gravity can affect a tall column of gas and will separate temperature, but there is no work being done in that process and there is no pump involved, just compression and expansion. As much material goes up as down. Just because there is a temperature gradient does not mean you can extract energy from it indefinitely because extraction will cool the whole system.
This is straightforward. A system at any given temperature above 0 K can be cooled. A system with a temperature gradient can be cooled using a heat engine that exploits the temperature gradient. But that is nothing like ‘free enery forever’ That is ‘stored energy being converted to a non-thermal form’.
A spherical planet with some gravity and an atmosphere isolated from the universe will have a temperature gradient from top to bottom, cold to hot, because of the universal gas law and Brownian motion. Energy can indeed be extracted from the temperature difference between the top and bottom of the system provided that device is not entirely made of gas – it has to not respond to the universal gas law. It will produce, say, electricity. That is a conversion of heat into electrical energy. The whole system cools in response. That is not free energy and ‘gravity’ will not make up for what is missing. That is a thermoelectric generator. Over time the bottom of atmosphere will cool until, in theory, the atmosphere condenses. Gravity does not ‘heat the atmosphere’ is causes the heat in the atmosphere to be unequally distributed according to the pressure gradient.
Bart says:
January 20, 2012 at 11:00 pm
davidmhoffer says:
January 20, 2012 at 8:10 pm
“There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt.”
I just realized your and Kevin’s thought experiment supports my hypothesis. As you say, all those photons backradiated into the Earth should just be spit back out again. So, why should there be gaps in the emissions at TOA in those bands?
Are they re-radiated back from the surface at even lower wavenumber? (There’s no way the Earth is re-radiating the backradiation at higher energy.) If so, shouldn’t we see a hump at lower wavenumbers in the emissions spectra? In fact, we see just the opposite: the emissions spectrum at TOA over Guam, for example (see here in Fig. 2), in the low wavenumber range appears to fit an isocline near 275K, while that at higher wavenumbers appears to follow a 300K or so isocline.
Curiouser and curiouser…
Now, consider my mechanism in which the IR absorbing gases actually pin down the spectrum in their radiative range, while the rest of the spectrum is trying to push itself up around it. That is a much better fit with what is actually measured!
DeWitt Payne says:
January 20, 2012 at 9:15 pm
“Because the surface temperature is so much higher, the 15 μm band of CO2 doesn’t contribute much. Even at 92 bar, there are still ranges of relatively low absorption for CO2 alone. However, those gaps are nearly filled by SO2 and H2O, even though the partial pressure of H2O is small. Replacing half the CO2 with nitrogen does cool the surface, but not by much.”
In light of my comment above, I note that the composite emissions spectrum of Venus is “pinned down” in the H2SO4 band, while tiny bits are taken out in the CO2 and other bands. How about giving it a go? Suppose, for the sake of argument, that I am right. Where does it lead you in interpreting Venus’ emissions spectrum?
In response to
Jordan says:
January 20, 2012 at 2:28 pm
Hans Jelbring says:
January 20, 2012 at 5:21 pm
Come on guys, you are both missing the point. The basic question is – Does increasing the amount of greenhouse gases in the atmosphere cause the atmosphere to get warmer (the consensus) or colder? In my paper, I suggest a atmosphere made from a hypothetical gas that does not absorb or emit IR radiation. This avoids the very valid exceptions that both of you have made. In this very hypothetical case, the atmosphere has no way to loose energy to space and, therefore, becomes isothermal.
At that point, it does not matter if you add salt particles, O2, or CO2, there is now a mechanism for heat to escape the atmosphere. (However, with O2 the loss is so small it might not be measurable.) Since some of that heat returns to the surface, it gets warmer as the atmosphere cools.
As long as the atmosphere is optically thin, the entire thickness of the atmosphere will cool about the same amount and an adiabatic lapse rate will form near the surface. If the atmosphere becomes optically thick (nearly opaque), then only the top will cool significantly and the lapse rate will not be very close to the adiabatic rate.
On the real Earth, the dry adiabatic lapse rate is about 9.8 K/km, the typical measured troposphere lapse rate is about 6.5 K/km. With an atmosphere lacking any IR emitters, the lapse rate will be 0 K/km (isothermal). Since the actual lapse rate is between 0 and 9.8, if follows that the atmosphere is optically thick in the frequencies that matter.
Beside the mechanisms you have suggested, there are other cooling scenarios caused by pressure gradients produced by the surface temperature variations between the day and night sides of the planet, as well as the pole vs equator differences. But remember, the point of this exercise is not to describe some hypothetical planet, but to determine the function of greenhouse gases.
DeWitt Payne says:
January 20, 2012 at 9:59 am
People, please pay attention to this interchange. Hans Jelbring asserted categorically that “Any surface radiation power exceeding 100 W/m^2 is bull.”
In response, DeWitt posted up an actual measurement from a surfrad station. It shows a 24 hour plot. The MEASURED, OBSERVED surface radiation swings between 300 (night) and 400 (day) watts per square metre. So what is “bull” is Hans’s crazy claim.
I want you all to consider the level of willful blindness about climate science that it would take to get to the year 2012 and still be making Hans’s claim that nowhere on the earth is there surface radiation exceeding 100 W/m2. I mean, to do that you would have to avert your eyes from half the scientific papers published about the climate. You’d have to never, ever google “upwelling longwave” to see if your 100 W/m2 claim is correct. You’d have to think that the Stefan-Boltzmann equation didn’t work. You’d have to never look at a global energy budget.
Tallbloke, upstream you claimed that I was making an “ad hominem” attack when I said:
Now … perhaps given Hans Jelbring’s demonstrated ignorance regarding upwelling radiation in this interchange, you might be willing to rethink your comment?
w.
PS—Please be clear why what I said is not an ad hominem attack. I am not saying that Hans’s scientific theories are wrong because Hans is ignorant of basic information about radiation. That would indeed be an ad hom attack, where you try to bring down the man’s scientific claims by attacking the man.
I am saying that Hans’s scientific theory are wrong, and that Hans is ignorant of basic information about radiation. The two are not linked. Hans’s theory fails on its own merits, with no reference to the author of the theory.
The same is true about my comment about how people who are generally ignorant of basic science believe in theories like Jelbrings. That’s not an ad-hom argument either. Saying that someone is ignorant of basic science does not mean that any given belief of theirs is wrong, that would also be an ad hominem attack.
You see, an ad hominem attack means that you are attacking the man rather than attacking the argument. I’m not saying that those folks arguments are all wrong, or are wrong because they don’t know science.
I’m just saying that a lot of them don’t understand basic science, which is a simple statement and not an ad hominem in any form.
Marc77 says:
January 20, 2012 at 12:00 pm
Once again, we are not discussing whether we can draw power from the Earth’s atmosphere. Everyone agrees we can do that. We are discussing whether we can draw power from either the atmosphere in Jelbring’s thought experiment, or (equivalently) from a tall, thermally insulated cylinder.
w.
Joe Born says:
January 20, 2012 at 12:35 pm
Joe, all considerations about the mechanism go nowhere. Consider the outcome. If it’s true that gravity can separate molecules by temperature, then we can pull energy out of tall insulated cylinders of air, Jelbring is right and we never have to worry about energy again.
Do you believe that we can do that? Really?
w.
Bart says:
January 20, 2012 at 3:35 pm
Thanks, Bart. You seem to have momentarily forgotten that in the Jelbring thought experiment, the planet and atmosphere are surrounded by an impermeable perfectly insulating sphere which doesn’t allow any energy in or out. It is a sealed-off portion of the universe, with a planet and an atmosphere.
It is in that world, which I have nicknamed “Stygia” for the stygian blackness inside the shell, than Hans says the atmosphere will be thermally stratified by gravity.
Since (as you say above) the inhabitants of Stygia could pull work out of such a temperature difference, then they could use it to light up their eternal darkness.
As you know, that would be a perpetual motion machine, so Jelbring’s hypothesis is falsified.
w.
Please can anyone point out any logical flaws in the following reasoning:
Consider two systems, one gravitating and one not. Each is surrounded by a perfectly insulating, perfectly reflective shell and contains a spherical, homogeneous, black-body planet with a homogeneous, transparent atmosphere.
In both cases the surface of the planet radiates energy which is reflected completely back onto the planet’s surface by the shell, yielding no net radiative transfer from the planet. There exists a boundary condition at the surface of the planet such that the planet’s temperature and the atmospheric temperature are the same.
In the non-gravitating case, any temperature differences in the atmosphere will be eliminated by conduction. Thus the atmosphere is a uniform temperature that matches the surface temperature of the planet and has uniform pressure and density at all altitudes.
Now add gravity to this system. By doing so we have added a considerable amount of total energy to the system in the form of gravitational potential energy. The mass distribution of the atmosphere changes to ensure that at each altitude the pressure is equal to the pressure exerted by the atmosphere above it. The compression of the atmosphere at lower altitudes causes it to heat up, as per the ideal gas law. We now have pressure, density and temperature gradients decreasing with altitude as we are all familiar with from Earth.
Now we must return to our boundary condition: the planet’s surface must be the same temperature as the atmospheric temperature at the boundary. The temperature of the atmosphere at the surface has risen as it was heated by compression. There is now a temperature difference between the planet’s surface and the surface layer of the atmosphere. Thus energy will flow from the hotter body to the cooler, from the atmosphere to the planet until our boundary condition is restored.
The gravitating planet’s surface is now hotter than the non-gravitating planet. The GPE released by redistributing the mass of the atmosphere was converted to thermal energy. Now we turn to our other boundary condition: the insulating, reflecting shell that makes this a closed system. No energy can leave the system and a gravitating system has greater total energy than a non-gravitating one. Since the atmosphere is transparent the increased radiation of the hotter black-body planet is perfectly reflected back by the shell to the planet. The gravitating system has higher total energy and higher planet temperature. The increase in total energy is proportional to the GPE of the atmosphere which is proportional the the mass of the atmosphere. More massive atmosphere yields higher temperatures.
A non-gravitating system is isothermal, a gravitating one is not.
So, unless someone wants to point out the flaw in the logic, we now have a stable system containing a temperature gradient that should persist indefinitely.
Now let’s turn to the perpetual motion issue with the heat engine. I think we all agree that the heat engine will operate (do work) as there is a temperature gradient. However in doing work to produce light, the heat engine must reduce the temperature gradient that makes it operate since this is a closed system. If thermal energy is converted to light, there’s less thermal energy to go round. Stygia is now lit but colder. The surface temperature boundary condition ensures that the planet cools.
Let’s suppose that the heat engine is a real one and not 100% efficient, it gives off heat. This will heat the atmosphere locally and the atmosphere will radiate the heat. Since the atmosphere is transparent and the reflective shell is perfect, the only place for this radiated heat to be absorbed is by the planet (which is now cooler thanks to the heat engine).
Stygia is lit but colder. With the perfectly reflecting shell the light will bounce around the system indefinitely until it is absorbed. The only absorber is the planet. As the planet absorbs the light, the light’s energy is converted back to heat and the planet warms. As the planet warms, so does the surface layer of the atmosphere and the thermal gradient in the atmosphere is increased once more.
We’ve ended up with a system where energy is converted by the heat engine to light and local heating of the atmosphere depleting the temperature gradient available to the heat engine. The heat engine would theoretically convert all energy in the system to light making the system uniformly cold and the heat engine stop working. However the local heating of the atmosphere and the energy in the light is reabsorbed by the planet as there is nowhere else in the closed system for it to go. This heats the planet and restores the temperature gradient which means the heat engine keeps running.
The entropy in the system can only increase OR STAY THE SAME. The theoretically perfect, lossless system described can operate because the specification of perfect losslessness means the entropy change for all the steps is zero. Thermodynamic laws are NOT violated.
I think where the column of air arguments are failing is that they do not account for the redistribution of energy in a perfect, closed system. Taking energy from the column of air must either deplete the energy in the air column or it must be replenished by taking energy from the rest of the system. Taking the energy from the rest of the system depletes the thermal gradient that is being used to generate the energy thus energy generation will eventually stop. Still no free, limitless energy for everyone.
Downdraft says:
January 20, 2012 at 3:37 pm
Forget about the mechanism, Downdraft, and start with what makes sense.
Suppose Jelbring were right and that a tall insulated cylinder of air would thermally stratify by gravity, with warm air at the bottom and cold air at the top. If so, we could use that temperature difference to do work.
Of course, the operation of our heat engine would cool the warm bottom air inside the tall insulated cylinder, and warm the cool air at the top. That’s what heat engines do. But Jelbring says no worries, gravity will separate the warm air and cold air again so the heat engine can continue running indefinitely.
Now Downdraft, be honest with me here … do you really think we can pull work forever out of a tall, perfectly insulated cylinder of air?
I didn’t think so. Which means that the air in the cylinder must be isothermal, and that Jelbring’s hypothesis cannot be right.
If you start thinking about mechanisms from there, from what you know to make sense, you’ll have more chance of sussing it out.
All the best.
w.
I don’t think Dr Brown appreciates the tiny, but significant, exchanges of kinetic for gravitational potential energy between his layers. Additionally, he treats these layers as being much more dense than they are. Even at the surface, 99.9% of a volume of air is vacuum. A conduction model doesn’t work well here, I think.
Let’s assume the atmosphere is made from mostly empty layers (it is mostly vacuum, afterall) that are 10^-5cm think. This is about the average distance a molecule in air travels before colliding with another molecule. Also assume a molecule traveling about 500 m/s. This is also the mean speed of an air molecule at the surface.
It has kinetic energy (ignoring rotational energy) of an N2 molecule then is
1/2 * 28 * 1.66 x 10^-27 kg * (500 m/s)^2 == 5.81×10^-21 J.
A molecule passing through the mostly empty space from the bottom of a layer to the top of the layer has kinetic energy converted to gravitational potential energy so that
28 * 1.66 x 10^-27 kg * (9.8 m/(s^2)) * 10^-5 cm = 4.56×10^-32 J of kinetic energy is lost (or gained if moving down from a higher layer).
This seems small but it is significant.
Imagine one molecule at the bottom of the atmosphere made energetic by contact with the surface of the Earth hitting another molecule 10^-5cm above it and so on through the layers as the atmosphere tries to reach equilibrium. Each time a lower molecule loses some kinetic energy to gravitational potential energy as it travels across the current layer before it strikes a molecule in the next layer above it. At 10km, 4.56×10^-21 J has been converted to gravitational potential energy — a significant amount when compared with the 5.81×10^-21 J at the start. The reverse is true, too. Molecules propelled upward are matched by molecules falling downward and each downward falling molecule gains 4.56×10^-32 J of kinetic energy as it falls through a 10^-5cm layer.
These differences in kinetic energy in each layer must translate into different temperatures in each layer.
Of course linear kinetic energy in the direction normal the surface of the Earth isn’t the only energy in a N2 molecule. There is kinetic energy parallel to the surface. There is also rotational energy, too. A thermometer will integrate all these to determine a temperature. This is one reason why the temperature of the atmosphere at 10km is higher than what is predicted. Still, much energy is in the form of vertical kinetic energy and this is converted from kinetic to gravitational as molecules move upwardly. The absence of this kinetic energy must be reflected in a lower measured temperature.
Looks like the very basics of “science” are still not settled 😉
Lets forget the gravity for a moment. Claim of many of us is, that the bulk atmosphere itself (N2+O2) is from a considerable, if not full part, responsible for warmer average than Moon (and much less diurnal variations, which is at least so important).
Earth day is much cooler than on the Moon. We have clouds, water surface evaporation, cloud/snow albedo and air thermals cooling our day. Without bulk atmosphere, none of it should work. The main “greenhouse gas” – water vapor – in its various forms causes net daytime cooling.
Earth night is much warmer than on the Moon, on average by tremendous 240° C. Temperature in 2m altitude, measured by thermometer, is equal to number of molecular collisions per given area per given time unit, and their speed of movement. Without bulk atmosphere, there are no molecular collisions to be measured. Mere presence of “greenhouse gases” is not sufficient. Mars black body temperature and actual temperature is the same – 310K – even there is 6,000 ppm of CO2 in its atmosphere. However, this is already 95% of the whole atmosphere, it means it is very thin. There is nothing, which should hold the warmth from the surface, heated by sunlight (or long-wave IR coming from that CO2, if we believe so). How thick would be the back radiation arrow in Kiehl-Trenberth diagram for Mars? How many Watts would be assigned to it? Net result is still ZERO. Still 310K.
The ultimate point of all that is to say, that the claim “greenhouse gases raise surface temperature by 33K” is totally wrong; 33K is wrong and the whole attribution to “GHG” only is wrong as well.
Jordan says:
January 20, 2012 at 3:48 pm
Thanks, Jordan. You and several others have said that practical problems in utilizing the heat to do work invalidate the proof. They do not. The existence of the temperature difference means we can extract work, whether or not we’ve invented a way to do so yet.
In fact, there is a lovely plan upthread to use two columns of different gases. As they have a different specific heat Cp, at every level but one they will be at different temperatures. This gets past the question of the distance between the ends of the thermocouples.
Also, your “rate of transfer” argument above doesn’t work. All I have to do is turn off the heat engine for a little while. Then gravity (says Jelbring) would re-sort the molecules by temperature. Tomorrow, I turn on my heat engine again. Perpetual motion.
w.
KevinK says:
January 20, 2012 at 4:04 pm
Dang, bro’, with all due respect, I didn’t say you were.
w.
Bart says:
January 20, 2012 at 4:23 pm
I don’t understand. Are you claiming that the temperature in a tall insulated cylinder of air will never reach equilibrium?
w.
Willis Eschenbach says:
January 20, 2012 at 11:48 pm
Are you certain? Suppose the engine were a Stirling engine. A Stirling engine could do no work as the gas inside the engine would stratify just as the gas in the insulated cylinder.
Having read Willis’ comments, and Jellbring’s comments about Willis’ comments, I am quite surprised. How on earth is it possible that Jellbring has such a weak grasp of physics? Yes, his paper was peer-reviewed, which is what is just so very surprising! And as Willis points out, this is not an ad hominem attack, I have read his comments and come to the conclusion he has some very basic concepts wrong. He says:
I have already pointed out that a “perpetuum mobile” is possible to construct in our real atmosphere since the cold air at the top of Mount Everest for sure is colder than air at the surface.
Yet he does not understand what is obviously wrong with that statement. That’s not a perpetual motion machine, that’s just another energy source like any other that we have today. The energy is derived from outside sources (the sun), which in turn causes the temperature gradient, which in turn allows you to extract work.
Willis is correct, if gravity were to provide a temperature gradient, then you could create a perpetual motion machine, you could do work from that gradient, which in theory would reduce the temperature gradient, but of course the theory states that gravity would restore the gradient once more (that’s the Jellbring theory!), thus enabling the perpetual extraction of work! This is energy creation from nothing other than a force (gravity), hence impossible.
And of course we have other commenters like fred berple pointing out that temperature gradients exist in the real world. How utterly irrelevant.
Meanwhile, tallbloke keeps making comments that exhibit a very weak understanding of basic physics, not even using the terminology correctly for basic concepts such as energy, work, and temperature.
I just don’t understand how people who clearly have a weak grasp of physics to be so utterly and totally oblivious to the fact that they have a weak grasp of physics.
Dewitt Payne: “The entropy of an isothermal atmosphere is higher than for an atmosphere with an adiabatic lapse rate.”
You believe that. Willis believes that. Robert Brown believes that. I believe that. But our believing it doesn’t make it true. Nor does Willis’s heat-engine “proof,” because it begs the question. I.e., by assuming that the heat engine will work, he’s assuming that the adiabatic-lapse-rate atmosphere’s entropy is lower than an isothermal one, which, ultimately, is what he set out to prove.
The only proof would be to show that the number of states that fall within the “isothermal” definition vastly exceeds the number that fall within the adiabatic-lapse-rate definition. And, of course, your response would be that you have indeed seen such proofs.
But have you really? Here I’ll help you The Coombes and Laue paper discussed at Tallbloke’s Talkshop’ Loshmidt thread, http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/ , does indeed purport to provide a proof that the maximum-entropy state is isothermal even in the presence of gravity.
However, the Velasco et al. paper also discussed at that site demonstrates that Coombes & Laue’s conclusion is only an approximation asymptotically approached as the number of molecules gets large; for any finite number of molecules, the maximum-entropy state has a non-zero lapse rate given implicitly (for a monatomic ideal gas) by the (altitude-dependent) expression I set forth above.
Does the mean that Helbring is right? No. The lapse rate at which Velasco et al. arrive is much less than the adiabatic lapse rate. Indeed, by most people’s standards, it is negligible for any number of molecules of which we’d take notice.
But what it does mean is that neither Willis’s nor Robert Brown’s nor your “proof” is valid; none of them does anything more than state a conclusion. (The same is true of that Science of Doom discussion the summer before last.) And, technically, that conclusion is wrong.
Or, at least it’s wrong if Velasco et al. are right. Now, I’m no physicist, and I found Velasco et al. (and the Román et al. paper on which it depends) tough sledding. But the relationship they derive, namely (3E/(5N-2))(1-mgz/E) for the mean single-molecule kinetic energy, where N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude, is clearly correct for N = 1, unlike the isothermal conclusion everyone thinks he remembers.
Willis, you said:
“It is in that world, which I have nicknamed “Stygia” for the stygian blackness inside the shell, than Hans says the atmosphere will be thermally stratified by gravity.”
This is correct.
“Since (as you say above) the inhabitants of Stygia could pull work out of such a temperature difference, then they could use it to light up their eternal darkness.”
Briefly. Extracting heat energy from the atmosphere would result in net cooling.
>As you know, that would be a perpetual motion machine, so Jelbring’s hypothesis is falsified.
Incorrect. It is falsified but not for the reason you state (that it will not re-stratify). If Jelbring says that the atmosphere will re-sort the temperature into a stratified one, he is correct. If he does not mention that the system will have cooled (net) he is missing something. If he thinks gravity will add heat to the atmosphere he is incorrect. The moment he says that, he has created a perpetual motion machine.
Re-stratification does not mean ‘re-heating’. There is no energy input. The energy extraction from the heat engine will work. The atmosphere will indeed re-stratify. It is not re-heated while doing so; no net gain in energy from gravity, there being no mechanism for it. Ergo your clear statement that re-stratification = perpetual motion is incorrect.
Suppose you employed a 100% efficient heat engine in a stratified atmosphere. Wherever it was placed, it would cool the system. Overall, the system would cool and eventually have no heat left. The fact that a less efficient heat engine would take longer, and involve re-stratification, does not change the physics: work can be done using the temperature difference. The system will wind down. How is that a perpetual motion machine? It is the extraction of energy from a heat battery. Nothing more.
If Jelbring says it will continue to generate energy indefinitely, that is a contradiction of the Law of Conservation. If you say that the atmosphere will not re-stratify, that is a contradiction of the Universal Gas Law. You are correct to say his claim creates a perpetual motion machine IF he states that the upper gas will gain energy from gravity as it drops lower. It will warm, but not to the previous temperature.
A ‘gas packet’ will increase in temperature when it goes down, but temperature is not a measure of energy. Energy is thermal mass times temperature. If you cram more mass into a unit volume, the temperature rises because there is more mass in there not because each unit making up that mass, on its own, gained energy. “Temperature” is a measure of how many bangings there are into the thermocouple. You can’t measure energy with a thermocouple.
It is akin to heating a spring v.s. compressing a spring. Both are reservoirs of energy. You cannot tell how much energy is in the spring by measuring the temperature only. Stacking springs vertically in a gravitational field will compress the bottom one the most and it will have more energy that the top ones, even if they are all the same temperature. You can extract work by decompressing the bottom springs to the same length as the top ones, but it reduces the total stored energy. Imperfect analogy but the point needs to be made – extracting energy will wind down the system. Gravity will not make it up.
jae says:
January 20, 2012 at 6:53 pm
First, I fear that what happens on other planets is of little interest to me. The planets are so different from the Earth, and we understand our own climate so poorly even though we live in it, that what is happening on Jupiter or Venus can provide little insight, falsification, or support for theories about the Earth.
Second, I’m still not clear (other than planetoids) what evidence you say exists in the “papers”.
Third, I don’t know what “papers” other than Huffman you are discussing. There is no evidence in Jelbring’s paper that I know of.
Fourth, Joel pointed out above that the planetary “evidence” in N&Z is a fit of an equation with many free parameters, viz:
I am not impressed by a fit with four visible free parameters plus selection parameters. That has no evidentiary value at all.
As to whether I “ignore” the question of the planets, no, I don’t. I’m just not impressed by arguments about what’s happening on Venus. Down at the bottom of the Venusian atmosphere, CO2 is neither a gas nor a liquid, but a supercritical liquid. If you think that has some relevance to the Earth’s climate, I don’t see it.
Finally, you ask why I “do not confront it/explain your position. Is it confirmation bias? Old age? Ego? What, Willis?”
I am under no obligation to either confront something, or explain why I have not done so, merely because you think it is important. Look at this thread. How many people have posted how many different theories and claims and explanations and evidence and insights and all the rest?
Some of it I reply to. Some I don’t. Some I follow up on. Some I don’t. Time is short, the road is long.
A fit of an equation with 4 free parameters to a bunch of carefully selected planetary data is not something I follow up on.
And if I do or don’t follow up on it, I do not owe you an explanation. In this case I didn’t discuss it, because, frankly, it is meaningless.
I am not amused by your accusations of confirmation bias and the rest. I advise you to leave them off the next post. Bear in mind that it may not be answered, for the reasons listed above. I have answered you this time in part so you will know why I may not answer you in the future.
w.
I have finally found time to re-run my empirical experiment into the N&Z hypothesis. This time I followed “Joules Verne’s” suggestion and regulated the pressure in the test chambers. I used an air bladder (hot water bottle) and weights (bricks) to maintain constant pressure in the high pressure chamber.
Due to thunderstorms and rain I was unable to use sunlight as a long wave source and had to use a flood lamp (too much IR). However the results were just as before. The chamber with the higher pressure always rises to a higher temperature when illuminated. I should point out that illumination is only started when the chambers have been pressurised and allowed to equalise temperatures.
With just 6 house bricks on the air bladder, chamber temperature differentials of over 4 degrees were observable. Again I reiterate that low and high pressure chambers were allowed to equalise in temperature before illumination.
When the weather clears I will re run the tests with sunlight instead of a floodlamp. In the meantime I am confident in claiming that Nicolov and Zeller are correct and that Willis and Joel are entirely wrong. Again.
scf says:
January 21, 2012 at 1:37 am
How can you be certain work could be done from that gradient? Gravity may prevent that.
The working substance in a heat engine, after doing work, must move to the cold sink. The cold sink in this hypothetical perpetual motion machine is high above the heat source. Energy must be used to raise it. I suspect the energy required to do that is equal to the energy that appears can be extracted from the gradient.
jae says:
January 20, 2012 at 8:22 pm
It was a suggestion offered in friendship, in support of you getting some traction, jae. No reason to be upset.
w.
Willis Eschenbach: “If it’s true that gravity can separate molecules by temperature, then we can pull energy out of tall insulated cylinders of air.”
Without realizing it, you’ve begged the question. “We can pull energy out of tall insulated cylinders of air” in which a non-zero lapse rate prevails, but only if that air is not at maximum entropy–and you have not shown that maximum entropy necessarily requires a zero lapse rate.
You think you know it’s true. And, as I mentioned above to DeWitt Payne, people like Robert Brown think they’ve proved it. But the only real way to prove it is to show that the number of states that exhibit isothermality greatly exceeds the number thereof that exhibit a non-zero lapse rate. And, to this non-physicist at least, the Velasco et al. paper, together with the Román et al. paper on which it relies, shows that it does not.
Now, as I mentioned above to DeWitt Payne, that doesn’t mean that Jelbring is right; in fact, it shows he’s wrong. But it also shows you haven’t proved he’s wrong.
Willis Eschenbach says:
January 20, 2012 at 11:41 pm
DeWitt Payne says:
January 20, 2012 at 9:59 am
Hans Jellbring,
“Any surface radiation power exceeding 100 W/m^2 is bull regardless if it is from equatorial, midlatitude or polar regions during days or night. Just show how this fantasy power radiation changes between day and night in polar regions as an exsample.”
Here’s a plot of upwelling IR radiation measured over 24 hours at Desert Rock, NV by a SURFRAD station there. It looks to be more than 100W/m² to me. Note that the time axis is UTC. Desert Rock is -8 hours from UTC so local noon would be 2000 on the time axis.
There are seven SURFRAD stations in the US. You can access the data here.
People, please pay attention to this interchange. Hans Jelbring asserted categorically that “Any surface radiation power exceeding 100 W/m^2 is bull.”
In response, DeWitt posted up an actual measurement from a surfrad station. It shows a 24 hour plot. The MEASURED, OBSERVED surface radiation swings between 300 (night) and 400 (day) watts per square metre
———————-
Here are all the radiation flows (In and out) for Table Mountain Co. SURFRAD station for the 24 hour period of Nov, 29, 2009. (I did this more than a year ago so that is why the date) .
http://img140.imageshack.us/img140/4109/tablemountainall.png
http://img12.imageshack.us/img12/3225/tablemountainnets.png
How you can dismiss the atmospheres of the other planets I find incredible. They all obey the same laws of physics. It is just that you GHGers are wedded to the claims that somehow a trace gas drives temperature.
You also fail to accept that compressing a gas will increase its kinetic energy which results in a higher temperature. That is what adiabatic means.
How about this for an add on to the compression theory. Convecting gas in the atmosphere will cool as it rises at the Saturated Adiabatic Lapse Rate, around 5C/Km rise, and form clouds at height. This rising air causes air at height to descend but this air is dry, having formed clouds, and warms at the Dry adiabatic Lapse Rate, 9.8C/KM, so will arrive back at the surface warmer than when it convected. (Like a vertically looped Chinook Wind). This warms the surface above the theoretical BB temperature not the GHG theory that violates the laws of thermodynamics.
I wonder if you can be bothered to get down this far in the reply list?
Willis Eschenbach says:
January 21, 2012 at 12:33 am
“You seem to have momentarily forgotten that in the Jelbring thought experiment, the planet and atmosphere are surrounded by an impermeable perfectly insulating sphere which doesn’t allow any energy in or out.”
Well, you can’t forget what you never knew. To be truthful, I haven’t looked at the Jelbring hypothesis much – it sounded fishy to me from the get-go. The setup you describe, and which I have now looked at the paper to see for myself, is IMHO very contrived. If you take away the insulation, and there’s no influx, then the atmosphere will freeze and even the usual caveats about the ideal gas law go out the window.
I have been looking at the standard greenhouse theory and trying to find loopholes in its physical basis. What I have found is that it depends very much on an assertion that a derived formula (SB) intended for matter in thermodynamic equilibrium holds even in situations which are far from that condition. This is a loophole.
In researching the topic, I have found relatively narrow experimental results existing for non-equilibrium radiative behavior, and there is evidently quite a bit of current research going on into the very question of how it works. To borrow a phrase, the science is not settled. It appears plausible to me that rapid conductance of heat could significantly reduce thermal radiation, as the heat gets conducted away before the particles can radiate away their energy.
I have also found that the standard explanation that IR emitters heat the surface has a mirror image explanation with the same resulting steady state behavior, but with very different non-equilibrium dynamics. Looking at things from this perspective gives an actual reason for why an equilibrium condition in an emitting atmosphere is even approached, and solves a couple of other dilemmas which I have pointed out now:
1) How did the Earth ever heat up enough to unfreeze the water vapor that presumably contributes most of the “greenhouse” warming to heat the planet to its present state?
2) Why is there a gap in IR emitter bands at TOA when the IR radiation broadcasted back to the Earth should simply re-radiate back out?
To those, I will add a third I only just started thinking about. We all know by now that CO2 and CH4 levels increase after emergence from an ice age, and not before. Why? Well, if IR emitters actually make things colder, then recovery should proceed after enough of them have been sequestered by a change of state, and the warming which releases them again should set into motion a new decline. And, that is what we see.
Willis Eschenbach says:
January 21, 2012 at 1:21 am
“Are you claiming that the temperature in a tall insulated cylinder of air will never reach equilibrium?”
The SB loophole allows this to be a physically realizable possibility.
jae says: “you did not really read his stuff and pay attention to the DATA (aka, EMPIRICAL EVIDENCE!). Same with the other papers. It seems that the atmospheres of other planetoids that have atmospheres have temperatures that are not correlated in any way to the amounts or types of GHGs present in those atmospheres. ”
I don’t understand how this is supposed to be evidence. You have a small sample set. You have a million other variables with each of those planets, You might as well be claiming that the colours of those planets are the cause of the unknown temperature differentials, rather than the size.
And, in another forehead slapping moment, I suddenly realized I do not even have to argue that SB violation is possible. That argument is tailored for a fictitious planet with a non-radiating atmosphere, which can never actually exist. All that is needed is to establish that there is a tendency for heat to accumulate in an atmosphere (which there trivially is).
As the heat rises, the emitters in the atmosphere will draw more and more of it off, increasing the allowable temperature rise of the surface to remain in radiative equilibrium according to SB. The rising temperature stops when the emitters have taken enough out such that the surface temperature induced radiation balances with the incoming flux from the Sun and the backradiation from the emitters.
Under this dynamic if, for example, you had no IR emitters, but emitters only say in the visible part of the spectrum, the temperature would climb until those emitters can establish an equilibrium. The total emissions still balance on the way up because the energy radiated by those emitters is orders of magnitude more energetic than IR emitters.
So, in this way, a planet’s equilibrium temperature is set by the minimum threshold at which available emitters can balance all the fluxes.
And, adding more IR emitters to the Earth’s atmosphere will tend to cool, rather than heat, it.
So, your temperature rises to the SB limit. The tails of the first major emitter back-radiate, which allows your temperature to rise more within the SB limit. So, you get more back-radiation, and your allowable temperature rises some more. And, so on, until you have reached the point where your surface is radiating significantly into the main lobe of the emitter, and you reach an equilibrium.
That’s it!
Thanks for pointing out my errors A Physicist.
scf says:
January 21, 2012 at 1:37 am
“Having read Willis’ comments, and Jelbring’s comments about Willis’ comments, I am quite surprised. How on earth is it possible that Jelbring has such a weak grasp of physics?”
Please, point out exactly what part is weak in my paper insteaed of using general degrading statments about my understanding in physics. Willis behavior in this respect has been outstanding and what I consider him to be is not printable. The results in the article stands and falls by applying 1:st law of thermodynamics and 2:nd law of thermodynamics. Are these 2 laws familiar to you? Just tell me which one is wrong or point to any other fault IN THE ARTICLE. Zeroth´s law is not applicable in this case since gravity is involved and it makes the constant energy per any equal mass within the insulated sphere constant. This is equivalent to the adiabatic temperatuer lapse rate of -g/Cp. To avoid the sill PM argumet I confirm that a system that has reached maximum entropy cannot include a PM so I agree with you on that point. The treated atmosphere only contain ideal gases and no perpetuum mobile.
Willis and you seem to share the same problem. You don´t understand my paper and still you know that it is wrong by adding unspecified circumstances. Just read it and comment line by line what is wrong. Don´t forget to comment on observational evidence that support the model results which primarily are observed temperture lapse rates on other planets that support the model result. That includes observed temperature lapse rates on Venus, Mars, Titan, Jupiter, Neptun and Uranus.
Willis Eschenbach;
I am not impressed by a fit with four visible free parameters plus selection parameters. That has no evidentiary value at all.>>>
I, on the other hand, am not impressed with an accusation that isn’t true. Go back and take a look at the equations yourself Willis and count the g_d d_mned parameters. Equation 7 is actually two equations on a single line, and further, it is an interim step. Each of the two equations expressed in Equation 7 has 2 variables. You cannot possibly be suggesting that we count up the variables in two different equations and accuse N&Z of having four variables in a single equation, can you? Oh wait. You just did.
The purpose of the TWO equations shown in Equation 7 (which should be expressed as Step 7 for clarity) are to explain the transformation that leads to Equation 8, which is their FINAL equation, and which contains TWO variables, from which the surface T of 8 planets is then calculated.
So, your harsh criticism of N&Z being nothing more than curve fitting based on four free parameters is falsified simply by reading what they said instead of reading what someone else SAYS they said. Are we skeptics seeking the truth? Or rabid confirmation biased critics stooping to half truths and omitted facts to support our position?
Willis Eschenbach;
People, please pay attention to this interchange. Hans Jelbring asserted categorically that “Any surface radiation power exceeding 100 W/m^2 is bull.”>>>>
People, please pay attention to Hans Jelbring’s entire comment where it becomes clear that English is not his first language and that he struggles in several places to articulate clearly what he means. Rather than going all attack dog on his choice of words, how about cutting the guy some slack and asking for clarification?
Hans Jelbring,
Any surface radiation power exceeding 100 W/m^2 is bull regardless if it is from equatorial, midlatitude or polar regions during days or night. Just show how this fantasy power radiation changes between day and night in polar regions as an exsample.>>>
Hans, could you expand on this statement for clarity? Your wording suggests that the surface of the earth never exceeds radiance of 100 w/m2 which just isn’t correct. Is this what you meant? Or were you referring to net radiation? Or some sort of averaged number over time? We don’t understand exactly what the 100 w/m2 you are speaking of here actually refers to.
OK, jae, let’s provide some empirical evidence (and in the process, hopefully we’ll depoliticize this argument).
We’ll consider a different system, that has NO politics associated to it, and yet has (essentially) similar physics. That system is a cryogenic tank, also called a “dewar”, also called a “thermos flask.”
In its simplest form, a cryogenic tank is insulated by a layer of vacuum (say 1 cm thick). Of course, heat can be carried across that gap by radiation. NASA engineers (among others) would like to reduce that radiative heat flow, so that spaceships can hold liquid oxygen/hydrogen/helium longer.
Weird-sounding engineering idea: Fill that one centimeter vacuum gap with ten (or more) ultrathin, aluminized, lightly crumpled, sheets of plastic film (typically ordinary mylar), with each mylar sheet having a thickness of only 0.001 cm (or less), such that the thickness of each sheet is tiny compared to the thickness of the vacuum gap, and in particular, such that the net thermal resistance of all the mylar layer’s together is negligible.
Now a photon traversing the vacuum gap is absorbed (by an ultrathin mylar sheet) and reemitted … absorbed (by a sheet) and reemitted … absorbed (by a sheet) and reemitted … (analogously, as infrared is absorbed (by CO2) and reemitted … absorbed (by CO2) and reemitted … absorbed (by CO2) and reemitted … ).
Question: What happens to the heat leak from the vacuum gap? Does the gap insulate the tank better, or worse? (analogously, does CO2 make our atmosphere insulate better, or worse?)
Answer: As discovered by NASA in the 1960s, the thin-layered vaccum tanks — called “multilayer superinsulation tanks” insulate heat flow orders of magnitude better than tanks without superinsulation. That is why every cryogenic tank manufacturer in the world now uses superinsulation.
So that is one more “Elevator Answer” to Willis’ question. Adding CO2 to the Earth atmosphere has effects similar to adding superinsulation to a cryogenic tank. In both cases, thermal conductivity is reduced, with the (desired) result that NASA’s spaceship tanks retain their cryogen longer, and with the (similar, but sobering) result that the Earth’s surface warms to a higher temperature from the sun’s solar input.
Ultrashort Elevator Answer: CO2 acts as a multilayer superinsulator.
The point being, there’s essentially zero doubt (among scientists and engineers) regarding multilayer superinsulation; both theory and empirical observation agree that multilayer superinsulation just plain works.
And this is yet another reason why there’s essentially zero doubt (among scientists and engineers) regarding the GHE in general: the GHE just plain works
I said early on in this thread
“The isothermal/adiabatic distribution for an isolated ideal gas in a gravitational field has long been debated.
For the isothermal distribution we have Maxwell, Boltzmann and Clausius.
For the adiabatic distribution we have Loschmidt, Laplace and Lagrange.
The smart money must be with the isothermal advocates but I would not regard this as a debate of which was settled and of historical interest only.
Here for instance is a member of the physics department of the University of California making a very up to date case for the adiabatic distribution.
http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4990v3.pdf”
Its a pity that Willis who last week was with the adiabatic camp, now thinks to hold such a position is only held by “ignorant” people.
This is an overreaction which has more place on a Deltoid thread than here.
However anyone who makes the adiabatic distribution for an thermally isolated ideal gas in a gravitational field a cornerstone of their theory about the real adiabatic atmosphere is needlessly going against the orthodox physics.
A bit like Claes Johnson and photons.
Its much better to stay within the orthodox framework of physics when addressing a general problem such as the Earths Climate.
Then separately argue the case for the existence or otherwise of photons or whether an adiabatic distribution for isolated ideal gas in a gravitational field is appropriate.
Bart;
Bart says:
January 21, 2012 at 3:21 am
And, in another forehead slapping moment, I suddenly realized I do not even have to argue that SB violation is possible.>>>
You NAILED it!
Well, until the last sentence…
Bart;
And, adding more IR emitters to the Earth’s atmosphere will tend to cool, rather than heat, it.>>>>
Nope. You should get to the same surface temperature, or close to it.
By opening up more pathways for emission to space, you don’t actually cool anything provided that you consider the whole surface and the average of T^4 of that surface. I think it easier to explain by going the other way. Take a system with a given number of pathways and shut some of them off. Does the temperature rise? One would think so, but not by nearly as much as one would think.
As soon as you shut some of the pathways off, that forces energy that otherwise would have escaped to be recirculated by other means. But the over all temperature does NOT need to rise (or more accurately, it doesn’t need to rise as much as one would think) to re-establish equilibrium. Because the energy is not being forced to re-circulate, it must also have the effect of re-distributing energy about the planet. Shut off some pathways, and there is no other possible result but that net energy flow from the tropics to the poles increases. The result is, because P varies with T^4, is that the coldest parts of the planet experience the most increase in T, which in turn results in a planet of more uniform temperature.
Equilibrium still requires that emission to space match absorption. By shutting off some channels, we force the emission to be spread around more than it would otherwise. Equilibrium however still arrives at the exact same w/m2 being emitted because absorption didn’t change. We get a cooler tropics and a warmer high latitudes, but the average of T^4 remains identical. If we fall into the trap of averaging T instead of T^4, we will get an average of T that is higher than it was before we shut some of those channels off.
Your main thought process howevere is correct. Shut down some channels, or create some new ones, and you redistribute flux in terms of what frequencies and how much escape from where. But the surface temperature simply “evens out” until equilibrium is established again. But change in net energy balance? There isn’t any.
A Physicist: “With regard to thermodynamics and transport theory (which is broadly what this WUWT topic is about), an historically recent and very broadly applicable framework regards thermodynamics and transport theory as (essentially) the study of the geometry of flow on manifolds, specifically the study of Hamiltonian dynamical flows on manifolds that are equipped with a symplectic structure.”
That’s beyond what vestigial mathematical skill remains from the last math course I took, over forty years ago. But, since you appear to be conversant in these matters, perhaps you could point out where I am wrong in understanding the above-identified Velasco et al. paper to reason from Hamiltonian dynamics to a conclusion that at equilibrium an ideal gas in a gravitational field will exhibit a non-zero temperature lapse rate. Or maybe you could point out the equation at which that paper or the Román et al. paper on which it depends went off track. Those papers are discussed in this thread: http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/.
In this context it is not enough to say that Velasco et al.’s lapse rate approaches zero as the number of molecules approaches infinity, because Willis’s argument is based on the lapse rate’s being zero at equilibrium even for a finite number of molecules.
Both the GHE and the N&Z paper try to solve a problem in explaining the current avg. temp
of 288K, the first starts from 255K, the other from 154,7K.
What both approaches are neglecting is that the earths surface consist mostly of oceans,
70% area, minimum 3 km deep and a temperature on average of +2C, already 20K higher
than the blackbody temperature the GHE uses when the sun has heated the blackbody.
Ocean surface temperature is on average 290K, just 15K higher than the temperature of the
deep oceans. This warmer layer is ~200m deep in the tropics, reducing to 0m near the
polar circles.
So instead of heating a blackbody from 0K to 255K (255K difference) all the sun has to do
is heat a smal part of the oceans from 275K to 290K, just 15K difference.
This warm ocean then heats the atmosphere, and results in our pleasant 288K average
temperature near the surface.
As background information may serve that the earth radius is ~6370 km, ocean depth
~3km, oceanbed 5 -10 km. The other 6350+ km are hot to very hot, (400K – >5000K)
although it is assumed that allmost no heat is flowing from the hot interior to the oceans
Ocean temps are basically steady since system earth is in radiative balance with the sun.
Ben Wouters
This thought experiment is similar to elevator music. It is there, but not particularly memorable.
I saw one commenter taking a correct approach. He(she) looked at the gas properties, including Prantl number and did an amazing thing, calculated the changing properties with temperature, viscosity, thermal capacity and pressure. I mentioned that isothermal conditions in the tall column of air would require changes in the mixed gas viscosity, perfect insulation and the velocity/energy of the upper most molecules would be greater than the lower. If you take the top and bottom off the tube, what would happen? So it is an unrealistic thought experiment, but maybe a start to a useful one. If that theoretical column was integrated over the entire surface, using the calculations to determine the final temperature, viscosity and pressure of each column, then we would be on to something, but not an elevator description or back of the envelop kinda thing.
There have been a lot of comments mentioning conductivity. About time. Now how about the change in conductivity? CO2 changes the conductivity of a mixed gas. H2O doesn’t have as much impact on the conductivity of a mixed gas. It takes the change in conductive hundreds of years to thousands of years to change climate, in the mean time, Ein to the oceans is never equal to Eout of the oceans. Shouldn’t that be where the problem starts? A ball of water in space over eons absorbing more energy that it emits? Then one fine day it reaches near equilibrium. Gravity does have an impact there. Then add layers of atmosphere one at a time.
The top down approach so far has been problematic, why not start from the bottom up?
I will frankly admit that I haven’t read all the comments, but I see that some people still insist that gravity would still raise the temperature at the bottom of the atmosphere as compared with the top, by means of compression.
If this is the case, the argument wouldn’t be confined to a gaseous amosphere. It would also apply to any solid column – say, a granite pillar – since all solid bodies are at least slightly elastic. To find an easily visualisible model, consider a tall hollow tube filled with tennis balls. The balls have weight, and the balls at the bottom of the tube have more weight above them than those near the top, so they are more compressed. The density of balls, and the pressure they exert on each other through their elasticity, is greater at the bottom.
We assume that the tube is perfectly insulated from the outside world (with respect to heat flow).
Now, does anyone suggest that in equilibrium the balls at the bottom will be hotter than those at the top? If so, what is to prevent heat being conducted from the hotter balls to the cooler ones until the temperature is equalised?
Note that I refer to the situation in equilibrium. If we disturb that equilibrium, say by putting in more balls at the top of the tube, there will be a temporary increase in temperature at the bottom, as the balls there are further compressed, but again the heat generated by compression (which is actually a form of kinetic energy) will eventually be diffused evenly throughout the tube. Does anyone disagree?
If they agree in the case of an elastic solid column, but disagree in the case of a gas, they need to explain the difference. Of course in the case of a gas there will be convection as well as conduction. If the gas were a perfect Newtonian fluid, I suppose that by adding more gas at the top of the column we might set up a perpetual circulation, doing no net work, just as if we drop a perfectly elastic tennis ball we can set up a perpetual bouncing motion. But there are no perfect fluids, any more than there are perectly elastic tennis balls. In any real gas, without any new input of energy, the convection current would eventually come to a halt as its energy is dissipated by friction and converted to heat.
davidmhoffer says:
January 20, 2012 at 8:10 pm
David, you’re pretty close to what I have been saying ever since Ira’s response to the K&Z hypothesis . It’s what I have been calling the maximum GHE. If you look at my comments over the last two weeks you will your description is very close. However, I don’t think your dam analogy is right. More water behind the dam would indicate a warmer climate.
It’s more like an overflow valve. The dam is full and any more energy just gets kicked out the overflow. It’s more like a V-shaped river. The initial flow of water increases the height behind the dam quickly. However, as the lake grows the same amount of water increases the height less and less. The overflow amount can match this rise and the water level does not increase.
In the atmosphere I think this is due to the fact that CO2 has both a warming and cooling effect. The GHE gets maximized once these two effects become equal. This occurs because the warming effect is a stronger one but diminishes more quickly due to saturation.
PS. I even pointed out that the warming from adding 100 ppm of CO2 would be .005% … which is the increase in mass when you convert O2 to CO2.
davidmhoffer says:
The above is the exact argument made by Alan Siddons, one of the “Slaying the Skydragon” crew. It is an extremely silly argument. Yes, the amount the Earth is emitting as seen from space has to be the same amount it is receiving from the sun in radiative balance…It has to be. However, the amount emitted by the surface of the Earth is not. The radiative greenhouse effect is the only thing that can explain how the surface of the Earth can be at an average temperature so high that it emits ~390 W/m^2 while the Earth as seen from space only emits ~240 W/m^2 and is thus still in radiative balance with what it receives from the sun.
Congratulations on now embracing the arguments of the “there is no greenhouse effect” crew who you once laughed at!
No…The water depth is the analog of the surface temperature. And, while the W/m^2 at the top-of-the-atmosphere is the same as it was, the W/m^2 leaving the surface is not.
The only way that the atmosphere can lower the probability of radiative energy from the surface from escaping to space is radiatively…i.e., by absorbing this energy. So, the only way that the mass of the atmosphere can come into it is in a way that changes the radiative absorption (which it can do to some extent by broadening of the absorption lines of the GHGs).
Response to http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871539 wherein David M Hoffer invents a new way to count parameters:
Great…Now all I have to do if I perform a fit of y vs. x with 100 free parameters is break it into two equations, one that defines some intermediate quantity as a function of x in terms of 99 of the parameters and the other that gives the final relation between this intermediate quantity and what I want (y) with 1 free parameter. Then I will have magically gone from a 100 parameter fit to a 1 parameter fit and everyone will be very impressed with my fit. I’ll have to remember that trick!
davidmhoffer says:
January 21, 2012 at 4:07 am
[Willis Eschenbach;
I am not impressed by a fit with four visible free parameters plus selection parameters. That has no evidentiary value at all.>>>]
I, on the other hand, am not impressed with an accusation that isn’t true. Go back and take a look at the equations yourself Willis and count the g_d d_mned parameters. Equation 7 is actually two equations on a single line, and further, it is an interim step. Each of the two equations expressed in Equation 7 has 2 variables. You cannot possibly be suggesting that we count up the variables in two different equations and accuse N&Z of having four variables in a single equation, can you? Oh wait. You just did.
The purpose of the TWO equations shown in Equation 7 (which should be expressed as Step 7 for clarity) are to explain the transformation that leads to Equation 8, which is their FINAL equation, and which contains TWO variables, from which the surface T of 8 planets is then calculated.
This was the same argument Joel tried to use against me back in Ira’s thread. It’s unfortunate that they are willing to discard what may very well be a valid correlation by simply waving their hands.
The two variable are atmospheric pressure and input energy. Those two define the GHE on every one of the planets. I think the non-GHG atmosphere is simply a red herring. GHGs are required, they just have a maximum effect in a gravity field.
This makes a lot of sense when you think about the effective radiation height. The incoming solar energy is what drives the atmosphere to be well mixed … to a point. Lower energy or higher pressure, less mixing. Higher energy or lower pressure, more mixing. The mixing drives the GHGs to a point in the atmosphere that ends up defining the effective radiation height. Once you’ve got enough GHGs to block almost all outgoing radiation you have maxed out the GHE.
The Title of this thread and the whole arguement by mr Eschenbach is a Strawman.
His hypothetical machine is quite impossible and he knows it. You cannot have a perfectly insulated column of air, just as he admits you cannot get any “work” out of downwelling IR.
He also admits that ice can form in the desert at night even though the temperatures are above freezing.
There have been many experiments that show that Cooling occurs, not warming from the night sky and not just in deserts. See Roy Spencer’s Experiment, where he thinks the Air in the Box should fall even lower than 5C lower than the surrounding air temperature and it is DIR that is keeping it from falling that much lower.
Talk about fool yourself with pre-conceived ideas.
http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/
Richard M says:
Yes, there are 2 variables but there are at least 4 free parameters in their fit.
And, they are not fitting to the GHG effect. They are fitting to their “surface temperature enhancement”. Only 3 of the 8 bodies that they consider have a very significant radiative greenhouse effect and only for one of those bodies (Venus) is the greenhouse effect that majority of what they define as the “surface enhancement effect”. (For the Earth, the radiative greenhouse effect is ~25% of their total surface temperature enhancement.) Most of the “surface temperature enhancement” is simply due to an evening out of the temperature distribution with no change in the W/m^2 emitted by the surface.
It is strange that people who have not understood the most basic facts about what they have done nonetheless seem to think that they have extremely wise and intelligent opinions about it!
Bart says:
January 20, 2012 at 6:38 pm
KevinK says:
January 20, 2012 at 4:04 pm
“When the energy returns to the surface from the “GHG” you cannot ADD it to the energy arriving from the Sun to produce an alleged “energy budget”.”
Energy is constantly coming in. If some of it is made to hang around longer than instantaneously, before the new batch arrives, then you will accumulate a net offset.
It’s not about increasing energy flow, which is always nearly constant. It’s about impeding that flow so that you keep more close to you.
Hi Bart. Do we not have a similar situation with the effect of gravity creating a gradient of pressure in an atmospheric mass? Because air is compressible, it means the density of the air will be greater near the surface, and energy hangs around longer in denser masses of a given composition than it does in less dense masses of that same composition due to the higher heat capacity.
Thank you, Joe, for this very reasonable question, which I will try to answer concretely.
From the viewpoint of geometric thermodynamics, every conserved quantity is associated to a thermodynamic potential. All systems conserve energy, and the thermodynamic potential associated to energy is called temperature. Some systems (but not all) conserve particle number, and the (less familiar) thermodynamic potential that is associated to particle number is called the chemical potential. Ideal gases are an example of a system that does conserve both energy and particle number.
If we regard the atmosphere as a stack of thin layers, we observe that each layer exchanges both energy and particles with the layers above and below it, and moreover the exchange of particles is associated to work done against a gravitational potential.
So a preliminary accounting of any thermodynamical theory of the atmosphere must ask:
(1) Does this theory account for the temperature potential?
(2) Does this theory account for the chemical potential?
(3) Does this theory account for work done against the gravitational potential?
Unless all three potentials — temperature, chemical, and gravitational — are accounted, the theory is thermodynamically wrong.
With reference to Tallbloke’s page, we find a mistaken implication in the second paragraph:
Mistakenly, the Loschmidt analysis goes on to completely neglect the chemical potential, on the grounds (presumably) that because there is no net exchange of particles, the thermodynamic effects associated to the exchange of individual particles can be neglected.
But since the exchange of individual particles really does occur (and is in fact the mechanism that sustains a pressure-and density gradient), Loschmidt’s analysis is wrong to neglect the chemical potential.
One way to get clear on these matters is to extend the analysis to thermodynamically account (explicitly) not only temperature and chemical potential gradients, but also pressure and density gradients. The result is simply that (at equilibrium) atmospheric pressure and density vary so as to ensure that both the temperature and the chemical potential are uniform.
Ultrashort Elevator Summary: Gravito-thermal theories wrongly neglect chemical potentials.
DavidB says:
January 21, 2012 at 5:34 am
I see that some people still insist that gravity would still raise the temperature at the bottom of the atmosphere as compared with the top, by means of compression.
consider a tall hollow tube filled with tennis balls. The balls have weight, and the balls at the bottom of the tube have more weight above them than those near the top, so they are more compressed. The density of balls, and the pressure they exert on each other through their elasticity, is greater at the bottom.
We assume that the tube is perfectly insulated from the outside world (with respect to heat flow).
Now, does anyone suggest that in equilibrium the balls at the bottom will be hotter than those at the top? If so, what is to prevent heat being conducted from the hotter balls to the cooler ones until the temperature is equalised?
This goes back to the proof I offered Duke.edu physicist Robert Brown up near the top of the thread. he said:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.”
And I pointed out:
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
But add in some adiabatic action – a nice little bit of vigorous vertical mixing – and the temperature gradient reappears. The elevator speech goes thus: “When air moves in a vertical airflow from the top to the bottom it is compressed and thus heats. When air moves in a vertical airflow from the bottom to the top it is decompressed and thus cools. In an atmosphere with a lot of vertical mixing you therefore will see a temperature gradient.”
Close. One has to ask what drives the “vigorous vertical mixing”. The answer is, temperature difference in the non-equilibrium, open system caused by differential heating. What does the heating? The Sun, of course. It isn’t so much air compressing and heating as it falls as it is air being heated at the bottom and rising (and displacing cooler denser air as it does so). The convective flow actually cools the bottom when it is being heated, without exception, although it does establish the lapse rate. Otherwise you have another of those processes that moves heat from colder to hotter.
The other point is that your elevator speech has nothing to do with either Jelbring or N&Z. The general thermodynamics of adiabatic expansion and its importance in the atmosphere is long known and completely absent from both of their papers. They both assert that something about the atmosphere differentially warms it due to gravity, not that the sun heats the ground, establishes convective flow that moves the heat around (cooling the ground in the process) and that radiative imbalance in GHGs cause differential cooling of the upper troposphere to maintain the energy flow.
Without the rapid cooling of the upper atmosphere that keeps it cold relative to the ground — cooling that is strictly radiation, because sooner or later the Earth has to lose the incoming heat from the sun — the adiabatic warming profile would not exist, and in parts of the atmosphere that profile inverts even as it is (for example, over the arctic in the long arctic night) as further proof that this isn’t an atmospheric compression effect, it is plain old convection.
rgb
Richard M says:
It wasn’t hand-waving at all! I am the ONLY ONE to my knowledge who has even bothered to replicate their fitting procedure. Then I showed how I could change the temperature of the three bodies that have a significant greenhouse effect to essentially eliminate that effect and still get almost as good a fit using their fitting form (even though the data for N_TE was now quite a bit squirrelier since Venus now had a value considerably smaller than Earth and Titan and I made no attempt to change the fitting form or definition of T_sb to optimize the fit)!
I have also explained that there is likely a generally positive correlation between P and their N_TE because as you add an atmosphere to a planet, you will even out the temperature distribution, raising the average temperature. (Higher pressure also tends to correlate with more greenhouse gases and it can also cause broadening of the GHG absorption lines, although I think this is a smaller issue in the positive correlation that they see because, for the most part, their surface temperature enhancement does not really reflect the radiative greenhouse effect.)
Here’s a shorter elevator speech:
If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.
Wow, so perfectly correct! Two laws of thermodynamics (0 and 2). Done.
rgb
Tallbloke:
Your last argument in response to my ‘tennis ball’ model seems to prove too much. Using this argument you could prove that heat can never be conducted along a metal rod. Suppose one end of the rod is in a fire, and the other end in a bowl of ice water. There is a continuous gradient of temperature along the rod. But if we conceptualise the rod as a collection of infinitely thin ‘slices’, each slice will be at the same temperature as the next one. Therefore, by your argument, ‘no heat will flow’. And yet it does!
Willis,
I’m just a bozo on the bus, so I post with trepidation. I was wondering about something that seems to have been addressed by Crispin above. Starting with a perfectly insulated column of non-GHG gasses initially at uniform temperature and pressure throughout, suppose you cause it to appear on the surface of a planet, and suppose that its gravity does indeed cause a temperature gradient to appear.
Then, suppose you take advantage of the temperature differential at the top and bottom to drive a heat engine, which reduces the temperature differential. And so on. That seems to me to say that at some point, the engine will stop.
My question is this: is Jelbring saying that gravity will somehow add heat to the system, thereby restoring initial heat differential? Is he actually explicitly saying that he is creating a perpetual motion machine, or is that something you have inferred? Could he simply be saying that gravity will cause a temperature differential?
I don’t know enough physics (hardly any, in fact), to venture an opinion and certainly an elevator speech is far beyond my capabilities. But I’d just like to know what Jelbring is explicitly saying and whether you might be addressing a straw man argument. I’m not being in any way antagonistic – this is a genuine and open question asked out of ignorance.
The problem is that GHGs and back radiation does not explain the vertical temperature of the atmosphere. The inescapable conclusion of this is that the GHG model is not capable of explaining our atmosphere and that there is more at ‘play’ than the GHG model would suggest.
Why not? I would have said that this is precisely what they predict, although the observed vertical temperature profile of the atmosphere is due to convection induced by differential heating as a part of the overall greenhouse process. In order to get the greenhouse effect, you need to radiate (in a frequency band) from the greenhouse gases up where the atmosphere is cold instead of down where it is as warm as the surface. That’s all it takes. That is experimentally observed to be the case in the actual IR spectrum as measured by NASA satellites.
The real question is: Why do you think one needs additional explanations at all?
Willis Eschenbach says:
January 20, 2012 at 11:56 pm
Joe Born says:
January 20, 2012 at 12:35 pm
… I have to confess that two weeks ago I was arguing over at tallbloke’s place for the same position you’re taking now: http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/#comment-12926. Currently I think I was wrong then and that you’re wrong now. Maybe by tomorrow you and Dr. Brown will have convinced me otherwise.
Joe, all considerations about the mechanism go nowhere. Consider the outcome. If it’s true that gravity can separate molecules by temperature, then we can pull energy out of tall insulated cylinders of air, Jelbring is right and we never have to worry about energy again.
Do you believe that we can do that? Really?
Willis, the rules for falsifying a proposition by appeal to the consequent theoretical constructability of a perpetual motion machine of the second kind are very clear. You have to specify the machine and demonstrate that it will produce work.
Armwaving is insufficient.
I made this point a lot earlier in this thread and you ignored it. Just as I pointed out to Robert brown why his appeal to “the zero’th law failed and he ignored that too.
Unresponsiveness speaks volumes.
A Physicist:
Multi layer insulation – Never knew that!
http://www.technifab.com/cryogenic-resource-library/cryogenic-insulation.html
==
RADIATION FROM THE SKY
From the excellent source of info:
Pat Arnott – thank you!
http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
Atmospheric longwave irradiance uncertainty: Pyrgeometers compared to an absolute sky-scanning radiometer, atmospheric emitted radiance interferometer, and radiative transfer model calculations Rolf Philipona, etal.
Please check out fig 3:
Figure 3. Longwave downward irradiance measured with all
pyrgeometers and the absolute sky-scanning radiometer from
September 22 to 29, 1999, at SGP. Field calibration and Albrecht
et al. formula with C, k2, and K is used for all pyrgeometers.
Nighttime and daytime slots are used for the analysis
of nighttime and daytime measurements.
If I read this correctly there is not much difference in LW IR between day and night.
Around 20 to 50%
Now note that this is LW IR so most of the solar sw input is not present. and solar IR is small
We know(?) that N2 and O2 and Argon do not emit much (zero mainly) IR..
But obviously the sw solar hits the ground gets re-emitted as cool LW ir. this gets reflected (re-emitted actually) back down by IR (or magic depending on your point of view). So this would explain the daytime level. Some of the GHGs will transfer warmth to the non ghgs by collision some will be radiated upwards.
At night there is NO source of solar input, and only a warm atmosphere. So where is the night time IR coming from? The system measures radiation not temperature, The non GHGs cannot emit radiation. So we are left with magic or GHGs.
Now combine this info with the down and up spectra shown in slide 9 of:
http://www.patarnott.com/atms749/powerpoint/ch6_GP.ppt
Where the GHG emissions are obvious.
they are reduced in TOA flux and increased in down welling flux.
Someone said but why is it miss from TOA if it goes down it must also go out. But of course some gets absorbed by the sea/land and provides heat which will be spread thinly over the BB radiation spectrum.
Can anyone explain these OBSERVED an recorde effects without GHGs?
The air moving up and down exchanges potential energy (PE) for kinetic energy (KE). The air moving down loses PE but gains KE, and vice versa for the air moving up. A higher KE means a higher temperature, a lower KE means a lower temperature. So the air moving down increases in temperature (KE), while the air moving up decreases in temperature (KE). This will maintain the adiabatic lapse rate, warmer air at the bottom of the column and cooler air at the top.
Your second point is wrong, there is no equalisation of temperature. Therefore, your conclusion in the third point of your elevator speech is also wrong.
Ah, so the air is moving, is it? What part of equilibrium is escaping you?
Now try again where the air is not moving in bulk because there is no differential heating or cooling of the air, which is trapped in an adiabatic container with no heat flow in or out. It quickly — for all practical purposes instantaneously — develops a pressure gradient and (in response to its bulk compressibility) density gradient, so that there is no bulk transport up or down the air column. At that point, heat moves from higher temperature to lower temperature within the gas because that is what heat does. In order for heat not to flow, the gas would have to not conduct, not thermally mix, not be a physical gas at all.
As you note, the temperature of the gas is directly proportional to its kinetic energy, but molecules of gas are not on average moving up and down the gas column so there is no net conversion of kinetic energy to potential or vice versa. They are however, colliding and exchanging kinetic energy. Those collisions, on average, share the kinetic energy equally among all of the molecules. Any time a molecule has more than the average, it is likely to (on average) lose it in collisions to others, any time one has less it is likely to gain it.
What maintains the adiabatic lapse rate is convection caused by differential heating of the gas column, specifically more heat being delivered to the bottom. Why is it so very difficult for you to see how having a spontaneous separation of temperature in a gas column due to gravity enables a perpetual motion machine of the second kind to be built, and thus is absolutely impossible?
rgb
DavidB says:
January 21, 2012 at 7:34 am
Tallbloke:
Your last argument in response to my ‘tennis ball’ model seems to prove too much. Using this argument you could prove that heat can never be conducted along a metal rod. Suppose one end of the rod is in a fire, and the other end in a bowl of ice water. There is a continuous gradient of temperature along the rod. But if we conceptualise the rod as a collection of infinitely thin ‘slices’, each slice will be at the same temperature as the next one. Therefore, by your argument, ‘no heat will flow’. And yet it does!
Hi David. My argument was assuming an energy equilibrium with a consequent thermal gradient as in Jelbring’s hypothesis. Clearly, if you start stacking extra energy in at one end, heat will flow. This doesn’t affect my argument.
Cheers
Rog
If we wanted gravity derived concepts for heat, we should look at the temperature profiles of planets like Jupiter to see if there is a corresponding gravity induced effect. A quick look at the temperature profile here:
Not the best choice of examples. Jupiter is radiating more heat than it receives because it is still slowly collapsing at the center. What you want is a planet that is in (near) thermal equilibrium, with an atmosphere too thick for the center to be differentially heated by sunlight. The other problem is that the figure you post only goes to the base of the troposphere. The troposphere is by definition the part of the atmosphere that is differentially warmed at the bottom and cooled at the top and exhibits (approximately) an adiabatic lapse rate maintained by vertical convection.
The interesting question is — what is the thermal profile inside the troposphere? The atmospheres of the gas giants do not stop at the bottom of the troposphere. It is difficult to maintain convection (of any depth) while heating from the top. It is not at all unreasonable that the interior atmosphere of the gas giants resembles nothing more than the oceans of the Earth, with a more or less constant temperature and very little convection.
Nobody argues that the troposphere isn’t warmer at the bottom and cooler at the top. That’s what a troposphere is — the part of the atmosphere that is warmer at the bottom and cooler at the top. If you look at the very same figures for Jupiter (and for that matter Uranus or other planets) you can also see the layering and composition of the greenhouse gases in those atmospheres that seem to establish the thermal gradient that drives the convection that establishes the adiabatic lapse rate that perpetuates the greenhouse effect and makes the troposphere’s thermal distribution dynamically stable.
rgb
Robert Brown says:
January 21, 2012 at 7:24 am
Here’s a shorter elevator speech:
If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.
Wow, so perfectly correct! Two laws of thermodynamics (0 and 2). Done.
I’ll add it to my collection of potentially conflicting formulations of the thermodynamics laws – thanks for endorsing it.
Tallbloke, your statement is wrong-on-the-facts: Willis (and I) did describe a perpetual motion of the second kind, namely a thermopile column, said column having its warm end at ground-level and its cold end at altitude.
And the mistake in “gravito-thermal” theories is evident too: these theories include a thermodynamic potential (namely the temperature) together with a gravitational potential, and yet they (wrongly) neglecting the chemical potential.
Just to mention, ideal gases do have a chemical potential, whose effects must be included in any thermodynamically consistent theory that involves particle exchange (in this case, the exchange of molecules between layers of the atmosphere). Because “gravito-thermal” theories mistakenly omit the ideal gas chemical potential, it is unsurprising that the predictions of “gravito-thermal” theory violate basic thermodynamical principles (precisely as Willis asserted at the beginning of this post).
Hopefully, further study of the theory of chemical potentials will clarify these issues for the authors and fans of “gravito-thermal” theories. Wikipedia’s article on chemical potentials is a good start, and includes many further references.
Joel Shore;
Great…Now all I have to do if I perform a fit of y vs. x with 100 free parameters is break it into two equations>>>
Stop Joel. You are making a fool of yourself. Their final equation is Equation 8 and it has two variables in it. Two. count ’em. Two.
Joel Shore;
The above is the exact argument made by Alan Siddons, one of the “Slaying the Skydragon” crew. It is an extremely silly argument. >>>
Unable to refute it you instead try to equate it to something else. Is it right or isn’t it?
Joel Shore;
The radiative greenhouse effect is the only thing that can explain how the surface of the Earth can be at an average temperature so high that it emits ~390 W/m^2 while the Earth as seen from space only emits ~240 W/m^2 and is thus still in radiative balance with what it receives from the sun.>>>
Really Joel? Really? You cling to math so ridiculously wrong that your attempt to justify it is just pathetic. You have no idea what the theoretical black body temperature of the earth is, you have no idea what the practical black body temperature of the earth is, and thus you have no idea what the difference between the two is, let alone what causes it. But do read BenAW’s comment upthread where he has a nice explanation of EXACTLY how it works. You insist on not only ignoring the falacies introduced into your conclusions by the misaplication of SB Law, you insist also on ignoring the heat capacity of the planet, a point that I have made to you many times and which you continue to ignore.
You earned my respect in a variety of threads a while back, you have now lost it entirely and I think it tragic that someone as supposedly intelligent as Willis Eschenbach has somehow been duped by your duplicity.
Robert Brown says:
January 21, 2012 at 7:50 am
Why is it so very difficult for you to see how having a spontaneous separation of temperature in a gas column due to gravity enables a perpetual motion machine of the second kind to be built, and thus is absolutely impossible
As I said to Willis, armwaving is insufficient.
The rules for falsifying a proposition by appeal to the consequent theoretical constructability of a perpetual motion machine of the second kind are very clear. You have to specify the machine and demonstrate that it will produce work continually forever.
I have become convinced that the isothermal hypothesis, although correct as an approximation, is theoretically true only in the limit….
(referring to the following from tallbloke):
It is from the thus-obtained Velasco et al. Equations 5 and 6 that they claim to obtain their Equation 8 for temperature as a function of altitude, and it is that equation that seems to me not to jibe with any significant lapse rate: although it does have temperature fall with altitude, the drop, if my calculations are correct, is negligible.
Does anyone see where the authors or I went wrong here?
Sure. You and the authors seem to fail to appreciate the fact that any lapse rate in equilibrium violates the second law of thermodynamics.
You obviously understand thermo and some stat mech, so surely you understand Willis (and my own, independent) argument. In your imagined atmosphere, with hotter temperatures anywhere that are maintained by e.g. gravity as a steady state, if you run a thermally insulated conductor from the hot to the cold, what will happen, Joe? How long will it happen?
That’s right, heat will flow (to equalize the temperature difference) because that’s what heat does — it flows from hot to cold through any material that can conduct heat. It doesn’t matter if the gas parcels in thermal contact with the ends of the “heat pipe” are at the top or bottom of a vertical column of air or running between two adiabatically isolated containers of air at different temperatures and pressures, heat will flow from the hot side to the cold side. If Velasco is predicting that the gas will spontaneously separate into hot and cold reservoirs, well, I don’t think it is possible to imagine a more pure example of a violation of the second law, can you?
The beauty of the laws of thermodynamics is that it no longer matters if one can find the precise error made by Velasco in a complicated stat mech analysis. In all probability it involves including momentum or failing to achieve local equilibrium in the first place — ultimately what he is asserting is that there is a failure of detailed balance (because detailed balance leads to isothermal equilibrium). But no matter — heat will not flow forever in an isolated system, will it? And yet, if the top and the bottom of the container maintain themselves at different temperatures, it will, the instant you provide an alternative pathway for heat to flow in the still isolated system.
If Velasco has somehow proven that gravity acts as Maxwell’s Demon — which is precisely what this is — then either he deserves a Nobel Prize, for proving that the second law is openly violated in the Universe and at the same time solving our energy crisis as we can start building 100% efficient perpetual motion machine of the second kind heat engines, engines that just recycle heat into work indefinitely, or he deserves to be instantly ignored until he goes back and finds his own mistake. Personally, I wouldn’t dare to publish a paper that asserted that the second law is violated, that gravity acts like Maxwell’s Demon in any atmosphere ideal or real, because I know that if I somehow got that as a conclusion, my work would almost certainly be incorrect. I’d be just as leery of publishing something with an error in units, or that violated energy conservation.
rgb
Richard M;
This was the same argument Joel tried to use against me back in Ira’s thread. It’s unfortunate that they are willing to discard what may very well be a valid correlation by simply waving their hands.>>>
Frankly I am disgusted. Equation 8 in N&Z’s article has two variables, and anyone who can read can see that. All their results are calculated using THAT equation. Trying to represent it as having four variables by pointing to an intermediate step is an outright lie.
Willis, please. Read the effing Equation 8 for yourself and set the record straight.
Robert Brown says:
January 21, 2012 at 7:24 am
“Here’s a shorter elevator speech:
If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.
Wow, so perfectly correct! Two laws of thermodynamics (0 and 2). Done./rgb”
Perfectly wrong since you refuse to include gravity in the system you are treating. Since you now are informed that the Zeroth Law doesn´t work in this situation it seems that you are practising “willful ignorance” which does not belong to scientific methods.
Best
Hans Jelbring
tallbloke says:
January 21, 2012 at 7:10 am
“Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.”
Tallbloke, please bear in mind that for any parcel that leaves an air layer, another parcel must enter to conserve mass, and the parcel that enters has an equal likelihood of expanding on entry as compressing. Thus on balance the number of parcels that expand when entering the layer must equal the number of parcels that compress when entering, leaving no net change in the energy content of the layer, and no opportunity for perpetual motion on account of the adiabatic process. However, there is still conduction, which will eventually remove any temperature gradient that exists in the absence of an external source of power. It should be clear that an external source of power is required to develop the adiabatic temperature profile and in our neighborhood the sun does the trick.
Can somebody please conceive this fact; Energy cabn be also employed, not just conserved.
Add kinetic energy to matter and the E does not equal mc/2.
OK, once again we have a case of “get an intro physics textbook and read it before stating things that you don’t understand and that are in any event incorrect”.
First of all, energy cannot be “employed”. Free energy can be employed, then it turns into heat. Heat cannot be employed to make work unless there are two thermal reservoirs at different temperatures. If you want to discuss this, please learn the laws of thermodynamics. If you want to assert that the laws of thermodynamics are incorrect, all I can say is don’t be silly! That may not stop you from being silly, but it should.
Second, E = \gamma m c^2 (for an isolated massive particle moving at constant speed in an inertial frame). This includes the kinetic energy. In fact, the low-velocity “usual” form of the KE is the first non-constant term in a binomial/Taylor series expansion of \gamma m c^2. A common way of asserting this is to include the \gamma in the m and say that particles with kinetic energy are more massive. This is, in fact, the case — a hot object is indeed more massive than the same object when cold.
rgb
Extend the column or cylinder to TOA and you have a mechanism to lose the top layer to space fairly rapidly, and by logical extension, the whole atmosphere.. Some disconnect with reality here?
Not at all. Learn about “escape velocity”. This is how the Earth does, gradually, lose atmosphere. It just takes a long, long time for the heavier molecules at the average temperature of the Earth.
rgb
In the event that anyone read my previous post, I should point out the following: Stygia is a perpetual motion machine. It was defined as such by Willis himself. A perfectly insulating, perfectly reflective shell defines a closed, perfectly lossless system: the very definition of a perpetual motion machine.
The average temperature of Stygia must drop as light is generated but there will be a point where the energy reabsorption (and heating) of the planet will exactly balance the energy extracted by the heat engine. Stygia will be light but colder than if it was dark.
In answer to the charge that we could build energy sources that extract energy from a temperature gradient in the atmosphere. The energy drained from the atmosphere is replenished by insolation. Well known parallels would be the photovoltaic cell or wind turbine. Extracting energy directly or indirectly from the energy of insolation. Not very efficient I grant you but that’s why we don’t want to rely on renewables.
BenAW;
What both approaches are neglecting is that the earths surface consist mostly of oceans,
70% area, minimum 3 km deep and a temperature on average of +2C, already 20K higher
than the blackbody temperature the GHE uses when the sun has heated the blackbody.>>>>
I recommend BenAW’s comment upthread to everyone who demands an explanation of surface temperature that is higher than blackbody via a means other that back radiation. The earth surface is ALREADY above the blackbody temp, all the sun has to do is MAINTAIN it, and there is sufficient energy from the sun to do that while also satisfying the laws of thermodynamics. you cannot arrive at that conclusion however unless you first properly apply SB Law instead of starting off with mind bogglingly WRONG averages of a set of variables that CANNOT be compared via averages.
BenAW, the one quibble I have with your comment is that N&Z missed this. My reading of N&Z is that it is founded upon this (amongst several other things).
Robert Brown says:
January 21, 2012 at 7:39 am
The problem is that GHGs and back radiation does not explain the vertical temperature of the atmosphere. The inescapable conclusion of this is that the GHG model is not capable of explaining our atmosphere and that there is more at ‘play’ than the GHG model would suggest.
Why not?
Since there is nothing more unphysical than 390 W/m^2 backradiation. Such an effect can easily be felt by the skin. IR radiation in dífferent directions are no vector quantities that can cancel each other. If you have two suns shining from opposite directions you will fell the heat from both directions. The same is true about any electromagnetic radiation and IR is of that type.
Since you don´t feel any heat fromn any direction during night time it is all bull. I would glacly examine all påredtended measurements if I get information of the instrument and their constructions. This fraud is just passing any limits and is degrading science to superstition.
It is good to know that you and willis are great IPCC supporters.
Robert Brown says:
January 21, 2012 at 8:28 am
This is starting to look like a bit like a battle between 19th century physics and 20th century physics.
The concept of heat doesn’t not rise superior to the concept of atoms. If the movement of atoms under the influence of gravity shows they lose kinetic energy as they rise then some heat disappears and becomes gravitational potential energy. What heat remains still flows, but that heat lost cannot flow.
Bart says:
January 21, 2012 at 2:43 am
Not plausible. The thermal conductivity of air is low. You need a very large temperature gradient to get rapid conductance. That sounds like the same argument that Bill Illis made on the other thread that the short time between molecular collisions could somehow prevent an excited molecule from radiating. But collisional energy transfer must be orders of magnitude faster than radiative transfer for LTE to apply. Emission is determined by the number of molecules or surface states that are excited. That number is a constant in any small volume where LTE applies. For every excited state that is de-excited by collision, another is created by another collision. That’s the how Maxwell-Boltzmann statistics work. LTE only breaks down at very high altitudes in the atmosphere.
You can hand wave all you want about non-equilibrium thermodynamics, but you have yet to demonstrate that they actually apply and give a better fit to observations. The observed emission spectra of surfaces and the atmosphere and spectra calculated assuming that LTE applies are the same within the experimental error of the measured (and calculated ab initio) line parameters and the experimental error of the observing spectrometer.
pochas says:
January 21, 2012 at 8:32 am
Tallbloke, please bear in mind that for any parcel that leaves an air layer, another parcel must enter to conserve mass, and the parcel that enters has an equal likelihood of expanding on entry as compressing. Thus on balance the number of parcels that expand when entering the layer must equal the number of parcels that compress when entering, leaving no net change in the energy content of the layer, and no opportunity for perpetual motion on account of the adiabatic process. However, there is still conduction, which will eventually remove any temperature gradient that exists in the absence of an external source of power.
Hi Pochas, my two line proof which Robert Brown continues to ignore shows that the thermal gradient in a gravitationally forced gas column which is at energy equilibrium will not conduct heat (KE).
As Trick put it:
[Any claim that] the column will be isothermal, meaning all at the same temperature top to bottom…. violates the 2nd law; KE + PE = constant at each h in the presence of an inexplicable gravity field in the gaseous cv of interest, namely an adiabatic (no gain or loss of heat from CV) GHG-free air column.
The proof again incase Robert wants to respond:
tallbloke says:
January 21, 2012 at 7:10 am
“Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.”
Robert Brown says:
January 21, 2012 at 7:50 am
“What maintains the adiabatic lapse rate is convection caused by differential heating of the gas column, specifically more heat being delivered to the bottom. Why is it so very difficult for you to see how having a spontaneous separation of temperature in a gas column due to gravity enables a perpetual motion machine of the second kind to be built, and thus is absolutely impossible?
What I prove in my E&E paper is that “dynamic” DALR is equal to the “static” DALR so convecting is not needed to reach the maximum entropy situation. All available processes will act to reach taht process. It is the content of the 2:nd law of thermodynamics applied in an insulated atmosphere.
I am interested how you are going to construct your perpetuum mobile out of ideal gases. You or anybody else has alread disingrated and your bound energy belongs to the total inclosed energy from start of the thought experiment.
Joe Born,
Quick and dirty. Meteorologists use a concept called potential temperature to look at atmospheric temperature profiles. It’s the temperature that would result if a packet of air were moved back to the surface adiabatically. Obviously an atmosphere with an adiabatic lapse rate has a potential temperature gradient of zero. Potential temperature is, in fact, a measure of entropy. An adiabatic lapse rate is isentropic. So now lets make the atmosphere isothermal. Potential temperature now increases with altitude so the average potential temperature, and therefore the entropy of the system, has increased. The entropy of the atmosphere could be increased even more by making the temperature increase with altitude. But that’s not stable. Heat would be conducted back to the surface because of the temperature gradient. Therefore, an isothermal atmosphere has the maximum stable entropy.
jelbring:
“If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.”
this is a definition. it is not wrong. get literate – then you can get logical – then you can get scientific. you aren’t close yet.
Robert Brown says
“What maintains the adiabatic lapse rate is convection caused by differential heating of the gas column, specifically more heat being delivered to the bottom. Why is it so very difficult for you to see how having a spontaneous separation of temperature in a gas column due to gravity enables a perpetual motion machine of the second kind to be built, and thus is absolutely impossible?”
Robert it seems that some within the Climate Science define convection as an UNSTABLE vigorous vertical exchange of air. .
See bottom of page 13.
The stable condition (hydrostatic approximation) is used to derive the DALR. See page 12
Air parcels moving up and down at constant speed (no unbalanced force) will track the DALR.
These air parcels are assumed not to exchange heat with their surroundings.
On going up expansion work PdV is stored by the surroundings(temperature dropping by 9.8K/km)
At TOA there will be a loss of heat by radiation to space causing the down phase
On going down the surroundings do work on the parcel (PdV) (temperature increasing by 9.8K/km)
Stationary parcels will not change temperature.
This atmospheric condition is known as the neutral atmosphere and can be stable particularly at night.
See the near neutral RESIDUAL LAYER page 31
These two idealised adiabatic processes (like the adiabatic stages in the Carnot Cycle) will result in the parcel returning to Earth with nearly the same temperature as leaving (the slight drop being accounted for by radiation at TOA).
Energy supply is from Sun heating the Earth surface and atmosphere during daylight and the Earth surface at night
After reading many comments, my elevator speech is:
The troposphere is a giant heat engine with energy input from the sun heating the lower part and the upper part being the condenser dumping energy to outer space.
Energy emission from the the top of the troposphere is a thermodynamic requirement for the engine to run.
Gravity is necessary for the function of the engine but does not in itself cause the thermal gradient.
Greenhouse gases must be present in the upper troposphere as this is the only significant mechanism available for energy to be removed from the upper troposphere.
The lapse rate is a thermodynamic manifestation of the heat engine at work.
In conclusion, I side with Willis and Dr. Brown on this thread.
Bart says:
January 20, 2012 at 11:00 pm
OK. The upper temperature limit is the surface temperature. Assuming an emissivity of 1 as specified, the surface temperature will be 255K. The surface temperature cannot increase without limit without violating the Second Law.
You’ve got to be kidding. That makes no sense at all. How is emission being suppressed? If the surface has a temperature above absolute zero and the emissivity is 1, it will emit radiation. Period. Molecular collisions with the surface cannot suppress emission, they can only raise or lower the temperature. You can’t keep pumping heat into the atmosphere and maintain a negative temperature gradient if there’s no heat sink at the top of the atmosphere. You can assert that this could happen all you want, but your assertions are unphysical, and quite frankly, absurd.
tallbloke:
I wouldn’t read anything nefarious into Brown & Eschenbach’s failure to engage on the issue of what proves that equilibrium implies isothermality. To them, one who questions the proposition that it does no doubt appears to be just as much a crank as someone who questions conservation of energy would appear to you or me. And have I proved conservation of energy or had it proved to me? I don’t think so. It’s just that I know all the technology around us was designed by people who counted on that law, and they seem never to have met a counterexample. Brown & Eschenbach no doubt accept the equilibrium -> isothermal thing for the same reason. And would you or I waste much time on someone who professes to believe that conservation of energy does not apply in a particular situation? I probably wouldn’t.
To them it no doubt seems pointless to take the time to look deeply into a “law” they think is supported by the experience of hundreds of thousands of engineers over centuries. And why should they listen to an uncredentialed layman like me about how likely it is that a serious paper actually does say that the “law” is not exactly what they think it is?
Still, it is frustrating.
Robert
I forgot to link the paper in my last post
http://www-as.harvard.edu/education/brasseur_jacob/ch2_brasseurjacob_Jan11.pdf
A physicist says:
January 21, 2012 at 8:12 am
tallbloke says: Willis, the rules for falsifying a proposition by appeal to the constructability of a perpetual motion machine of the second kind are very clear. You have to specify the machine and demonstrate that it will produce work.
Armwaving is insufficient. I made this point a lot earlier in this thread and you ignored it.
Tallbloke, your statement is wrong-on-the-facts: Willis (and I) did describe a perpetual motion of the second kind, namely a thermopile column, said column having its warm end at ground-level and its cold end at altitude.
In comparison the the machine described in email to me by Peter Berenyi, this is an incomplete and inadequate description. Try harder.
And the mistake in “gravito-thermal” theories is evident too: these theories include a thermodynamic potential (namely the temperature) together with a gravitational potential, and yet they (wrongly) neglecting the chemical potential.
Yes, I read your reply to anther comment about this. However, the argument fails because the chemical potential is at equilibrium in the Jelbring thought experiment. The Gibbs free energy is zero.
gnomish says:
January 21, 2012 at 9:13 am
“If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.”
this is a definition.
It is a definition. One that is severely lacking in universality. It fails as a general law as soon as it is placed orthogonally to a gravitational field.
We know that gravity exists but not one acedemic has managed to explain with any confidence why our boots are attracted to the Earth’s core.
Your point being? Or if you prefer, exactly what would you consider an answer?
Science consists of observing the world, noting regularities, proposing hypotheses to explain those regularities, and fitting all of the competing sets of hypotheses into a coherent, mostly consistent whole. The whole set of your beliefs about the world is called your worldview. Nothing in your worldview except the empirical ongoing experience of your own existence is certain (see e.g. the movie The Matrix, or Plato’s parable of The Cave for ways — however implausible or plausible you might think them — that we could be mistaken even about the external reality we infer on the basis of our sense data including our memory).
In that context, knowing why is impossible. The best we can do is have a set of well-founded beliefs, the beliefs that are in some sense the best things to believe, given the data and our experience.
Gravity is very much in that category. We know that all the mass we can see appears to move as if Newton’s Law of Gravitation is correct, subject to minor tweaks associated with relativity (which we believe for a variety of equally good reasons). There are a very few astronomical observations where we cannot verify gravity — but where the assumption of Newton’s Law requires an “odd” distribution of mass we cannot see, so called “dark matter”. We do not yet know enough to be certain what the explanation of the deviation is. We also do not yet have a completely consistent theory of everything yet, and the transition between quantum theory and general relativity is not yet well-understood.
So what? Gravity could be caused by invisible fairies and it would still be the case that the invisible fairies make masses accelerate pretty much as if Newton’s Law of Gravitation is true. Or are you just trying to assert something childish, like “you physics guys don’t know everything, look, you don’t even know why gravity exists, so that means when you say thermal equilibrium is isothermal you could be wrong, you just don’t know”?
The number of logical fallacies in such an assertion is left as an exercise for the reader. Of course, you didn’t make that assertion, so I don’t really know what your point is. If you have one, please express it clearly.
rgb
The kinetic energy of mass is the mechanism that enhances its employment of energy.
Which means, would that be, nothing at all?
It would.
rgb
Joe Born says:
January 21, 2012 at 9:23 am
tallbloke:
I wouldn’t read anything nefarious into Brown & Eschenbach’s failure to engage on the issue of what proves that equilibrium implies isothermality. To them, one who questions the proposition that it does no doubt appears to be just as much a crank as someone who questions conservation of energy would appear to you or me. And have I proved conservation of energy or had it proved to me? I don’t think so. It’s just that I know all the technology around us was designed by people who counted on that law, and they seem never to have met a counterexample. Brown & Eschenbach no doubt accept the equilibrium -> isothermal thing for the same reason. And would you or I waste much time on someone who professes to believe that conservation of energy does not apply in a particular situation? I probably wouldn’t.
To them it no doubt seems pointless to take the time to look deeply into a “law” they think is supported by the experience of hundreds of thousands of engineers over centuries. And why should they listen to an uncredentialed layman like me about how likely it is that a serious paper actually does say that the “law” is not exactly what they think it is?
Still, it is frustrating.
Joe, I think you have got to the heart of the matter.
I doubt that the physicists ever got the engineers to build anything tall enough or sufficiently sensitive or well insulated enough for them to spot the gradient and wonder what caused it.
It’s easy enough to derive from formulations of the thermodynamics laws which concentrate on energy rather than heat. I guess we’ll have to let the old school labour under their two dimensional delusions and just get on with testing predictions of planetary surface temperatures against empirical measurements. It seems Nikolov and Zeller are streets ahead using the gravito-thermal theory on that front already.
I’ll abandon attempts to get Dr Brown and Willis to engage for now and write up the unanswered proofs for the blog.
All sounds like my first year university Physics lectures.
Perhaps a few more Physicists need to read and understand and comment.
Where’s Richard Feynman when we really need him?
Dead. I’m doing my best, but it is difficult to teach a full year of intro physics in a blog, and between people who are clueless about work, energy, gravity, Newton’s Laws, fluids, and thermodynamics my hands are full.
What would really help would be more people going back to that first year physics textbook and reading it. Then if they would simply stop making posts that are the equivalent of “introductory physics is all wrong”. It just makes them sound very silly. It’s fine to discuss, debate, and learn, but if one wants the debate to be useful, it really helps if both sides do their homework.
I’ve done my homework, so to speak, from 1973 up to the present. Climate physics per se I’m working on, but physics itself from intro stuff that I teach a full course of twice a year these days through graduate level E&M and quantum and research level condensed matter physics. I make no pretense to being as good as Feynman — he was truly exceptional — but as the author of an online physics textbook and teacher of all this stuff to hundreds of very smart students a year, I’m not terrible — if anyone will listen to me.
rgb
Ref Earth’s Heat
Don’t forget that we still quite a bit left from the accretion of the planet.
Nicely locked up inside an insulating crust. The average outward flux of heat has been posted repeatedly. It hasn’t been forgotten, it is just negligible, completely inadequate as an explanation of “warming”. If it weren’t for the sun, oceans and the atmosphere, the Earth would be warm like the moon is warm, which is pretty much “not”. And even the heat that is in there is at least partially due to tidal heating from the moon.
I’m happy to hear a quantitative assertion of a significant contribution from geothermal energy, but it would have to be backed by references and data.
rgb
100 kg of gas 100 kms from Earth will e have 98,000,000 less joules of energy than 100 kg of gas at the surface.
Gravitation potential energy actually turns into real thermal energy for a mass that is falling through a gravity field or objects with mass that are closer/farther from the centre of the gravity field.
GPE = Mass * Gravity * Height
100 kg of gas 100 kms from Earth at 100C will have the exact same kinetic energy per molecule, independent of the volume it occupies (as long as it remains at least dense enough to be in thermal equilibrium).
As you drop the container of gas from the upper height to the lower, the temperature inside will remain the same. Sure, the gas gains KE, but not heat — all of the KE is organized and associated with one very non-thermal direction. The kinetic energy per molecule in the gas in container-centered coordinates will make only tiny changes as the fluid density redistributes during the fall.
The gas gets hot when it hits something and abruptly comes to rest. This basically slams all the molecules into the walls of the container, and transfers all that KE they accumulated into random motion, that is, “heat”.
The problem with a static column of gas is that it isn’t falling. Get it? It’s not moving up or down. The density/pressure is constant as a function of height. Gravity isn’t releasing more energy or absorbing more energy. Gravity is doing no net work on the system. Consequently, the molecules can share their kinetic energy around and come to thermal equilibrium without moving up or down and involving gravity.
rgb
The isothermal atmosphere is not the maximum entropy state. If you start with an isothermal atmosphere and exchange two parcels of air in the vertical direction, the one you bring down becomes warmer than its environment (due to compression at a higher pressure) and the one you bring up becomes cooler. Now mix those exchanged parcels at their new level and keep doing this, and you are making lower levels warmer and upper levels cooler.
In an isentropic atmosphere (meaning constant potential temperature and adiabatic temperature lapse rate), exchanging two parcels in this way results in no temperature change because the lapse rate cancels the compression effect. However much you mix adiabatic-lapse-rate air you won’t alter its temperature profile. This is the well-mixed maximum entropy state.
Robert Brown says:
January 21, 2012 at 10:01 am
But IT is made of particles that are moving up and down within the column.
Imagine one molecule at the bottom of the column made energetic by contact with the surface of the Earth hitting another molecule 10^-5cm above it (average distance between particles) and so on through the layers as the atmosphere tries to reach equilibrium. Each time a lower molecule loses some kinetic energy to gravitational potential energy as it travels across the current layer before it strikes a molecule in the next layer above it. At 10km, 4.56×10^-21 J has been converted to gravitational potential energy — a significant amount when compared with the 5.81×10^-21 J at the start. The reverse is true, too. Molecules propelled upward are matched by molecules falling downward and each downward falling molecule gains 4.56×10^-32 J of kinetic energy as it falls through a 10^-5cm layer.
These differences in kinetic energy in each layer must translate into different temperatures in each layer.
Hans Jelbring says:
January 21, 2012 at 8:53 am
It’s already been described up thread, but I’ll do it again. Take two vertical cylinders 100 m tall. Fill one with helium and the other with xenon at 1 atmosphere. Thermally couple the bottom of each cylinder to the other and insulate the rest of the cylinders. The DALR for helium is 0.001888 K/m and 0.061 K/m for xenon. If gravity causes the DALR to be established in both cylinders, the temperature difference at the top of the cylinders would be 6K. If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top. The cylinders will then be isothermal. We disconnect the heat engine. If gravity then re-establishes the DALR, we can run the heat engine again, and again, and again. When the DALR is re-established, the heat that was moved from the helium to the xenon will move back again. The total energy content of the two cylinders won’t change. Energy conservation is violated. But of course, there won’t actually be a temperature difference between the tops of the two cylinders.
Hans Jelbring says: January 21, 2012 at 8:42 am
Since there is nothing more unphysical than 390 W/m^2 backradiation. … I would glacly examine all påredtended measurements if I get information of the instrument and their constructions. This fraud is just passing any limits and is degrading science to superstition.
It is good to know that you and willis are great IPCC supporters.
=======================
Please see my post above:
jjthoms says: January 21, 2012 at 7:49 am
http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
Atmospheric longwave irradiance uncertainty: Pyrgeometers compared to an absolute sky-scanning radiometer, atmospheric emitted radiance interferometer, and radiative transfer model calculations Rolf Philipona, etal.
Please check out fig 3:
Figure 3. Longwave downward irradiance measured with all
pyrgeometers and the absolute sky-scanning radiometer from
September 22 to 29, 1999, at SGP. Field calibration and Albrecht
et al. formula with C, k2, and K is used for all pyrgeometers.
Nighttime and daytime slots are used for the analysis
of nighttime and daytime measurements.
If I read this correctly there is not much difference in LW IR between day and night.
Around 20 to 50%
…
Only downward radiation here BUT in this document slide 9 toa and ground are compared.
http://www.patarnott.com/atms749/powerpoint/ch6_GP.ppt
Where the GHG emissions are obvious.
they are reduced in TOA flux and increased in down welling flux.
Crispin in Waterloo says:
January 21, 2012 at 1:58 am
There are 3 ways to move heat – radiation, convection, conduction
In an atmosphere without any IR active substances (such as greenhouse gases), radiation is not a factor. Using only convection, the Universal Gas Law applies and, in a gravitational field, the dry adiabatic lapse rate (DALR) is produced. However, conduction must also be considered. It is conduction (much slower than convection) that produces the isothermal atmosphere.
Basically, the Universal Gas Law assumes that no energy enters or leaves the system. This is referred to as adiabatic. Conduction allows energy to move from one part of the system to another and, therefore, allows the atmosphere to have a lapse rate different than described by the DALR.
“It is a definition. One that is severely lacking in universality. It fails as a general law as soon as it is placed orthogonally to a gravitational field.”
t.b – the definition of a definition is something for you to understand. definitions don’t possess the characteristic of universality. they limit. the latin ‘fin’ does not mean universal.
if you redefine the context, you are changing the definition.
the laws of thermodynamics do NOT require ‘gravity’. mass is not a property of heat. you may as well speak of age, length and color effects on temperature. it’s not even good nonsense.
density is not heat. god, man – temperature is not heat. i can have gas at any temperature and density i please. there is no state equation of gas laws that infers, implies or has anything to do with gravity or density.
bad physics is a direct consequence of the abuse of language.
dude- take out one of the several torches you own and heat some things. if you heat a spot of anything at all, you have a hot spot. if the whole thing is not hot, therefore you have a gradient because things have a property called ‘thermal conductivity’. every single bit of mass in the universal universe does that. the dynamic gradient is produced it is not static. it has nothing to do with gravity.
if i understand the proposition, gravity is supposed to increase density which increases thermal capacity. this is as much sense as i can extract so far from the gibberish.
in fact, the take home lesson i am getting from this is that people who can’t use language properly can’t think properly and are busily truncating the ‘iens’ offa our species name.
DeWitt Payne says:
January 21, 2012 at 10:21 am
Take two vertical cylinders 100 m tall. Fill one with helium and the other with xenon at 1 atmosphere. Thermally couple the bottom of each cylinder to the other and insulate the rest of the cylinders. The DALR for helium is 0.001888 K/m and 0.061 K/m for xenon. If gravity causes the DALR to be established in both cylinders, the temperature difference at the top of the cylinders would be 6K. If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top. The cylinders will then be isothermal. We disconnect the heat engine. If gravity then re-establishes the DALR, we can run the heat engine again, and again, and again.
Thanks DeWitt, this sounds more promising. So for a 1m^2 column we could generate how much power? According to Graeff’s experiments, the gradient was re-established within 36 hours with water diffusing through ground glass particles when he turned the columns end for end. I expect the gas would re-organise itself more quickly.
But wait, the extraction of energy will cool the Helium gas overall. Where is the energy going to come from to warm it up again? Can’t come from outside the machine or it’s not a perpetuum mobile, so it has to be perfectly insulated. Or if you think that all that is required is to have a differential, how is the heat engine going to continue running once the temperature approaches absolute zero? Or if you are going to use the generated energy to warm the gas again, how do we tell it is working at all?
Tallbloke, simpler it better!
You didn’t tell us what amazing, complicated machine your friend imagined, but the simple machine that has been described (the thermopile) does exactly break the 2nd law of thermodynamics if a lapse rate is the equilibrium condition.
Thinking about DeWitt’s machine some more, I think I’ll think about DeWitt’s machine some more, before I try to evaluate it in a summary, so scrub that second paragraph for now.
Good puzzle! Thanks DeWitt.
Tim,
Peter’s machine wasn’t complicated, just better specified than the previous attempts I’d seen on this thread. I missed DeWitt’s earlier post.
The atmosphere ALWAYS reconfigures its energy distribution so that the Atmospheric Thermal Effect/Adiabatic Lapse Rate is maintained.
Wilde’s Law.
Tallbloke, this is a case where “handwaving is not enough”, and in fact your assertion is flat-out wrong.
Please consult a kinetic theory textbook that gives the (simple) expression for the chemical potential of an ideal gas in a gravitational potential. Like this one, for example.
davidmhoffer says:
January 21, 2012 at 8:41 am
“BenAW, the one quibble I have with your comment is that N&Z missed this. My reading of N&Z is that it is founded upon this (amongst several other things)”.
I’m trying to engage in a discussion, but they seem to busy at the moment.
See eg.
http://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/#comment-13485
or
http://tallbloke.wordpress.com/2012/01/17/nikolov-and-zeller-reply-to-comments-on-the-utc-part-1/#comment-14747
or
http://tallbloke.wordpress.com/2012/01/17/nikolov-and-zeller-reply-to-comments-on-the-utc-part-1/#comment-14848
Would be interested in at least some reaction.
Ben Wouters
John Marshall says:
January 21, 2012 at 2:34 am
John, I don’t “dismiss them” I do note that:
1. We understand the mysteries of our own climate only poorly, so our knowledge of the inner workings of e.g. Neptune’s climate must perforce be orders of magnitude worse. This makes it hard to draw any solid conclusions.
2. The atmospheres of the other planets are very, very, very different from the Earth. For example Venus, which is in some ways is the most similar, has temperatures and pressures at the surface that make the CO2 into a supercritical fluid:
Venusian atmosphere and spectroscopic properties of CO2
As can be seen from above link, partial pressure of CO2 in the Venusian ground atmosphere amounts to ~89 bar and ground temperature to 737 K on average. The “critical point” data of CO2: p(crit.) = 72.9 bar; T(crit.) = 304 K (handbook of physics). A substance beyond the critical point is neither gas nor fluid, though in general referred to as supercritical fluid. It surely is not a gas and very surely not a trace gas. SOURCE
John, perhaps you could explain to us all what an atmosphere where CO2 is not a trace gas, not a gas at all, not a liquid, but a supercritcal fluid can teach us about the Earth’s climate?
Yes, as you point out, planetary atmospheres all obey the same laws of physics as they apply to gases, but so does my grandma’s dying breath … so what?
w.
Never mind all that detail.
“The atmosphere ALWAYS reconfigures its energy distribution so that the Atmospheric Thermal Effect/Adiabatic Lapse Rate is maintained.”
Wilde’s Law.
On the WUWT threads many are doing all they can to ignore the implications of the Ideal Gas Law and N & Z’s ATE ( formerly widely accepted as the Adiabatic Lapse Rate or ALR).
Just look at Wikipedia here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas.”
and:
“the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature normal with increasing distance from the conductive source.”
which together show that we are dealing here with long established science although recent shenanigans have made it necessary for N & Z to try to redefine the adiabatic lapse rate (ALR) as ATE for the current uneducated generation.
So the issue isn’t whether the phenomenon is the ATE (Atmospheric Thermal Effect) or the ALR (Adiabatic Lapse Rate)
We can see that they are one and the same.
The issue is whether increasing GHGs can make a difference to the ATE/ALR by virtue of their radiative characteristics.
Well what Ned is doing is simply refining the data and the S-B calculations to show that the effect of more GHGs is zero as against the AGW theory that more GHGs can ADD to the ATE/ALR.
Where I propose a novel idea is in showing WHY the extra GHGs have zero or near zero effect on the ATE/ALR. Ned demonstrates the fact that there is a zero effect and I say WHY there is a zero effect.
The fact is that the entire atmosphere responds to ANY forcing that attempts to disrupt ATE/ALR by altering the tropopause height and that change in height then allows the surface pressure distribution to slide poleward or equatorward beneath the tropopause to negate the thermal effect of any change in tropopause height either at equator or poles.
The same process occurs on EVERY planet that has an atmosphere which tries to disrupt ATE/ALR. The atmosphere ALWAYS reconfigures so that ATE/ALR is maintained.
There is no other possible solution. IMHO.
To my mind the jigsaw is complete.
Stephen Wilde says:
January 21, 2012 at 11:11 am
Wilde’s Law.
With a little help from his friends. 😉
It may turn out to be consistent with Willis’ favourite the Constructal Law too….
Tallbloke says:
>So for a 1m^2 column we could generate how much power?
That is immaterial. ANY violation is a violation.
>According to Graeff’s experiments, the gradient was re-established within 36 hours
You don;t descibe the experiment in detail or provide a link, but only tiny imbalances are need to create a lapse rate of 10 K/km in the atmosphere (less that 1 mW/m^2 out of the top of the column. It would take VERY careful experimentation to get close enough to “perfect insulation” to see the effects we are discussing.
>Where is the energy going to come from to warm it up again?
From the ground in this case. The bigger question is where will the energy go? We have a single “hot reservoir”. You would take energy from this hot reservoir, create a machine (a pair of tall cylinders with a thermopile at the top), and turn thermal energy directly into usable work with no energy rejected to any “cold reservoir”.
How about we start with the definition of free energy as described in Chapter 21 (Systems Subject to a Gravitational Field) of Klotz/Rosenberg’s Chemical Thermodynamics 5th Ed:
G = f(T, P, n1, n2, …, ni, x) Eqn 21-6
The total differential is therefore:
dG = (dG/dT)*dT + (dG/dP)*dP + dG/dn1*dn1 + dG/dn2*dn2 + … + dG/dni*dni + dG/dx*dx Eqn 21-7
Now let:
dG/dx = mg Eqn 21-2
dG/dT = -S Eqn 11-3
dG/dP = V Eqn 11-4
with the assumption that the composition of the atmosphere consists of indistinguishable particles such that:
dG/ni = 0
Equation 21-7 becomes:
dG = -SdT + VdP + mgdx
or in terms of molar quantities:
dG = -Sm*dT + VmdP + Mgdx
The molar volume of an ideal gas is given by:
Vm = V/n = RT/P
The molar entropy of an ideal gas with a constant specific heat can be taken as:
Sm = CvlnT + RlnVm + Sm0 Eqn 6-111
After making the appropriate substitutions, we’ll get:
dG = -(CvlnT + Rln(RT/P) + Sm0)*dT + (RT/P)*dP + Mgdx
So, all one needs to do is solve this differential equation to answer the original question. At equilibrium, dG = 0 by definition, thus:
0 = -(CvlnT + Rln(RT/P) + Sm0)*dT + (RT/P)*dP + Mgdx
The solution is trivial if dT = 0, which is consistent with one formulation of the zeroth law of thermodynamics. It is of interest that Klotz & Rosenberg specify that the column is isothermal in their example.
Michael Larkin says:
January 21, 2012 at 7:38 am
Michael, the only stupid questions are the ones you don’t ask. And in this case, you’ve asked good ones that go to the heart of the discussion.
I don’t know what mechanism Jelbring thinks will restore the initial heat differential, but he definitely thinks it will be restored.
He does not say that, and in fact vigorously denies it. It is true nonetheless.
Saying as Jelbring does that gravity will cause an on-going, restorable temperature differential in an energetically isolated column of air is the same as claiming a perpetual motion machine. Here’s why.
IF gravity could cause such a temperature differential, we could use that differential to produce work. Producing work from heat has to cool the warm end and warm the cool end of the system. This is because you have removed heat from the warm end to do the work, and rejected the waste heat to the cool end of the system.
Jelbring claims that after such extraction of energy or any other disturbance, gravity will restore the prior temperature differential. This means that we could pull energy out of it forever, which is a perpetual motion machine.
And this means, of course, that gravity cannot create a temperature differential in an energetically isolated system such as Jelbring describes.
This is the part that all of the folks who are arguing about the mechanism are missing. The problem isn’t in the mechanism. It is in the outcome, which violates conservation of energy, so considerations about possible mechanisms by which gravity could/should/must cause a temperature differential are meaningless.
w.
On reading the discussion, I feel drawn to the conclusion that DALR defines a set of physical conditions where no useful work can be extracted from a compressible fluid using a heat engine.
Temperature difference alone is not decisive.
Consider a heat engine with a high altitude heat sink and low altitude heat source. Energy transfer is by physical contact only (conduction). For now, assume molecules which can transfer a discrete amount of energy on contact with the source or sink.
Can we get useful work from temperature difference at DALR ?
Firstly, consider a single molecule which has just gained additional energy at high altitude from the engine’s heat sink (high potential energy + increased temperature (kinetic energy)). The molecule then drops to the heat source where potential energy is lower but kinetic energy is high (boosted by the energy it had gained at the sink). The engine appears to have found a way to boost its own performance by transferring heat back from its own sink to its source.
This violates thermodynamic principles and the safest conclusion is that something is wrong with this modern-day version of the well-crafted pile of wood at the top of this thread.
I believe the mistake is to assume that only one physical property (temperature) is decisive in the potential to do work from conditions at the source and sink.
Second case: a heat engine with no capacity to store energy and therefore operates due to simultaneous events at the source and sink (single molecule doesn’t do anything here, we need to consider many molecules).
If a molecule is available to transfer-in energy at the source, the engine needs a molecule at the sink to transfer energy out. More generally, if there are ‘n’ molecules at the sink, the source can only accept the energy from ‘n’ molecules despite there being many more molecules available at the source.
It is not necessary to stick to the above constraints. My contention is that a DALR heat engine doesn’t work because the conditions at the sink limits the potential to receive energy at the source.
I would respectfully suggest is a mistake to postulate machines which assume only temperature difference is relevant to operation.
If number of molecules is equated to the density of a compressible fluid and density is a function of pressure, the above arguments suggest that pressure difference cancels temperature difference at DALR.
The capacity to do work is determined by total potential at different altitudes (pressure and temperature) for compressible fluid.
And this guides me to the view that there is nothing thermodynamically wrong with a sustained (permanent) temperature gradient in a gravity field.
OK guys – shoot me down
😉
Dr Brown says:
“Without the rapid cooling of the upper atmosphere that keeps it cold relative to the ground — cooling that is strictly radiation, because sooner or later the Earth has to lose the incoming heat from the sun — the adiabatic warming profile would not exist, and in parts of the atmosphere that profile inverts even as it is (for example, over the arctic in the long arctic night) as further proof that this isn’t an atmospheric compression effect, it is plain old convection.”
I would suggest that the arctic atmosphere does not “invert” in the sense of the action verb as a result of seeking an equilibrium but instead inverts as a consequence of differential cooling between the atmosphere and the surface. To claim that as evidence against Jelbring’s world one has to acknowledge the differential cooling, subtract that out and then demonstrate that whats left is something seeking an isothermic equilibrium. Jelbring’s world was defined in such a way that one cannot use inversion layers caused by differential surface cooling as evidence.
So I am blowing the whistle on that play. . . .out of bounds!!!!
Digging down a little further, just what influence does gravity have on creating heat further down into the earth’s crust? IF it does have a participatory influence, then why not the atmosphere where closer to the earth the greater the pressure?
A Physicist & DeWitt Payne: Thank you both for your efforts.
A Physicist, I appreciate your heroic effort to reach a level of concreteness that a layman like me can comprehend. Unfortunately, even that much concreteness left your explanation at a level of abstraction I cannot reliably navigate; in the corn and soybean fields of Indiana, where I live, we think “chemical potential” has something to do with batteries.
And, in any event, you addressed Loschmidt, who apparently argued for the adiabatic lapse rate at equilibrium, whereas my question was instead about Levasco et al., who agreed that isothermality prevails, but only in the limit: they found a small but finite lapse rate for finite numbers N of molecules, and their formula yields a value for N = 1 that I would have thought hard to argue with.
DeWitt Payne, you I could understand, and your explanation sounds admirably plausible. Although you, too, addressed Loschmidt (or, equivalently, Jelbring) instead of Velasco et al., I think I can adapt your argument to Velasco et al. well enough. You would begin by recognizing that, unlike in the adiabatic-lapse-rate configuration for which I understand Loschmidt argued, the hypothetical rising packet in a configuration where the atmosphere has Velasco et al.’s (less-than-adiabatic) lapse rate would absorb heat from its surroundings. But you would point out that the dQ/T difference prevailing in that heat transfer would not be as great as in the zero-lapse-rate, i.e., isothermal, situation. Quite understandably, therefore, you would conclude that the isothermal configuration is the higher-entropy one, and you would pronounce your proof successful. Q.E.D. Game over, thanks for playing. Exeunt omnes.
But here’s the problem. People who know this stuff tell me that, from a statistical-mechanics point of view, entropy is the log of the number of microstates. And, as I understand it, when Velasco et al count those up, they find that the configuration with the most microstates actually is one that has a slight but non-zero temperature lapse rate: it’s not isothermal–although their lapse rate does approach zero asymptotically as the number of molecules approaches infinity.
Or at least that’s how this layman understands what they say. And, if what they say is correct, the arguments you all are making for why Jelbring got it wrong have to be adjusted.
So I’m left to guess at who’s right. And I have to say that the approach Velasco et al. took appears from where I sit to be the more rigorous one. Couldn’t one of you smart guys out there actually engage with the Velasco et al. and Martin et al. papers tallbloke provides links to in his Loschmidt thread–and show us where they or I went wrong?
Ok, energy from the surface that would otherwise escape to space is returned to the surface by the action of GHG, where it warms the surface temperature over what it would otherwise be. We call this back-radiation.
By the same principle, energy from the surface that would otherwise escape to space is conducted back to the surface (eg: at night and towards the poles) by N2 and O2, where it warms the surface temperature over what it would otherwise be, We call this back-conduction.
If GHG can heat the planet through back-radiation, then non GHG must heat the planet as well through back-conduction.
http://tallbloke.wordpress.com/2012/01/20/greg-elliott-use-of-flow-diagrams-in-understanding-energy-balance/
DeWitt Payne said @ January 20, 2012 at 8:54 pm
I stand corrected; it was but a single lecture. The Goodsteins’ book has considerable supplementary material and that made it seem to me much longer. I thoroughly enjoyed working through the reasoning, though his students were less enthralled. They would have preferred the “shorthand” of calculus. Of course Newton had to present his argument in the language of the day, rather than the language of calculus that he (and Liebnitz) were developing.
Also out of bounds on the former play is the contention the earth has to eventually lose the incoming heat from the sun. Thats not necessarily the case in Jelbring’s world as there is no incoming solar energy and both that and its losses are out of bounds.
We can all understand that jelbring’s world is impossible but Jelbring is in essence asserting that none of that is relevant to the question at hand as he asserts he was fair handed in drawing the boundaries of the issue by removing what is a balanced equation.
So we either have to criticize Jelbring’s fairness in defining the imaginary world we are dealing with or keep the play inside the bounds.
My apologies for being a self-annoited moderator but I find all this extremely thought provoking.
tallbloke says:
January 21, 2012 at 7:39 am
Thank you for the ideas and the comment, TB. Is there a book somewhere that contains that rule? Because it’s a curious rule if so. You claim that to falsify a perpetual motion machine I must first “demonstrate that it will produce work”????
In any case, let’s follow your logic. I have specified the machine, a tall insulated cylinder of air. Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.
As to your claim that I have failed to demonstrate that work can be produced out of a temperature differential … surely you are kidding. That’s how heat engines work. Are you claiming that a temperature differential cannot be used to produce work?
So as far as I can see, I have done what you have asked. I have specified the machine. Jelbring claims that it will produce a temperature differential, which means it will produce work.
Finally, if people do not respond to you, there are two possibilities you seem to have overlooked:
a) They didn’t notice the post in a thread that now has over 500 comments, or
w) They thought there was no point in answering, that the debate wasn’t worth it, either because the science expressed in the post was so colossally bad or because they reckoned your mind is made up and you don’t want to be bothered with further facts.
I fall into “Group W”, tallbloke. From my perspective, I’d done what you ask above. I’d specified the machine as a tall insulated cylinder of air, and Jelbring says gravity will thermally stratify it, cold at the top and warm at the bottom. I never even considered the possibility that you would think work could not be extracted from a temperature differential. It can. So Jelbring’s hypothesis is falsified.
So perhaps my silence “speaks volumes” … but it’s not saying what you think.
w.
PS—Please note that Robert Brown may be in “Group W” as well regarding your claim, viz:
He may not be all that impressed by some random guy on the internet claiming to find a violation of the zeroth law …
To me, the problem with your claim is the idea that there is a temperature gradient across A, so that each microscopically thin slice of A is a bit cooler than the previous slice … but there is no difference between the last slice in A and the first slice in B, you claim that they are the same temperature. That’s not a temperature gradient.
So you do not have a “thermal gradient” across them as you claim. You have a gradient everywhere but at the interface between A and B.
If that is the case and the last slice of A is at the same temperature as the first slice of B, then momentarily no heat will flow from A to B. As a result, heat will run down the thermal gradient from the next-to-last slice in A to the last slice in A, warming it slightly. In the same way, heat will flow down the thermal gradient from the first slice in B to the second slice, cooling the first slice slightly.
At that point you will have a true thermal gradient across the stack, where every slice is warmer than the previous one, including from the last slice in A to the first slice in B. And of course, as the zeroth law states, then heat will flow from A to B and on down the line.
However, the idea that Robert Brown is somehow afraid to confront your “proof” that the zeroth law of thermodynamics is wrong is … well … like I said elsewhere, I’m reforming, so let me just call it “less than probable”. Robert doesn’t have time to address every claim, and a claim that you’ve overthrown centuries of science and shown that one of the laws of thermodynamics is wrong, well, that is one claim that I can see he might not want to get involved in, that one is well out in the fruit-loops zone … if you are right, you’re up for a Nobel Prize, and Jimmy the Greek is not giving good odds in Las Vegas on that happening.
All the best,
w.
Willis and Dr Brown are completely right here of course.
The system proposed by Hans Jelbring would allow the construction of a PM machine.
Unless Jelbring is willing to state that the Thermodynamic laws are wrong or incomplete then his hypothesis must be rejected.
There is a problem with Gedanken experiments though, they can allow you to construct something that seems viable and yet breaches agreed physical laws.
I could invent a closed system that was initially composed of a diffuse cloud of particles in
thermodynamic equilibrium. Now that system would be near maximum entropy. However the addition of gravity starts to cause the particles to compress and voila I now have a system like the solar system and I have reduced the entropy of the system without the addition of energy or increasing the entropy of another system.
Perhaps someone could say that this proves gravity can reverse the flow of entropy in a closed system and the Thermodynamic laws need revising.
However, in the real universe we inhabit we cannot create such a system in such an inital state. Perhaps a supreme, all powerful being could but I am not holding my breath! The existence of such a being would invalidate all known physical laws in any event.
So we have to be careful with gedanken experiments. I tend to the view that if such a system is proposed, that is in breach of thermodynamic laws, that it is either in error or could not ever be created in the universe we inhabit.
Alan
“Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.”
Only when energy is being added from an external source.
Gravity stratifies by mass not temperature.
Read up on the Ideal Gas Law and adiabatic lapse rate.
Just look at Wikipedia here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas.”
and:
“the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.”
You, Willis appear to be trying to rebut 150 years of established science in favour of some 20 years during which some bozos failed to appreciate that radiative physics is a minor player and readily offset by the atmospheric response.
All your recent threads have been corrupted by that misapprehension which is sad given the quality of your Thermostat Hypothesis (even though you really should have extended it globally whereupon you would have realised your current errors).
If you had extended your Thermostat Hypothesis to the entire planet you would have realised exactly why you are so wrong now.
davidmhoffer says:
January 21, 2012 at 8:29 am
Equation 8 does indeed contain two variables. One is So, the surface pressure. The other is NTE(Ps).
But NTE(Ps) is in turn a function with three variables, Ts, Ps, and Tgb. As a result, if we expand equation 8, it is obvious that there are FOUR FREE PARAMETERS in equation 8—So, Ts, Ps, and Tgb. You sure you understand this “algebra” deal? Let me go over it.
IF
a = b + c + d, and
final_answer = f + a
then we can expand the second equation by substituting b + c + d for “a” to give us
final_answer = f + b + c + d
In other words, although (as your claim states) there are only two variables listed in final_answer = f + a, in fact there are four variables that make up the final_answer. The exact same thing is true of equation 8.
Save your disgust, your snark, and your accusations of lying for when you are right, david. They don’t taste so good when you have to swallow your words.
w.
Thank you, Joe!
Chemical potential is a pretty dang tough concept to grasp, and arguably it’s best for everyone to find their own path to it. Because in thermodynamics, kinetic theory, and transport theory there is no “royal road”, and the struggle is a necessary part of the learning process.
For folks who like to see their physics come-to-life concrete numerical simulations (as I do), one good free-as-in-freedom reference is the article “Understanding the temperature and the chemical potential using computer simulations” (2004), which includes both links to free software and a very physical explanation of temperature, chemical potential, free energy, and other mysterious topics.
Joe Born says:
January 21, 2012 at 9:23 am
“Brown and Eschenbach” have been engaging you, tallbloke and a host of others on these very questions for several threads now. The idea that we are “failing to engage” fails the laugh test. Yes, at times we might not answer a post for some reason, mainly (in my case) raw incredulity about the claims made … sometimes I just shake my head and move along, I can’t answer every wacko theory about how the world works.
But indeed, the noble firm of “Brown and Eschenbach” and a host of other good folks here are doing all we can to fight the amazing ignorance of thermodynamics so ably demonstrated by you, tallbloke, Jelbring, and others …
w.
This a comment addressed to all that believe a non-zero temperature gradient develops in an atmosphere in a gravitational field at equilibrium. This builds on upon Dewitt’s two cylinders and Rasey’s two tubes comments above. We will build a two gas, double loop system that according to believers in a steady-state non-zero lapse rate, must work as a perpetual motion machine.
Construct a rectangular pipe loop, height H, width W. The pipes are thermally insulated. Whether they are optically transparent will not matter for now, but later we might want them to be so. Arbitrarily, we call upper right point P, and clockwise the corners are Q, R, S, the base being QR, verticals PQ and RS. On the length PQ, we drill holes and insert thermally conductive plugs (C) along its length, then wrap all pipes in insulation. Have a thermal conductive plug at Q. This is Loop 1. Copy it and make Loop 2. Arrange them in mirror images with PQ of Loop 1 nearest PQ of Loop 2. The Bases QR(i) are on an isothermal ground and horizontal. Final configuration Loop 1, clockwise from upper right, PQRS, Loop 2 CounterClockwise from upper LEFT, PQRS. QR(1) and QR(2) are at the same gravitational potential. PS(1) and P2(2) are also horizontal and at the same gravitational potential. There is an arbitrary separation between PQ(loop2) and PQ(loop 2), insulated, so no thermal exchange between them…. Yet.
Fill Loop 1 with Helium. Fill Loop 2 with Argon. Expose the plugs at Q on an isotheral ground. Let them come to thermal equilibrium and via the plugs at Q
Now, to be clear, I am in the isothermal camp. I believe (along with Willis, Dr. Brown, Dewitt) that the equilibrium state is an atmosphere on a dead planet is isothermal.
But let’s play by the theory of Jelbring that an DALR will be the equilibrium state. If so, at equibrium, temperature T at Q Loop 1, T(q1) = T(q2). Then T(r1)=T(r2). By Jelbring’s reasoning, T(p1) > T(p2) because the lapse rate in PQ(1) helium is lower (closer to zero) than it is in PQ(2) argon. T(sp1) is hotter than T(sp2). T(s1) = T(p1) > T(p2) = T(s2). PQ(1) has a different temperature. If it was isothermal T(q1)=T(p1)=T(p2)=T(q2).
Now, rip off the thermal insulation from PQ1 and PQ2, slap them together so the conductors are in contact and reapply the insulation, including at Q to decouple from the ground. The two loops that were in isolated equilibrium are no-longer in equilibrium because heat if flowing from P1 to P2. The Helium at P1 is cooling, the Argon at P2 is warming. Al along PQ1 the conductors are moving heat to the colder PQ2 except at point Q where they are the same temperature. As the Helium PQ1 cools, it becomes more dense. As the argon PQ2 heats, it becomes less dense. Both loops become unstable. Helium flows clockwise PQRS, Argon flows CLOCKWISE, PSRQ. We have started the system in motion and that should be disturbing, but it doesn’t prove perpetual motion yet. Sooner or later, the system will reestablish equibrium. If you believe as I, then the equibrium will be isothermal, all points in loops 1 and 2 at the same temperature. But if you believe as Jelbring, the system will have to find some equilibrium where T(p1) < T(q1) and T(p2) < T(q2), but T(P1) = T(P2) and T(q1) = T(q2) via the conductors.
But we face a great big problem. At equilibrium RS(1) lapse rate should approach the ALR for Helium and RS(2) should approach the ALR for Argon. But PQ(1) an PQ(2) must be at the same real lapse rate because of the shared conductors. That means the verticals of loop 1 have different lapse rates, which means different densities, so they are pneumatically out of balance and must flow with greater density toward the bottom. The same goes for Loop 2.
The only way PQ(1), RS(1), can stay in balance is if they have the same lapse rate. Likewise for P2(2), RS(2). And PQ(1) must have the same real lapse rate as PQ(2) since they are thermally conducted. The only solution is for the real lapse to be zero — Isothermal.
Jim D says:
January 21, 2012 at 10:18 am
Since it takes energy to do the mixing that you propose, please come back when you’ve considered what would happen if you start out isothermal and you don’t add mixing energy. Remember, we’re discussing the atmosphere on Jelbring’s thought experiment planet of Stygia (or equivalently the air in a tall thermally insulated column). In such a situation you can’t add work to do the mixing you describe.
Will the air move away from the isothermal state if there is no energy to mix it?
w.
gnomish says:
January 21, 2012 at 9:13 am
jelbring:
“If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.”
this is a definition. it is not wrong. get literate – then you can get logical – then you can get scientific. you aren’t close yet.”
Agree. Thank You. My misstake reading the opposite and being in a hurry late after midnight.
DeWitt says:
If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top. The cylinders will then be isothermal. We disconnect the heat engine. If gravity then re-establishes the DALR,
I agree the helium column will cool, but disagree that the gas in the cylinders will become isothermal, because convection would occur which would maintain the adiabatic lapse rate, as the second law demands.
Once the helium column has cooled by 3C and the Xenon column has warmed by 3C, the heat engine stops.
What is going to warm the helium column up again so the heat engine can be run a second time?
This is not a perpetual motion machine. All you have done is exploit the Gibbs free energy between the two dissimilar gases until it’s gone.
How about we start with the definition of free energy as described in Chapter 21 (Systems Subject to a Gravitational Field) of Klotz/Rosenberg’s Chemical Thermodynamics 5th Ed:
Very clear, and pretty much inarguable. BUT, anybody that is silly enough to still believe that gravity is the long-sought Maxwell’s Demon, capable of spontaneously sorting out fast moving and slow moving molecules in thermal equilibrium after the discussion and law-of-thermodynamics-violating counterexamples presented so far will simply allow their eyes to glaze over and pretend that your textbook presentation is bad science, while Jelbring or N&Z — since they state things that support their fantasy that greenhouse gases have nothing to do with the average temperature of the Earth — are all “good science”.
I swear, I’m not a “warmist” or a “coolist” — I’m a physicist, and all I care about is objective truth as best we can make it out with good science — but it’s getting to where I can see where some of the “warmists” are coming from.
Look — the Earth’s climate isn’t driven by the things you want it to be driven by. That’s “confirmation bias”, visible in spades as people continue to try to defend the indefensible (because things that openly and obviously violate the laws of thermodynamics are indefensible). If you ever have spontaneous, stable separation of an adiabatic (thermally isolated) system in good thermal contact into hot and cold reservoirs, you have already violated the second law of thermodynamics. I don’t care what internal mechanism you try to invent to act as Maxwell’s Demon to accomplish it. One of the first things you learn about in stat mech is detailed balance — it is the basis, in a manner of speaking, of stat mech.
It is this sort of nonsense that permits the warmists to loudly ignore its challengers. The greenhouse effect is established science. Get used to it. Get over it. It’s experimentally verified by IR spectroscopy from the top of the atmosphere. Claiming that there is no such thing in the face of direct experimental evidence is just plain stupid. Claiming that the warming of the Earth is accomplished by a mechanism that creates a thermal gradient in an isolated atmosphere is just plain stupid. Claiming that “gravity” is in any sense whatsoever directly responsible for the warming is just stupid.
None of these observations of things that are stupid suggest that there isn’t a thermal gradient to the atmosphere, only that it exists because the atmosphere is differentially heated at the bottom and cooled primarily at the top. This creates several mechanisms that establish the gradient — convection, radiation, conduction — where convection described as adiabatic expansion of a parcel of dry air provides not a law or universal rule, but rather a useful heuristic baseline for understanding the bottom to top lapse rate.
The bottom to top cooling not only isn’t a law, it isn’t even universally maintained in the troposphere! There are plenty of times the air overhead is warmer than the air on the ground. That’s when things like freezing rain happens. It happens as standard practice over Antarctica during the long winter night — the ground layer of air is often colder — sometimes much colder — than most if not all of the rest of the troposphere. The atmosphere is no longer being heated at the bottom, you see.
I will point out that I can keep this up indefinitely. Thermodynamics isn’t the easiest physics in the world, but this particular issue isn’t subtle. It’s one of the first things you learn in any introductory class in thermodynamics — heat flows from hot to cold, never the other way around, no matter what kind of system you have on either end, as long as there is any interaction between the hot and cold ends that can carry energy. It can be conduction, convection, or radiation. Energy transfer can be mediated by electromagnetic forces, gravity, whatever you like. If you watch the system over times long compared to the thermalization interaction time, it will come into thermal equilibrium at the same temperature.
Will the air move away from the isothermal state if there is no energy to mix it?
The answer, in case you didn’t get it from the way he asked the question, is “no”. It will move towards it. Otherwise, you violate the second law and can trivially create any of a number of perpetual motion machines of the second kind between the top and the bottom of the air column.
rgb
Mr. Brown:
I commend you on your continued posts here.
Thank you.
If sheer volume of words carried scientific weight, Robert and Willis would be winners by a mile.
However, Robert still hasn’t addressed my short proof that no heat has to flow in a gravity induced thermal gradient at energetic equilibrium.
Here it is again if he feels like getting round to it.
Robert says:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
“… all considerations about the mechanism go nowhere. Consider the outcome. If it’s true that gravity can separate molecules by temperature, then we can pull energy out of tall insulated cylinders of air …”
But whether the lapse is enabled by gravity or just thermal gradients, we do already have a lapse rate in our current atmosphere, so of these magical vertical lapse rate generators that you and Dr. Brown keep speaking of, why have these not already been built?? Sound so simple. I’ll take some of that forever free energy Willis no matter what forms the lapse, just show us the schematics…. PLEASE? ☺
Willis, I am pointing out why the isothermal state is not the maximum entropy state, which you have not disputed yet. An isothermal or any stable stratification with a linear profile is stable against diffusion. There are multiple stable states possible when only local diffusion is permitted, but the only one stable to eddy mixing is the isentropic one. If you start isothermal, it will stay isothermal. If you start isentropic it will stay isentropic, and if you start half way between these, it will stay like that too.
Robert Brown says at 1/21 10:01am
“..the molecules can share their kinetic energy around and come to thermal equilibrium without moving up or down and involving gravity.”
Certainly the delays in many conversations are irritating especially stopping in and out like me. But it is interesting and fun pastime to engage on this particular thermo topic.
To regroup: control volume is an adiabatic GHG-free tall air column in the presence of inexplicable gravity. I continue to be on the side of those thinking this column must stratify while at the same time it cannot generate energy forever. No energy must be harmed in this thought experiment. Or created.
Have I detected at least a little backing off by Robert Brown no-stratify position by above seeking to take out the n*g*h PE term by molecules exchanging KE limited to lateral movements only? Or is this somehow out of context?
Remember in the top post Willis’ quotes Robert Brown writing: “On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down…”
This last statement seems to be a violation of conservation of energy in a gravity field where each molecule PE + KE = constant since in my view: each molecule moving up in the presence of an inexplicable gravity field cannot have exactly the same velocity thus kinetic energy as those moving down in presence of gravity field PE (= n*g*h) since KE changes with h for total energy to be constant.
That stratifying seems to square with 0th,1st, 2nd Law and ideal gas law. I don’t think Maxwell-Boltzmann distribution can be invoked in a gravity field since it deals with statistics of molecular movements w/o external forces. Here gravity is an external force. M-B can only be invoked if gravity is turned off.
In this 1 system cv with gravity no heat can flow (there is no system 2), cannot reduce the KE + PE of any h > 0 upper molecule, total is always constant. Cannot increase the KE +PE of any lower h~0 molecule, total is always constant. Thus temperature has got to vary because KE varies as P*V varies with h (via PE = ngh).
My view this cv has to stratify by temp. consistent with all the laws and therefore by density thru PV=nRT. If it doesn’t, total energy of each molecule is not conserved. Same reason it can’t generate energy forever. There is no hot & cold source within the CV maintained forever if molecules always move with same constant energy.
Like Tallbloke said, build a Perpetuum Mobile if you can – let us know your design but energy conservation will always and everywhere stop it.
Oh wait, I see now Willis says at 1/21 12:34pm:
“I have specified the machine, a tall insulated cylinder of air. Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.
As to your claim that I have failed to demonstrate that work can be produced out of a temperature differential … surely you are kidding. That’s how heat engines work. Are you claiming that a temperature differential cannot be used to produce work?”
Willis’ should read the 2nd Law v-e-r-r-y slowly, there has to be two systems – one hot and one cold for heat to flow when they touch. In this cv there is only 1 system, heat cannot flow. There IS no temperature differential between two systems here, only one system. Or show me the second system. Otherwise, Willis’ specified Perpetuum Mobile machine is not demonstrated. One system cv: no heat flows or work produced. Can’t even be said the insulated cylinder of air system contains heat (does contain energy) if I read my Maxwell definition right (heat only flows).
Oh Willis, one more thing, if this is another 5km high wire gizmo, thermocouple or not, I’ve already tried that… no thanks… burned down the shed it was anchored to and I’m sure it was not from the lapse rate but more from the electro-static differential found in our atmoshere at times! ☺
The perpetuum mobile won’t happen because we talk about an hypothetical planet with no radiative transfer. Try to build a thermocouple with no emissivity and then we’ll talk. If your thermocouple emits heat, eventually the planet will cool.
The zero adiabatic lapse rate cannot happen without a Maxwell demon. Remember, the number of molecules traveling upward is equal to the number traveling downward if the atmosphere is in balance. Now, if you have the same temperature everywhere, it means that molecules at higher altitudes are traveling faster. So, at a certain height, the molecules traveling downward should be warmer than the molecules traveling upward. Except if you have a Maxwell demon blocking the faster molecules traveling downward.
The only way PQ(1), RS(1), can stay in balance is if they have the same lapse rate. Likewise for P2(2), RS(2). And PQ(1) must have the same real lapse rate as PQ(2) since they are thermally conducted. The only solution is for the real lapse to be zero — Isothermal.
Well done, but way too much work. The easiest PPM2 one can build is a simple thermocouple that lights a light bulb or turns an electrical motor. Place the upper contact in the supposedly cold air at the top of the static column. Place the lower contact in the warm air at the bottom. Insulate the heat transfer medium (e.g. silver) all the way down to the actual junction, so that heat is conducted from the hot bottom to the cold top, through the thermoelectric junction.
The motor turns, doing work as heat is conducted from the warm bottom to the cold top. But the top never gets any warmer, as the DALR is supposedly the stable equilibrium state of the gas. No matter how much heat we remove from the bottom and deliver to the top or middle, it somehow “falls” back to the bottom to re-establish the DALR.
Even simpler, don’t worry about the actual motor — simply connect the top and bottom with a silver rod insulated on the sides. Bottom is hot, top is cold, so heat flows along the rod — forever. As fast as you deliver heat to the top, it is sorted out by gravity and returned to the bottom to keep it from actually cooling. The system never reaches equilibrium! Which violates so much simple common sense that maybe — although I doubt it — even Jelbring and Tallbloke with concede that this cannot happen, and that the assertion that any spontaneous stable process can create a permanent thermal gradient in an adiabatic system is false (since this argument catches them all, which is why it is used to show the equivalence of the various forms of the second law in intro thermodynamics textbooks)..
rgb
Willis Eschenbach says:
January 21, 2012 at 12:34 pm
PS—Please note that Robert Brown may be in “Group W” as well regarding your claim, viz:
This goes back to the proof I [tallbloke] offered Duke.edu physicist Robert Brown up near the top of the thread. he said:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.”
And I [again tallbloke, not me —w.]] pointed out:
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
He may not be all that impressed by some random guy on the internet claiming to find a violation of the zeroth law …
To me, the problem with your claim is the idea that there is a temperature gradient across A, so that each microscopically thin slice of A is a bit cooler than the previous slice … but there is no difference between the last slice in A and the first slice in B, you claim that they are the same temperature. That’s not a temperature gradient.
So you do not have a “thermal gradient” across them as you claim. You have a gradient everywhere but at the interface between A and B.
If that is the case and the last slice of A is at the same temperature as the first slice of B, then momentarily no heat will flow from A to B. As a result, heat will run down the thermal gradient from the next-to-last slice in A to the last slice in A, warming it slightly. In the same way, heat will flow down the thermal gradient from the first slice in B to the second slice, cooling the first slice slightly.
At that point you will have a true thermal gradient across the stack, where every slice is warmer than the previous one, including from the last slice in A to the first slice in B. And of course, as the zeroth law states, then heat will flow from A to B and on down the line.
I’ll be interested to see if Robert endorses this laughable response to my proof. Clue, no heat is going to run down the thermal gradient, because KE+PE at any altitude h is in equilibrium over the height of the column.
The rest of your longer than elevator length reply is more ad hominem attack. You seem to be all out of science, so I’ll leave you to it and go do my write up for the blog.
Cheers
TB.
Smokey said @ January 20, 2012 at 6:32 pm
Actually, no. Magic means something is caused supernaturally. Quoting Feynman; he asked rhetorically “What is behind the Law?” and answered, “There is nothing behind the Law”. Many (most?) of us accept that. Some believe that God is behind the Law. I have no quibble with that belief (I just don’t share it), but that belief is magical (supernatural).
DeWitt Payne says:
January 21, 2012 at 8:45 am
“You can hand wave all you want about non-equilibrium thermodynamics…”
I am no longer handwaving. The question is moot. All the other side discussions are moot. All the rest of this blog thread is moot. I am now proclaiming Bart’s Law as settled:
Bart’s Law:
It follows that surface temperature sensitivity to the addition of radiative gases is negative: the addition of more radiative gases will tend to lower the surface temperature. This is a sensitivity – it does not exclude the possibility of feedback from other processes tending to resist the change.
Bart’s Law is justified by the following line of reasoning:
Convergence is initially superlinear, because the more you bite into the areas of the tails, the faster you increase the backradiation and hence surface temperature. Eventually, you reach a point where successive bites approach zero, and there you stabilize.
Willis Eschenbach:
Please take note of the above. Here is a physically viable alternative to the standard greenhouse theory.
Trick says:
January 21, 2012 at 1:42 pm
Excellent comment Trick.
See my attempted disproof of DeWitt Payne’s twin cyclinder version of Willis’ one lunger above at
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871872
Thoughts?
Willis wrote;
(In response to my expressed umbrage at allusions to ignorance amongst those of use that do not believe in the GHE)
“Dang, bro’, with all due respect, I didn’t say you were.”
Willis, my sincere apology, I believe I may have read another comment about those of us that do not believe in the GHE as being too ignorant to understand it. I then read your post about being deluged with ignorant people and mingled the two posts together in error.
But those that think that all people who do not accept the GHE are ignorant may learn otherwise in the next few years when we all finally figure this thing out.
Please keep trying to explain things, and eventually we all might reach a better understanding about what is going on.
Cheers, Kevin
Tallbloke, may I suggest that you (and other skeptics) simulate that precise experiment, using the methods and software of Understanding the temperature and the chemical potential using computer simulations (2004).
To handle the reservoir coupling, have the simulation’s “demon” exchange energy and atoms with *both* reservoirs A and B. Furthermore, have that demon exchange particles and energy only with the top 1/10 of reservoir A (the lower reservoir) and the lower 1/10 of reservoir B (the upper reservoir). And finally, don’t forget to assign the particles in reservoir B a higher gravitational potential energy than those in reservoir A.
Start the simulation in some arbitrary atomic configuration, and let it run for awhile. You will find that it approaches an equilibrium in which:
(1) Reservoirs A and B are at the same temperature
(as mediated by the exchange of energy)
(2) Reservoirs A and B are at the same chemical potential
(as mediated by the exchange of particles)
(3) Reservoir A (the lower reservoir) has a higher density and pressure than B
(in consequence of B’s higher gravitational potential)
Then after thinking some more, you will appreciate that microscopic atomic simulations will always respect the laws of thermodynamics, in the sense that their “demons” will always act so to equalize temperature and chemical potential, in accord with orthodox thermodynamical theory (naturally), and entirely at variance with “gravito-thermal theory.”
Speaking personally, I have mixed feelings about claims that the GHE is “established science.”
Skepticism about established science is good, to the extent that this skepticism motivates people to study these questions seriously and in-depth. And from the point of view that skepticism can be an excellent motivation for learning, this entire WUWT discussion has been terrific!
And yet what Robert Brown says is 100% correct too: “The greenhouse effect is established science. Get used to it. Get over it.” In particular, the “gravito-thermal” theorists have been making extraordinary claims, but they have not brought forth any of the extraordinary evidence — mathematical, theoretical, computational, experimental, or observational — that is needed to support their claims.
So perhaps the best balance for rational skepticism is this: “The greenhouse effect is established science. Get used to it. Get over it — after you have checked its key ideas for yourself.”
tallbloke said @ January 21, 2012 at 9:31 am
I am truly gobsmacked! A definition is a tautology. A tautology is a proposition that is unconditionally true. But not unconditionally true according to TB. Quoting Aristotle (translated):
“one cannot say of something that it is and that it is not in the same respect and at the same time”.
Post-modernists do this all the time of course, thus excluding themselves from rational discourse.
Robert Brown says at 1/21 1:52pm
“The easiest PPM2 one can build is a simple thermocouple that lights a light bulb or turns an electrical motor. Place the upper contact in the supposedly cold air at the top of the static column.”
You are mixing metaphors. The cv of interest allows only one air system. This PPM2 is more than one system. And things are going across the cv. Try again.
However, Robert still hasn’t addressed my short proof that no heat has to flow in a gravity induced thermal gradient at energetic equilibrium.
Here it is again if he feels like getting round to it.
Robert says:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B.
Sure, and in return you can address the actual topic of the thread, which is how any system that spontaneously generates a thermal gradient violates the second law of thermodynamics and thereby enables the construction of PPM2K’s, including the trivial thermocouple example in the previous post.
First of all, what you offered is not a proof. You simple restate your conclusion. You postulate two systems in thermal contact, one at a higher average temperature than the other, and then say “no heat will flow” as long as the boundary in between them is at some (presumably intermediate) temperature.
The problem is that temperature is a meaningless concept for a two dimensional surface. What you’re saying, in a nutshell, is that you have two volumes neither of which is in thermal equilibrium (if they were, they’d have the same temperature throughout each volume.
But let’s see if I can come up with a simple, simple argument that will convince you.
Go into your column of air with its imagined lapse rate. Get perfectly insulating jar and fill it with air from the top. Get a second jar and fill it with air from the bottom. By hypothesis, these jars are in thermal equilibrium, but have different temperatures. Note that once you fill the jars, you can move them up or down all you like — the gravitational field doesn’t vary (by enough to matter). You’ve simply replaced the fluid that was supporting the air with the jar walls — nothing else changes. Which, incidentally, is why your entire argument is wrong — no gravitational potential energy is involved in heat exchange or energy balance. You move the jar slowly up, move it slowly down, nothing in the jar knows that it has been moved at all!
Now put the two jars in thermal contact. I don’t care how — move them together, use an insulated silver conductor of heat to couple their otherwise adiabatic walls without moving them up or down — it doesn’t matter, does it. What happens? Well, they are at different temperatures, so heat flows. If you change the temperature across any jar heat flows. There is nothing special about the heat flow. It will flow up or down or side to side to neutralize a thermal gradient no matter what you do to the otherwise unaltered jars of air.
The existence of the jar itself doesn’t matter. Put a side-insulated silver rod into the air column and heat will flow in it forever. Put a heat engine into the air column and you can run it off of that temperature difference forever.
The point is that your air column isn’t in equilibrium. Heat can easily be conducted through the air — the equilibrium distribution of the air (and its temperature) isn’t just determined by convection. Once it settles down into a density profile, heat will flow throughout that profile until the temperature is the same because the air at the top cannot tell any difference between heat that arrives through the intermediary air and heat that arrives through a silver rod. The point is that heat spontaneously flows from hot to cold, never the other way around, in the absence of something that does continuous work to push it the other way.
Nothing about the density of air in a jar tells you its temperature. Heat is perfectly happy being conducted uphill, downhill, or sideways. Warmth has no mass, and objects placed in gravitational fields do not spontaneously separate into warm side down. Indeed, in general it is rather the other way, at least at first! But equilibrium is always isothermal in a case like this with no meaningful constraints.
rgb
I’m getting frustrated by the doziness of some of the contributors here, including some Phds.
The column only becomes graded by temperature when energy is added from an external source.
Until there is an external source of energy the column gets graded by MASS not temperature.
Gravity works only on mass. It ensures that the greatest density is at the bottom of the column.
Apart from that gravity does no work and supplies no energy.
In the absence of an external energy source the column will become isothermal i.e. the same temperature from top to bottom.
But
there will be a higher energy content at the bottom because there will be more mass per unit volume at the bottom due to the effect of gravity.
Then when the whole column is irradiated by an external source of energysuch as the sun the temperature will rise most at the bottom because the incoming radiation has more molecules to react with where the density is greater.
This is well established physics of more that 150 years provenance as here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas”
and
“temperature decreases with altitude at a fairly uniform rate. Because the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.”
This is all well proven basic physics.
What is going on here and in other threads that seem to deny basic well demonstrated physical principles ?
DeWitt Payne says:
January 20, 2012 at 6:51 pm
If as you say the TOA temp is the same as at the surface, then there would be an abrupt stepdown to the near-zero K of space. That’s not a satisfactory explanation. I fear that something essential is being missed by positing thermal equilibrium, which preordains the isothermal conclusion. In other words, it becomes a tautology. Space cannot store energy, only transmit it radiatively. Molecules, on the other hand, can. After all, temperature is the manifestation of the kinetic energy of molecules,and depends upon their DENSITY in any given volume. Don’t have time to say anything more.
Robert Brown says:
January 21, 2012 at 1:52 pm
Even simpler, don’t worry about the actual motor — simply connect the top and bottom with a silver rod insulated on the sides. Bottom is hot, top is cold, so heat flows along the rod — forever. — even Jelbring and Tallbloke with concede that this cannot happen.
Excellent. another suggested experiment. My old dad is a fine qualified civil engineer with a wealth of fluid dynamics experience. He told me earlier today that he once did a set of calcs which showed that a water pipe of infinite length with infinite pressure at one end would have no flow at the other due to frictional loss. I suspect the heat flow in your 10km silver wire might have similar conduction issues due to atomic interstice losses. By all means give it a try though.
Robert Brown says:
January 21, 2012 at 2:22 pm
Robert,
Thank you. I will give your reply careful consideration before responding. I hope we can sort this out together.
Cheers
TB.
RE: Robert Brown says: 1:52 pm about Rasey’s 1:08 Helium-Argon Loops
Well done, but way too much work.
Thank you. You and Willis and others had already done the thermocouple PPM2 refutation. I decided to play in Jelbring’s ballpark and stay strictly in lapse rates and pneumatics. Let’s let him find non-zero real lapse rate in the argon and helium loops that work at equilibrium. I’m not even trying to extract work out of the system. What’s is the system’s equilibrium state in a gravitational field if not isothermal?
On the planet Stygian, you have one house on the (hot) beach, and your sister has another house on the (cold) mountaintop.
Fortunately, your family possesses two one-meter^3 blocks of copper. You keep one (enormously heavy!) copper block at the top of the mountain, the other at the bottom.
Every morning you and your sister exchange these two copper blocks (this requires zero work, as the two blocks weight the same—a simple pulley raises the beach-house block at the same rate that the mountain-house block descends).
You use the mountain-cooled copper block to air-condition your beach house all day and even (via a thermoelectric generator) to power your beach-house computer and beer cooler all night.
Similarly, your sister uses the beach-warmed copper block to heat her mountain house, and to power her computer and hot-chocolate-maker.
Next morning, exchange the now-warmed beach block for the now-cooled mountain block, and do it again. Which amounts to this: free heating, air-conditioning, and electrical power for you and your sister, forever, with zero work input! Thank you, thermo-gravitic theory!
Willis said
“I realized after posting this that some people might say “well, maybe the isothermal case is the high energy state”. There’s an easy way to settle that.
If an energetically isolated system is in its lowest energy state, it cannot perform work.
If the isolated atmosphere in Jelbring’s thought experiment is warm at the bottom and cold at the top, I can stick a thermocouple into it and use the temperature differential to generate electricity to perform work.
Therefore, the isothermal state (same temperature everywhere) is the lower of the two energy states, since I cannot use it to do work. ”
Actually this might not be a proof.
One at least has to reflect on the [miniscule] gravitational effect on the squillions of electrons in the thermocouple wires .Or to make it simpler consider the electrons in one wire subject to the Seebeck effect. Maybe, just maybe, the gravitational effect on the electrons in the wire exactly balances the gravitational effect on the air columns temperature gradient.
-In a thermocouple the gravitational effect in each wires material might be proportional to its Seebeck coefficent thus no work could be done 😉
Robert Brown says:
January 21, 2012 at 1:52 pm
How do you know this? Heat must move from bottom to top mediated by silver atoms. Agitated silver atoms at the bottom are at times displaced upwardly, striking atoms higher up and so on transferring some of their energy, right? The trouble is that they must do this in the presence of gravity. A silver atom moving up loses a little energy to gravitational potential energy. This can’t be conducted. As the atom moves back down it accelerates and its kinetic energy increases. Lower kinetic energies mean lower temperatures. Higher kinetic energies mean higher temperatures. A temperature gradient must exist.
Tallbloke says 1/21 at 2:06:
“Thoughts?”
It was obvious DeWitt had two cv DALR (note says DALR – must think stratified) particular gas systems at different T which as Robert Brown says can always run a heat engine when touching from hot source, outputting work, to cold sink. Stops at 3C as all agree.
The connection of two cylinders is a different problem than Willis’ cv of interest in a gravity field, like connecting his to earth surface; it is not by adiabatic definition. After each cylinder is disconnected & made adiabatic in a gravity field, each is one system again & in my view has to be stratified to obey all thermo laws being discussed, harm or create no energy.
Robert Brown said @ January 21, 2012 at 1:20 pm
Substitute “philosopher/historian” for “physicist” and it’s ditto for the rest. Further, no warmist has managed to budge my convictions by a single iota (so far), but many of the “sceptics” here may well achieve what the warmists have failed to do. It’s a strange world…
More insight into Bart’s Law:
The equilibrium temperature will be established by the local maximum of the most substantial peak of the radiative response of the atmosphere. If the peak is in the IR, it will be slightly above the peak emission for the IR gases alone, because the total includes emissions from higher energy radiative sources.
If you add more IR radiating gas, that should tend to pull the peak lower in wavenumber and/or broaden the distribution below, resulting in surface cooling. If you add more gases which radiate at higher frequency than the peak, that will tend to push the peak to a higher wavenumber and/or broaden the distribution above, and you should get surface heating.
The key to all of this is stability. In Bart’s Law, it is stated that the temperature will “rise” until equilibrium is reached, and the atmospheric emissions keep it from rising further. The greenhouse theory implicitly says it will fall until equilibrium is reached, and the atmospheric emissions will keep it from falling further.
But, the situation is not symmetric. More energy is emitted than retained. Therefore Bart’s paradigm meets increasing resistance going up, but the GHG paradigm meets decreasing resistance to going down. Therefore, Bart’s paradigm is stable and bounded, but the GHG paradigm is not.
sky says:
January 21, 2012 at 2:27 pm
“In other words, it becomes a tautology.”
Glad someone else recognizes that. But, I have found what I believe to be the solution to the conundrum which slices through the Gordian knot. In any real atmosphere, there are always emitters, and that makes it possible for temperatures to increase within standard SB limitations until such as time as balance is achieved.
Robert Brown says 1/21 2:22pm:
“…You move the jar slowly up, move it slowly down, nothing in the jar knows that it has been moved at all!..”
Robert please slow down along with Willis, I am sure you can think this thru way better s-l-o-w-l-y.
Here you now have three systems & F=ma; everything in those jars & even the jars themselves know they have been moved. Mixing metaphors again. Just slow down. Please stay with Willis’ original cv of one system GHG-free air in a tall column in the presence of inexplicable gravity field.
We can learn from you, no kidding. I’m glad you’re engaging. My & other’s view is gas in this cv has to stratify, M-B can’t be invoked due to the ngh term; your view it does not stratify & invoke M-B. Let’s engage on that cv – Willis’ premise. Please.
Robert continues:
“Now put the two jars in thermal contact. I don’t care how — move them together, use an insulated silver conductor of heat to couple their otherwise adiabatic walls without moving them up or down — it doesn’t matter, does it. What happens? Well, they are at different temperatures, so heat flows.”
Yes, yes we get this Robert, you have two systems now, hot jar connected to cold jar works in two systems. Now get into Willis’ one system cv again. Put down the jars. Then do not cross the cv. Help me et.al.
“As to your claim that I have failed to demonstrate that work can be produced out of a temperature differential … surely you are kidding. That’s how heat engines work. Are you claiming that a temperature differential cannot be used to produce work? ”
=====
Willis, Tallbloke etc:
Off the shelf Thermal engines (and vortex tubes)
stirling engines running of ice/hand/coffee with videos.
http://gyroscope.com/default.asp
As Willis would put it, I’m just a “random guy.” And, frankly, I have almost no knowledge of thermodynamics. For that matter, I didn’t understand all of the paper by Velasco, Roman, and White, whom I’d never heard of. So, how can I have the chutzpah to credit that paper over the opinions of Robert Brown, a Duke professor, and Willis Eschenbach, who regularly demonstrates more physical-science insights in a month than I’ve probably had in my life–and over the opinions of many other undoubtedly well-credentialed people at this site who have patiently explained to me that I’m ignoring one of the most basic laws of physics?
Four reasons.
Three of the reasons are two monatomic molecules and the following equation:
= (3E/(5N-2))(1-mgz/E),
where is mean single-molecule kinetic energy, N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude.
Place the two monatomic molecules by themselves into the isolated thought-experiment column we’ve belabored. Have them share a total (potential + kinetic) energy of 2mgz_mid. Let them equilibrate. When we plug this little system’s values into the equation above, which is adapted from Velasco et al.’s paper, we get (3/4) mgz_mid(1 – z/z_mid) for the mean per-molecule kinetic energy in our gas column.
According to Velasco et al., that is, the mean per-molecule kinetic energy in our gas column is 0 at 2z_mid, (3/8) mgz_mid at z_mid, and (3/4) mgz_mid at the bottom of the column. On the other hand, Brown, Eschenbach, and the many physicists whom I hereby thank for having patiently given their time here to “fight the amazing ignorance of thermodynamics [that I’ve] so ably demonstrated” say that the mean kinetic energy at equilibrium is independent of altitude: it’s the same at z=z_mid as at z=0>
A thermodynamics innocent, I cannot understand how what Brown et al. say can be so. On the other hand, my simple mind can indeed believe the story the Velasco et al. equation tells.
So, yes, my ignorance of thermodynamics is indeed amazing. But until the experts can explain to me in terms I can understand just which of Velasco et al.’s equations is wrong and why, Ill have to believe Velasco et al.
Oh, I said there was a fourth reason why I credited the unknown Velasco et al. over the obviously brainy people who opine on this site. It’s that I spent my career having such folks as clients, and my experience is that things tended to go badly when I took their word on complicated technical matters instead of forcing them to make me understand.
“Tallbloke
January 21, 2012 at 2:27 pm
Robert Brown says:
January 21, 2012 at 1:52 pm
Even simpler, don’t worry about the actual motor — simply connect the top and bottom with a silver rod insulated on the sides. Bottom is hot, top is cold, so heat flows along the rod — forever. — even Jelbring and Tallbloke with concede that this cannot happen.
Excellent. another suggested experiment. My old dad is a fine qualified civil engineer with a wealth of fluid dynamics experience. He told me earlier today that he once did a set o f calcs which showed that a water pipe of infinite length with infinite pressure at one end would have no flow at the other due to frictional loss. I suspect the heat flow in your 10km silver wire might have similar conduction issues due to atomic interstice losses. By all means give it a try though.”
Frankly, Tallbloke, I am dissapointed with your responses to the science posts on here.
Initially I thought there was a whiff of reflection about the science being presented to you. But now there is a positive stench of desperation in your responses.
Give it up man, we are all wrong at some point in our lives, no shame in that.
I am very very skeptical about the CAGW hypothesis. My skeptism does not involve trying to deny that changing CO2 concentrations in the atmosphere has no effect, I think it does, it is the magnitude of these effects, given all the other connected feedbacks, where my doubts lie. However I am open to evidence which will show that I am wrong.
You are giving the cause of CAGW skeptism a bad name by this ludicrous defense of a flawed hypothesis. An hypothesis so flawed it is clear to people who have not completed a tertiary education.
Alan
Scot Allen says at 1/21 2:49pm:
Fine silver atom thoughts. I figured to eventually go there. When Robert Brown drops an insulated silver rod into Willis’cv tall air column, remember these thoughts. To make progress in understanding Willis’ original top post premise.
My view the silver rod, insulator and ideal gas all equilibrate, no heat flows, no energy forever b/c even though we have 1 stratified gas system, and 2 solid bodies they are all in energy equilibrium & 1 cv system obeys all the laws we’ve discussed. M-B not applicable since have ngh PE term. No energy harmed or created.
Willis Eschenbach;
Save your disgust, your snark, and your accusations of lying for when you are right, david. They don’t taste so good when you have to swallow your words.>>>
From N&Z:
Equation (7) allows us to derive a simple yet robust formula for predicting a planet’s mean surface temperature as a function of only two variables – TOA solar irradiance and mean atmospheric surface pressure, i.e
Ts=25.3966(So+0.0001325)^0.25*NtePs
Two variables.
Yes, you need multiple variables to calculate NtePs in the first place because mean surface pressure is not uniform. If you understood N&Z (which you state that you don’t) you would understand that a big part of what they have done is provide a mechanism to calculate NtePs. It could just as easily be measured in which case it would indeed represent a single variable. The problem is that measuring it is just as complex as measuring the temperature of the earth compared to the insolation because they don’t vary with each other. P varies with T^4. All N&Z are showing you in this part of their equation is how they got to the correct value in the first place, which is not straight forward. But if we could just stick a pressure meter into the earth and get that number, it would indeed be a single variable.
My accusation and snark stands.
Hi Robert, and thanks again for your reply. I’m going to split my response into two halves. the first will deal with your statements and claims about my position. The second half will deal with the argument you feels sufficiently addresses the matter in hand.
You say that:
“The problem is that temperature is a meaningless concept for a two dimensional surface. ”
Well, I was only trying to address your ‘slices’ argument from the original post, so I humbly submit that you are the one who started it with the surfaces malarkey.
and you say that I should:
“address the actual topic of the thread, which is how any system that spontaneously generates a thermal gradient violates the second law of thermodynamics”
Well you can’t define us out of existence. Total energy = KE+PE. The second law I’m working from (feel free to define the one you’re using) tells me that in a gravitational field, the higher up you are, the less Kinetic Energy you’ll have and the more Potential Energy you’ll have. Total energy must be conserved. Everyone is agreed on that one, so, less KE = less heat high up.
Now, I’m aware that there are some very sophisticated arguments about this stuff which use statistical mechanics, but as Joe Born will tell you, he’s been through the math in the Coombes and Laue paper, and the Valesco et al paper, and he has found the reason C&L come to the isothermal conclusion. Joe says it doesn’t work. They are playing with limit cases which don’t relate to the mechanical behaviours of actual atmospheric gases.
“and thereby enables the construction of PPM2K’s, including the trivial thermocouple example in the previous post.”
The only viably well specified machine I’ve seen in this thread fails the logic test quite quickly. keep trying though.
“What you’re saying, in a nutshell, is that you have two volumes neither of which is in thermal equilibrium (if they were, they’d have the same temperature throughout each volume.”
Your problem here is that you keep going back to your own preconceptions instead of following our argument. KE+PE is constant, so as we go up, PE increases. Therefore KE (and temperature) falls. So an air packet is in energetic equilibrium when the temperature gradient across it is such that it matches the lapse rate defined by gravity and it’s specific heat at constant pressure. Loschmidt held it should be that way for solids too by the way, so your silver rod may not behave as you’d expect.
Now, your capturing of gases in jars argument seems hopelessly confusing to me. That may be my fault or yours or just that we’re seeing things differently. Try this explanation of why we have ended up with differing interpretations and see if it makes sense to you.
H/T Frank
“In a laboratory setting, we say that the pressure is the same in all directions and therefore that the kinetic energy associated with movement of molecules (thermal diffusion) is the same in all directions. On a planetary scale, we say that pressure change is associated with the weight of gases overhead, and forget to consider what this means about the behavior of molecules. If there is a pressure gradient with height, it can only be caused by molecules moving upward more slowly than they move downward. The obvious reason for this is that the molecules moving upward are being slowed down by gravity and converting kinetic energy to potential energy (and cooling when we consider a large enough group to treat mean kinetic energy as temperature). The opposite is true for molecules moving downward.”
Now from what I read in an earlier comment, someone claimed that you’ve said that the KE of molecules won’t be affected by gravity. If that’s so, please could you explain why they are exempt from the law of gravity. What goes up must lose velocity. If it’s mass stays the same, it ends up with less KE. QED
Thanks and I hope we can continue discussion.
TB.
Robert Brown says:
January 21, 2012 at 8:39 am
“get an intro physics textbook and read it…….”
=============
I’ve always wanted one, does Amazon sell a good one ?
Do you sell a good one ?
Don’t be bashful, just tell me which book to buy.
Thanks.
respect.
Alan Millar says:
January 21, 2012 at 3:32 pm
Frankly, Tallbloke, I am dissapointed with your responses to the science posts on here.
The feeling is mutual Alan. Except I’m disappointed, rather than dissapointed.
thepompousgit says:
January 21, 2012 at 2:17 pm
tallbloke said @ January 21, 2012 at 9:31 am
It is a definition. One that is severely lacking in universality. It fails as a general law as soon as it is placed orthogonally to a gravitational field.
I am truly gobsmacked! A definition is a tautology. A tautology is a proposition that is unconditionally true. But not unconditionally true according to TB.
Wipe your windows pompous. I didn’t criticise it as a definition, I pointed out it failed as a general law. And while your smug logico-philosophical point is trivially true, the problem we face is people running riot with limited definitions of the thermodynamics laws which are useful heuristics in limited settings and pushing them to the status of generally applicable ‘laws of nature’.
This is the reason for most of the confusion in this thread.
Sorry, folks, the web site ate part of the equation in my last post; it must not have liked the mean-indicating angle brackets
Here it is without the mean-indicating brackets:
K = (3E/(5N-2))(1-mgz/E),
where K is mean single-molecule kinetic energy, N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude.
As I said before: Place the two monatomic molecules by themselves into the isolated thought-experiment column we’ve belabored. Have them share a total (potential + kinetic) energy of 2mgz_mid. Let them equilibrate. When we plug this little system’s values into the equation above, which is adapted from Velasco et al.’s paper, we get (3/4) mgz_mid(1 – z/z_mid) for the mean per-molecule kinetic energy in our gas column.
According to Velasco et al., that is, the mean per-molecule kinetic energy in our gas column is 0 at 2z_mid, (3/8) mgz_mid at z_mid, and (3/4) mgz_mid at the bottom of the column. On the other hand, Brown, Eschenbach, and the many physicists whom I hereby thank for having patiently given their time here to “fight the amazing ignorance of thermodynamics [that I’ve] so ably demonstrated” say that the mean kinetic energy at equilibrium is independent of altitude: it’s the same at z=z_mid as at z=0.
Tallbloke your reply to Robert is just another obfuscation.
You have been shown that Jelbring’s system allows heat to flow indefinitly. Robert’s silver rod shows that. You have not attempted to disprove that by the use of the current thermodynamic laws. No, you have attempted to disprove that by trying to say that whist the laws apply in general they do not apply in the particular.
You have said, in refutation, that ‘something’ in atomic structiures will prevent connected heat sources, at different temperatures, exchanging heat. That something is presumably due to the distance between the heat sources and/or the nature of the connection or perhaps it is something else.
You have to publish the details man, a Nobel prize awaits. Nobody could deny you it. You would go to Norway garlanded, as the man who has added to and improved, the laws of thermodynamics.
That is assuming you have some evidence or mathematics to back up this claim!!!
Do you?
Alan
Would an experiment resolve these issues?
For example, Take a sealed cone filled with non GWG and spin it about so that the pointed end is on the extremity and the flat end is nearest the centre. Measure the temperatures near the pointed end and near the flat end.
tallbloke says:
January 21, 2012 at 1:17 pm
You didn’t pay close attention to the specification of my device. The two cylinders are thermally coupled at the bottom. So the gas at the bottom of the column in contact with, say, a copper plate that is otherwise insulated must be at the same temperature. So the cooled helium that falls to the bottom of the cylinder causes heat transfer from the bottom of the xenon cylinder cooling the xenon at the bottom while the heat engine (or a silver rod) warms the top of the xenon and cools the helium. In the end, both cylinders are isothermal at the same temperature. So now we have two isothermal cylinders. But if Jelbring is correct and we thermally disconnect the top of the cylinders, the lapse rates will be re-established and more work can be done. Or, if we extract work slowly enough, the engine will run forever. A classic perpetual motion machine of the first kind. Or possibly it’s the second kind. I haven’t actually done the calculations to see if the heat content of the cylinders decreases over time.
tallbloke said @ January 21, 2012 at 4:27 pm
Wipe your own windows TB. The Laws of Thermodynamics are called Laws just because they are universal, show no sign of being contradicted by any repeatable experiment, are unchanging etc. There is no chance of any of these Laws being repealed by those posting to this thread who view them as “limited heuristics in limited settings”. They have been accepted as Laws for over a century not just because they make sound theoretical sense, but also because they have withstood every test of them so far.
Please note that I am not saying that they will remain as Immutable Laws for Ever and Ever Amen. Paradigm shifts do occur. But every paradigm shift I have studied was enabled by people who clearly understood the paradigm they were overturning.
Most of the people on this thread exhibiting confusion are doing so because they do not understand basic physics.Reading Feynman’s The Character of Physical Law might help remedy this.
BTW Tautologies are far from trivial; they are the building blocks of the edifice we call mathematics, science and logic.
Alan Millar says:
January 21, 2012 at 4:43 pm
Robert’s silver rod does not show that. The atoms in the silver rod are affected by gravity in the same way molecules of air are affected. Hotter silver atoms at the bottom deliver their energy to those above, and those to more atoms above and so on during conduction. Each tiny temporary displacement carries kinetic energy, but some of that energy is converted to gravitational potential energy. This stored potential energy can’t be conducted. It is converted back to kinetic energy as the atoms displace downward. These differences in kinetic energies between layers must create a temperature gradient.
There are really only two possibilities.
1) The second law works and thus the temperature must be uniform.
2) KE + PE is constant, so the particles cool as they go up, and thus the temperature drops.
I can give the “elevator speech” for why #2 is suspect. For any given trip, the KE + PE is indeed constant, but for different trips, the value is different (ie the boltzman distribution). If you look near the top, the “low energy tail” never gets that high, so you are only looking at self-selected particles that started in the high energy tail. These originally-high-energy-particles have indeed lost some KE on the way up, but they started with extra on average. This can (and does) leave this SUBSET of particle with the right average energy to be at the same temperature as the WHOLE set was at the bottom.”
I certainly haven’t proven this rigorously in this non-mathematical paragraph. However, it is clear the “lose KE and lose temperature” argument has a huge hole in it. Those who want to pursue this line of reasoning IN THE FACE OF STRONG COUNTER-ARGUMENTS, would need to determine the distribution of energies of particles at any altitude, and show that those remaining particles are indeed lower KE then the whole set was at the bottom. I am sure they can’t do this because I am sure they are wrong.
But hey, that Nobel Prize is still waiting to be claimed….
sky says:
January 21, 2012 at 2:27 pm
But the gas density decreases to zero with altitude. For an isothermal atmosphere, the volumetric energy density decreases with altitude exponentially because the volumetric kinetic energy density decreases much faster than the volumetric gravitational potential energy increases. There is no kinetic energy temperature for space. The pressure is too low and the mean free path is too long for a Maxwell-Boltzmann kinetic energy distribution to form. Local thermodynamic equilibrium doesn’t exist, so a gas kinetic temperature can’t be defined. That’s not to say there isn’t an effective radiative temperature. There is, but it would have no effect on the kinetic energy of an individual molecule. Any molecule that might be excited would radiate long before it could transfer energy to another molecule and change the kinetic energy of either molecule. Similarly, collisions between molecules that resulted in transferring enough energy to excite one of the molecules so it could radiate and result in a loss of total kinetic energy would be extremely rare.
Also, at low density, the thermal diffusivity is extremely high, leading to a rapid decrease in any existing temperature gradient if there is no other source of energy, which there isn’t by the definition of the problem.
DeWitt Payne, where are you getting your lapse rates for helium and xenon? Are they empirical or calculated?
Alan Millar says:
January 21, 2012 at 4:43 pm
Tallbloke your reply to Robert is just another obfuscation.
You have been shown that Jelbring’s system allows heat to flow indefinitly.
=====
Isn’t the confusion here that between perpetual motion and recycling? The Water Cycle is recycling, no one thinks of it a perpetual motion..
Taking energy out of a system is going to have some effect on the system, how that plays out if taking energy from the gravity piston is what has to be calculated, (gravity doing the work of a piston by default as it were, the gases changing density by temp using gravitational potential by being in the different levels by weight/density, rising and falling, so energy extracted from anywhere along that is going to have to be replaced).
For example, extracting wind energy to power must be having some effect, altering wind patterns perhaps?
“DeWitt Payne says:
January 21, 2012 at 5:02 pm
You didn’t pay close attention to the specification of my device.”
You are wasting your time DeWitt.
It is clear that Tallbloke is no longer in a rational thinking mode.
He is in fullblown ‘La La La’ denial mode.
He has invested so much of his standing and prestige in this position that any evidence that contradicts his position is just going to induce the ‘La La La’ response.
We have seen what happens on the ‘other’ side of the CAGW debate. First they get defensive, then they get angry and then they start censoring and then they stop listening.
Sad really!
Alan
Why are you all wasting time on inconsequentials? The problem is solved.
Scot Allen,
Calculated. The heat capacity of a monatomic gas is nearly independent of the atomic weight on a per mole basis and is equal to about 21 J/K^-1 mole^-1. But Cp must be in the form J kg^-1 K^-1 to calculate the lapse rate. So take g and divide by Cp on a per mole basis and multiply by the atomic weight expressed as kg/mole. That’s 0.004 for helium and 0.13129 for xenon.
Tim, the reasoning of your post is fine … except for the final conclusion follow “and thus” … which (upon careful atom-counting and energy accounting) turns out to be incorrect, per a previous post titled “a WUWT Puzzler”.
As often happens in thermodynamics and statistical mechanics, careful counting and accounting lead us to surprising conclusions!
Bart,
Might I suggest you do yourself a favor and not name Physical Laws after yourself. It is bad form, first you have to PROVE your law and then others have to acknowledge that you are in fact correct. Sorry, but that’s the way it works.
This comes from someone who has not had the high honor of having a LAW named for me, perhaps someday I might have that honor. But, frankly it is not very high on my “bucket list”.
Is there a specific point in my hypothesis (it was clearly enumerated if I recall correctly) that you dispute ?
Cheers, Kevin.
Alan Millar,
With respect to Tallbloke, I’m quite aware that it’s futile. I have the same problem with Nasif Nahle when he insists that partial pressure is identical to partial pressure times path length. That causes him to think that the emissivity of CO2 in the atmosphere is low. It’s impossible to convince him otherwise because he has invested too much of his reputation in his hypothesis. Note that Tallbloke seems to think Nasif is a reliable source.
But I’m retired so I have the time. And I refine my own understanding by having to use different explanations and examples. Also, for people lurking on this thread, I think it’s important to not leave false statements unanswered.
Scot Allen says:
January 21, 2012 at 5:14 pm
And we’re back to perpetual motion again. What you’re saying is that a 100 m vertical insulated silver rod will have the same difference in temperature from top to bottom as a 100 m insulated column of air. That could only be true if the two are isothermal. Otherwise, the lapse rate of a silver rod must be less than a column of air and you can extract energy from the system.
“Scot Allen says:
January 21, 2012 at 5:14 pm
Robert’s silver rod does not show that. The atoms in the silver rod are affected by gravity in the same way molecules of air are affected. Hotter silver atoms at the bottom deliver their energy to those above, and those to more atoms above and so on during conduction. Each tiny temporary displacement carries kinetic energy, but some of that energy is converted to gravitational potential energy. This stored potential energy can’t be conducted. It is converted back to kinetic energy as the atoms displace downward. These differences in kinetic energies between layers must create a temperature gradient.”
So what are you are trying to indicate here? That two connected heat sources can be prevented from exchanging heat?
If no, then Jelbring’s hypothesis is a bust. If yes, then publish with evidence and/or mathematics and get your Nobel prize.
Which is it going to be?
Alan
An isothermal atmosphere has a vertical gradient of total energy since enthalpy plus potential energy increases upwards (enthalpy is proportional to T). As such, if left to itself, there is a downward (downgradient) flux of energy, which is only sustainable if you drain the excess from the bottom and add it at the top. If you impose top and bottom insulated boundaries that don’t allow an energy flux, it ends up warming the bottom and cooling the top towards an isentropic state that has no energy flux because, in this adiabatic state, enthalpy plus potential energy is a constant.. So an isothermal state is only sustainable with the correct flux boundary conditions.
There is another basis for saying that in the absence of all energy sources, a column of gas must be isothermal, with or without gravity. A temperature gradient necessarily creates entropy, continuously. It takes energy (and heat pumping) to counter this and maintain a steady state gradient. I’ve blogged about this here.
DeWitt Payne says at 1/21 at 10:21:
“If gravity then re-establishes the DALR, we can run the heat engine again, and again, and again. When the DALR is re-established, the heat that was moved from the helium to the xenon will move back again. The total energy content of the two cylinders won’t change.”
If I get this, you have taken the cv of interest which is Willis’ GHG-free tall gas column in presence of inexplicable gravity and modified the bottom of his cv so it can touch to a simulated earth, in particular with xenon canister. Now heat can flow out/in of Willis’ original 1 system premise, things have changed. This is fair of course and a logical extension of the adiabatic original discourse. Not adiabatic anymore.
To begin, let me just discuss the one DeWitt cylinder on top as the 1st system.
My view would be DeWitt top system immediately starts with helium T stratified. Physical plate at bottom touches to enable heat flow out of helium in order to equilibrate with the 2nd xenon system at lower T. This heat flow lowers the DeWitt top system internal P & internal T field but keeps first (helium) system V and n unchanged.
Equilibrium with bottom cylinder is reached and top DeWitt cylinder is disconnected (untouched). We are back to Willis’ original cv except ideal gas PV=nRT is lower at every h. Since KE is lower each molecule is speeding around less fast. The system stratifies as before but molecules total energy have lower KE (since P is lower, V constant) with same PE above the h=0 gravity field ref. point.
Don’t see how you can write “the total energy content of the top cylinder won’t change.” Fill me in. Maybe I didn’t get it.
KevinK says:
January 21, 2012 at 5:49 pm
“Might I suggest you do yourself a favor and not name Physical Laws after yourself. “
I haven’t. Figure out that riddle – it’s an easy one.
“Is there a specific point in my hypothesis … that you dispute ?”
Yes. I have disputed it. I am not going to discuss it anymore. I have an actually physically viable hypothesis I am trying to get people to notice and start thinking about.
A physicist says “Tim, the reasoning of your post is fine … except for the final conclusion follow “and thus” … which (upon careful atom-counting and energy accounting) turns out to be incorrect”
I am a bit confused, since I think we both came to the same conclusion — that the second camp (who believe the declining temperature profile is the equilibrium condition) are wrong. My post actually follows the same line of reasoning that yours did. I stated specifically that “it is clear the “lose KE and lose temperature” argument has a huge hole in it. ”
I just said it much later in the thread after missing your post earlier. 🙂
I think I see a logic here with convecting air. The relationship with kinetic energy and temperature has been confusing to say the least.
Perhaps the problem is in the way its been explained.
OK so as a gas rises it does not lose kinetic energy thats whats been said about the adiabatic “process”.
But at the same time we are told it loses temperature.
One logic would be the kinetic energy always equates to a certain temperature in a single kind of material.
And when air rises it also mixes with other air. The mixing lowers the average temperature of the mixed gas, which is all we can read with measuring devices; but it takes a while for the kinetic energy of individual molecules to come to equilibrium as you need collisions or radiation or something of that nature for them to dump their kinetic energy.
If I can adopt that idea then I can see my way through this morass a bit more easily.
Here I can see an isothermal atmosphere and a lapse rate created by the fact the convection/conduction system fails to keep up.
It really fails to keep up during the cooling cycle as it might require a lot of years for conduction to work its way through the atmosphere. Convection on the other hand might only take a few days or a week or so to catch up (though that last nth of a degree could be long extended as convection slows to a crawl so slow a snail looks like a speeding bullet.)
So since this equalization in effect is in fact determined by a combination of gravity and atmosphere mass that makes Jelbring basics right (just his equilibrium is wrong). And to prove Jelbring was wrong about the equilibrium, his critics were required to resort to an argument that proved the essence of this theory was right that convection and conduction would eventually catch up and equalize the temperature of the atmosphere. (or at least I can think of no other means for the surface to be normal temperature and have a need for it to be the source of warming the upper atmosphere. So in the end Jelbring’s model proves beautifully useful.
Now one more step! If the reason we have a lapse rate is this delay in equalization of the atmosphere it follows that the longer the delay the hotter the surface and the cooler the upper atmosphere and viola we are back in business with a lapse rate and an explanation for why the surface is warmer than its expected blackbody temperature. Or at a minimum we have a concrete theory it at least partially explains it. Then the correlations across planets with disregard to GHG begins to explain why it might be a very large part of it.
I am feeling pretty comfortable with that. Does somebody have a pin to puncture my balloon?
Tim Folkerts says 1/21 at 5:15pm:
“..but for different trips, the value is different (ie the boltzman distribution).”
Tim – if you mean Maxwell-Boltzmann, double check a couple thermo texts or ref.s on that – I find a main assumption for M-B distrib. to hold is no external forces (electrostatic, gravity) on the molecules.
Willis’ cv of interet has inexplicable gravity field which is an external force on the molecules – it introduces the PE (ngh) into molecular total energy term that M-B doesn’t consider (M-B KE = total molecule energy for M-B). Maybe find a ref. that does consider PE in modified M-B and fill us in. There might be a cool answer in this line of reasoning.
“DeWitt Payne says:
January 21, 2012 at 5:52 pm
But I’m retired so I have the time. And I refine my own understanding by having to use different explanations and examples. Also, for people lurking on this thread, I think it’s important to not leave false statements unanswered.”
Ahh… I see, we must be of a similar age. I am semi-retired and got my first job as a scientist in the late 1960’s. Changed career path though, subsequently.
I now try to contribute to these sort of blogs in logical layman terms. One, because I think that is what is best for the general readership and two, because I don’t fancy having to brush up on my Mathematics after nearly 40 years!
Alan
Bart’s Law:
It follows that surface temperature sensitivity to the addition of radiative gases is negative: the addition of more IR radiative gases will tend to lower the surface temperature. This is a sensitivity – it does not preclude the possibility of feedback from other processes tending to resist the change, e.g., cloud albedo effects.
Bart’s Law is justified by the following line of reasoning:
The foregoing is, I think eminently reasonable and appealing. The following is my current thinking about how it all fits together:
Convergence is initially superlinear, because the more you bite into the areas of the tails, the faster you increase the back-radiation and hence surface temperature. Eventually, you reach the peak of the emissions spectrum and you can go no higher, because each additional increment of surface temperature produces too little extra back-radiation to sustain it in equilibrium with SB.
The other frequencies outside the limiting emissions spread will still attempt to reach the higher temperature emissions levels, but they are constrained by a requirement of continuity in the total emissions, and the dominant emissions spectrum constrains them. The result is a “pincushion” spectrum at TOA which dips in the region of the dominant emissions spectrum.
“…Eventually, you reach the peak of the emissions spectrum and you can go no higher…”
Or, maybe you can go a little higher, at any rate until the rate of increase in back radiation is insufficient to sustain the surface temperature above the SB equilibrium limit.
The way I see this ‘gravity’ theory explained in the comments is-
1/ gravity can be easily made to work. eg hydroelectric plants convert gravity into electricity.
2/ atmospheric pressure is caused by the weight of the air. ie gravity working on the atmospheric mass to pin it to the source of the gravity, earth. other gravitational fields also alter this eg tides. this pressure is the expression of the gravitational force.
3/ the earths surface is warmed directly by sw radiation from the sun. the surface temperature will be higher than space as long as the atmosphere has mass. a temperature gradient is created.
4/ the portion of sw radiation converted to lw radiation by the blackbody earth will come across no absorbers or few, so will pass freely to space.
5/ air (not classed as a greenhouse gas, but capable of retaining heat.. just!) is a very poor thermal conductor, but with time will eventually form a gradient due to the fact that both ends remain at constant temperatures.
to add bits that have been mentioned further-
6/ one end of the gradient changes within a 12hr period. ground and air cools at night.
7/ changes in gravity from various tides also change/shift pressure.
the way i see it though is that while conduction and convection surpass the radiative factor at the lower altitudes, where the gas is thin in the upper atmosphere the radiative exchange becomes more important.
About the the WUWT puzzler : http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-870583
If the shells have the same speed at all altitude but there are fewer of them at higher altitude. Then the sum of momentum is lower at high altitude. So the gas analogy would give us a lower temperature at a higher altitude.
You are entirely correct Tim — and also, my post should have been more clear that were agreeing (for which confusion I hereby apologize)
By the way, it’s fun to see that folks actually care about the GHE, and are following (and contributing to) the discussion. And it is mighty impressive to see how closely the discussion here on WUWT have evolved in parallel with the historical evolution of GHE theory.
These ideas most definitely are *NOT* simple or trivial. One wonders, at what point can it be fairly said, that among skeptics and nonskeptics alike, a consensus is emerging that the GHE is real?
Willis, you reply to me with this DUMB statement and some other nonsense:
“First, I fear that what happens on other planets is of little interest to me. The planets are so different from the Earth, and we understand our own climate so poorly even though we live in it, that what is happening on Jupiter or Venus can provide little insight, falsification, or support for theories about the Earth.”
Willis, we are kindred spirits in many ways, and I really, really respect you and enjoy your posts, but I’m concerned that you are putting your head under the sand here! Just because you have “no interest” in something has nothing to do with the truth and science, no? STRAWOMAN??? Your statements are nothing but a total denial of FACTS, sir! You HAVE NOT ADDRESSED THE EMPIRICAL EVIDENCE THAT CHALLENGES YOUR THEORY, WILLIS!!!! IN FACT, YOU CONTINUE TO IGNORE ALL THE EMPIRICAL EVIDENCE!!* Are you afraid of the empirical evidence?
Until you address these concerns, I remain unconvinced of some magic “radiative GHE effect.
And the current average global temps are not helping you, either, sir.
[SNIP: Swearing at the moderators and calling them bad names is bad form and makes them cry. -REP]
Willis Eschenbach says:
davidmhoffer says:
Ackk! Now, you guys are both getting me confused because you are confusing the terms “parameters” and “variables”. What bothers me is not so much the variables as the free parameters. To define N_TE in terms of P_S, there are 4 free parameters…Those numbers in Equation (7). How do we know they are “free parameters”. Well, for one reason, they say, “However, we discovered that NTE was strongly related to total surface pressure through a nearly perfect regression fit via the following nonlinear function” which tells you that they did a fit; the other is that the numbers are not simple values like 1 or 2 or 3…They are clearly numbers that were obtained through a fitting procedure.
Now, to write T_S in terms of N_TE, there are two further numbers, although I think one can argue that those are not “free parameters” at least once their definition of T_gb is chosen (and one is just to correct for the 3 deg background of space…which is numerically irrelevant anyway). However, even discounting these, to express T_S in terms of P_S (and S_0 and T_gb), there are 4 free parameters…the ones that appear in Equation (7).
So, like I said, their fit involves 4 free parameters, plus some freedom in their choice of T_sb. And, given the different values that are floating around for the average temperature of the moon and, I imagine might be similarly true for the other nearly airless bodies, there may have even been some “play” in the data itself.
It takes an amazing absence of skepticism to look at their fit and be that impressed by it. Nobody who does so can seriously call himself a “skeptic” in anything but a very Orwellian sense of the word! In fact, nobody who takes their paper seriously at all after the sort of gargantuan errors we have found can by called “skeptics”. It is simply a ridiculous use of the word.
…at any rate until the additional increment in back radiation is insufficient to sustain the additional increment in surface temperature above the SB no-atmosphere equilibrium limit.
Robert Brown,
I have been following your posts both here and at Tallbloke’s blog. Unfortunately I have not read all of the posts here at WUWT (although I have covered a lot of them) so I hope what I say is not a repeat of what others may have said.
First, you talk extensively about “equilibrium” and “thermal equilibrium”. But I believe we should be talking specifically about “thermodynamic equilibrium”. Thermal equilibrium is but a subset of thermodynamic equilibrium. A system is in “thermodynamic equilibrium” when it is in thermal equilibrium, mechanical equilibrium, radiative equilibrium and chemical equilibrium. Thermodynamic equilibrium equals thermal equilibrium only when a system’s internal energy can be described by
U = CvT (1)
In this case the system’s internal energy is thermal energy and thermal equilibrium is all that matters. Your statement “Thermal equilibrium is isothermal, period” only applies to such a system. But that is not the system of a planetary atmosphere under the influence of a gravitational field.
Second, we need to better define where the classical laws of thermodynamics and equilibrium are valid. They are valid with a homogeneous system where all the locally defined intensive (e.g., per unit mass) variables are spatially invariant. But a system is not homogeneous if it is also affected by a time invariant externally imposed field of force, such as gravity. Thus in a gravitational field the laws of thermodynamics have to be applied in a manner that reflects the external field. This is a paradigm buster in itself.
Third, the thermodynamics of an atmosphere cannot be described wholly through considerations of heat transfer only (Trenberth diagram, anyone?). The atmosphere must be treated as a system undergoing both heat and mass transfer. Mass transfer is the reason that gravity is so important in understanding atmospheric thermodynamics. This gets to the mechanical equilibrium part of thermodynamic equilibrium.
Fourth, chemical equilibrium is also very important in atmospheric thermodynamics. But if you assume uniform molecular diffusivity in a horizontal layer (gravitational potential energy handles the vertical diffusion) we are left with latent heat and that is beyond our discussion here.
Thus, as I explained in a previous Tallbloke post,
http://tallbloke.wordpress.com/2012/01/16/the-gravity-of-some-matter/#comment-14236
the proper formulation of the first law of thermodynamics for a “dry” atmosphere under the influence of a gravitational field is:
dU = CvdT +gdz – PdV (2)
The first energy term, CvdT, is the only variable that deals with thermal energy (temperature). The other two energy terms, gdz (gravitational potential energy) and PdV (mechanical work energy), deal with mass transfer. Thus temperature reflects only one variable in the energy composition of the atmosphere. As Tallbloke has pointed out, for the first law to apply energy can be transformed from one form of energy to another, but total energy content must remain constant (dU = 0). The temperature profile through the atmosphere is dependent on the mix of energies at any given point.
The temperature profile resulting from a dry adiabatic lapse rate is covered in my previous post above. And that profile reflects an isentropic system in steady state dynamic equilibrium described by:
CpdT + gdz = 0 (3)
The same system in static equilibrium can be described by the equation of state:
CpT + gz = constant (4)
But for an isothermal temperature profile to exist in an atmosphere in a gravitational field both CpT and gz must increase with altitude. Thus internal energy (U) is not a constant but must also increase with altitude. That violates the first law. The only way that could be achieved is if work is being done to the system. And such a system would not be in thermodynamic equilibrium.
The zeroth law only applies to a homogeneous system and the atmosphere under the influence of a gravitational field is not a homogeneous thermodynamic system (as defined above). And equation (3) does not violate the second law since the process equation reflects an isentropic (constant entropy) process.
The Jelbring hypothesis holds up perfectly from a thermodynamics standpoint.
Your insertion of an insulated silver wire into the discussion makes for an interesting mental exercise, but it has nothing to do with the thermodynamics of a planetary atmosphere. You have just added a non-existent subsystem to the existing system. And the wire cannot be a metaphor for conduction since vertical conduction is almost non-existent in a gaseous system in a gravitational field.
As for the N&Z hypothesis, I too have no firm opinion as yet. I am awaiting Part 2 of their submission that deals with the detailed thermodynamics. But if that is based on a premise similar to the Jelbring hypothesis, I will be pleased.
I would like to provide one quote from Thomas Kuhn’s book “The Structure of Scientific Revolutions” to wrap things up:
“Because the unit of scientific achievement is the solved problem and because the group knows well which problems have been solved, few scientists will easily be persuaded to adopt a viewpoint that again opens to question many problems that have previously been solved.” (Page 169)
Bill
IMPOSSIBLE ….. Willis in a nutshell…..
It’s amazing, but if we ignore the Puzzler shells that fall short of a given altitude, and calculate the mean kinetic energy of the Puzzler shells that *do* reach a given altitude, then we find that these Puzzler shells have a greater mean kinetic energy (at altitude) than the mean energy of the shells leaving the gun barrel.
In fact, in the limit of very large altitude, the vanishing small fraction of Puzzler shells that reach these far-higher-than-average altitudes turn out to retain twice mean the kinetic energy at-altitude as the shells leaving the gun barrel (showing that the energy-gain is a factor of two requires a tedious calculation).
So in this statistical sense, the Puzzler shells become “hotter” as they climb!
Joe Born says:
January 21, 2012 at 3:26 pm
I just read Velasco. Unless I missed something, Velasco agrees with Robert Brown and myself.
First, they say:
So Velasco et al say that Coombes and Laue conclude correctly that the temperature is the same throughout, i.e. isothermal.
Then they analyze the same question for an infinite rather than a finite system. Their conclusion is:
So they are distinguishing between finite and infinite systems, and they clearly say that the equilibrium situation in both is not choice (2) above, but choice (1), that temperature is the same throughout because the system is in equilibrium.
So I’m unclear why you think Velasco disagrees with me. If I understand it, which is always an open question, it seem they agree that the final distribution is isothermal. What am I missing?
w.
William, that’s why I constructed the WUWT Beach House story … to show in concrete, practical terms that Willis Eschenbach is entirely correct in asserting that “gravito-thermal” theories do violate thermodynamical principles.
tallbloke says: January 21, 2012 at 1:17 pm
“This is not a perpetual motion machine. All you have done is exploit the Gibbs free energy between the two dissimilar gases until it’s gone.”
OK, what’s the end state? Suppose you connect heat engines across both top and bottom. If there is a temperature difference, heat flows and work is done. If not, then both top and bottom are at the same temperature, and no heat is flowing. This recreates the original conditions – sealed cylinders. But they can’t both be at their lapse rate. What will the gradient be?
Willis:
Bless you; you’ve finally engaged. And, as I said, I’m happy to be set straight; I don’t contend I know the answer. I was just trying to attract some eyes to those papers to see if others see what I thought I did.
The part you quote is highly ambiguous; I couldn’t be sure what they were saying. But I think the key is Equation 8, which I read as the mean single-molecule kinetic energy as a function of altitude. In my last post I applied it, with f=3, to an N=2 system for which E=2mgz_mid. The result appealed to my intuition.
When you put realistically large values of N to it, you get a lapse rate that, as I said above, is negligible for practical purposes (to the extent that our hypothetical gas column is practical) but nonetheless non-zero. That is why over the last couple of weeks I came to question the equilibrium-implies-isothermal result everyone swears by.
“It’s already been described up thread, but I’ll do it again. Take two vertical cylinders 100 m tall. Fill one with helium and the other with xenon at 1 atmosphere. Thermally couple the bottom of each cylinder to the other and insulate the rest of the cylinders. The DALR for helium is 0.001888 K/m and 0.061 K/m for xenon. If gravity causes the DALR to be established in both cylinders, the temperature difference at the top of the cylinders would be 6K. If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top.”
There minor things like, your lapse rate will affected by having pressure greater and less then the atmosphere- bottom of both will have more than 1 atm. Top of helium will be greater then ambient air pressure, and Xenon one at top will have less then ambient air.
One could have your lapse by having helium at or below ambient pressure at top- this also means less pressure at the bottom- so all of helium cylinder will have slight vacuum. With Xenon just have pressure at bottom 1 atm and top will have slight vacuum.
This wouldn’t make much energy. A sunny sidewalk and shaded sidewalk has far more temperature difference.
And insulating these cylinders would difficult- gases do not conduct heat well. Air is great insulator if you inhibit convection. This machine seems to generally inhibit convection- on purpose, you are shielding it from the energy of sun.
So this would make some tiny amount of energy powered by gravity.
There other ways of powering it by gravity. You use a balloon using buoyancy of helium. Inflate balloon from bottom of cylinder. Lift water with balloon. Deflate balloon at top into lower pressured helium, float balloon down with less inflated balloon, fill balloon up again, lift water, etc.
Probably get more energy. But still a lousy way to make energy.
Hydro dams are better way to make energy from gravity.
No gravity- no energy from hydro dams. Less gravity, less power from a hydro dam. More gravity and more energy from hydro dam.
Energy from gravity is commonly used.
Now, back radiation is something like 200 watts per square meter? If so this would dwarf all hydro dams or all nuclear energy if you get 5% of it. The only source of energy greater that back radiation is solar energy, but alas no one has figured out how to harvest this huge source of energy.
Robert Brown says:
Exactly. There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave. It is not. Whenever one finds the actual lapse rate to be close to the adiabatic lapse rate, the reason likely is that the lapse rate would even be higher given how the atmosphere is being differentially heated by radiation (or conduction from the Earth’s surface)…But a lapse rate above the adiabatic lapse rate is unstable to convection, which then transports heat up into the atmosphere until the lapse rate drops down to the adiabatic lapse rate. The adiabatic lapse rate is just a stability limit…Think of it as a ceiling on the lapse rate. If the lapse rate is naturally below that ceiling (as it is in the stratosphere and, as Robert has noted, in certain cases even in the troposphere), then such a lapse rate is perfectly acceptable. If the lapse rate exceeds that ceiling, then it is pushed down to the adiabatic lapse rate.
It is really no more magical than that.
Tallbloke to Willis:
“the rules for falsifying a proposition by appeal to the consequent theoretical constructability of a perpetual motion machine of the second kind are very clear. You have to specify the machine and demonstrate that it will produce work.”
Willis may have considered this unworthy of a response. Or he responded and I missed it. But one simple answer is to place an infra-red sensitive photocell at the top of the atmosphere and generate power with the i-r emitted from the surface due to the thermo-gravitational effect.
And the reason, in case it has been forgotten, why the supposed Jelbring effect makes a sunless planet luminous is, as stated by Willis with abbreviation, as follows:
“…assume that we have the airless [sunless] blackbody planet … The temperature of this theoretical planet is, of course, the theoretical S-B temperature.
Now suppose we add an atmosphere to the planet, a transparent GHG-free atmosphere. If the theories of N&K and Jelbring are correct, the temperature of the planet will rise.
But when the temperature of a perfect blackbody planet rises … the surface radiation of that planet must rise as well.
And because the atmosphere is transparent, this means that the planet is radiating to space more energy [i.e., micro-wave background] than it receives. This is an obvious violation of conservation of energy, so any theories proposing such a warming must be incorrect.”
I think this is worth repeating, since it seems to eliminate all need for further discussion.
Joel Shore says:
January 21, 2012 at 7:39 pm
You are right, upon re-reading my explanation doesn’t track at all. Let me try again in a simpler manner.
The value of the Nte(Ps) is given in equation 7 as:
exp(0.233001 Ps ^0.0651203 + 0.0015393 Ps ^ 0.385232)
This has four numbers in it that were used to fit the value of Nte to the surface pressures of the various planets. These are the four “free parameters”. The fact that they are fits is evidenced by the number of decimals. This kind of fitting can be easily done in Excel using the “Solve…” menu item.
There are also variables which are somewhat arbitrarily chosen although as you say this is the lesser problem. One issue is that on earth the surface pressure is defined as being at sea level. But where is the reference level on a planet with no ocean? At the lowest point? At the average elevation? RMS average elevation?
In any case, the point remains. They have used a very suspect equation (a single variable appears twice) to fit the surface pressure of eight planets and moons to the “Nte”, which in turn is the amount the planet is above the “greybody” temperature. This kind of 4-free-parameter fit, particularly with only 8 datapoints, and pressures that vary by 15 orders of magnitude, carries absolutely no weight with me. With apologies to those who have heard it, the apposite story is as follows, by Freeman Dyson:
Four free parameters, eight data points, that adds up to nothing at all in my world. Those who think that this is “evidence” of anything other than rampant curve-fitting are fooling themselves.
w.
William Gilbert says:
The problem with using Thomas Kuhn to support your notions is that people quoting Thomas Kuhn have correctly predicted about 10,000 of the last 10 paradigm shifts (although that first number is probably too conservative an estimate). [This sort of makes economists, for which this sort of quip was originally applied, look good by comparison!]
And, the the supporters of the notion that the surface of the Earth can emit 390 W/m^2 without some of this radiation being absorbed by greenhouse gases in the atmosphere have still not produced a simple explanation of how this can be done using CORRECT physics principles.
Besides which, we know for a fact that the reason why the Earth’s surface is emitting 390 W/m^2 is not because there is some magical new source of energy but because there is absorption occurring in the atmosphere such that only ~240 W/m^2 is escaping to space.
Hence, what N&Z and Jelbring have exposed is how many otherwise intelligent people will apparently believe utter nonsense if it supports what they really strongly want to believe. I keep thinking that one of these guys is going to step up and admit that their hypothesis is all a Sokal-style hoax ( http://en.wikipedia.org/wiki/Sokal_affair ). These “theories” really have revealed some very embarrassing things about the AGW-skeptic movement…which, to their credit, some skeptical folks like Willis, Robert Brown, and Roy Spencer are trying desperately to correct, although it makes it all the more amazing that such crazy nonsense continues in light of their best efforts to squelch it.
As an example of the arbitrary nature of the N&Z choice of values for surface pressure in their paper, they say the pressure of the earth is 98,888.20 pascals. I particularly liked the fact that it is allegedly accurate to the nearest hundredth of a pascal.
The oddity is that standard sea level pressure for the earth is generally taken to be 101,325 pascals (no tenths or decimals) … so their number is off by a couple of percent. They describe their number as “mean atmospheric surface pressure” and “mean total surface pressure”, so it is unclear just what they think they are measuring. I don’t even know what “total surface pressure” might be.
Finally, whenever you see parameters given to seven significant figures, and the earth’s surface pressure given to two decimals, you know you are dealing with amateurs. That kind of fake precision is a big red flag at any time.
w.
The best discussion of the perpetual motion paradox that I have seen on this thread is the one by Robany here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871446
But I did not see any responses to him. Anyone have any comments?
Personally, I am having a lot of trouble with the whole PM concept for the Jelbring system. Gravity is a time invariant externally imposed field of force acting on mass. When did Newton’s second law of motion become a perpetual mobile of any kind?
Bill
Willis Eschenbach says:
January 19, 2012 at 10:32 pm
You are right and I was wrong. If the g force were constant, it would be isothermic. The g force is inverse square with distance, so the temperatures are a little different, but >99% of the difference from 0-30 km is pressure, and <1% is temperature. I don't have the exact equations, but I can do orders of magnitude.
“…assume that we have the airless [sunless] blackbody planet … The temperature of this theoretical planet is, of course, the theoretical S-B temperature.”
Which is what temperature? 2 K
“Now suppose we add an atmosphere to the planet, a transparent GHG-free atmosphere. If the theories of N&K and Jelbring are correct, the temperature of the planet will rise.”
So atmosphere would certainly be 2 K.
N&K and Jelbring are saying the atmosphere warms the planet?
“But when the temperature of a perfect blackbody planet rises … the surface radiation of that planet must rise as well.”
Well, a blackbody is imaginary.
And it’s suppose absorb all radiant energy of whatever wavelength and emit all energy at all wavelengths.
So blackbody which similar to earth wouldn’t have internal heat, because blackbody is suppose to be a perfect conductor of all energy- unlike rock. So if black surface is 2 K, it’s core planetary temperature is also 2 K. And so now a dinky atmosphere is suppose to warm the entire planet. A million hydrogen bomb would not warm that planet- does that mean nuclear bombs don’t create heat?
William Gilbert says: 7:42 pm
It seems to me that you have given insufficient consideration to the mechanical energy changes in Pressure P as a function of z and T. And doesn’t this formula apply to a constant mass? Therefore the volume is a function of pressure which is a function of z.
Some other people like Joel Shore have commented: There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave. It is not. I agree. The Adiabatic Lapse Rate only defines the critical threshold above which convection becomes possible.
The ALR is far more like the Critical Angle of Repose of a Pile. This defines the maximum angle (slope) of a pile before it becomes unstable and sloughs off.
ALR does not require that the temperatures reach this lapse rate. They only define a maximum of the lapse rate before the system will be unstable. A column of gas can exist at ½ the ALR and not convect, but it will still thermally conduct. So what will happen to this column if no energy is applied and it is thermally insulated from the outside? Will the temperature gradient increase making the bottom hotter and the top colder? No. It will become isothermal and its pressure distribution will change accordingly.
A physicist says:
January 21, 2012 at 2:36 pm
Tiring of of the Sysyphean task with the copper blocks, you lay a pair of insulated pipes up the mountain between your houses. You fill the pipes with a carefully selected refrigerant. At the top, the gaseous refrigerant is exposed to the cool mountain air in a condenser. This condenses the gas, yielding liquid refrigerant, with the waste heat warming her house.
The liquid refrigerant then flows by gravity down to the bottom of the hill in the first pipe. At the bottom it is evaporated in a standard air conditioner and cools your house.
At that point it is again a gas, and so it flows up the second pipe to the condenser at the top, driven by the counterbalancing weight of the liquid refrigerant in the other pipe that is pressurizing the expansion valve.
So now you have a modern refrigeration cycle operating without any effort necessary on your part, no need to move those 19th century copper blocks … and folks, that’s just one of many 21st century technology advances brought to you by TGC USA, a wholly owned subsidiary of ThermoGravitiCorp Ltd, a Cayman Islands Corporation. Invest now in the energy source of the future! Pick up the phone and dial 1-(800)-472-8489, that’s 1-(800)-GRAVITY, call right now for full details on where to send your checks, and remember, this opportunity won’t last!
w.
pegged that one, willis.
now we can see that there is a difference between static and dynamic.
tallbloke says:
January 21, 2012 at 3:59 pm
I partly agree. There is also rotational and vibrational energy.
That is true for a block of atmosphere that changes altitude without gaining or loosing energy.
However, thermal conduction from warm to cold means that energy will move from one block of atmosphere to another. As a result, the actual lapse rate will be much closer to zero than it is to the DALR. The question is – How much closer?
Let’s assume that the average speed of a molecule is extremely high. When a molecule moves either toward or away from the center of the Earth, it will undergo an acceleration for some time (dt). If the change in velocity due to gravity is a significant part of the average speed, then there should be a measurable induced lapse rate. However, I suspect (I have not computed this) that the gravitationally induced change in velocity will be at least nine orders of magnitude less than the average velocity (mainly because dt is extremely small). As a result, the gravitationally induced lapse rate will not be significant with respect to the current models. In other words, it would be indistinguishable from zero. (Of course, when the pressure gets low enough, dt increases and the lapse rate should become noticeable.)
More to the point, the DALR and the gravitationally induced lapse rate are created via totally different mechanisms. As a result, their magnitudes are not even close.
Stephen Rasey says: January 21, 2012 at 10:06 pm
“ALR does not require that the temperatures reach this lapse rate. They only define a maximum of the lapse rate before the system will be unstable. A column of gas can exist at ½ the ALR and not convect, but it will still thermally conduct.”
I agree with almost everything you said, but not this. I explained here how forced vertical motion in the stable regime acts as a heat pump, driving heat downward, and the gradient toward the ALR. And there’s plenty of wind to do the forcing.
Nick Stokes, hold on there. There is nothing forcing a vertical motion in my very short example. It is a closed, insolated system, a simple sealed column that we heat to 1/2 the ALR rate, seal it up, and then see what happens.
I agree with you, if you pump any system, especially with a day-night cycle, you will set up a non-isothermal atmosphere with a non-zero lapse rate. That is all the more reason I think “divide by 4” climate models are worse than useless; they lead you down a blind alley.
Anthony and Willis, I would love to see a post that looks into matter of how the suns radiation budget is used up from space to ground to space.
The title of the post can be: Two sides of the Albedo debate
Re: Joel Shore 9:25 pm
The question that has been bugging me for a year is that discussions seem to universally take the suns energy 1367 W/m^2, divide by 4 to remove night and day and make a thermally dead planet (which we now know has an isothermal atmosphere !! ) and get 342 W/m^2. We give the earth a 30% albedo, so only 70% make it way to the ground, about 240 W/m^2.
No, Willis, the planet Earth receives 342 W/m^2. Some of it bounces immediately off the Albedo. Most of it goes through and some of it bounces around a bit – because the Albedo layer has two sides: a top and a bottom.
I asked this question in a Jan 12 “Earth’s Baseline…” comment. Why do we account for albedo only on the direct-from-sun energy, but everyone seems to ignore it on surface-to-sky path? I set up a general formula where the top Albedo was A= 30% as accepted, but I allowed for an unknown Albedo reflectivity from the bottom as “a”. Using infinite series of decaying reflections…
So if A=a, then the heat at the ground would be 340 W/m^2 with 240W/m^2 of heat leaking through the layer. Add to that the 30% originally reflected and you have the original 340 W/m^2. Maybe “a” isn’t equal to A. But where is the proof that “a = 0”?
There is really no such thing as a “one-way mirror”. It really only a partially reflective two-way mirror. The Albedo layer over the earth acts the same way. But I fear many people treat it as a one-way valve, a diode, a mirror with only one side. There are two sides.
If I am wrong, then I want to establish prior claim to a new invention: “The Portable Backpackers Liquid Oxygen Generator.” Get a cardboard box, cover it in 99% reflecting Mylar (one way of course). Let’s see, 340 W/m^2 * (1 %) leaves 3.4 W/m2 inside the box. Yep, 3.4 W/m^2 equates to 88 deg K, and Oxygen liquefies at 90 deg K. (Oh, that was easy! Why hasn’t anyone thought of this before…)
[Blockquoting fixed. —w.]
Anthony and Willis, I would love to see a post that looks into matter of how the suns radiation budget is used up from space to ground to space.
The title of the post can be: Two sides of the Albedo debate
Re: Joel Shore 9:25 pm
The question that has been bugging me for a year is that discussions seem to universally take the suns energy 1367 W/m^2, divide by 4 to remove night and day and make a thermally dead planet (which we now know has an isothermal atmosphere !! ) and get 342 W/m^2. We give the earth a 30% albedo, so only 70% make it way to the ground, about 240 W/m^2.
No, Willis, the planet Earth receives 342 W/m^2. Some of it bounces immediately off the Albedo. Most of it goes through and some of it bounces around a bit – because the Albedo layer has two sides: a top and a bottom.
I asked this question in a Jan 12 “Earth’s Baseline…” comment. Why do we account for albedo only on the direct-from-sun energy, but everyone seems to ignore it on surface-to-sky path? I set up a general formula where the top Albedo was A= 30% as accepted, but I allowed for an unknown Albedo reflectivity from the bottom as “a”. Using infinite series of decaying reflections…
So if A=a, then the heat at the ground would be 340 W/m^2 with 240W/m^2 of heat leaking through the layer. Add to that the 30% originally reflected and you have the original 340 W/m^2. Maybe “a” isn’t equal to A. But where is the proof that “a = 0”?
There is really no such thing as a “one-way mirror”. It is really only a partially reflective two-way mirror. The Albedo layer over the earth acts the same way. But I fear many people treat it as a one-way valve, a diode, a mirror with only one side. There are two sides.
If I am wrong, then I want to establish claim to a new invention: “The Portable Backpackers Liquid Oxygen Generator.” Get a cardboard box, cover it in 99% reflecting Mylar (one way of course). Let’s see, 340 W/m^2 * (1 %) leaves 3.4 W/m2 inside the box. Yep, 3.4 W/m^2 equates to 88 deg K, and Oxygen liquefies at 90 deg K. (Oh, that was easy! Why hasn’t anyone thought of this before…)
A physicist says: January 21, 2012 at 2:36 pm
On the planet Stygian …
Next morning, exchange the now-warmed beach block for the now-cooled mountain block, and do it again. Which amounts to this: free heating, air-conditioning, and electrical power for you and your sister, forever, with zero work input! Thank you, thermo-gravitic theory!
Well, of course, you could do this on Earth. Which emphasises my point that it takes work to maintain the ALR. Energy that comes from the wind.
I should also respond to Joel Shore’s
“There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave.
…
If the lapse rate is naturally below that ceiling …, then such a lapse rate is perfectly acceptable. If the lapse rate exceeds that ceiling, then it is pushed down to the adiabatic lapse rate.”
No, there is a wind driven heat pump, which by forced motion in the stable regime, pushes the gradient up toward the ALR, exactly analogous to the convective engine operating when the lapse rate exceeds the ALR. It’s not a fallacy.
Stephen Rasey says:
January 21, 2012 at 11:40 pm
Stephen, the measured albedo of the planet is about 0.3. Because this is a measured figure, perforce it includes all “double reflections”. It includes reflections from the surface plus reflections from the clouds and aerosols minus any double reflections of the type you have noted.
As you point out, double reflectors will be present whenever there are clouds. As a seaman, I have seen this in the form of what we used to call “atoll blink”. This is where from over the horizon you can see the reflection of unseen lagoons on the underside of the clouds. The sun strikes the shallow water in the blue lagoon. That lovely turquoise blue light makes a curious glow on the underside of any clouds hanging over the lagoon. That light can guide the canny mariner home when the atoll can’t be seen.
But to answer your question, yes, it’s included.
Having said that, there’s a couple kinds of albedo. One is what is reflected back to the eyes of a viewer between the earth and the sun. Another, the bond albedo is what is reflected not just back into the observers eyes but in all directions. The bond albedo cannot be practically measured, you’d need a swarm of hundreds of satellites in a sphere around the earth to measure scattering in all possible directions. So it’s done with a big pile of computer modeling of the reflection. Do they include double reflections? I’d be shocked if they didn’t. I say this because to calculate bond albedo you have to calculate what’s happening at low grazing angles to the planet. At those angles, tall clouds shade large areas, including other clouds. In addition much more light is reflected from the surface, and has to fight through a lot of cloud to get out.
So it’s not a simple problem, and I suspect the scientists have not ignored the issue. I don’t have the references to hand, but if you google something like “double reflection” “bond albedo” “cloud” on google scholar and look for PDF papers, you’ll find something.
w.
Willis – you just described a solar powered version of the gas refrigerator.
http://home.howstuffworks.com/refrigerator5.htm
KevinK – Think about it for a second, there are lots of very obscure physical effects that have practical applications. For example, the Bernoulli effect is what makes a plane fly..
~~~
Most of the upward thrust comes from the downward deflection of air particles by the wing. That is why stunt planes can fly upside down:
http://www.cap-ny153.org/forceslift.htm
Robert Clemenzi says:
January 21, 2012 at 10:43 pm
Average speed of N2 at surface is 500m/s.
It has kinetic energy (ignoring rotational energy)
1/2 * 28 * 1.66 x 10^-27 kg * (500 m/s)^2 = 5.81×10^-21 J.
Average distance between molecules at surface is 10^-5 cm.
28 * 1.66 x 10^-27 kg * (9.8 m/(s^2)) * 10^-5 cm = 4.56×10^-32 J of kinetic energy change over 10^-5 cm.
Seems small, but a molecule at the bottom of the atmosphere transferring energy to another above and so on up to 10km gives 4.56×10^-21 J KE to PE change. That’s significant.
I think it doesn’t matter to the base gravity/mass theory if the atmosphere at equilibrium isothermic or stablizes with a lapse rate.
But I think Jelbring’s theory is cleaner and more intuitive if the atmosphere does stabilize at an isothermic condition.
If incoming and outgoing radiation is prohibited from the model world that means the equalizing of the atmosphere has to be internal. Thus the lower atmosphere has to be the source of warming the upper atmosphere proving that at equilibrium the lower atmosphere would be cooler.
Thus gravity and atmosphere mass are the main functions. The delayed response to equilibrium that breaks down the lapse rate via taking heat from a cooling lower atmosphere to a warming upper atmosphere provides the mechanism that gravity uses to raise the surface temperature.
The isothermic delay claimed by the isothermal crowd is evidence for the gravity/mass/friction relationship that pegs the lapse rate as its signature when the atmosphere is in motion. When you remove the external inputs it becomes apparent. The fact that conduction and convection are behind schedule is evidence of this relationship.
Jelbring’s model is ideal for teasing this out whether or not the atmosphere stabilizes at the lapse rate or continues on to isothermic equilibrium. If it just abruptly stopped you would have hopeless continuous perpetual motion mouth movement and perhaps its deserved. But when it continues to via an allegedly well understood physical process to an equilibrium the mechanism has not changed it just continues to work to obtain the equilibrium the physicists claim as fact. So convection and conduction continues its work albiet at a greatly lower rate of change now that the big 174 petawatt driver has been unplugged so it can catch up on its homework.
Tim Folkerts says:
January 21, 2012 at 5:15 pm
There are really only two possibilities.
1) The second law works and thus the temperature must be uniform.
2) KE + PE is constant, so the particles cool as they go up, and thus the temperature drops.
The laws of thermodynamics have been stated in terms of energy rather than heat for over a hundred years. The news hasn’t reached Duke University it seems.
I can give the “elevator speech” for why #2 is suspect. For any given trip, the KE + PE is indeed constant, but for different trips, the value is different (ie the boltzman distribution). If you look near the top, the “low energy tail” never gets that high, so you are only looking at self-selected particles that started in the high energy tail. These originally-high-energy-particles have indeed lost some KE on the way up, but they started with extra on average. This can (and does) leave this SUBSET of particle with the right average energy to be at the same temperature as the WHOLE set was at the bottom.”
I think this is wrong because of the very short mean free path length between collisions. Maxwell did a fair bit of work on that when he was considering the collisions and distributions of various sized gravel (analogous to differing KE in molecules) in the rings of Saturn (also sorted by gravity) and this is what led him to develop Clausius’ statistical mechanics. I’ve been reading up on this at the library in my University and will be writing an article sometime in the next few weeks.
I certainly haven’t proven this rigorously in this non-mathematical paragraph. However, it is clear the “lose KE and lose temperature” argument has a huge hole in it. Those who want to pursue this line of reasoning IN THE FACE OF STRONG COUNTER-ARGUMENTS, would need to determine the distribution of energies of particles at any altitude, and show that those remaining particles are indeed lower KE then the whole set was at the bottom. I am sure they can’t do this because I am sure they are wrong.
Well, do the maths and see if you get the same answer as Joe Born and Valesco. I’m certainly willing to concede the possibility that the gravito-thermal effect we are outlining might not account for the whole of the dry adiabatic lapse rate, but a substantial part of it. There may be some room in there for radiative effects too. It’s current ongoing work. Real science in real time.
I have to say I’m a bit surprised to see you using categorical statements like “This can (and does)” when you haven’t done the maths. Mind you, you are not alone…
Remember the path lengths when you get around to it. It’s important. Maxwell knew that. It may be why he never attempted a mathematical refutation of Loschmidt, but appealed to his own second law formulation. Boltzmann did attempt it, but never got a satifactory result. The controversy continues.
Willis,
Thanks a lot for your reply to this particular bozo. You say:
“I don’t know what mechanism Jelbring thinks will restore the initial heat differential, but he definitely thinks it will be restored.”
That’s the money quote. If Jelbring wants to contradict that, I’m all ears. Otherwise, I lean to the view (we non-physicists can only lean! :-)) that he is effectively proposing a perpetual motion machine is possible.
Sorry that you replied that way, Willis.
I think that if you put the two-sided Albedo up as a post, the ruckus following “Perpetuum Mobile” will be a “tempest in a teapot” compared to “Two-sided Albedo”.
In all seriousness, I think there are significant issues regarding reflectance from the cloud layers that deal with solar and reflected (converted) spectrums. That is why “A” is not necessarily equal to “a”?
And why does the backpacking liquid oxygen generator not work? (Yes! I know why it doesn’t, but what are we doing differently when we say the ground receives 240 W/m^2 ?)
Willis, I think your answer was just to a higher (google) authority. Hence, there is value in a separate discussion on a “Two Sided Albedo” to marshal the facts.
Re: Willis 12:17am. Stephen, the measured albedo of the planet is about 0.3. Because this is a measured figure, perforce it includes all “double reflections”. “Perforce?”
Do to the math, Willis! You will always measure from satelite 340 W/m^2 for any value of “a”. But I can drive the energy trapped in a ground-cloud wave guide to a lot of different levels based upon different values of “a”. I’m not even excluding “a” might be greater than “A” when you take into account converted spectrum from visible light into more infrared energy. I do not know what “a” is, but I have no reason to believe it is zero.
Surely the potential energy of the gas in the bottom is contained in pressure? Take gas from the top, lots of PE due to gravity, no pressure. Take an enclosed sample from the bottom, take it to a vacuum, and you can get work from its expansion, even though its start temperature is the same as the gas from the top. No temp difference required, and the energy sums the same.
DeWitt Payne says:
January 21, 2012 at 5:52 pm
Alan Millar,
With respect to Tallbloke, I’m quite aware that it’s futile. I have the same problem with Nasif Nahle when he insists that partial pressure is identical to partial pressure times path length. That causes him to think that the emissivity of CO2 in the atmosphere is low. It’s impossible to convince him otherwise because he has invested too much of his reputation in his hypothesis. Note that Tallbloke seems to think Nasif is a reliable source.
Thanks for the mind reading DeWitt.
I’m painfully aware that since the near fatal accident I was in I often get things wrong, and I’m less able to do complex algebra than I used to be. So instead I now concentrate my efforts on providing a platform for others to have their ideas discussed. Nasif’s idea was discussed. It was strongly criticised by John Eggert, who knows a thing or two, though no doubt you have your disagreements with him too. That’s science.
As an engineer, I was taught thermodynamics from a kinetics and molecular mechanics point of view. It seems Robert Brown is more a heat is work kinda guy. Nikolov and Zeller, Bill Gilbert and Joe Born do the maths. As a historian of science, I record their output and act as librarian ferreting out relevant old papers. As a blogger, I decide how to hold my space in the way I see best. No doubt there is subjectivity and personal bias involved, but I do my best to strike the balance I want to in the partisan atmosphere of the climate debating blogosphere. Sometimes I find myself in agreement with the direction WUWT thrusts in, sometimes not. Same goes for other science blogs.
Let the chips fall where they may, and the best science win.
Robert Clemenzi: “The DALR and the gravitationally induced lapse rate are created via totally different mechanisms. As a result, their magnitudes are not even close.”
If you put numbers into Velasco et al.’s Equation 8 for mean single-molecule kinetic energy as a function of altitude, the magnitude of the (non-zero) lapse rate there implied is indeed much less than the dry adiabatic lapse rate if the number of molecules is on the order of those we usually encounter.
Robert Clemenzi says:
January 21, 2012 at 10:43 pm
tallbloke says:
January 21, 2012 at 3:59 pm
Total energy = KE+PE.
I partly agree. There is also rotational and vibrational energy.
However, thermal conduction from warm to cold means that energy will move from one block of atmosphere to another. As a result, the actual lapse rate will be much closer to zero than it is to the DALR. The question is – How much closer?
Let’s assume that the average speed of a molecule is extremely high. When a molecule moves either toward or away from the center of the Earth, it will undergo an acceleration for some time (dt). If the change in velocity due to gravity is a significant part of the average speed, then there should be a measurable induced lapse rate. However, I suspect (I have not computed this) that the gravitationally induced change in velocity will be at least nine orders of magnitude less than the average velocity (mainly because dt is extremely small). As a result, the gravitationally induced lapse rate will not be significant with respect to the current models. In other words, it would be indistinguishable from zero. (Of course, when the pressure gets low enough, dt increases and the lapse rate should become noticeable.)
Thank you Robert Clemenzi.
I agree that as path lengths get longer, the effect will increase. However I disagree that conduction would take place at equilibrium for the reason I gave Robert Brown. In the Jelbring model at energetic equilibrium the proposed thermal gradient will not produce a flow of heat between contiguous air packets because the gradient passes through those air packets. Therefore the upper ‘surface’ of the lower packet will be at the same temperature (Same KE+PE) as the lower ‘surface’ (arbitrarily chosen measuring point) of the air packet above it. The two packets would have different average temperatures commensurate with the g/Cp relationship, but since the adjacent ‘surfaces’ are at the same temperature, no heat will flow because energy is in equilibrium.
in the real atmosphere, the input from the Sun leads to energetic heat transfer, conduction, convection and radiation. However, the underlying gravitational pressure gradient induced thermal effect means the dry adiabatic lapse rate will tend towards the gravity-pressure induced apse rate. This is indeed what observational evidence shows.
Robert Brown says
“Well done, but way too much work. The easiest PPM2 one can build is a simple thermocouple that lights a light bulb or turns an electrical motor. Place the upper contact in the supposedly cold air at the top of the static column. Place the lower contact in the warm air at the bottom. Insulate the heat transfer medium (e.g. silver) all the way down to the actual junction, so that heat is conducted from the hot bottom to the cold top, through the thermoelectric junction.”
Using DALR of 9.8K/km work out the figures for a real thermocouple connected by real copper leads using real resistance and heat loss figures and you will see why this proposed experiment is imposible.
If there is no outside influence on the planet, and no internal heat source, all atoms would reach absolute zero, so there would be no heat no matter where in the atmosphere atoms were situated, and therefore no difference in temperature top to bottom.
If there is a source of heat -the sun or from the planet core, the pressure of atoms would make temperatures higher at the bottom, fueling circulation which equalizes temperatures, but never quite gets there.
In the last case a heat engine would be possible, but in reality fueled by the heat source. So no perpetuum mobile.
Robert Brown:
If I understand your position, it’s that at equilibrium the laws of thermodynamics require heat to flow from hot to cold no matter how low the ratio of temperature gradient to potential gradient is. To me that is not a self-evident result of those laws. Maybe it is a result. But you have not demonstrated that it is. You have just argued as if the result were itself the law.
In contrast, Velasco et al. have provided a mathematical proof that it is not. They may be wrong, but at least they showed their reasoning, whereas you have not. Perhaps you don’t have the time. Fine. But know that this leaves the intelligent layman no choice but to credit Velasco et al., at least until he can find the error in that proof on his own.
Willis, your comment about supercritical gasses does not negate compressive heating which occurs in liquids and solids. It is certain that the liquid outer core of our planet is liquid due in part by compressive heating, the bulk heat is from radioactive decay of potassium, thorium and uranium in order of importance, and the lithosphere has areas of compressive melting at subduction zones that produce granite It is compressive adiabatic heat that enables Jupiter to radiate more heat than it receives despite the atmosphere being hydrogen and helium mix not known for their GHG credentials. This is not a PMM but a natural process gaining heat energy from gravitic work done. (water on the high pressure side of a pump is warmer than that in the low pressure side and water is ‘incompressible’).
As far as Venus is concerned there is such a violent atmosphere that mixing might change the supercritical properties of the mainly CO2 atmosphere. But compressive heating will still occur and more severely due to the high surface atmospheric pressure. If it were not for the presence of water on Earth we would have an atmosphere with similar properties to Venus. But this is the subject of a different discussion completely.
William Gilbert says:
January 21, 2012 at 9:27 pm
The best discussion of the perpetual motion paradox that I have seen on this thread is the one by Robany here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871446
But I did not see any responses to him. Anyone have any comments?
Personally, I am having a lot of trouble with the whole PM concept for the Jelbring system. Gravity is a time invariant externally imposed field of force acting on mass. When did Newton’s second law of motion become a perpetual mobile of any kind?
Bill
====
I hadn’t read it, my eyes glaze over fairly rapidly trying to read this thread, but it is what I was trying to say (not a scientist) about the gravity not being perpetual motion as energy being taken out of a natural cycle as we have it is bound to affect the cycle (and if this changes the component gases it could impact on the gravity field by changing it, by changing the potential gravitational energy available, say – can the enclosed system be scaled down?
Willis Eschenbach says:
January 21, 2012 at 9:26 pm
As an example of the arbitrary nature of the N&Z choice of values for surface pressure in their paper, they say the pressure of the earth is 98,888.20 pascals. I particularly liked the fact that it is allegedly accurate to the nearest hundredth of a pascal….
Finally, whenever you see parameters given to seven significant figures, and the earth’s surface pressure given to two decimals, you know you are dealing with amateurs. That kind of fake precision is a big red flag at any time.
At least they are using their arms to operate calculators instead of waving them around furiously.
Joel Shore says:
January 21, 2012 at 9:25 pm
These “theories” really have revealed some very embarrassing things about the AGW-skeptic movement…which, to their credit, some skeptical folks like Willis, Robert Brown, and Roy Spencer are trying desperately to correct, although it makes it all the more amazing that such crazy nonsense continues in light of their best efforts to squelch it.
Classic Joel Shore.
When you can’t successfully refute the theory, smear the proponents.
We know who is trying to squelch what Joel.
Independent of the question about the effect of gravity on temperature, N&Z have shown using empirical data and a better application of S-B that the average temperature of the surface of the Moon is ~90K cooler than previously thought. This is a mortal blow to Joel’s scientific beliefs, and he is desperate to get N&Z buried on any pretext.
Willis Eschenbach: “So I’m unclear why you think Velasco disagrees with me. If I understand it, which is always an open question, it seem they agree that the final distribution is isothermal. What am I missing?”
Let me address this more completely than I did above, where I said the Velasco et al. passage you quoted is ambiguous. As I said above, I resolved the ambiguity in favor of a non-zero lapse rate for finite numbers of molecules because that is what I read Velasco et al.’s Equation 8 to say.
Before I parse that quoted passage, though, I’ll briefly review what preceded it. Their Equation 5 gave state density as a function of velocity and altitude, Equation 6 gave it as a function of altitude only, and Equation 7 gave it as a function of velocity only. Since Equation 5’s left side was not the product of Equation 6’s and Equation 7’s left sides, Velasco et al. concluded in the first bullet point after Equation 7 that velocity and altitude are not statistically independent. They then presented Equation 8 to show the dependence of mean kinetic energy on altitude. This is why I resolved the ambiguity the way I did.
Now to the passage in question, against which we are to assess my interpretation of Velasco et al., which is that in an isolated system the lapse rate approaches zero as the number of molecules approaches infinity–but is non-zero for finite numbers of molecules. As you say, the Coombes & Laue paper that Velasco et al. were critiquing came to the conclusion that you and Robert Brown have, namely, that the equilibrium configuration would be isothermal, independently of the number of molecules. Here is how they characterized Coombes & Laue:
“The problem proposed and analysed by these authors is the following:
If a vertical column of an adiabatically enclosed ideal gas is in thermal equilibrium, is the temperature the same throughout the column or is there a temperature gradient along the direction of the gravitational field?
According to Coombes and Laue, there are two conflicting answers to the above question:
(1) The temperature is the same throughout because the system is in equilibrium.
(2) The temperature decreases with the height because of the following two reasons.
(a) Energy conservation implies that every molecule loses kinetic energy as it travels upward, so that the average kinetic energy of all molecules decreases with height.
(b) Temperature is proportional to the average molecular kinetic energy.
Coombes and Laue concluded that answer (1) is the correct one and answer (2) is wrong. They reached this conclusion after finding that statement (2a) is wrong, i.e., the average kinetic energy of all molecules does not decrease with the height even though the kinetic energy of each individual molecule does decrease with height.”
And here is the conclusion they drew:
“In conclusion, in our opinion a full explanation about why answer (2) to the paradox formulated by Coombes and Laue is wrong must discern between the cases of a finite system and an infinite system. In the former case, statement (2) is wrong because the assumption in statement (2b) is wrong. In the latter case, statement (2) is wrong because the conclusion in statement (2a) is wrong (as it has been established by Coombes and Laue).”
Until I had more carefully considered the equations that preceded it, I interpreted this passages as you did: I thought it meant that the finite and the infinite system are both isothermal.
But, after considering the preceding equations, I realize that when they say “statement (2)” they mean to include not just the lapse-rate conclusion but also the reasoning behind it; in the finite case, statement (2) is wrong not because its conclusion is wrong but because its reasoning is wrong. Note that for the finite case they don’t say that statement (2a), namely, that “the average kinetic energy of all molecules decreases with height,” is wrong. To do so would be inconsistent with their Equation 8. They say that only for the infinite case.
Willis Eschenbach;
As an example of the arbitrary nature of the N&Z choice of values for surface pressure in their paper, they say the pressure of the earth is 98,888.20 pascals. I particularly liked the fact that it is allegedly accurate to the nearest hundredth of a pascal.
The oddity is that standard sea level pressure for the earth is generally taken to be 101,325 pascals (no tenths or decimals) … so their number is off by a couple of percent.>>>
Until I really thought through N&Z in detail, I accepted 255K as the blackbody temperature of earth, and now realize I was wrong. That number is too high. I accepted that the average temperature of earth surface was 288K, and now realize I was wrong. That number is too low. I accepted that there has been a warming trend in the average temperature for the last 150 years or so, and now realize that some portion of that warming trend is due to the temperature of the earth becoming more uniform rather than from a net increase in energy.
So is the number of decimal points they’ve use a fair criticism? It is. Is the fact that the generaly accepted value is 101,325 rather than the 98,888 they calculate falsify their calculation? It does not.
Your cheer leader on this issue, Joel Shore, has already admitted that 288K is too low, though he maintains that is is by only a couple of degrees. The same logic applied to the calculation of 255K however, yields a value about 100K lower, a matter that Joel Shore refuses to engage in regard to. Bottom line however is that for two long accepted numbers that N&Z have refuted, Joel Shore has defacto stipulated to the innacuracy of both.
Did N&Z get the mean pressure calculation correct? Here is the important point on this, and why they only have two variables in their equation, not four. It doesn’t matter. If their method for deriving mean surface pressure is innacurate, then by all means substitute a better one. But at day’s end their work rests primarily upon correcting the misaplication of SB LAw to calculate effective blackbody temperature of earth, misaplication of averaging measured surface temperature to arrive at the blackbody radiance of earth surface, and the misaplication of trended average surface temperature data to conclude that there has been a long term energy net increase in energy retained on earth.
If they’ve erred in their calculation of mean surface pressure, then by all means, point out the error and suggest a fix. But don’t throw the baby out with the bath water.
The only problem with Perpetuum mobile is that nobody wants to buy it. I know how it should looks like : simple, cheap, small, very powerfull, very usefull. I also know how to build it for use in ships, for use in cars, for use for homes, etc … I believe there are several types of this engine, but it is not enough just that it works, it is also important how powerfull and usefull it could be. 100 kw at weight 100kg would make this engine usefull and comparable with Diesel and Otto engine. OK, 10kw at weight 100kg would be also usefull for cars, because it could produce the electricity 24 hours per day.
Davidmhoffer, I am happy to oblige, and you are entirely correct that “baby cherishing” and “bathwater discarding” both are necessary to good science and to rational skepticism.
(1) Nikolov & Zeller erred in neglecting radiative transport in the atmosphere.
(2) The symptom of the Nikolov & Zeller error is that (as is generically true of “gravito-thermal” formalisms) their model exhibits one of the following two flaws (depending upon details):
(a) the model either predicts an isothermal atmosphere
(which at odds with observation), or else
(b) the model violates the second law of thermodynamics
(3) The fix for the Nikolov & Zeller error is to incorporate radiation absorption and emission.
(4) Working through the details of this change, the standard GHE mechanisms are recovered.
Elevator Summary: The “baby to be cherished” is the practice of working through the details. The “bathwater to be discarded” is gravito-thermal formalisms that neglect radiative heat transport in the atmosphere: these theories are just plain wrong.
WHOA! GUYS!
You guys are so intent on arguing about the “end state” of Jelbring’s imaginary world you are missing the point.
The contention the atmosphere would be isothermal if it were isolated from any external heat source or sink is evidence that the earth’s lower atmosphere is warmer than it should be. (period)
That so precisely because of the gravity physics acting on expanding gases and a resistance to that response that causes world’s with solar input to display a lapse rate.
The disappearance of that lapse rate in Jelbring’s stygian world is irrelevant because it can only happen if you turn off the sun and if it doesn’t happen nothing is different except the precise mechanism of how gravity does it.
Thus the Perptuum Mobile argument is irrelevant to Jelbring’s theory, its only picking around the edges of the Jelbring paper.
davidmhoffer says:
Well, we don’t know what it means: It could be something as innocuous as trying to correct for the fact that not all of the “ground” is at sea level. On the other hand, another possibility is that they “tuned” the data in order to improve their fit. It is not a big issue, but it is a bit of a curiosity and I agree with Willis that there are too many sig figs there…which admittedly is pretty small on their list of sins.
That is not exactly true…You seem to switch between wanting the average T to be computed as the fourth root of the average of T^4 or as just a direct average according to how it suits your purpose. As far as I know, 288K is correct for the average of T. I have said that the fourth root of the average of T^4 is probably AT MOST 2K higher. Furthermore, since we are interested in the average of epsilon*sigma*T^4 to compute the average power and the effects of the non-uniform temperature distribution and of epsilon not being exactly 1 go in opposite directions, it is likely that the combined effect is even smaller.
That is even more untrue. I have engaged in discussing this in considerable detail, including why I believe your “100 K lower” claim is pure fantasy for any reasonable definition of and Earth without greenhouse gases but otherwise similar.
N&Z have not refuted anything. We never said that 33K was perfect, i.e., that it was 33.0000 K. And, we have always pointed out that the fundamental issue is the difference between the amount of power the Earth absorbs (~240 W/m^2) and the amount that its surface emits (~390 W/m^2) and that for a blackbody emitter absorbing 240 W/m^2 from the sun, the MAXIMUM average temperature that it can have without greenhouse gases is 255 K. It could be less if the temperature distribution is uneven.
tallbloke says:
Pure silliness. N&Z wave their hands around quite a bit. And, when they do operate calculators, they do so with silly, embarrassing errors like putting convection into a radiative model in a way that drives the troposphere to an isothermal state.
Furthermore, those of us who are refuting N&Z have done plenty of calculations themselves. I have not only replicated their fitting procedure but have gone on to investigate how well that procedure would fit altered data.
It is not smearing opponents. It is trying to make you guys understand how incredibly ignorant and silly you are making the “skeptic” community look in the eyes of the scientific community. Hey, but if that’s your goal, I’m okay with it!
This is not a mortal blow to any belief of mine. When people started arguing about the average temperature of the moon in Ira’s thread, I pretty much couldn’t care less and noted that this argument was really just a distraction from the real issues.
It is completely understood that Holder’s Inequality only sets a maximum limit on the average temperature that a blackbody can have given the average amount of power it emits. And, one consequence of this is that the “average temperature” one measures for the surface of an airless body like the moon is going to depend strongly on the exact definition of “surface” that one uses, which is why you see so many different numbers flying around. (This is a fact that N&Z have not come to grips with because they don’t understand how to correctly apply Conservation of Energy, as they made clear in part 1 of their reply. They think that “Conservation of Energy” says that a planet’s average temperature can’t change if its temperature distribution becomes more or less uniform, because they have failed to account for the fact that the planet is not an isolated system but is actually receiving energy from the sun. In fact, what Conservation of Energy correctly applied says is that any temperature distribution compatible with the condition that the planet is radiating back out into space as much power as it receives from the sun will be a steady-state condition.)
davidmhoffer says:
January 22, 2012 at 4:57 am
If they’ve erred in their calculation of mean surface pressure, then by all means, point out the error and suggest a fix. But don’t throw the baby out with the bath water.
I don’t think this is relevant. It’s not the variables in equations 7 and 8 that are the free parameters. It is the constant values they specified (those values could be anything). I also didn’t look close enough at the equation to see what was being done. So, on this point Willis and Joel are right.
However, if we look closely at the curve and also at Figures 5 and 6, we see something interesting. The curves match fairly closely. This means the so-called free parameters did not have to vary over a wide range to fit the curve. I think N&Z did themselves a disservice by trying to make a perfect fit. There is clearly a relationship that bears further investigation. This is the baby that needs to be viewed with more scutiny.
Nick Stokes says:
Nick,
Interesting…but I am still having trouble completely coming to grips with your argument. I understand how it is possible in principle for parts of the atmosphere to act as a heat pump…but I am having trouble understanding why you think that this drives the atmosphere up from a lower lapse rate to the adiabatic one. What is the driving force? (For an engine, I tend to think that the driving force is simply that heat wants to flow from hot to cold but it seems to me that a heat pump doesn’t have anything driving a certain temperature distribution…I turn on my air conditioner and what determines the final temperature in my house is simply the amount of electricity I run it with, e.g., the power of the air conditioner, its efficiency, and how long it runs.)
Also, how does this jive with the observed fact that the stratosphere is not being driven to the ALR? Or, are you saying that it is necessary to have things, such as in the troposphere, where part of it would be working as a heat engine (because convection is driving the lapse rate down) in order to have the part where it is acting as a heat pump?
John Marshall says:
Are you suggesting that coconuts migrate? (http://thinkexist.com/quotation/are-you-suggesting-that-coconuts-migrate/761764.html )
Srtioudly, are you saying that the Earth and its atmosphere are undergoing gravitational collapse and that this is producing 150 W/m^2? (And, that somehow all the satellite measurements that show that the Earth, as seen from space, is only emitting ~240 W/m^2 are wrong?)
tallbloke says:
January 22, 2012 at 3:41 am
“Independent of the question about the effect of gravity on temperature, N&Z have shown using empirical data and a better application of S-B that the average temperature of the surface of the Moon is ~90K cooler than previously thought. This is a mortal blow to Joel’s scientific beliefs, and he is desperate to get N&Z buried on any pretext.”
The avarage surface temperature of the moon was measured by two different Apollo missions in mid-latitude locations. It’s exactly what everyone thought it would be and what theory still predicts it should be which is 250K.
How do you explain your apparent discounting of experimental data? Do you believe both Apollo experiments were somehow flawed or what? In a nutshell – two different Apollo missions to the moon bored 3 meter deep holes in the regolith and then placed thermocouples at various depths from surface to bottom of the hole. The intent was to measure thermal conductivity of the regolith which is uber-important in the design of any habitats that may be constructed on the moon as it tells them how deep the structure needs to go to be immune from the large diurnal temperature swings and what the temperature of the regolith is once you’re deep enough so that it becomes constant.
As it turns out the constant temperature is 250K which is precisely what S-B predicts the average surface of the moon should be and the depth at which the temperature becomes constant is anything greater than 50 centimeters. The experiments continued transmitting temperature data to the earth for 4 years.
When some new hypothesis comes along from an undistinguished source that defies both long established laws of thermodynamics as well as repeated experimental data there’s invariably an explanation and that explanation is that the new hypothesis is wrong. This is crank science. Please stop dallying with it. It’s making you look bad.
That stratifying seems to square with 0th,1st, 2nd Law and ideal gas law. I don’t think Maxwell-Boltzmann distribution can be invoked in a gravity field since it deals with statistics of molecular movements w/o external forces. Here gravity is an external force. M-B can only be invoked if gravity is turned off.
That would be fine except that click through references to one straight up algebraic proof that you are incorrect has been offered, one reference to a textbook (Caballero) has been offered that both derives/explains the adiabatic lapse rate and has as an explicit end of section homework problem to prove otherwise, and finally your answer is inconsistent. If MB statistics were not valid in gravitational fields, how would they ever have been discovered? Is there somewhere Maxwell or Boltzmann could go where they were absent?
Therefore consider the “jar” argument once again. Bear in mind that the jar in question is one that is differentially small but large enough that it contains enough molecules that they can achieve “thermal equilibrium” (large compared to the mean free path). Grab a jar of air at the bottom of your “equilibrium” room with a DALR. Its pressure is a bit higher and temperature is a bit higher than air at the top. Grab a second jar of air at the top, where pressure and temperature are both a bit lower.
So far, we know nothing about the density of said air. Nothing in fluid dynamics requires a fluid to be compressible. Real fluids range from nearly incompressible water to highly compressible air — I don’t know how you want to idealize your “air” but let’s assume that it is moderately compressible. Even ideal fluids will almost certainly have a positive thermal expansion coefficient across the temperature range as well (a factor in the DALR) so we can assume safely enough that even an idealized fluid with a temperature lapse will have a density lapse as well.
So in the end, J_b from the bottom has T_b, P_b and N_b in volume V, where jar J_t from the top has T_t, P_t and N_t. T_b > T_t, P_b > P_t and N_b > N_t.
Gravity, however, is the same at the top and at the bottom. Moving the jar at the top to the bottom (remember, the fluid is in jars with rigid sides) does not change anything about the state of the fluid. Moving them anywhere does not change their state, as long as it is done gently enough that one doesn’t slam the fluid molecules around inside their jars (this is called “quasi-static” motion in thermodynamics, and is assumed for anything like an adiabatic process so we haven’t really introduced new assumptions there).
The fluid in the two jars is not in thermal equilibrium. Surely you agree? They are at two different temperatures. If the two jars have any sort of thermal pathway” opened between them, they will come to thermal equilibrium at a temperature in between T_b and T_t, one we can actually compute as it will depend on their heat capacities at constant volume which depend on N only (and in fact are equal to e.g. 3/2 Nk or 5/2 Nk for ideal monatomic or diatomic gases respectively).
Note well — extremely well, if you please — that the heat capacity of the gas gets no contribution from gravity. Moving a jar of air up or down in gravity does not change its ability to store heat. From this alone you should be able to see that gravity is decoupled from the problem, because your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect. dT/dh = 0 (where the derivative should be a partial derivative btw). Adiabatic lapse is a non-equilibrium state.
We surely agree that the two jars are not in equilibrium. Even if we leave them where they were in the first place and simply run a heat-superconducting wire between them, we expect heat to flow in the wire because they are not in thermal equilibrium and thermal equilibrium does not depend on where you are! The whole point of the zeroth law is that if I carry a thermometer with me it will predict whether or not heat would flow between any two reservoirs whose temperature I measure no matter where they are and no matter what the state is of each reservoir and whether or not the two systems are at all similar. I can measure air temperature in the Swiss Alps and water temperature in the North Carolina ocean, and if they are the same then I know that if I placed an “ideal” thermal conduit between the Alps and the Ocean no heat would flow.
Once again, you are stuck. You have a closed system that — you claim — is in thermal equilibrium with two different temperatures, one at the top and one at the bottom. Yet that means that an ordinary thermometer carried from the bottom to the top would read two values, and, just as would be the case for any two systems whose temperature is measured, means that heat would flow between them if it could.
a) It can. Air is a conductor of heat and heat will flow from the bottom to the top until the system is in equilibrium.
b) If you persist in arguing that it cannot, then you have established gravity as a Maxwell’s Demon for air. Please read about Maxwell’s Demon as it would greatly improve your understanding of detailed balance and why this argument ultimately microscopically fails.
c) Given the stable thermal gradient in this case, one can trivially construct a heat engine that does perpetual work moving heat from the bottom to the top and then “re-using” this energy after gravity sends it back to the bottom. That’s again a simple fact. One could run one of those little bobbing ducky things with a fluid with a high expansion coefficient in glass inside your container forever. Head down, it warms and squirts fluid through a tube to the other end (which is higher, hence cooler). Eventually the tail is heavier than the head, so it falls. Now the head cools and the tail warms to squirt the fluid back the other way. It falls again. Each cycle carries heat from bottom to top, but no matter, gravity will re-partition the heat again so you can use it over.
rgb
Drinking Bird:
http://en.wikipedia.org/wiki/Drinking_bird
rgb
@Robert Brown
In your isothermal arguments you are conflating temperture and energy. Shame on you. In the absence of gravity the column would be isothermal. However gravity induces a pressure gradient and this introduces a temperature gradient along with it. Total energy in the column however does not change and neither does the total energy in any horizontal layer. Total energy is a mix of gravitational potential energy and kinetic energy. As you move further up the column total kinetic energy declines and gravitational potential energy increases commensurately. “Heat” is energy of motion, measured by thermometers (“sensible”) but it isn’t the only kind of energy. The books balance just fine with a thermal gradient produced by a gravitational field. Gravity does not produce or reduce total energy, it doesn’t change the distribution of energy, it merely changes the form the energy takes on at different altitudes.
The isothermal/adiabatic distribution of a thermally isolated gas in a gravitational field has never been tested experimentally.
I was aware that the usual suggestions (such as a thermocouple or several thousand thermocouples with their real copper or silver connectors) don’t stack up when real values are calculated.
Stephen Rasey proposed to use two columns each with a different gas with widely different values of Cp.
I suggested two and realised that you could get a 2K difference between the two 100m tubes.
DeWitt Payne improved the model by suggesting another two gases that gave 7Kdifference.
It seemed all very practical and measurable.
I was convinced and did not need much persuading (as I tend toward the isothermal camp).
Why did someone not think of this test before, I thought?
However on reflection the test is questionable .
If the gases are allowed to mix at the top you have a one off heat transfer which cannot be repeated.
If the gases are kept separate by say copper plugs with a thermocouple in between I would like to see practical figures for gas to copper heat transfer before the matter becomes part of “settled science”.
William Gilberts post to Robert illustrated the wider context where he says:
“First, you talk extensively about “equilibrium” and “thermal equilibrium”. But I believe we should be talking specifically about “thermodynamic equilibrium”. Thermal equilibrium is but a subset of thermodynamic equilibrium. A system is in “thermodynamic equilibrium” when it is in thermal equilibrium, mechanical equilibrium, radiative equilibrium and chemical equilibrium. Thermodynamic equilibrium equals thermal equilibrium only when a system’s internal energy can be described by
U = CvT (1)
In this case the system’s internal energy is thermal energy and thermal equilibrium is all that matters. Your statement “Thermal equilibrium is isothermal, period” only applies to such a system. But that is not the system of a planetary atmosphere under the influence of a gravitational field.
Second, we need to better define where the classical laws of thermodynamics and equilibrium are valid. They are valid with a homogeneous system where all the locally defined intensive (e.g., per unit mass) variables are spatially invariant. But a system is not homogeneous if it is also affected by a time invariant externally imposed field of force, such as gravity. Thus in a gravitational field the laws of thermodynamics have to be applied in a manner that reflects the external field. This is a paradigm buster in itself.
Third, the thermodynamics of an atmosphere cannot be described wholly through considerations of heat transfer only (Trenberth diagram, anyone?). The atmosphere must be treated as a system undergoing both heat and mass transfer. Mass transfer is the reason that gravity is so important in understanding atmospheric thermodynamics. This gets to the mechanical equilibrium part of thermodynamic equilibrium.
Fourth, chemical equilibrium is also very important in atmospheric thermodynamics. But if you assume uniform molecular diffusivity in a horizontal layer (gravitational potential energy handles the vertical diffusion) we are left with latent heat and that is beyond our discussion here.”
Robert Brown and Joel Shore both pointed out that at times of an atmospheric inversion the atmosphere above the earth surface can be at a higher temperature than the surface.
But what happens then?
the radiative flux from the surface is more intense than the radiative flux from the atmosphere and so the atmosphere fails to stop the surface cooling further.
The heat flow is still from the surface to the atmosphere.
Radiation is not heat.
Temperature is not heat.
Until a fully implemented experiment or at least fully calculated presentation with real figures is supplied I for one will keep an open mind.
A Physicist;
(1) Nikolov & Zeller erred in neglecting radiative transport in the atmosphere.
(2) The symptom of the Nikolov & Zeller error is that (as is generically true of “gravito-thermal” formalisms) their model exhibits one of the following two flaws (depending upon details):
(a) the model either predicts an isothermal atmosphere
(which at odds with observation), or else
(b) the model violates the second law of thermodynamics
(3) The fix for the Nikolov & Zeller error is to incorporate radiation absorption and emission.
(4) Working through the details of this change, the standard GHE mechanisms are recovered.
——————
A grandiose case of circular reasoning. Your position is pretty simple. In your mind it is impossible for the earth surface to be higher than the effective blackbody temperature without back radiation (GHE). You conclude from your belief system that any formula that doesn’t attribute the higher than effective blackbody temperature of earth to GHE must therefore by default be wrong and to violate the laws of thermodynamics. Wrong.
Consider a nice hot day in the tropics. The sun comes up at dawn and the temperature is already at 20C or 293K. Over the course of the day, the temperature increases by 20K to 313K. After it peaks, it begins to cool again, falling back to 293K by the following morning. Do we need back radiation to explain this? No. Note that I’m not saying that back radiation isn’t part of the equation, but I am saying that you do not need back radiation to explain this. If you try an stick to “average” insolation and “average” temperature to understand what happens every single day, you will get large numbers to attribute to GHE and they will be wrong. The following is illustrative.
If we assume an average insolation of 240 w/m2, then insolation couldn’t possibly maintain the temperature in the scenario above at 20C,let alone increase it during the course of the day to 40C. 240 w/m2 equates to only -18C, so if all we were relying upon was “average” insolation, the earth would actually cool, not warm, even during the day. Do we need back radiation or GHE to explain this rise in temperature that occurs every day on earth despite this? We do not.
The insolation that the tropics is exposed to every day ranges from 0 w/m2 at night, to 1,000 w/m2 at noon. What is the blackbody equilibrium temperature of earth at 1,000 w/m2?
Answer: 364K or 91C
Question: Does the earth surface ever get to 91C?
Answer: No
Question: If GHE raises the temperature of earth surface to a temperature higher than the blackbody temperature expected from insolation, why doesn’t the temperature go even higher than 91C?
Answer: I know what you are thinking right now. You can blame this conundrum on conduction and convection cooling the earth surface. Back to circular reasoning again. You can’t claim that the surface of the earth is at a temperature higher than blackbody because of GHE and also claim that it doesn’t get to blackbody at all DESPITE GHE. Further, there is a far simpler explanation: heat capacity.
Explanation:
The earth surface does not and cannot respond to changes in insoltion instantaneously. The earth surface has a heat capacity, and for any given change in insolation, there must be a time constant applied to determing how long it will take to warm the earth surface to equilibrium.
The temperature of the earth surface in the tropics when exposed to 1,000 w/m2 never gets to 91C because it doesn’t have TIME to get to 91C. Further, at night when insolation is zero, it doesn’t cool off to absolute zero because it doesn’t have TIME to cool that much. The “average” daily temperature can easily be maintained at well above the non existant imaginary 255K without a single watt from GHE provided that we STOP averaging insolation and treating it like a fixed number and instead accept that it varies wildly every single day and when combined with the heat capacity and time constant that governs warming and cooling, explains most if not all the the surface temperature over and above the supposed 255K.
Do convection, conduction, and back radiation get involved in the process? Of course they do. But is GHE required to raise the temperature of the earth above 255K?
No.
N&Z did not break the laws of thermodynamics, and what they showed is NOT that they ignored GHE, but that properly calculated, GHE is insignificant.
Joel Shore says:
January 22, 2012 at 6:56 am
“It is not smearing opponents. It is trying to make you guys understand how incredibly ignorant and silly you are making the “skeptic” community look in the eyes of the scientific community.”
Really. And who exactly is this “scientific” community and how does its defintion manage to exclude the “skeptic” community. There are many scientifically literate people who are skeptical of the alarmist narrative. Some of those skeptics even have the politically correct credentials that inbred academic bigots regard as credibility markers. I find you characterization insulting and a disservice science everywhere. Science is a discipline that may be practiced by anyone at any time and it is also a discipline which can be abandoned by anyone at anytime especially when power and tribal politics are a factor in the practioner’s behavior.
RichardM;
I don’t think this is relevant. It’s not the variables in equations 7 and 8 that are the free parameters. It is the constant values they specified (those values could be anything). I also didn’t look close enough at the equation to see what was being done. So, on this point Willis and Joel are right.>>>
The point is that if you arrive at the mean surface pressure of the various planets by another means, and plug those values into N&Z, you should either get the same results at they did, or, if you get different results, then either the way you calculated mean surface pressure is wrong, or the way they did is. Or I suppose, both could be wrong.
But I don’t see anyone jumping up and showing that the mean surface pressures of the various planets are appreciably different from what they calculated. Oddly, would that not be the easiest way to falsify at least that part of their work?
But allow me to suggest an alternative. If you understand what they are trying to get at over all, is there any reason they could not have used atmospheric weight (note that I said weight, not mass) instead of mean surface pressure?
Joel Shore;
The same logic applied to the calculation of 255K however, yields a value about 100K lower, a matter that Joel Shore refuses to engage in regard to.
That is even more untrue. I have engaged in discussing this in considerable detail, including why I believe your “100 K lower” claim is pure fantasy for any reasonable definition of and Earth without greenhouse gases but otherwise similar.>>>
Sir, I have provided you with sample insolation curves and the blackbody calculations derived from them by doing the calculation over time and shown quite conclusively that it is easily possible to have a daily fluctuation of insolation from 0 to 850 w/m2 that averages 240 w/m2 and yields an average temperature over time of 140K. Your rebuttal amounted to screaming “that’s impossible”. Hardly a refutation. Read my comment upthread about heat capacity and time constants.
Robert Brown says at 1/22 8:04am:
“We surely agree that the two jars are not in equilibrium.”
Robert – Thank you for the well prepared & thought out reply. It is excellent & consistent with your earlier discussion. It will take a bit to compose a prepared response with the given reading assignments – I will have to double check them.
My view continues that the jars are in energy equilibrium with Willis’ gas system 1 before they are sealed, T stratified & no energy flows in the equilibrium cv up to the instant of sealing – after sealing their energy equilibrium becomes a second & third body equilibrium w/gas & as such energy can be made to flow thereafter. Basically, count the control volumes, there are two more cv.s after jar sealing. Willis’ premise only allows 1 cv, the answer has to be consistent with 1 cv and nothing ever crosses the 1 cv whilst forming conclusions. This is always hard.
This thread made me aware that apparently even the thermo grand masters had different views of Willis’ premise prior to the sealing of the jars. As Tallbloke wrote – that’s science. Might be a l-o-o-o-ng thread for good reason – “Waiting for Godot “ Act II quote: “In the meantime let us try and converse calmly, since we are incapable of keeping silent.”
Heat can be emitted by the ground has IR or has a transfer of momentum to the gas above. IRs can be back-radiated by certain gases. The upward momentum of a particle will be back-radiated by gravity. In the end, the heat emitted by the ground in both forms will be back-radiated to the ground. Can anyone tell me the difference between the 2 kinds of back-radiation?
Also, if a gas particle is accelerated downward and hits a CO2 particle, this CO2 particle is more likely to emit an IR photon.
The way to calculate the adiabatic lapse rate in a no emissivity atmosphere is to define to invisible surface at a slightly different height, and calculate the amount of momentum that goes through these layers. You have 4 cases:
1- A particle passes through the lower layer, does not interact in between, and then goes through the top layer.
2- A particle passes through the top layer, does not interact in between, and then goes through the lower layer.
3- A particle passes through the lower layer, does not interact in between, does not have enough momentum to make it to the top layer, so it goes back through the lower layer again.
4- Multiple particles hit each others in between.
If you calculate the amount of momentum that goes through those 2 layers. You will find that 1, 2 and 3 show a higher sum of momentum at the lower layer. 4 is more tricky, but I believe it can be simplified by taking the center of mass the multiple particles. But in reality, you also have to consider the individual amount of momentum in relation to this center of mass. But I think it is inconsequential, I might be wrong.
In the case of a planet with no sun and a no emissivity atmosphere. The planet would slowly cool to space. The top of the atmosphere would always be cooler than the surface of the planet. Eventually, the top of the atmosphere would rain down to the ground. In the end, the planet would get to a very low temperature. Using a thermocouple in this atmosphere could potentially accelerate the cooling because most thermocouple are made of materials with a non-null emissivity.
tallbloke says:
January 22, 2012 at 2:22 am
Think of a metal rod with a heat source at one end and something very cold at the other. If you divide this rod into slices, you could make the same argument – that heat can not flow because “adjacent ‘surfaces’ are at the same temperature”. Yet, we know that heat does flow. This is because each “slice” is not at some temperature, but because all slices have a temperature gradient.
Using the numbers provided by Scot Allen, January 22, 2012 at 12:27 am
Average speed surface is 500m/s
Average distance between molecules at surface is 10^-5 cm.
10^-5 cm * (1 m/ 100 cm) / (500 m/s) = 2E-10 seconds between collisions
2E-10 s * 9.8 m/s2 = 1.96E-9 m/s — the maximum change in speed due to gravity
(1.96E-9 m/s) / (500 m/s) = 3.92E-12
(500 m/s + 1.96E-9 m/s) ^2 – (500 m/s) ^2 ~= 2 * 500 * 1.96E-9 = 1.96E-6
Assuming 1/2 kT = 1/2 mv^2, then T = m/k * v^2 and
T2/T1 = v2^2 / v1^2
T2 = T1 * (v2^2 / v1^2)
So the question becomes – What is v2? A linear sum of the maximum velocity difference is obviously wrong since more energy will be transferred by molecules traveling parallel to the surface than by those traveling perpendicular to it.
Assuming a linear sum of the maximum values (the wrong approach)
(1.96E-9 m/s) / (1,000 m / 1E-7 m) = 19.6 m/s change every kilometer
T2 = 300K * (500 + 19.6)^2 / 500^2 = 324K – whoa, that’s humungous
Some other examples, just moving the decimal point
T2 = 300K * (500 + 1.96)^2 / 500^2 = 302.4K
T2 = 300K * (500 + 0.196)^2 / 500^2 = 300.2K
Arcsin(0.10) = 5.74 degrees
Arcsin(0.01) = 0.573 degrees
Granted, half a degree is a pretty small skim angle, but this at least puts numbers to how I understand the problem. Over a long enough period of time, an even shallower angle would transfer enough energy to create an isothermal atmosphere (at least in the limit of infinite time and no turbulence).
Bryan says on January 22, 2012 at 8:21 am:
“The isothermal/adiabatic distribution of a thermally isolated gas in a gravitational field has never been tested experimentally.”
Hmmm. Lets see what history tell us:
“Fourier (1824, p. 153)
It is difficult to know how far the atmosphere influences the mean temperature of the globe; and in this examination we are no longer guided by a regular mathematical theory. It is to the celebrated traveller, M. de Saussure, that we are indebted for a capital experiment, which appears to throw some light on this question. The experiment consists in exposing to the rays of the sun, a vessel covered with one or more plates of glass, very transparent, and placed at some distance one above the other. The interior of the vessel is furnished with a thick covering of black cork, proper for receiving and preserving heat. The heated air is contained in all parts, both in the interior of the vessel and in the spaces between the plates. Thermometers placed in the vessel itself and in the intervals above, mark the degree of heat in each space. This instrument was placed in the sun about noon, and the thermometer in the vessel was seen to rise to 70°, 80°, 100°, 110°, (Reaumur,) and upwards. The thermometers placed in the intervals between the glass plates indicated much lower degrees of heat, and the heat decreased from the bottom of the vessel to the highest interval.The effect of solar heat upon air confined within transparent coverings, has long since been observed. The object of the apparatus we have just described, is to carry the acquired heat to its maximum; and especially to compare the effect of the solar ray upon very high mountains, with what is observed in plains below. This experiment is chiefly worthy of remark on account of the just and extensive inferences drawn
Fourier (1824, p. 154)
from it by the inventor. It has been repeated several times at Paris and Edinburgh, and with analogous results”
By the way Tallbloke has done us all a big favour by posting “Fourier 1824 as translated by Burgess 1837” on his blog, so go read it and perhaps learn – a lot.
His (Fourier 1824) pagenumbers may not match mine, but that should not be a big issue
davidmhoffer says: 1/12 at 8:30 am
Keep it up, David. One very minor correction: “what happens every single day” isn’t really correct because in the average insolation, there is no night and there is no day. 😉
All this analysis based upon a “divide by 4”, average insolation, must stop. It is a dead end on a dead planet. Heat capacity is vital to the study of the temperature history of the earth. Here is a link to arguments I have made before on WUWT with traceback to climateetc since summer 2011.
It is time to bury the Dead Planet Model.
David, the short answer is “Yes, only back-radiation can explain this.”
Because if we miraculously turned-off all back-radiation in the atmosphere, then the first night would be very cold (as it is in the dry desert at night), then during the following days and nights each new cumulus cloud, thunderstorm, or hurricane, would dump heat into the upper atmosphere that could *never* radiate away, with the result that within a few weeks all such storms would cease, with new clouds formed during the day at ground-level unable to rise through the now-heated upper air.
New patterns of global air circulation would then emerge, between the (now-cold) tropic latitudes and the (now even colder) polar latitudes … the resulting global weather patterns on such a non-GHE planet might be very interesting, but they would not much resemble the weather patterns of our planet.
The weather, I think, would end up looking very much like the weather of Snowball Earth a world of cold foggy days and colder nights even at the equator, a world without hurricanes or thunderstorms, a world with ever-growing polar ice caps, progressing until the entire world was locked in ice.
Robert Clemenzi says:
January 22, 2012 at 9:03 am
tallbloke says:
January 22, 2012 at 2:22 am
The two packets would have different average temperatures commensurate with the g/Cp relationship, but since the adjacent ‘surfaces’ are at the same temperature, no heat will flow because energy is in equilibrium.
Think of a metal rod with a heat source at one end and something very cold at the other. If you divide this rod into slices, you could make the same argument – that heat can not flow because “adjacent ‘surfaces’ are at the same temperature”. Yet, we know that heat does flow. This is because each “slice” is not at some temperature, but because all slices have a temperature gradient.
Sure, if you create an energy differential between the ends, heat will flow. My point is that Jelbring’s hypothesis (and Loschmidt and Laplace and Lagrange and all of classical mechanics) say that there will be a thermal gradient at energy equilibrium. If energy is in equilibrium, then no heat will flow, despite the fact that there is a thermal gradient.
dynamic equilibrium means there is a flux.
static equilibrium means there is no flux.
this simple device called a ‘word’ can really help sort out thing if you don’t post normally contort it at whim.
a rubber ruler can’t be a useful standard.
davidmhoffer says:
January 22, 2012 at 8:30 am
Skin temperature measured by satellite reaches 70C. Nice global maps in the link below. Soil temperature can get much warmer than that below the surface and reaches a maximum about a quarter inch deep in dry sand. FYI.
http://www.cienciayclima.es/files-pdf/Hot-places-on-earth.pdf
O H Dahlsveen
I said
“The isothermal/adiabatic distribution of a thermally isolated gas in a gravitational field has never been tested experimentally.”
You said
“Hmmm. Lets see what history tell us:
“Fourier (1824, p. 153)”
Your interesting passage from Fourier did not deal with a thermally isolated gas .
Which means no heat enters or leaves the gas sample.
Re: Joules Verne -9:34 am Mildrexler 2011 map.
Cool Map! Uh.. Hot Map!… Uh.. A keeper in any case.
Joules Verne says:
January 22, 2012 at 7:59 am
The avarage surface temperature of the moon was measured by two different Apollo missions in mid-latitude locations. It’s exactly what everyone thought it would be and what theory still predicts it should be which is 250K.
So it wasn’t the average temperature of the Moon which was measured, it was two mid-latitude locations. Extrapolating from that using a misapplication of the S-B law doesn’t give you the average temperature.
How do you explain your apparent discounting of experimental data? Do you believe both Apollo experiments were somehow flawed or what?
Not at all. N&Z use the old apollo data as well as the new data which shows colder than expected high latitudes. Have you read their paper?
“Data obtained during the LRO commissioning phase reveal that the Moon has one of the most
thermal environments in the solar system. Surface temperatures at low latitudes soar to 390K
(+117C) around noon while plummeting to 90-95K (-181C), i.e. almost to the boiling point of liquid
oxygen, during the long lunar night (Fig. 2). Remotely sensed temperatures in the equatorial region
agree very well with direct measurement conducted on the lunar surface at 26.1o N by the Apollo 15
mission in early 1970s (see Huang 2008). In the polar regions, within permanently shadowed areas
of large impact craters, Diviner has measured some of the coldest temperatures ever observed on a
celestial body, i.e. down to 25K-35K (-238C to -248C). ”
When some new hypothesis comes along from an undistinguished source that defies both long established laws of thermodynamics as well as repeated experimental data there’s invariably an explanation and that explanation is that the new hypothesis is wrong. This is crank science. Please stop dallying with it. It’s making you look bad.
They bullied people who supported the discoverers of the ulcer causing stomach bacteria too. The discoverer of plate tectonics was told he was a crank as well.
I’ll be providing a mathematical proof of the misapplication of S-B on my site soon. Watch this space.
Joel Shore says: January 22, 2012 at 7:18 am
“What is the driving force?”
Joel, the driving force is the wind, which generates isotropic turbulence and forces air to move up and down, even against buoyancy resistance in the stable regime. Air forced down warms faster than ambient, acquiring an upward buoyancy. Its momentum (and local pressure gradients) keep it moving for a while (taking KE from the air). When it arrives it
diffuses heat to ambient; this is heat that has been pumped down against the gradient. The amount of heat is proportional (for a given motion) to the difference between the LR and the DALR. The effect is to drive the gradient toward the DALR, and the strength of the pump is proportional to (DALR-LR).
Upward motion pumps heat in the same direction (down).
When LR exceeds DALR, signs change; heat is pumped up, and KE is added to the air. This is the restoring mechanism you identified there. They are dual processes.
It’s like shovelling sand up a slope where it’s sliding down. If you can shovel hard enough you can maintain the slope. If not, the sand spreads until the slope reaches a lower value where the sliding matches your shovelling rate. Since the pump has finite power and its rate drops as LR ⇒ DALR, it can’t force LR all the way to DALR.
As the air thins, the effectiveness of the pump diminishes relative to conduction, so the actual LR gets further from DALR. The perturbing role of the heat source of absorbed insolation (and the sink of IR emission to space) also becomes relatively greater.
I can’t believe you guys are just continuing merrily to cruise along when I have ripped the fundamental foundation for Anthropogenic Global Warming right out of the ground. My latest amended proposal:
Bart’s Law:
It follows that surface temperature sensitivity to the addition of radiative gases is negative: the addition of more IR radiative gases will tend to lower the surface temperature. This is a sensitivity holding everything else equal – it does not preclude the possibility of feedback from other processes tending to resist the change, e.g., cloud albedo effects.
Bart’s Law is justified by the following line of reasoning:
What happens when you add additional IR gases, which are the lowest energy emitters, to the atmosphere? Without them, you would have continued increasing surface temperatures until higher energy emitters limited the surface temperature. Therefore, they have limited the temperature to less than it would have been. It follows that, if you add more of them, you will pin the surface temperatures down to a lower level.
IR emitters in the atmosphere do not heat the surface. They limit the surface to a lower temperature than otherwise would be the case. Adding more of them will further cool the surface.
Gravity can also concentrate heat with the use of photons. A photon gains energy when it travels downward in a gravity well. In fact, a black hole can prevent heat from escaping completely. It seems that gravity will concentrate energy more easily using particles with a high ratio of momentum to kinetic energy.
And other question, when you measure the radiation from a CO2 particle, how do you know if it participates to warming? The energy could have been previously absorbed from above. So this energy could be half of a reemission associated with a back-radiation to space(cooling). An example is the condensation of water vapor. If you have a lot of greenhouse gases between the condensation layer and the ground. Those gases with back radiate the energy to space. So the energy emitted to the ground is associated with a cooling by greenhouse gases.
To reinforce what David Hoffer Stephen Rasey are saying about what the Earths heat capacity and rotation does to maximumum and minimum temperatures.
The rotational effect is very pronounced here.
This is a new paper paper by Nikolov and Zeller
Posted on Tallbloke.
Have a look at the temperature /time graph of various locations on the Moon.
Eyeballing the graph and by removing the middle 27 Earth day periods we get an almost horizontal (little temperature variation) plot.
Pick say the temperature at the equator.
The explanation of the Holder inequality is worth a look
http://tallbloke.files.wordpress.com/2012/01/utc_blog_reply_part-1.pdf
@Nick Stokes: Nick, your heat pump idea made for something good to ponder while I was playing ultimate frisbee today…and I think I have a better understanding of it now. Let me put it into my own words and see if you agree that this is what you are talking about:
A heat pump / refrigerator / air conditioner pumps heat from cold to hot using an expansion-compression cycle. So, what you are saying is that the variation with pressure with height that occurs for an atmosphere in a gravitational field provides the setup for such a heat pump, with air at higher pressure at the ground and lower pressure further up. All that is needed is something to provide the actual “pumping” action. And, your point is that the pumping action will in practice be provided by any differential heating that occurs due to the diurnal cycle or the variation in insolation with latitude, presumably coupled with something like orographic effects to turn the horizontal motions into vertical motion.
Do I have the basic picture? If so, that’s pretty cool…and I have definitely learned something new! Have you thought about how important this effect is likely to be…E.g., it seems to me that, in the absence of greenhouse gases, an atmosphere will have a lapse rate somewhere between the isothermal distribution that is its equilibrium and the adiabatic lapse rate that is favored by this “heat pump” effect, but it is not clear to me where on that scale it is actually likely to fall.
By the way, for the benefit of those who want to violate the Laws of Thermodynamics: I will just emphasize a point that I am quite confident Nick will agree with, which is that whether or not there is a lapse rate is irrelevant to the average temperature (or really the average power emission) that a surface would have to have in the absence of a greenhouse effect. I.e., the constraint on the temperature has to be at the surface in this case because that is what is communicating (via radiation) with the rest of the universe.
Tallbloke;
I’ll be providing a mathematical proof of the misapplication of S-B on my site soon. Watch this space.>>>
To limit my misuse of time on the internet I long ago made a deal with myself to participate in one an one only forum. Please provide that mathematical proof as soon as you can and I will break my three year old deal with myself on the spot. My initial reaction to N&Z was that they were in dreamland. I had to let go of some long held beliefs, but I did so because their arguments make a ton of sense, and if the ONLY think that comes out of this discussion is that the misapplication of SB Law gets exposed and discredited once and for all, science will have made a giant leap forward.
Accepted Views of Their Time:
o the earth is flat
o the sun circles the earth
o bumps on your head determine your mental health
o disease can be cured by letting some of your blood out
o witches exist
o witches float
o warts are caused by touching toads
o volcanoes will cease erupting if your throw enough virgins into them
o the blackbody temperature of the earth is 255K
o the surface temperature of the earth is 288K
o the only mechanism by which a higher than “average” blackbody surface temperature of the earth can be achieved is through back radiation of GHG’s.
They all have one thing in common. They are wrong. Well…maybe not the part about witches because I believe I have a personal relationship with one and she does indeed float, though she disputes being a witch.
On a rotating uneven sphere under a single sun with a non GHG atmosphere there would be huge dayside and nightside temperature differentials producing very strong winds.All the energy exchanges between surface and atmosphere would be via conduction and convection involving ALL the molecules of the atmosphere whatever their radiative characteristics.
There would be enough mixing to bring warmed upper air down to the cold ground on the night side which would smooth out both heating and cooling around the planet.
The pressure at the surface would still combine with solar input to give the bog standard adiabatic lapse rate (or Atmospheric Thermal Effect) for a planet of that mass and atmospheric pressure. The air circulation of the planet would simply restructure itself around the ATE.
Just as every planet with an atmosphere of any composition always has done and always will do.
You see, an atmosphere structured around ATE is the only possible stable structure. Unless it achieves ATE then the situation is unstable and the atmosphere either boils off or congeals on the ground eventually.
The Ideal Gas Laws have never been falsified.
GHGs wholly unnecessary.
Robert Brown says at 1/22 8:04am:
“…one reference to a textbook (Caballero) has been offered that both derives/explains the adiabatic lapse rate…”
Replying succinctly as I chug thru my reading assignments, these are exact Caballero quotes you can text search for the rest of the context:
Caballero: 2.1: “Note that pressure is due to only to the local properties of the gas and not to anything going on far away.”, 2.1: “..if the gas as a whole is at rest…”, 2.11: “..an ideal gas with an isotropic velocity Distribution..”, in 2.2: “…we assume that the gas has no mean motion… yields the Maxwell-Boltzmann distribution…”
And so on. Under above assumptions, find non-g field gas column is isothermal. Don’t remember I ever wrote Willis’ premise cv gas would be stratified without presence of that inexplicable gravity (g) field. The gravity field is not consistent w/all the assumptions above, because now Willis’ GHG-free adiabatic tall gas column is not at rest, it is accelerating at g w/o an isotropic velocity distribution.
As I advised Willis upthread, I read down thru Cabellero Chapter 4 s-l-o-w-l-y. I scanned thru the rest of the chapters. Do not think I missed Caballero extending calculations to add explicit Willis premise including g unless Caballero was writing on the full atmosphere which he says must stratify in T (i.e. must have a lapse rate). If I missed it, you are more familiar with his work, please advise. It makes sense the full atmosphere should collapse to Willis’ premise with stratification of T inside his 1 cv of interest.
Robert continues:
“If MB statistics were not valid in gravitational fields, how would they ever have been discovered? Is there somewhere Maxwell or Boltzmann could go where they were absent?”
Thought problems like Willis’ premise have worked before! Thought problems did work pretty well for Einstein also. Here, Caballero tells us above that M-B distribution was discovered when “we assume that the gas has no mean motion… yields the Maxwell-Boltzmann distribution.” M-B has not gone on to apply to gas in gravity field in this work, maybe Tallbloke’s trip to the library will find a reference.
Of course, an accelerated gas will have a mean motion, special unaccelerated M-B is thus not applicable though it might be a subset of more general M-B and maybe that is where N&Z are going. Willis’ premise is extending at least my science understanding & in part, causing this thread length.
Robert continues:
“…and has as an explicit end of section homework problem to prove otherwise, and finally your answer is inconsistent.”
Found the jet & scuba diver, but need a more precise pointer to Willis’ cv problem here Robert and I’ll dig into my homework assignment.
And thanks again Robert, that Caballero ref. was great on 1st read and today’s second read – should be very useful again down thread.
Bart says:
Bart: Here is the elevator speech as to why your reasoning is wrong —
Greenhouse gases both absorb and emit radiation. However, for an atmosphere where the temperature decreases with increasing height (which is what the atmosphere will tend to when populated by greenhouse gases…and which convection can only partially cancel), the amount absorbed will be less than the amount emitted back out into space. This means that the net effect will be to allow the average surface temperature to exceed any possible average temperature that would produce radiative balance in the absence of an IR-absorbing atmosphere.
This is all well-understood stuff. You may be very smart but you are not quite as smart as you think you are (a statement alas true about a lot of people around here).
Bart,
I’ve got it in a sightly different form:
The atmosphere of any planet will restructure itself regardless of composition to produce the ATE determined by mass and solar input.
The critical thing to note about our hypothetical atmosphere is that it is at equilibrium. By definition, we cannot extract any additional energy out of this system provided that it is the lowest possible energy state. If it was not in the lowest possible energy state, we could be envisioning an atmosphere that is actually metastable. To clarify, consider a metastable atmosphere (dG = 0) where G = G1 and an alternative, attainable, and stable atmosphere (dG = 0) where G = G2. If G1 > G2, we could envision building a device that could extract energy from the system until it reached the lower energy state. Bear in mind, our hypothetical atmospheric system was initially at equilibrium dG1 = 0. And, at the end of the process, it is once again at equilibrium dG2 = 0. The key thing to note is that our ability to extract energy from the original system will cease once it reached the new equilibrium position. Hence, arguments for an isothermal atmosphere that are based soley on the ability to build a perpetual motion machine are non sequitur.
The challenge that we have is that our equation describing the equilibrium free energy of our system simply reads:
dG = -(CvlnT + Rln(RT/P) + Sm0)*dT + (RT/P)*dP + Mgdx = 0
In and of itself, this equation does not provide us with a definitive answer to the question. It does tell us that the free energy of the atmosphere must be uniform. However, there is no unique solution to this equation. There are a class of solutions if we (perhaps arbitrarily as a mathematician might view it) set dT = 0, dP = 0, or dT = dP = 0. We may even have a valid class of solutions where dT = dP or dT ~= dP. Of these five cases, the only absolutely safe bet is that dT = dP = 0 will not be a useful solution, since this trivial solution would imply the absence of an atmosphere. The other possible solutions are what I think form the basis for claims of the potential of a stable atmosphere which is not isothermal. From a purely mathematical point of view, we only know that the energy distribution must be uniform. Our equation does not dictate that the temperature must be uniform. This is about the point where I would anticipate a mathematician saying, “a solution to your question exists,” and immediately walk away.
As physical scientists, we immediately apply the temperature formulation of the zeroth law, because every system that we know of approaches an equivalent temperature state (e.g. Newton’s and Fourier’s law). In such a case, a valid solution to the equilibrium equation would be given as: P = P0*exp[-Mgx/(RT)]. This solution suggests that there will be a pressure gradient within an isothermal atmosphere. Paradoxically, every gaseous system not subject to a gravitational field also approaches an equivalent pressure state (refer to Figure 6-10 (b) of Klotz & Rosenberg:
Klotz & Rosenberg go on to describe a way to build a machine that will extract energy out of such a system. Hence, we could envision building a perpetual motion machine of the second kind to exploit the pressure differential in our hypothetical atmosphere. The resolution to this apparent paradox was realized by Clausius, who provided the most rigorous definition of the second law: “The energy of the universe is constant; the entropy tends toward a maximum.” In short, the application of the gravitational field may confound our intuition.
My scientific approach is best described as that of an empiricist. To quote Lord Kelvin, “when you measure you know.” It sounds like there are no definitive (non-thought) experiments addressing this question; if anyone knows of such experiments, please provide a link (or reference) to the original work. If not, then I would be intrigued by the results of the following experiment:
1. Place a gas in a thermally insulated cylinder of sufficient length,
2. Place this cylinder into a centrifuge,
3. Ensure that the gas is both isothermal and isobaric such that it has reached a state of equilibrium,
4. Turn on the centrifuge,
5. Allow a sufficient amount of time to pass to ensure that the system has achieved a new state of equilibrium.
The tricky part will getting an accurate measurement of the temperature distribution along the cylinder without disrupting the equilibrium state. We would also want to approach the equilibrium state from multiple directions to ensure that we did not happen to hit upon a metastable equilibrium. Assuming it is possible to devise such a system, the results will provide not only the answer the original question, but also validate (or refute) a fundamental law of thermodynamics when applied to systems subject to a gravitational field. The null hypothesis for this experiment is dT = 0. Any statistically significant deviation from dT = 0 would imply that the temperature formulation of the zeroth law is a limiting case. In short, a statically significant result would require a major paradigm shift. I’m not quite ready to discard what has been accepted for more than a century, so I’ll anticipate a result of dT = 0, even though it has a 1/inf (purely mathematical) probability of being the most thermodynamically stable energy state.
Bryan says:
January 22, 2012 at 2:25 am
I love folks like you, Bryan. They ask people to do their work for them.
But what the heck. I’ll bite let’s suppose the thermocouple has a resistance of Rt, and the connecting wire has a resistance of Rw. Let’s suppose that the thermocouple puts out a voltage V for a temperature difference ∆T
Now, the total resistance of a series circuit is R, + R2 + … + Rn. In this case the resistance is Rt + Rw.
Now Mr. Ohm comes to our aid with his famous E = I R equation. That tells us that Voltage = Current times Resistance.
The current flowing in the circuit under discussion is therefore
Now Bryan, I want you to notice something. As long as those variable have real values, A CURRENT WILL FLOW. It doesn’t matter if the wire is made out of steel and the voltage is tiny. Doesn’t matter if the silver wire doesn’t conduct much heat. If the components have real, non-zero values, a current will flow.
So the experiment is not “impossible” as you claim. There’s no reason to use real values, because the values don’t matter. It doesn’t produce much power, but that wasn’t Robert said. So your claim, that the experiment is “impossible” is falsified.
w.
Willis Eschenbach says
“Now Bryan, I want you to notice something. As long as those variable have real values, A CURRENT WILL FLOW. It doesn’t matter if the wire is made out of steel and the voltage is tiny. A current will flow. So the experiment is not “impossible” as you claim. There’s no reason to use real values, because the values don’t matter. It doesn’t produce much power, but that wasn’t Robert said. So your claim, that the experiment is “impossible” is falsified.”
Well Willis, I think you realise that if you put real values in, the current could not be measured.
So as far as Modern Physics is concerned if it cannot be measured you cannot say it exists.
So all you are left with is a conjecture..
Stephen Wilde says:
For the 1 millionth time: the lapse rate does not uniquely determine the surface temperature. You also need the temperature at one particular height. That height is the height from which radiation can successfully escape into space. For a planet without a greenhouse effect, said height is necessarily the surface of the planet.
Now, I have it. If the atmosphere is very thick in IR frequencies that interacts with CO2. Those photons do not travel a long distance. Let’s say this distance is 100m. It means a CO2 particle absorbs IR from 100m below and 100m above. Since the temperature is nearly the same over 200m, the emission of this CO2 particle is nearly as much a back radiation to the ground as a back radiation to space. It would not hold true at the top of the atmosphere because there is little radiation from above in the IR.
Trick, a coffee cup on a table is experiencing the force of gravity, but it is not moving in the table’s frame of reference, therefore it is not accelerating. (I’m not going to get into the Earth as a rotational frame of reference.)
A gas column will be stratified by density in a gravitational field, but at equibrium it is a statics problem where all forces cancel out and bulk acceleration is zero. Gravity can cause acceleration, but it doesn’t require it.
davidmhoffer says:
Sir, I have provided you with sample insolation curves and the blackbody calculations derived from them by doing the calculation over time and shown quite conclusively that it is easily possible to have a daily fluctuation of insolation from 0 to 850 w/m2 that averages 240 w/m2 and yields an average temperature over time of 140K. Your rebuttal amounted to screaming “that’s impossible”. Hardly a refutation.
You are attacking strawman arguments of your own devising. I have not said that it is impossible. I have said that the approximation that local temperature is determined by demanding radiative balance with the local insolation is a very poor approximation for the temperature of a planet with any significant atmosphere, rotation, etc.
But…It is also largely irrelevant. Let’s say I am wrong on this point and that an Earth without a greenhouse effect really would have such a dramatically broader temperature distribution that it would have an average temperature of 140 K. What does that prove? It proves nothing. You seem to think that we now have some much larger temperature discrepancy to explain, but the explanation for it is no big mystery to anyone but N&Z, you, tallbloke, and their other groupies: It is simply that lots of different temperature distributions with average surface temperature less than 255 K are compatible with the constraint that the surface must emit an average of 240 W/m^2.
Where? I am not going to search this whole thread.
O H Dahlsveen says:
January 22, 2012 at 9:17 am
I’d love to … but I’m banned from the talkshop for my evil ways. Is it posted anywhere else?
w.
davidmhoffer says:
January 22, 2012 at 8:39 am
The point is that if you arrive at the mean surface pressure of the various planets by another means, and plug those values into N&Z, you should either get the same results at they did, or, if you get different results, then either the way you calculated mean surface pressure is wrong, or the way they did is. Or I suppose, both could be wrong.
But I don’t see anyone jumping up and showing that the mean surface pressures of the various planets are appreciably different from what they calculated. Oddly, would that not be the easiest way to falsify at least that part of their work?
But allow me to suggest an alternative. If you understand what they are trying to get at over all, is there any reason they could not have used atmospheric weight (note that I said weight, not mass) instead of mean surface pressure?
We’re in agreement that there is a relationship of surface pressure to temperature. That was the point of the rest of my comment that you didn’t copy. However, it’s not useful to argue against the fact that they did use multiple free parameters or exact values. Instead, it’s better to look at those free parameters and argue that they really weren’t completely free and that the result of most of them should be testable (as you indicated).
Also, I still feel the presence of GHGs is most likely required. I feel it’s a non-issue as GHGs were present in all the atmospheres that were compared. The pressure-temperature relationship is the important one.
Read Nick Stokes comment above about turbulence and the relationship to the lapse rate. I think he’s on the right track and this is why the N&Z relationship exists.
@Joules Verne and tallbloke: The average surface temperature of the moon is the perfect debating point if your goal is to never resolve anything. You can debate this all day but unless you carefully define how the average is taken (e.g., are you looking at the first fraction of a mm or at the first meter of the surface), you’ll get all sorts of different answers!
And, this is precisely because there are lots of different temperature distributions having lots of different average temperatures that are all compatible with the moon being in radiative balance.
Misaplication of SB Law
Conventional calculation for “average” insolation of 240 w/m2 = 255K
Practical Application of SB Law in hourly increments over 24 hours beginning at midnight, the following insolation curve in w/m2 is illustrative:
0, 0, 0, 0, 0, 0, 100, 150, 350, 650, 800, 850, 850, 800, 650, 350, 150, 100, 0, 0, 0, 0, 0, 0
Applying SB Law to the hourly increments we arrive at equilibrium temperatures in degrees K of:
0, 0, 0, 0, 0, 0, 205, 227, 280, 327, 345, 350, 350, 345, 327, 280, 227, 205, 0, 0, 0, 0, 0, 0
Averaging both these series OVER TIME we arrive at:
“Average” insolation = 242 w/m2
“Average” SB Law equilibrium temperature = 144K
REBUTTAL
Since I already know the tune, I’ll play the next verse in advance. My detractors will immediately scream something to the effect that this is unrealistic and use the moon as an example where, even with a 14 “day” night and a 14 “day” daytime, extremes of either O or 350K are not achieved, hence there must be something wrong with my model as presented.
RESPONSE
Why of course! Did you think the heat capacity of the moon was zero?
Joel Shore;
Let’s say I am wrong on this point and that an Earth without a greenhouse effect really would have such a dramatically broader temperature distribution that it would have an average temperature of 140 K. What does that prove? It proves nothing. >>>
You mean other than attributing 33K to the GHE is totally and completely wrong? Other than demonstrating that the observed temperatures cannot possibly be achieved by GHE alone and that leaving heat capacity and time constant of the earth out of the equation is totally and completely wrong? Other than showing that much of the observed positive trend in temperatures globally can be attributed to the temperature of the earth becoming more uniform rather than the existence of an actual energy imbalance due to increases in GHG’s?
well, I guess if you skip all of those, then yup, it proves nothing.
“For a planet without a greenhouse effect, said height is necessarily the surface of the planet”
Irrelevant because energy still gets into the air via conduction and convection and back to the surface via convection and conduction for then radiating out. So there will still be a lapse rate and it must match ATE otherwise the system is unstable. If it doesn’t match ATE then the atmosphere will accumulate energy via conduction and convection indefinitely until it boiled away or lose energy to the ground via conduction and convection indefinitely until it congealed on the surface.
The true Perpetuum Mobile is the concept of a planet with an atmosphere that is not precisely in equilibrium with its ATE.
Any disequilibrium will either boil off the atmosphere or congeal it on the ground.Once congealed on the ground it would be lost via sublimation.
The radiative GHG theory is itself a Perpetuum Mobile because it proposes that changing the composition changes ATE. Thus a bit more human GHGs are amplified by more water vapour and that gives more GHGs which amplifies again ad infinitum.
The atmosphere and oceans would get hotter and hotter until they boiled away.
We would see lots more planet sized bodies with no atmospheres at all because a little change in atmospheric composition would have been enough to destabilise it.
IIf we were to reduce GHGs so that they changed the ATE lapse rate the other way then the Earth would get steadily colder until the oceans and air congealed on the ground.
The system won’t allow it. Any change in composition that might introduce a disequilibrium with ATE is neutralised by a reconfiguring of the circulation pattern.
If you could find one planet where the Gas Laws do not apply then you would have me. Where is it ?
davidmhoffer says:
Various things essentially borrowed from Postma and then…
You are wrong. Indeed, GLOBAL radiative balance requires that you do need GHGs. Giving local examples to illustrate why you think you don’t have to does not work…Yes, on a local scale, we can have energy storage and energy moved from one place to another. However, globally, we can’t have the Earth emitting 76.5 TerraWatts (150 W/m^2 times the surface area of the Earth) more than it is emitting day-in and day-out. If it were doing so, it would be rapidly cooling until it was no longer doing so. End of story.
davidmhoffer says:
No…It is not wrong. It is correct as a result of Holder’s Inequality.
We’re not leaving them out. Do you understand what an inequality is? It tells you that something can’t be bigger than something else. The temperature of a blackbody emitter receiving and average of 240 W/m^2 cannot be greater than 255 K. It can be less, depending on the temperature distribution. Is that such a hard concept to understand?
Joel,
Yes, I think that’s it. You can make a cycle – suppose hydrostatic atmosphere, LR=.5*DALR. Take a balloon with 1kg air, and pull it down 100m. T rises to ambient+0.5, and work had to be done against buoyancy. Hold it until T is back to ambient. 500J moved against gradient. Then pull it back up. It cools to ambient-0.5. Hold till ambient. 500J flows in.
I’ve done some quantification here, in entropy terms. The gradient produces a rate of entropy increase
E_V = (k/T^2) ∇T • ∇T
k=conductivity. The heat pump has to counter that. As the LR rises toward DALR, the pumping diminishes to zero, and the entropy production increases. Somewhere there is a balance point.
And yes, I do agree that without GHG the lapse rate is the same but irrelevant to heat balance and surface temperature. The air does not then interact with the outgoing heat flux. Though there is the possibility of liquefaction 🙁
Joel Shore says:
January 22, 2012 at 10:46 am
“However, for an atmosphere where the temperature decreases with increasing height (which is what the atmosphere will tend to when populated by greenhouse gases…and which convection can only partially cancel), the amount absorbed will be less than the amount emitted back out into space.”
It does not matter. Back-radiation allows the surface temperature to rise, which begets more back-radiation. This is classic positive feedback. Stability ensues when the overarching negative T^4 feedback balances the positive feedback.
“This means that the net effect will be to allow the average surface temperature to exceed any possible average temperature that would produce radiative balance in the absence of an IR-absorbing atmosphere.”
This statement is in no way in conflict with my Law. In the steady state, there is no difference between what I am saying, and what the greenhouse hypothesis says.
But, the greenhouse hypothesis is unstable, because reducing the back-radiation reduces the temperature, which reduces the back-radiation, and so on. There is nothing to prevent reversion to the ground state.
Stephen Wilde says:
January 22, 2012 at 10:46 am
“The atmosphere of any planet will restructure itself regardless of composition to produce the ATE determined by mass and solar input.”
I don’t think that assertion has adequate support at this time. But, my logic above is compelling. If we could establish a common cause, I would advise we bridge the moat before we assault the castle walls.
A Stirling engine, with the hot side a sea level, and the cold side at altitude is the Perpetuum Mobile being discussed. Clearly such a device could be constructed today. The question is then, what is the mechanism that makes this possible?
We all agree that solar energy warms the surface air, which expands and rises. The reason it rises is because the atmosphere has mass and because of gravity. If there was no gravity, there would be no “up” direction for the air to rise.
As the air rises, it cools because of gravity, as kinetic energy is transformed into potential energy. There are of course other mechanisms of cooling, but the specific mechanism by which (dry) rising air cools must be due to conversion of kinetic energy to potential energy. This is inherent in the formula for lapse rate, which relies on the gravitational constant.
Now, as the air cools it contracts and becomes more dense. It will descent as a result. The reason it descents is because of its mass and gravity. If there was no gravity, there would be no “down” direction for the air to descent.
Thus, all that is required for our Stirling engine to operate as a Perpetuum Mobile is an atmosphere with mass, solar energy, and Gravity and one other ingredient: TURBULENCE.
The model described above will not work in a work without turbulence. The air would simply rise until it reached a steady state equilibrium and there would sit. Convection would come to an end. What is required to set the Perpetuum Mobile in motion is a disturbing force. On earth we have this in abundance, due to rotation, latitude, topography, etc.
Something that does not exists in the cylinder of air model.
Joules Verne said @ January 22, 2012 at 8:38 am
If we accept that the scientific community consists of just those who accept the laws of science as true, then sceptics are certainly not excluded from the scientific community. Sceptics who refuse to accept the validity of the laws of science are excluding themselves from the scientific community. I find it’s best to avoid attributing malice to someone’s statements, when a charitable interpretation of the statements can be made.
[snip – already? That’s a separate paper and separate post – confine your discussion to this one – Anthony]
Stephen Wilde says:
No…It is not irrelevant. It is what determines the temperature of the planet’s surface, not some magical “ATE” that violates the laws of physics.
Strawman argument. The ideal gas law does not imply your result. Not even close.
I must catch a plane for a conference, and will be out of pocket for the next several days. Despite Joel’s misunderstanding of feedback dynamics, my logic is ineluctable. Temperatures will rise until radiative emissions stop them from rising. More IR gases will reduce the level they are allowed to rise.
If, when I am able to check back, you guys still haven’t gotten it, and this thread has gone stale, I will have to post it again in another thread until serious people begin to take my challenge seriously. DeWitt? Tim? Where are you guys?
davidmhoffer says: January 22, 2012 at 11:24 am
“You mean other than attributing 33K to the GHE is totally and completely wrong?”
People sometimes say that Earth with the same atmosphere free of GHG would be 33K cooler. What you and N&Z claim is that Earth with no atmosphere at all would be some different temperature. This is a completely different proposition and, in my view, of very little interest.
REPLY: Nick and I find ourselves in rare agreement – Anthony
davidmhoffer says on January 22. 2012 at 8:30 am.
“Further, at night when insolation is zero, it doesn’t cool off to absolute zero because it doesn’t have TIME to cool that much. The “average” daily temperature can easily be maintained at well above the non existant imaginary 255K without a single ——-”
=========
I like your answer and have no “gripes” but would like to add,- if I may:
People who build “Climate Models” by putting their imaginary “Model Planet” in a situation that exists nowhere in this Solar System or anywhere else, as far as I know, i.e. surrounded by “multiple little suns” – or for any other reason are giving their planet 24 hour sunshine, say by averaging “The Solar Constant” to cover the complete sphere 24/7 -, do not have any chance of getting the correct answers. The two main reasons for me saying this are:
1) Earths “Solar Irradiadiation” is halved – or all attempts to find all the values necessary for correct calculations – are abandoned.
2) There is no night in their models. Therefore the efforts made, by earlier scientists, to establish the “Earth’s Surface Cooling Rate” at various latitudes are lost. –
As you say, the Earth turns and for every second that passes – the left, or eastern, part of its surface presents a “new”, small part of itself that is far warmer than zero Kelvin (0 K) to the Sun’s rays.
thepompousgit says:
January 22, 2012 at 12:02 pm
If we accept that the scientific community consists of just those who accept the laws of science
Theory and hypothesis are not scientific law. Even the laws are not absolute. They exist only so long as there is not contradictory information.
Those that believe the laws of science are written in stone are not practicing science, they are worshiping science.
Fourier said in 1824:
“The presence of the atmosphere and bodies of water, has the general effect to render the distribution of heat more uniform. In the ocean and in the lakes, the coldest particles, or rather those whose density is the greatest, are continually tending downwards, and the motion of heat depending on this cause is much more rapid than that which takes place in solid masses in consequence of their connecting power. The mathematical examination of this effect would require exact and numerous observations. These would enable us to understand how this internal motion prevents the internal heat of the globe from becoming sensible in deep waters.”
It looks to me like:
Attributing 33K to the GHE is a bit like creating enough fuel for all “Perpetuum Mobile Machines”
“Applying SB Law to the hourly increments we arrive at equilibrium temperatures in degrees K of:
0, 0, 0, 0, 0, 0, 205, 227, 280, 327, 345, 350, 350, 345, 327, 280, 227, 205, 0, 0, 0, 0, 0, 0
Averaging both these series OVER TIME we arrive at:
“Average” insolation = 242 w/m2
“Average” SB Law equilibrium temperature = 144K
….
RESPONSE
Why of course! Did you think the heat capacity of the moon was zero?”
So what is the average heat capacity of earth per square meter?
And once you add an atmosphere and ocean this heat capacity increases enormously.
Instead of: 0, 0, 0, 0, 0, 0, 205, 227, 280, 327, 345, 350, 350, 345, 327, 280, 227, 205, 0, 0, 0, 0, 0, 0 one gets something like 285 K repeated 24 times.
And all this is talking about a average spot on earth, and I would guess that is somewhere around sun light and air temperature of 38 degree latitude.
I excited about averaging, but once you start down road, one continue and average everything.
A post not posted:
“So average amount sunlight energy on earth is said to be 240 watts per square meter, and average temperature of earth is said to be 15 C. Using this data, where is the average place on earth?
So need place that averages 15 C, nite and day, summer and winter. And it gets 240 watts times 24 hrs. Which is 5760 watts per square meter in 24 hrs.
I haven’t looked. Any guesses?
So, we look at solar map. Google: solar map:
So basically southwestern states in US- southern Spain, middle east. southern part Australia, Northern countries in South America, much of India.
And I live in this region. I live somewhat close to LA. LA downtown over last century yearly average temperature have been as high as near 69 F and near 62 F as lowest:
http://www.climatestations.com/images/stories/los-angeles/lacvann2.gif
Or 69 F is 20 C and 62 F is 17 C. So it’s a bit too warm for average.
Sacramento: 60.8 F [16 C]
San Francisco: 57.1 [13.9 C]
Norfolk – Virginia: 59.6 °F [15 C]
The average temperature in Melbourne, Australia is 14.8 °C (59 °F).
Anyway average of 15 C is near or lower than 40 degree latitude.”
Nick Stokes says:
January 22, 2012 at 12:12 pm
This is a completely different proposition and, in my view, of very little interest.
I disagree. What we have in our discussion of climate is a single data point on an unknown function. That data point is planet earth. The more data points we can add, the more likely we are to arrive at the correct description of climate. By necessity this involves resolving the effects of no atmosphere versus atmosphere, as not all world have atmospheres.
ferd berple says:
January 22, 2012 at 12:04 pm
[snip – already? That’s a separate paper and separate post – confine your discussion to this one – Anthony]
I’m sorry Anthony but what I have I missed? I made reference to the “Jelbring hypothesis” which was the subject discuss in this post. Here is the sentence in the post:
“I was reading the Jelbring hypothesis this afternoon, which claims that greenhouse gases (GHGs) “
davidmhoffer said @ January 22, 2012 at 10:26 am
re Stefan and Boltzmann
I find myself saddened that there are some in this debate who will believe this farrago of nonsense.
We’re not leaving them out. Do you understand what an inequality is? It tells you that something can’t be bigger than something else. The temperature of a blackbody emitter receiving and average of 240 W/m^2 cannot be greater than 255 K. It can be less, depending on the temperature distribution. Is that such a hard concept to understand?>>>>
Sigh. No, that’s not what it means. It means the AVERAGE cannot be higher. The actual temperatures can vary substantially both higher and lower. The average of those temperatures cannot be higher than 255K.
As seen from space, the outbound LW radiance cannot average more than 255K. BUT, the outbound LW radiance does not and cannot originate wholy from earth surface, or from any single altitude. Assuming the existance of GHG’s which, including water vapour, ozone, and so on we have plenty of, a small amount of radiance must originate at earth surface and an increasingly large amount of radiance must originate from increasing altitudes. The “average” of these must equal 240 w/m2 in order to be in equilibrium with the “average” 240 w/m2 being absorbed. In other words, some of the photons originate from a surface temperature higher than 240 w/m2 and some from a layer lower than 240 w/m2.
A quick glance at AMSU shows this to be the case. At an elevation of 14,000 feet, the temperature is about 253K. At altitudes lower than that, the temperature is higher. At altitudes higher than that the temperature decreases until about 60,000 feet and then rises again to about 253K at 135,000 feet. Since each and every layer MUST radiate, and SOME portion of the radiance from each layer MUST exit the atmosphere to space, and since the amount radiated MUST equal 240 w/m2 in order to achieve equilibrium, the fact that MOST of the air column is well BELOW 255K, and that upward bound LW has the highest possibility of escaping the earth’s atmosphere from the HIGH altitudes (because there is less in the road to potentially intercept and absorb them) the only possible conclusion is that the earth surface MUST be warmer than the average in order to balance the much colder than 255K temps in the upper atmosphere.
Add in heat capacity and you have all you need to account for the disparity. Does that mean the effects of GHG’s are zero? No. But it also means you CANNOT take 255 subtracted from 288 and attribute it to GHG’s alone.
Ferd berple said @ January 22, 2012 at 12:17 pm
Please read what I wrote Ferd; do not put words in my mouth. I said nothing about theory and hypothesis. Nor did I say that the laws of science are absolute. I said earlier in this thread: “Please note that I am not saying that they will remain as Immutable Laws for Ever and Ever Amen. Paradigm shifts do occur. But every paradigm shift I have studied was enabled by people who clearly understood the paradigm they were overturning.”
Nick Stokes says:
January 22, 2012 at 12:12 pm
davidmhoffer says: January 22, 2012 at 11:24 am
“You mean other than attributing 33K to the GHE is totally and completely wrong?”
People sometimes say that Earth with the same atmosphere free of GHG would be 33K cooler. What you and N&Z claim is that Earth with no atmosphere at all would be some different temperature. This is a completely different proposition and, in my view, of very little interest.
REPLY: Nick and I find ourselves in rare agreement – Anthony
***********************
Then help me out guys. If we don’t know what the temperature of the earth is with no atmosphere, and we don’t adjust the temperature profile that results for heat capacity and latency (time constant), then on what basis do we quantify atmopsheric effects?
Willis Eschenbach says on January 22, 2012 at 11:01 am:
“O H Dahlsveen says:
January 22, 2012 at 9:17 am
… By the way Tallbloke has done us all a big favour by posting “Fourier 1824 as translated by Burgess 1837″ on his blog, so go read it and perhaps learn – a lot.
I’d love to … but I’m banned from the talkshop for my evil ways. Is it posted anywhere else?
w.”
=====
YesWillis,. You can either cut and either copy, cut and paste Fourier (1824) as translated by Burgess (1837) into your search engine or click on: http://geologist-1011.mobi.
Willis – My compliments.
It’s never easy to debunk a popular, yet physically incorrect argument – JoNova had much the same experience over the greenhouse effect and the 2nd law of thermodynamics. But it’s vital – if folks want to be part of the discussion, their arguments must make sense.
Again, my compliments.
davidmhoffer says: January 22, 2012 at 1:03 pm
“on what basis do we quantify atmopsheric effects?”
We don’t need to quantify the effect of atmosphere removal for anything to do with climate. We have an atmosphere, and it isn’t going to go away. But the GHG fraction has changed, and may change further. That’s what the 33K refers to, not this silly arithmetic about moonscapes.
davidmhoffer says:
Yes…You can (modulo the point that you need to replace “GHGs” with “greenhouse effect” since technically clouds are not a GHG since they are not a gas but they are part of the greenhouse effect). Because in the absence of any greenhouse effect, the effective radiating layer is at the surface and the highest blackbody surface temperature that can support the emission of an amount of power equal to 24W/m^2 times the earth’s surface area is a temperature of 255 K.
So, in the absence of the greenhouse effect, the Earth’s surface temperature has to be at least ~33 K less.
Stephen Rasey says at 10:58am:
“Trick, a coffee cup on a table is experiencing the force of gravity, but it is not moving in the table’s frame of reference, therefore it is not accelerating.”
Stephen – The cup IS accelerating at g wrt earth surface/table in their rest frame or cup would have no weight. In this table frame of reference, the coffee cup has mass m & inexplicable gravity generated weight force = mg down on the table at rest wrt surface of earth.
Equivalently in the cup rest reference frame, the table would be seen to be accelerating at a (=g) and the cup has no way of knowing whether (F=weight=mg=ma) is from table accelerating or from an inexplicable gravity field.
Wait for my maybe imperfect language to be corrected below…but that’s generally ok I think.
In your isothermal arguments you are conflating temperture and energy. Shame on you. In the absence of gravity the column would be isothermal. However gravity induces a pressure gradient and this introduces a temperature gradient along with it. Total energy in the column however does not change and neither does the total energy in any horizontal layer. Total energy is a mix of gravitational potential energy and kinetic energy. As you move further up the column total kinetic energy declines and gravitational potential energy increases commensurately. “Heat” is energy of motion, measured by thermometers (“sensible”) but it isn’t the only kind of energy. The books balance just fine with a thermal gradient produced by a gravitational field. Gravity does not produce or reduce total energy, it doesn’t change the distribution of energy, it merely changes the form the energy takes on at different altitudes.
Dearest J.V.
I’m doing no such thing. I’m showing quite clearly that in the presence of gravity a non-isothermal column of air is not in thermal equilibrium. Temperature is (kinetic) energy per molecule, not total energy in the gas or total energy per molecules. Surely my example regarding jars of air made that clear. You can move a jar up or down all you like. You can fill it with as many molecules as you like (as long as there are enough for thermal equilibrium to be a valid concept, so putting a single molecule in the jar won’t work on the low side, and putting so many into the jar that they form a solid or liquid won’t work). You can give it any pressure that you like. At the end of the day, after you’ve waited for the molecules in the jar to come into thermal equilibrium, the temperature of the air in the jar is a direct measure of whether or not it would be in thermal equilibrium with any other system, anywhere, in or out of a gravitational field, whether the second system is a solid, liquid or gas.
Having the “books balance” is a matter of the first law. There are many, many possible ways the energy in the box can distribute with the first law books balanced. The way that it will distribute, in equilibrium, is a matter of the second law. The second law requires that the equilibrium distribution in any undisturbed/isolated environment, including one in a gravitational field, be isothermal because heat will flow from hotter to colder if it can, otherwise.
Temperature does not depend on potential energy, it depends on degrees of freedom. That’s why a jar full of air at 300K riding in the space shuttle will have precisely the same velocity/kinetic energy distribution as a jar full of air at 300K sitting on the surface of the earth. It does not depend on pressure. A jar full of air at 2 atmospheres and at 300K has the same velocity/KE distribution as a jar at 1 atmosphere. It does not depend on the number of molecules (although you can get into trouble for very low numbers of molecules because your jar can get to where it only has one or a very few molecules in it, in which case “distribution” of velocities and “temperature” become shaky concepts). A jar with N molecules at pressure P and temperature T has the same average kinetic energy per molecule as a jar with 2N molecules at pressure P/2 and temperature T, whether or not one of them is floating in intergalactic space and the other is sitting (well insulated) in a balloon floating at rest in the atmosphere of Jupiter.
Until you and all of the many other people who seem to want to insist that the laws of physics, specifically the laws of thermodynamics in the context of gases, are somehow suspended learn these simple ideas out of any book on thermodynamics, you will continue to contribute — ahem — “hot air” to what should otherwise be a very cool discussion.
Let me state it clearly and categorically. Under no circumstances can any gas, ideal or not, including a gas sitting over the same gas, liquified, in a container, whether or not the gas and/or liquid are floating freely in space or sitting at rest on the surface of a planet, sit there for a very long time and end up with a separation of temperature in equilibrium. If that could ever happen, anywhere, we could build perpetual motion machines of the second kind, because heat in such a system spontaneously flows from cold to hot in order to form it from a temperature neutral initial state. Gravity can transiently do work to heat a gas as it collapses and produce a non-uniform temperature distribution, but conduction and convection will quickly thermalize it to a single temperature.
Seriously, J.V. — you seem well-read. Surely you have read and understand the arguments associated with Maxwell’s Demon, enough to realize that gravity isn’t one.
How about you, or anybody, actually try to address the perpetual motion problem? Explain how one can run a heat engine between any two reservoirs that are at different temperature unless they happen to be the top and bottom of an “equilibrium” column with a stable, gravitationally produced lapse rate, because if they could do it then they would run forever.
rgb
Nick Stokes says:
January 22, 2012 at 1:24 pm
davidmhoffer says: January 22, 2012 at 1:03 pm
“on what basis do we quantify atmopsheric effects?”
We don’t need to quantify the effect of atmosphere removal for anything to do with climate. We have an atmosphere, and it isn’t going to go away. But the GHG fraction has changed, and may change further. That’s what the 33K refers to, not this silly arithmetic about moonscapes.>>>
If one does not know what the atmospheric effects are, then by default one also does not know what the GHG effects are as they are simply a component of atmospheric effects. Further, I find it interesting that people are quite content to calculate 33K based on the theoretical blackbody temperature of earth as derived from the “average” insolation, but when I point out that the method used to arrive at that number is flawed, the come back is, well, you’re talking theoretical numbers not practical numbers.
Well Willis, I think you realise that if you put real values in, the current could not be measured.
So as far as Modern Physics is concerned if it cannot be measured you cannot say it exists.
Excuse me? Quite aside from the fact that we are discussing a Gedanken experiment rife with “ideal” things, why in the world do you think that one could be unable to measure the heat flow or electric current? Are you trying to say that the top and the bottom would be so close together in temperature that the difference would not be measurable and therefore could not be said to exist? Or do you think that there is some sort of lower bound to the potential difference a thermoelectric junction can produce?
Or do you have something against dippy ducks?
We can make the air column as large as we like. We can make g as large as we like. It only exists in our mind, after all. We can make it a kilometer high and place it in gravitational field of 100g. For an ideal gas, we can simply scale any lapse rate up to where the temperature difference is 100K. We can use not just one thermocouple but 1000, adding their tiny little voltages in series until it is large enough to measure.
The Carnot Cycle engine is a textbook example of an “ideal” heat engine used in discussions of thermodynamics. One can run an ideal Carnot Cycle heat engine between any two reservoirs at any difference in temperature. It won’t be very efficient, and it will take a long time to go through a cycle (in part because the adiabatic expansion has to be quasi-static to be reversible) and it would be perfectly happy running between the top and bottom of the air column if they were at different temperatures. And then there is the even simpler problem of perpetual heat flow if you place a simple insulated silver rod so that it connects the top and bottom of the reservoir. Then the system never reaches equilibrium — you have heat flow to the top through the silver (surely you don’t want to propose an adiabatic lapse rate in silver that is identical to that of air, do you?) as long as there is a temperature difference, and the heat somehow “flows” back to the bottom due to gravity as fast as it gets there.
A neat trick if you could do it. But you can’t. There ain’t no such thing as a free lunch, and a spontaneous thermal gradient is the potential provider of an eternity of free lunches.
rgb
Robert Brown says at 1/22 8:04am:
“Therefore consider the “jar” argument once again.”
Ok. This may be long, fair warning. Pour a libation of choice, may only interest Robert.
“Grab a jar of air at the bottom of your “equilibrium” room with a DALR. Its pressure is a bit higher and temperature is a bit higher than air at the top.”
Ok. Willis’ cv equilibrium not mine; I haven’t added anything to Willis’ adiabatic GHG-free equilibrated tall air column under presence of inexplicable gravity field. I add for clarification assuming your permission: an open Bjar, b for bottom, and its separate lid actually. Assume here both jar & lid were inside Willis 1 cv and in energy equilibrium (i.e. no “hot” jars or “cold” lids suddenly cross cv.)
“Grab a second jar of air at the top, where pressure and temperature are both a bit lower.”
Ok. Tjar, t for top. With same clarification as Bjar above.
“So far, we know nothing about the density of said air.”
In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).
“Nothing in fluid dynamics requires a fluid to be compressible….I don’t know how you want to idealize your “air”…. Gravity, however, is the same at the top and at the bottom.”
Ok. Lotsa’ thought experiments exist using incompressible fluids. Not my air – I’ll use Willis’ GHG-free atmospheric non-idealized air.
“Moving the jar at the top to the bottom… gently enough that one doesn’t slam the fluid molecules…”
Ok. Before Tjar is moved down to a decreased h, I’ll assume with Robert’s permission, the lid has been closed completely and we have a second body (Tjar + gas) within the cv, nothing across Willis’ 1 cv.
“The fluid in the two jars is not in thermal equilibrium. Surely you agree?”
YES for Tjar moved down (deliberately –w/no time to re-establish equilibrium) and “gently” thus heat flows before re-establishing cv equilibrium since Tjar is touching Willis air.
NO for 3rd body closed Bjar still presumably at the bottom, Robert hasn’t moved it yet. BJar is still in cv energy equilibrium near h=0.
“…the two jars have any sort of thermal pathway” opened between them, they will come to thermal equilibrium at a temperature in between T_b and T_t, one we can actually compute as it will depend on their heat capacities at constant volume which depend on N only.”
Robert didn’t write it, but I presume just a trivial oversight, that closed completely Bjar was then raised, deliberately and gently as Tjar was lowered. So YES, the thermal path way being established between T&Bjars at some h >0, establishes the jars new equilibrium (Caballero’s mean KE) temperature (of h) thru heat flow that ceases eventually.
“Note well — extremely well, if you please — that the heat capacity of the gas gets no contribution from gravity.”
Ok. Noted extremely well.
“From this alone you should be able to see that gravity is decoupled from the problem, because your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect.”
No. Robert & I part views here. Total Tjar energy = its KE+PE & is not ever changed; it is always conserved & constant. No total energy can be harmed or created by moving the jars. Robert is incorrect to write gravity field can be decoupled (i.e. ngh term ignored). Accepting that means no conservation of total energy & then a thermo law is broken.
Since gravity potential energy cannot be decoupled & conserve TE, the PE of Tjar is reduced (of h), the PE of Bjar is increased (of h). When the established thermal pathway allows Tjar and Bjar to reach energy equilibrium, the KE of the molecules inside each will be the same as the KE of the molecules in Willis’ cv air at h. (After Caballero, we are talking KE mean here, actually each jar has a parcel of molecules.)
After Caballero, each of those jar KE means are the respective jar equilibrium temperatures (of h). Tjar (KE +PE) = constant = Bjar (KE + PE) = Willis’ air (KE + PE) (all of h). In the Tjar case, mean KE reduces as PE increases. In the Bjar case, mean KE decreases as PE increases.
“…your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect. dT/dh = 0 (where the derivative should be a partial derivative btw). Adiabatic lapse is a non-equilibrium state.”
In my view, my belief is correct since no thermo law is broken. My view/belief invokes immutable conservation of energy in the 1 cv (as others have stated in this thread). dT/dh is not = 0 which it cannot be and keep total energy = constant. (as an aside, not a partial derivative I think, I could be wrong – trivial matter). Here in presence of inexplicable gravity field, adiabatic lapse IS an equilibrium state. Presence of this g field results in the ngh = PE term necessarily and always materializing. Curiously, this stratification also enables buoyancy – a matter for another day.
“Even if we leave them where they were in the first place and simply run a heat-superconducting wire between them, we expect heat to flow in the wire because they are not in thermal equilibrium and thermal equilibrium does not depend on where you are!”
This gets a little more interesting. Note extremely well here, if Robert runs his heat-superconducting wire between T&Bjars in their original positions (at hugely different h), in my view nothing of interest happens. (NB: don’t need exotic super-conductor wire here, an ordinary works, but maybe SC enables an easier-to-see solution). This (now 4th body) wire is in thermal contact with Willis’ cv gas all the way from top to bottom. At system equilibrium in the cv, that wire will be in thermal contact with the KE of Willis gas at each continuous h so the T of that SC wire will necessarily vary at each h. By immutable thermo law!
The bottom terminal end of the superconducting wire (ending in Bjar) will be at thermal equilibrium of Bjar KE and vice versa all along its length (of h) in Willis air KE to Tjar KE which is also in energy equilibrium. There is never any 1 body in the system cv that is not in equilibrium thus no energy can flow between any of the4 bodies forever – esp. with conservation of energy invoked.
“Once again, you are stuck.”
No, conservation of total energy rescued me. As it has for many of my exploits.
“You have a closed system that — you claim — is in thermal equilibrium with two different temperatures, one at the top and one at the bottom. Yet that means that an ordinary thermometer carried from the bottom to the top would read two values, and, just as would be the case for any two systems whose temperature is measured, means that heat would flow between them if it could.”
Yes, two thermometers will read two different temperatures (KE) even in the equilibrated CV because of conservation of energy in a gravity field. Heat will now NOT flow in Willis’ cv, since all 4 bodies is – at every continuous h in stratified T- in energy equilibrium. No irreversibility. Entropy is constant. No heat flow.
“Air is a conductor of heat and heat will flow from the bottom to the top until the system is in equilibrium.”
Yes initially & stop flowing heat (no energy flow forever here) when balanced – just don’t go across Willis cv, Robert. Same system, different h’s. There are NOT two separate cv.s, here one hot and one cold. Only Willis’ original 1 cv, which in energy equilibrium & touching everywhere.
Yes GHG-free air conducts KE (i.e. T) to the wire from each h at molecular mean KE (Caballero’s temperature defn.), thanks for pointing out. Willis cv can achieve stratified T equilibrium, all 4 bodies touch in energy equilibrium at continuous h’s thru the wire, no Perpetuum Mobile. Nothing crosses Willis’ cv. A drinking bird (got a laugh out of me) could be added but it stops drinking energy at equilibrium.
Courtesy of Robert, I’ve had a fun Sunday afternoon whilst nursing a head cold & refilling hot lemon tea. Now I’ll fill a Bjar w/scotch OTR, turn up the fire, & go re-read Maxwell’s Demon thing and report back (don’t cringe), maybe even w/an elevator speech. Geez, I badly need to employ an editor but this site is inefficient for conversations, my apologies, I gave fair warning.
“Trick, a coffee cup on a table is experiencing the force of gravity, but it is not moving in the table’s frame of reference, therefore it is not accelerating.”
Stephen – The cup IS accelerating at g wrt earth surface/table in their rest frame or cup would have no weight. In this table frame of reference, the coffee cup has mass m & inexplicable gravity generated weight force = mg down on the table at rest wrt surface of earth.
Equivalently in the cup rest reference frame, the table would be seen to be accelerating at a (=g) and the cup has no way of knowing whether (F=weight=mg=ma) is from table accelerating or from an inexplicable gravity field.
Wait for my maybe imperfect language to be corrected below…but that’s generally ok I think.
If this is an honest plea for correction, say no more.
Forget the fact that Earth is rotating and the small non-inertial correction (Coriolis force) as it is quite negligible.
The cup sits on the table. Gravity exerts a force down of mg. The table exerts a “normal force” up on it of N. The cup is at rest. F = N – mg = ma = 0. Therefore the correct statement is that the forces acting on the cup are balanced, with gravity pulling it down and a column of electrostatic and pauli exclusion forces that stretch from the center of the earth to the top of the table pushing it up. It is indeed not accelerating.
The “weight” of the cup is mg, and of course this doesn’t change whether or not the cup is sitting at rest or falling. If we put the cup on a scale, a scale measures N, the normal force applied across the scale, and at rest N = mg.
Put the cup in an elevator (resting on a scale) and as the elevator accelerates up
N – mg = ma > 0
so that N > mg, the cup “weighs” more (not really, but in the accelerating frame it appears to). Accelerate it down and N – mg = ma < 0 and it weighs less. Drop it (so it is in free fall) and mg = ma and N = 0, so as far as the scale is concerned it is "weightless", although the weight of course has been mg throughout.
All of this is in my online intro physics book, with pictures and considerable explanatory text, if you care to learn it.
http://www.phy.duke.edu/~rgb/Class/intro_physics_1.php
rgb
@Trick 1:37 pm The cup IS accelerating at g wrt earth surface/table in their rest frame or cup would have no weight.
No, there is a FORCE due to gravity Fg = G(m1*m2)/r^2 There is no acceleration in this formula.
As I said, this is a statis problem. there is an equal and opposite force exerted from the table. No net force. F=(mass)(acceleration) if net force = 0, mass > 0, then acceleration = 0.
If the table broke, The upwards force from the table would drop and Fg would then cause the cup to accelerate downward.
Robert Brown says
“Excuse me? Quite aside from the fact that we are discussing a Gedanken experiment rife with “ideal” things, why in the world do you think that one could be unable to measure the heat flow or electric current?”
I said as far as I know there has never been an EXPERIMENT to test which of these conjectures are true for a thermally isolated column of gas in a gravitational field.
1. That is has an isothermal distribution
2. That is has an adiabatic distribution
Someone tries the experiment thermocouple with very long copper/silver leads and the most sensitive ammeter but finds no measurable current.
After consulting the physical tables and numbers he finds that the apparatus was not sensitive enough to detect a current.
I think we are at that stage with this particular method.
The two tubes with gases of widely differing Cp seems a better route but there are problems there too.
Joel Shore,
I am referring back to your post of 1/21/12 here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872192
You say:
” Besides which, we know for a fact that the reason why the Earth’s surface is emitting 390 W/m^2 is not because there is some magical new source of energy but because there is absorption occurring in the atmosphere such that only ~240 W/m^2 is escaping to space.
Hence, what N&Z and Jelbring have exposed is how many otherwise intelligent people will apparently believe utter nonsense if it supports what they really strongly want to believe.”
You are confusing two different things (and insulting good scientists). You have either not read or do not understand Jelbring’s paper. (I have no final opinion on the N&Z paper since I have not yet seen their Part 2 which explains their thermodynamic reasoning). You are confusing the adiabatic lapse rate derivation and the total heat content of the atmosphere. They are separate thermodynamic entities.
For a “dry” atmosphere in an electromagnetic and gravitational field, the “first law” can be written:
dU = CpdT + gdz (1)
This equation is also known in meteorology as “dry static energy” and is closely related to “potential temperature”.
For a system (atmosphere) in steady state, dynamic equilibrium, the internal energy (U) of the system is constant (conservation of energy) and dU = 0. This means that at any point in the system (atmosphere):
CpdT + gdz = 0 (2)
And for a system in static equilibrium
CpT + gz = constant (3)
This is the focal point of the Jelbring paper. As you ascend vertically in the atmosphere the thermal energy (and therefore temperature) decreases and the gravitational potential energy increases. Energy is conserved. Thus there is a vertical thermal gradient in an atmosphere that is under the influence of both an electromagnetic and gravitational field. The gradient can be represented by rearranging equation (2) to give:
dT/dz = -g/Cp (4)
Thus the vertical temperature profile in an ideal steady state atmosphere is solely a function of the gravitational acceleration and the specific heat capacity (at constant pressure) of the atmospheric gas. It does not matter if the gases are GHG’s or not, it does not matter what the down welling radiation is, it does not matter what the pressure is (as long as it is above 0.2 atm.), it does not matter what the density is, it does not matter what the absolute temperature is at any given point. The dry adiabatic temperature gradient is a constant, is a function of gravity and heat capacity only, and is a direct result of the first law and the conservation of energy (yes, there is a law defining the DALR).
Now let’s clear up your confusion as to how this applies to the surface temperature. The temperature at any given point is a function of the total heat content of the atmosphere. The heat content is a function of the incoming solar radiation and the outgoing long wave radiation. The total heat content can change (due to a change in the effective emission height for instance), and therefore the absolute temperature at any point can change, but once steady state is reached, the temperature gradient remains a constant. In all cases, the temperature at the surface will be warmer than the temperature at any altitude.
The “green house effect” is simply the difference between the surface temperature and the effective temperature at the effective emission altitude (z) (Jelbring’s outer sphere and diameter “d”). It is controlled by the adiabatic lapse rate, not by “back radiation”. “Back radiation” is a function of the surface temperature (and the lapse rate), not the other way around.
In Summary, 1) the GHE is simply a function of the effective emission height (i.e., effective emission temperature) and the lapse rate. 2) Surface temperature is simply a function of the total heat content of the atmosphere.
This is the Jelbring hypothesis and it is very straight forward. Where is the “utter nonsense”?
By the way, in your hypothetical, “isothermal” atmosphere, there would be no GHE (Ts = Te). This result would be true no matter the GHG concentration, whether it is zero or >1000 ppm. It would also be very cold!
I must also emphasize that this system (Jelbring) does not represent a perpetuum mobile – it is strictly a result of elementary thermodynamics. If this is a perpetuum mobile, then the rotation of the planets around the sun in the solar system is a perpetuum mobile – I don’t think Newton would approve! He understood gravity.
Bill
P.S., You also commented on my Thomas Kuhn quote in your post:
”“Because the unit of scientific achievement is the solved problem and because the group knows well which problems have been solved, few scientists will easily be persuaded to adopt a viewpoint that again opens to question many problems that have previously been solved.” (Page 169)
The problem with using Thomas Kuhn to support your notions is that people quoting Thomas Kuhn have correctly predicted about 10,000 of the last 10 paradigm shifts (although that first number is probably too conservative an estimate). [This sort of makes economists, for which this sort of quip was originally applied, look good by comparison!]”
You miss the point that Kuhn is trying to make. There would be many more paradigm shifts in science if the group practicing “normal” science were not so protective of their territory and utterly resistant to new ideas. Science would advance at a much faster rate. That 10,000 number contains many real paradigm shifts that were denied. Here is another Kuhn quote for you and Donald Brown:
“When examining normal science….we shall want finally to describe that research as a strenuous and devoted attempt to force nature into the conceptual boxes supplied by professional education.” (Page 5)
Bill
By the way, I have said it before, but I will repeat it here:
“I am confident that the complete Water-cycle is important to the surface- and near surface- temperatures. Therefore Water Vapor (WV) does also have a better capability than other “dry” gases of keeping heat in. The “Moist” or “Saturated Adiabatic Lapse Rate” is proof of that. None of the “lapse-rates” however, dry or moist, have changed since they were first established back in the 19th century.
Yet – CO2 has increased by 40 – 45 % since that time and if a pocket of ascending air contains 40 to 45 % more of a gas that is constantly being warmed by radiation from the surface below – and we are certain that must be the case because Kiehl & Trenberth (K&T) say that 40 watts per square meter are escaping from the surface out through “The Atmospheric Window.” – Therefore all radiation from the surface must be capable of reaching out to space – and must, by definition warm all gases in its path that absorb it.”
So, why does it not show up in the lapse rates?
In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).
I don’t know whether to do an itemized reply all at once or reply as you make mistakes. I’m also not sure what “cv in gravity field” stands for. What is cv? Also, if you look at the explicit context of my remark, it was fluids, and I later said that if one presumes an ideal gas… The point being that real gases have compressibilities that are not idea.
But either way, if you look at section 8.1.9 in my online physics textbook (the section titled “variation of pressure in compressible fluids” you will see the derivation of the density of an ideal gas (equal 711). So I think I’m familiar with this and you are quite correct — for an ideal gas;-)
Indeed, in the space immediately below this I derive the standard textbook isothermal variation in pressure that is a self-consistent solution to the buoyancy problem:
P(z) = P_0 exp(- Mg/RT)
where M is the molar mass of the ideal gas, T is the temperature, and R is the ideal gas constant (or factor in N_A to turn R into k if you prefer).
Note that this solution:
a) certainly exists, for an ideal gas;
b) is clearly in equilibrium. T is the temperature throughout the air column in question. No second law violations here.
This is a column that is clearly in static force equilibrium. The gas perfectly balances weight, buoyancy, and density throughout. That was the point of the \rho g z term used in the derivation (which if you read back in the text is the basis of fluid pressure in a static fluid). It is also in thermal equilibrium. Heat content will not be conducted up or down the column because it is all at the same temperature. This column does not violate the second law.
Note also that this variation of pressure with height is very, very close — within around 3% — to what is observed in the atmosphere, even though the atmosphere is better described by an adiabatic lapse rate than an isothermal one because it is not in equilibrium. And even the adiabatic lapse rate isn’t a very good approximation to the actual variation. It certainly doesn’t describe the air column from top to bottom, with troposphere, stratosphere, thermosphere and so on. It doesn’t even work all that well only within the troposphere. It’s just a useful baseline for a gas system heated at the bottom and cooled at the top — e.g. the atmosphere.
rgb
Robert Brown says at 1/22 8:04am:
“b) If you persist in arguing that it cannot, then you have established gravity as a Maxwell’s Demon for air. Please read about Maxwell’s Demon as it would greatly improve your understanding of detailed balance and why this argument ultimately microscopically fails.”
Ok, I’m back quickly (admittedly scotch OTR is only half gone) from reading Maxwell’s demon for air: this applies to a container of air not in a gravity field where the demon decreases the entropy by opening & closing a door selectively letting thru certain molecules. Here, Willis’ cv keeps entropy constant so gravity & energy conservation solution can stand w/o establishing gravity as a demon.
Could apply Maxwell’s demon to Willis’ premise of GHG-free air container in a gravity field for some additional fun I suppose. May even discover a PE demon that arises since Maxwell didn’t have gravity in his container. BTW – some think the demon when added to the cv adds enough entropy to keep it constant or increasing. Don’t let any demons cross the cv!
I will not persist in arguing same ground or hijacking a thread, ceptin’ what sneaks in later. My view stands on energy conservation, Robert’s view stands on writing “gravity is decoupled from the problem” which necessarily introduces ignoring some of molecule’s total energy in a gravity field. My view can certainly change but only in keeping w/conservation of energy in an inexplicable gravity field.
“c) Given the stable thermal gradient in this case, one can trivially construct a heat engine that does perpetual work moving heat from the bottom to the top and then “re-using” this energy after gravity sends it back to the bottom. That’s again a simple fact.”
No, one cannot construct a perpetual motion machine in Willis’ cv + gravity: the jars, lids and wires, all reach energy equilibrium exactly & since they are all touching, heat stops flowing eventually. I’m at end of Robert’s 8:04am post finally.
Robert Brown says at 1/22 3:03pm:
“What is cv?”
Control volume.
Joel Shore;
Yes…You can (modulo the point that you need to replace “GHGs” with “greenhouse effect” since technically clouds are not a GHG since they are not a gas but they are part of the greenhouse effect). Because in the absence of any greenhouse effect, the effective radiating layer is at the surface and the highest blackbody surface temperature that can support the emission of an amount of power equal to 24W/m^2 times the earth’s surface area is a temperature of 255 K.
So, in the absence of the greenhouse effect, the Earth’s surface temperature has to be at least ~33 K less.>>>>
1. To be correct your statement would have to say that in the absence of GHE, conduction, convection, heat capacity, oceanic and atmospheric currents, the earth’s surface temperature has to be at least ~33K less.
2. To be correct, you would also have to note that ~33k less could only occurr in one scenario, and one scenario only, which would be an earth with a uniform temperature across all latitudes, from day to night, season to season, and exposed to uniform radiance. All other instances result in a differential that exceeds ~33K and my simple math model demonstrates that this number is CONSIDERABLY higher than 33K.
Robert didn’t write it, but I presume just a trivial oversight, that closed completely Bjar was then raised, deliberately and gently as Tjar was lowered. So YES, the thermal path way being established between T&Bjars at some h >0, establishes the jars new equilibrium (Caballero’s mean KE) temperature (of h) thru heat flow that ceases eventually.
You still don’t understand. Yes, by “jar” I mean that you adiabatically isolate the fluid samples at the top and the bottom, in jars with lids that do not conduct heat. The jars contain fluid at some pressure and density. I’ve constrained them so that their pressure and density profile will not change — it supported their weight before, it still supports their weight. Molecules that hit the walls bounce elastically off of them, precisely. Note that this elastic collision process is precisely similar in ever respect to having actual parcels of fluid at all of the walls that instead transfer molecules in equilibrium back into the “jar” at the same rate that random motion transfers them out. After all, equilibrium is adiabatic, right? No net transfer of energy, momentum, or particle number across the imagined boundaries of parcels of fluid in equilibrium. A perfectly elastic boundary that reflects particles back into the volume has the exact same effect, within the usual thermal noise that vanishes in the large particle number limit.
Now that the fluid is in the jars, however, we can see the irrelevance of gravity.
We can move the jars anywhere and the temperature inside of them doesn’t change. You seem to agree, although you completely miss the point about “thermal pathway” and how the invariance of the temperature inside of those jars as they are moved up and down matters.
I don’t care where the jars are when we establish a thermal pathway between them. I didn’t say that we have to bring them together and make them touch. It doesn’t, after all, matter where they are or how the pathway is established. They now just jars containing gases at two different temperatures and pressures that do not vary no matter where you carry them to — they are inside perfect thermos containers with infinitely rigid elastic walls. Drill a hole in the thermos and put a perfectly insulated silver wire so that it runs in between them, and heat will flow from the hotter to the colder, whether or not the jars have been moved at all.
The point is that there is no difference between the perfectly adiabatic walls of the jar and the gas that otherwise surrounds and uplifts the jarful of the gas. Remove the jar, put the silver wire into the air column, and heat will flow because the parcels of air at the top and bottom are not in equilibrium.
There is, as previously shown, nothing at all contradictory about a column of ideal gas supporting its own weight isothermally. In fact, I provide a reference for where I derive its description literally as a textbook example. This is the only static distribution of gas in which heat cannot flow due to conduction. There is no bulk transport or convection, either. It is not only a self-consistent solution, it is the only self-consistent solution to the problem of evaluating pressure and density in which heat will not flow from hotter to colder regions, because you don’t need a silver rod — the gas conducts heat all by itself.
rgb
So, why does it not show up in the lapse rates?
Because CO_2 is 0.03% of the atmosphere? And because the atmosphere is already optically “thick” from the CO_2 that is already there? And because the things that determine the troposphere-stratosphere boundary height probably have nothing serious to do with CO_2 per se?
Not disagreeing with your argument, by the way. I’m just pointing out that it is quite reasonable to expect the GHE to vary extremely slowly with CO_2 concentration on all of these accounts. And it does — without positive feedback, CO_2 forcing due to the additional CO_2 is probably 0.3-1.0C, although some might argue that it should be a bit higher than that — still certainly less than 2C and most of the estimates I’ve read are around 0.6C.
The argument is over the feedback. The GHE is quite real and almost certainly is responsible for a large chunk of the warming of the earth compared to an ideal (if not terribly realistic) blackbody. In fact, it is probably responsible for part of more than the 33C or whatever that is estimated compared to this, because the initial state for a rotating Earth with a heat capacity would be cooler than the ideal blackbody in the first place.
However, it is largely a saturated effect straight up. Doubling it doesn’t give us 66K warming — nobody but maybe Hansen with his “boiling seas” asserts that. The question is whether or not there is additional forcing, in particular from increased water vapor that will presumably be added to the atmosphere as it heats. But that in turn neglects the many negative feedbacks that self-organize into things like the global oscillations and increased albedo, that all increase cooling as water vapor and temperature rise.
Although the data are still inconclusive, the evidence is mounting that a) climate variation from non-GH sources is at least as large as GH related variation; b) overall feedback is somewhere in the weakly negative to weakly positive range, excluding the “catastrophic” part of CAGW already in the sense that it is already more likely than not that there will be no catastrophic warming, and the boundaries are actively being lowered by every year in the current stretch where the Earth seems to be failing to follow the “catastrophic” prediction curves.
This is skepticism at its best, quite frankly. It isn’t crank science, openly violating laws of thermodynamics. It is simply observational evidence that isn’t agreeing with the predictions of the more extreme models with high feedback. The other place where really good skeptical science is being done is in the examination of the sun and the various oscillations. Solar-albedo feedback is quite plausible and represents an additional driver that can easily erase almost any amount of GHE warming at \Delta T \approx 91 \Delta \alpha — a variation of 0.011 in albedo is enough to completely cancel — or explain — the warming of the twentieth century. Not just some of it, all of it. The oscillations are all self-organized structures that exist because they improve long term heat loss and keep the Earth in balance. Their variation is again on the same order of \Delta T as that observed, at least in any short run 30-50 year stretch. Both of these have to be better understood and incorporated before making rash conclusions about “catastrophe” at our doorstep. The science isn’t really settled, not at all, however much the greenhouse effect per se is reasonably well understood and observationally supported.
rgb
No, one cannot construct a perpetual motion machine in Willis’ cv + gravity: the jars, lids and wires, all reach energy equilibrium exactly & since they are all touching, heat stops flowing eventually. I’m at end of Robert’s 8:04am post finally.
In other words, they weren’t in thermal equilibrium before. Exactly. Because the air is all touching and heat will flow until everything is the same temperature, because the “jars” are a convenient fiction to help you isolate your point of view, something you are having a hard time doing.
Thermal equilibrium is not “energy equilibrium” — it is detailed balance, not equal energy density everywhere. Gravity cannot act as a heat source in a static unmoving fluid, it only does work on fluid that is rising or falling in bulk. There is no fluid rising or falling in a static column of air. That was also the bit of how raising and lowering the jars has no effect on the temperature of the air inside. If you want to talk about the adiabatic expansion of parcels of air, they have to move up or down. In equilibrium they do neither. Even if an adiabatic thermal profile is reached first as you take air in a sealed container and “turn on” gravity, as it sits there it will quickly become isothermal because air conducts heat and heat flows from hot to cold until they are the same temperature. Your proposed thermal gradient is not static thermal equilibrium.
rgb
Okay, I give up trying to elicit a serious discussion from the many physicists here about why the Velasco et al. conclusion is or is not valid. But just in case there exists among them a spark of intellectual curiosity none has so far betrayed, I leave my layman’s (no doubt inaccurate) description of the technique the Velasco et al. and Román et al. papers employ to demonstrate that such a system would exhibit a non-zero temperature lapse rate.
Román et al. deal with a vessel that is disposed in a gravitational field and contains N identical-mass monatomic molecules (i.e., molecules whose kinetic energy is all translational; none of that energy resides in tumbling or vibration). The vessel so isolates the molecules that their aggregate total (kinetic plus potential) energy E is constant. To arrive at the molecular-velocity (and therefore temperature) distribution as a function of height, the authors use a technique they characterize as “counting microstates.” Rather than count discrete states, though, they integrate through a continuous “phase space.” That is, since each molecule’s state can be completely described by its (three-dimensional) position and (three-dimensional) momentum, the total system state can be described by an ordered set of 6N scalars, and the system state at any instance can therefore be thought of as a position in a 6N-space, a “phase space.”
Their technique is based on the assumption, which I take it is well accepted in statistical mechanics, that the probability of the system state’s falling within any region of that phase space is proportional to the volume of that phase-space region. Now, the constraint that energy is fixed restricts the system to a (zero-volume) hyperparaboloid in that phase space, so the probability that the (energy-constrained) system will fall within a given range of states having that energy becomes the ratio of that range’s area on the hyperparaboloid to the total hyperparaboloid area. Equation 2 adumbrates this approach, giving a function that is zero at all places in the phase space except those that have an energy E, where its Dirac delta function gives it an infinite value that results in a finite integral of that function.
.
To find the position and momentum distribution for a single one of those molecules, Román et al.’s Equation 9 integrates all the other molecules’ positions and momenta out of that function and restates the result into a form that lends itself to use of a magical identity they introduced in Equation 5 as proved by a pay-walled paper. This is how they perform the integration that leads to their Equations 12, 14, and 15, which Velasco et al. adopt as their paper’s Equations 5, 6, and 7 for the distributions of molecular position and velocity.
It is from the thus-obtained Velasco et al. Equations 5 and 6 that they claim to obtain their Equation 8 for temperature as a function of altitude. It is that equation that I interpret as saying that, for any finite number of molecules, there would be a non-zero (but for significant number of molecules very small) temperature lapse rate.
Bryan says on January 22, 2012 at 9:42 am:
“O H Dahlsveen
Your interesting passage from Fourier did not deal with a thermally isolated gas .
Which means no heat enters or leaves the gas sample.”
==========
Sorry Bryan, I have obviously misunderstood your statement – but I have thought a lot about it and – I have come to the conclusion that my answer should have been:
“Hmmmm, maybe it is because it is impossible to thermally isolate a gas, at least I have never heard of a way to do it. – You can isolate against light and electromagnetic energy radiation. But heat I am afraid only conducts and will therefore find its way through any container. – Just look at a thermos flask.” —
Actually lots of research, that we never hear of, is going on to find the “Perfect Gas/Liquid Isolator” So if you can improve the thermos flask to achieve 0 heat loss I shall be sending you begging letters.” However if you are successful my guess is that you will find the gas inside your “Super Thermos” will never change its temperature.
William Gilbert said @ January 22, 2012 at 2:53 pm
It seems to take a “strenuous & devoted attempt to force” the Received View (currently accepted paradigm) into minds here. How much more difficult to teach if there were no Received View and every institution taught its own version of Physics. I would agree that if the group practicing “normal” science were not so protective of their territory and utterly resistant to new ideas we would have more progress. OTOH, if they were utterly unprotective of their territory and accepted every new idea willy-nilly, there would be close to nil, or at best, glacial progress. No standing on the shoulders of giants allowed as it were. Just continual reinvention of the wheel (perpetuum mobile?) Kuhn was cognisant of this.
Oddly, The Character of Scientific Revolutions was the first book I purchased for my new Kindle. The second is a book called The Book of Lilith highly recommended by a friend who knows I rarely read fiction. You’ve gotta laugh 🙂
Forgot to put the link in to The Book of Lilith:
http://www.phy.duke.edu/~rgb/Lilith/Lilith.php
Stephen Rasey says at 1/22 2:23pm:
“No,… There is no acceleration…”
Merely trying to point out (me – poorly) whether the cup is accelerating or not depends on the reference frame. Cannot merely say the cup is not accelerating. Consider all ref. frames. Not just the familiar one.
My point: Einstein’s master insight was that the familiar “pull” of the Earth’s gravitational field is fundamentally the same as constant acceleration. The star position deviated w/sun gravity.
Table ref. frame: Force on Table = +F, cup mass m summation of forces F-ma = 0. F = ma. Same cup, same table, same situation, so same cup mass has to be accelerating at a (=g). Wish I could have written that simple deal 100 yrs ago. Nothing here about force of earth’s attraction. Now: Is the table accelerating?
This applies to what happens to the molecules in Robert Brown’s Bjar – so on topic, loosely. Need to account for total energy in a gravity field or the equiv. acceleration.
Robert Brown says:
January 22, 2012 at 3:48 pm
Agreed
Indeed
Let me suggest that you are mistaking the successful operation of a governing system for the absence of feedback. If the active cloud-based tropical thermo-regulatory system were to fail for only a few weeks, temperatures would soar to unimaginable heights.
This implies, not a “weakly negative to weakly positive” feedback, but the active application of both strong positive and strong negative feedbacks, in amounts that balance each other to maintain a temperature range..
The catastrophic part has always been much more spin than science.
The regulation of incoming radiation through changes in albedo (primarily in the tropics) is consistently neglected in climate science. It is the single largest factor. When the clouds form in the late tropical mornings around the ITCZ, incoming solar is cut by 40 W/m2. This implies that the variation of the time of tropical onset is a major regulatory factor, overwhelming any small changes from
Thank you very kindly, I have been one of few voices crying in the wilderness on this question, saying both that the greenhouse effect is real and that it doesn’t run the variations at our current temperatures.
All the best to you, thank you for your patience and perseverance, Oh, and thanks for the link to your online textbook. I can’t imagine what a pleasure it must be to say “see equation number seven hundred and freakin’ eleven in the physics textbook I wrote” …
w.
w.
Bart,
Even I don’t spend all my time reading this blog, which is one reason I haven’t commented on ‘Bart’s Law’. The other reason is that it makes no physical sense to me. That makes it hard to offer constructive criticism. One thing I will say is that you may be right about one minor point. Making the atmosphere over an isothermal planet absorb and emit in the thermal IR, may indeed lower the average temperature of the atmosphere compared to a transparent isothermal atmosphere. It would cause the surface to warm and the upper atmosphere to cool. Off the top of my head, I suspect that the warming of the near surface atmosphere might not be enough to offset the cooling of the upper atmosphere.
Joe Born says:
January 22, 2012 at 4:02 pm
Sirrah, that is vile mendacity. I read the paper at your behest, and I responded to you in detail. My opinion was, you didn’t understand Velasco. You can take your ball and go home with my blessing, but by gosh, you can’t blame me for it after I did exactly what you are now saying no-ne has done.
w.
Willis, don’t you not have anything to say about the empirical evidence? Why do other planetoids with atmospheres show about the same amount of “warming, DESPITE the amounts of GHGs present? “Why is it not getting warmer these days, when CO2 emissions are increasing drastically? The “bottom line” is always empirical evidence, but you just seem to shake it off!
I keep wondering just WHY you will not address this issue. You have no problem addressing other issues. Come on!
Could you have a case of confirmation bias?
William Gilbert,
Your argument that the atmosphere must show an adiabatic lapse rate is circular. Your requirement that KE + PE be constant with altitude is the definition of adiabatic expansion. Obviously you will get an adiabatic lapse rate with that condition. But there is, in fact, no requirement that KE+PE be constant with altitude. In the real atmosphere, it almost never is. Potential temperature, and thus KE+PE usually increases with altitude.
davidmhoffer says:
The point is that the greenhouse effect alone accounts for about 33 K or more…and it is the only thing that explains the important part of the warming…i.e., the part associated with the radiation emitted from the surface being 390 W/m^2 instead of 240 W/m^2. The rest is mainly what I would call “illusionary warming”. I mean, do we really want to base an estimate of the average temperature of the moon that depends radically on whether we average the temperature over the first micron, first mm, or first meter of its surface?
As for convection, it is actually what prevents the greenhouse effect from being about twice as large as it is.
That is because your model is too simple. It is probably not at all realistic for the Earth…even an Earth absent a greenhouse effect but otherwise similar.
I don’t think either of us would describe the current Earth as being having “a uniform temperature across all latitudes, from day to night, season to season, and exposed to uniform radiance” and, yet, for all intents and purposes, the error in assuming this in regards to distinguishing between averaging T^4 (and taking the 4th root) and averaging T is small.”
jae says:
January 22, 2012 at 7:09 pm
jae, don’t you not have following the thread?
w.
jae said @ January 22, 2012 at 7:09 pm
You ask: “Why is it not getting warmer these days, when CO2 emissions are increasing drastically?”
Answer: Because Willis invented this wonderful “regulation of incoming radiation through changes in albedo (primarily in the tropics) is consistently neglected in climate science. It is the single largest factor. When the clouds form in the late tropical mornings around the ITCZ, incoming solar is cut by 40 W/m2. This implies that the variation of the time of tropical onset is a major regulatory factor, overwhelming any small changes from” CO2. Just be grateful he invented it otherwise we would all be dead from malaria, heat stroke, increases in prostitution, obesity, decreasing biodiversity etc.
You also ask: “Why do other planetoids with atmospheres show about the same amount of “warming, DESPITE the amounts of GHGs present?”
Maybe, just maybe, I’m surmising here, Willis isn’t interested in the issue, or lacks the time, would prefer to go fishing, hitch-hiking, paying attention to the missus… Why don’t you prepare a paper on this and ask Anthony to post it? Or would that be too much like work?
Willis Eschenbach: “Sirrah, that is vile mendacity. I read the paper at your behest, and I responded to you in detail. My opinion was, you didn’t understand Velasco. You can take your ball and go home with my blessing, but by gosh, you can’t blame me for it after I did exactly what you are now saying no-ne has done.”
Willis, believe me, I appreciate your looking at the paper. And I recognize that the passage you seized upon does not lend itself to ready comprehension. But you’re forcing me to be blunt here. One could arrive at the conclusion you did only by failing to read or understand the equations that preceded it.
I quite understand how someone hurriedly skimming the paper might do that. But your perception of reality is distorted if thus merely tossing off the first comment that comes to your head after such a superficial review–and then ignoring the explanation I gave in response–is what passes in your mind for “serious discussion from the many physicists here.”
“Temperature does not depend on potential energy, it depends on degrees of freedom. ”
I don’t see why anyone would think temperature is related to potential energy. A rock on a hill has nothing to do with it’s temperature. Or the temperature gasoline and oxygen [potential energy] isn’t related to temperature.
“That’s why a jar full of air at 300K riding in the space shuttle will have precisely the same velocity/kinetic energy distribution as a jar full of air at 300K sitting on the surface of the earth. It does not depend on pressure.”
Well the air molecules are changing vector every nanosecond- or they are “pulling *huge* gees”
so 3 gees from a space shuttle, is rather trivial to a single molecule. Though 3 gees will affect all the molecules- creating pressure
“A jar full of air at 2 atmospheres and at 300K has the same velocity/KE distribution as a jar at 1 atmosphere. ”
Not sure what you mean by “velocity/KE distribution”.
But you seem to be suggesting that if you 1/2 volume/double pressure you would not effect the molecules average velocity.
It seems to me the initially molecules velocities are not affected, but with more molecules in a smaller volume with molecules travel the same velocity- it is higher temperature.
So, two cubic meters of 1 atm gas at room temperature forced into 1 cubic meter, initially retain their velocity [and are hotter than room temperature] and has it cools to room temperature the average velocity of gas decreases.
Obviously the 2 atm container of gas could have more Potential Energy, but that’s not important in regards to it temperature- If container is place in environment of 2 atm- it has no Potential Energy, if put in a vacuum it more Potential energy as compared to 1 atm.
It seems to me if you talking about x amount of gas molecule- the temperature is directly related to the velocity of the molecules. Increase their speed, warmer temperature. Decrease their speed
lower temperature. Hence a hot quantity of gas which cools, has the molecules of gas slowing down- or one changing the volume. With the atmosphere which isn’t in a container one can have expansion or contraction of volume of the atmosphere.
As for a jar on space shuttle- the air has little mass/weight and will not be not affected significantly from 3-4 gees [or 50 gees].
“It does not depend on the number of molecules (although you can get into trouble for very low numbers of molecules because your jar can get to where it only has one or a very few molecules in it, in which case “distribution” of velocities and “temperature” become shaky concepts). A jar with N molecules at pressure P and temperature T has the same average kinetic energy per molecule as a jar with 2N molecules at pressure P/2 and temperature T, whether or not one of them is floating in intergalactic space and the other is sitting (well insulated) in a balloon floating at rest in the atmosphere of Jupiter.”
The KE of molecules gas is 1/2 mass and velocity squared. Gravity only changes weight not mass.
But if double the mass of molecules- double the pressure you will increase temperature of gas and after it’s cooled to ambient temperature, the molecules would seem to have to have lost velocity from the point in time when the gas was heated from the added pressure.
Bart says:
January 20, 2012 at 6:04 pm
Phil. says:
January 20, 2012 at 5:02 pm
“…the only way that conduction can reduce the radiation loss from the surface is by cooling the surface.”
No, that is not the only way. Stefan-Boltzmann is an equilibrium relationship. You do not know the radiation loss from the surface in non-equilibrium conditions.
Amazing, you keep coming back with the same nonsense!
You have a situation of constant heat input to a surface in contact with a finite atmosphere heat capacity and the only method of heating that atmosphere is by conduction. Eventually the atmosphere will reach the surface temperature and the atmosphere cannot exceed that temperature. At that point the situation is equilibrium so your repeated objection doesn’t hold!
Explain to us how the atmosphere and surface temperatures exceed the equilibrium surface temperature, let’s have some equations and facts instead of the mumbo-jumbo hand waving that we’ve had from you hitherto.
You know, it is really annoying when someone proclaims they know everything about a phenomenon which is actively under investigation.
I agree it’s really annoying when someone asserts a non-physical model without any justification and pontificates about laws named after himself. Try using accepted physical laws and explain how the presence of a non-absorbing gas can increase the surface temperature above the equilibrium value.
Stephen Wilde says:
January 22, 2012 at 10:35 am
The Ideal Gas Laws have never been falsified.
If that were the case then Van der Waals wouldn’t have needed to develop his equation:
http://en.wikipedia.org/wiki/Van_der_Waals_equation
Joe Bom, I note that above you say:
Joe Born says:
January 22, 2012 at 8:13 pm (Edit)
Joe, you claimed above that you misunderstood it until you read the equations that preceded it. I went back and read them. I could not see how they changed the conclusion in any way, much less in the way that you claimed.
You quoted Velasco’s conclusion, which was:
Here are the two possible answers as stated in the paper:
Answer 2 was the adiabatic case. Answer 1 is the isothermal case. THEY SAID THE ANSWER WAS “ISOTHERMAL”. You went on to to argue that they did not say what they plainly said, for reasons that are still not clear to me, and made no sense.
So I did not answer your post, Joe. You are unable to read clear english. Read the conclusion above. In neither case is answer (2) right. Answer (2) is adiabatic lapse rate. They said three times in their conclusion that the final rate is isothermal, and you want to tap-dance around that.
Sorry, not interested. You want a rational discussion, but I can only provide half of the rationality required.
w.
@ Trick 1/22 at 5:56 pm
“ Rasey: No,… There is no acceleration…”
Merely trying to point out (me – poorly) whether the cup is accelerating or not depends on the reference frame. Cannot merely say the cup is not accelerating. Consider all ref. frames. Not just the familiar one.
Trick, the definition of acceleration is the second derivative with respect to time of the equation of position. Define Positions P in some coordinate system as a function of time. Velocity V is the first derivative, dP/dt. Acceleration is the second derivative of position, d^2P/d^2t, the first derivative of velocity, dV/dt. Gravity acts as a force, one of many forces at play, not an acceleration.
If you choose a reference frame of the table or corner stone of your building or the nearest USCGS Benchmark, the Position of the coffee cup on the table is constant with time, its velocity is zero, its acceleration is zero.
If you choose a reference frame based upon the Moon, a Celestial frame, a non-inertial frame, Sun-Galactic Coordinate System, or the SupergalacticCordinate system, then, yes, the coffee cup has non-zero acceleration that is hopelessly difficult to calculate, but at least you know the USCGS benchmark and the center of the Earth has the same acceleration difficult to calculate acceleration. Feel better?
Your life is a lot less complicated as a scientist or engineer by picking the frame of reference and the coordinate system that makes the math the easiest. Zeros are good things in math. That’s why we have multiple frames of reference and coordinate systems. That’s why tensor analysis was developed. Just ask Einstein.
Stephen Rasey says at 9:35pm:
“…yes, the coffee cup has non-zero acceleration… Feel better?”
Yes! It is all relative. Thanks, Stephen
PS: I’m just hanging out here last few days while sporting a head cold, interested in the science conversation. But I do feel better now, LOL.
Joel Shore;
The point is that the greenhouse effect alone accounts for about 33 K or more…and it is the only thing that explains the important part of the warming>>>
Since you don’t even know what the actual baseline temperature sans GHE is in the first place, you cannot possibly know how much of the observed temperature is due to other effects, nor, without being able to quantify each and every one of them, can you possibly calculate how much is GHE.
Here is the gold-standard of reference frames: the JPL ephemeris. http://www.cv.nrao.edu/~rfisher/Ephemerides/ephem_use.html
Prediction and reduction of pulsar pulse times of arrival… on the earth’s changing distance to a pulsar as it moves around the sun. … working at the level of a few hundred nanoseconds over long periods of time…. implies that the observer’s position with respect to the solar system barycenter be known to about 100 meters. Necessary if you want to shoot Cassini through the gap between Saturn and its rings seven years from launch. That’s science and engineering!
Keeping this climate related, here is the North Polar hexagonally shaped vortex of Saturn. Its been there for decades. A deep mystery.
Say you want to make Mars, have warm atmosphere as earth does?
The N&Zers and the CAGW could have different approaches.
Suppose what is wanted is ability for a human operate on Mars
without a spacesuit/pressure suit.
At the present, if given a oxygen mask, you can’t breathe on Mars-
there isn’t enough pressure. At 45000′ on earth, you can’t breathe
that air- well you can breathe it, it won’t give you enough oxygen to
remain conscious. If you use an oxygen mask you can breathe the air and
remain functional. But at 100,000′ an oxygen mask won’t help you.
And Mars atmosphere is similar to earth at 100,000′.
So on Mars or earth starting above 45,000 you need a pressure suit.
So what wanted is something as good as Mt Everest in term of pressure
and preferable a bit warmer than Mt Everest.
Now, for N&Zer, we can’t increase Mars gravity, but could use heavier
gas. And CAGWer would probably want some super greenhouse gas.
“Following these first proposals a variety of other partially even more ambitious concepts to terraform Mars were published. The include (1) changing the orbital eccentricity of Mars’ orbit around the Sun and (2) changing the obliquity of Mars’s spin, (3) channelling of volatile-roch cometary nuclei into the Martian atmosphere, (4) seeding of Martian atmosphere with heat-absorbing, cloud-forming particles, (5) heating the polar caps using large spaceborne mirrors, (6) devolatising of the carbon within the Martian crust, (7) inducing large-scale drainages of potential Martian aquifiers, (8) the introduction of microbes, bioengineered to survive the harsh environment on the Martian surface, (9) the addition of bioengineered plants to lower the surface albedo and (10.) the introduction of super-greenhouse gases (GHGs)”
http://journalofcosmology.com/Mars149.html
And:
“According to the Intergovernmental Panel on Climate Change, SF6 is the most potent greenhouse gas that it has evaluated, with a global warming potential of 22,800 times that of CO2 when compared over a 100-year period. Measurements of SF6 show that its global average mixing ratio has increased by about 0.2 ppt per year to over 7 ppt. Sulfur hexafluoride is also extremely long-lived, is inert in the troposphere and stratosphere and has an estimated atmospheric lifetime of 800–3200 years.”
http://en.wikipedia.org/wiki/Sulfur_hexafluoride#Greenhouse_gas
And Sulfur hexafluoride is 5 times denser than air, and non-toxic- you can watch Y-tube of people breathing it to get a deep voice [opposite of Helium].
Fluorine and sulfur are fairly abundant. There is more Fluorine than carbon or sulfur in earth crust:
http://en.wikipedia.org/wiki/Abundance_of_elements_in_Earth%27s_crust
Existing Mars atmosphere is 2.5 x 10^16 kg or about 25 trillion tonnes- mostly CO2.
One would need to do this economically. Assume less the 1 trillion dollars.
Assume costs are about $100 per kg for anything made, oppose to naturally occurring- such as water or CO2 in comets. Assume naturally occurring could be $1-10 per kg. One could get cheaper costs if whatever needed is “waste product” or by product of some other needed commodity.
With this constraint, one get about 10 billion kg of Sulfur hexafluoride. Let’s give a round number of 25 billion kg- or 1/millionth of Mars atmosphere.
Would such a small trace gas have a significant effect upon Mars temperature?
Good thread!
Having ignored your 100 km column height initially, I had a stupid hang-up about constituent stratification of air in a full atmosphere height column. I see now that would make no difference, the entire column must still be isothermal. Thanks Willis, Dr Brown, Anthony et al, I learned a lot of detail new to me.
davidmhoffer says:
January 22, 2012 at 9:58 pm
Actually the facts we have do allow us to draw some conclusions. Let me define the “theoretical S-B temperature” as the temperature of a blackbody with a uniform surface temperature corresponding to the amount of incoming radiation. Here’s what we know.
1. For a given incoming radiation, any variation in surface temperature from the isothermal state will lower the mean surface temperature.
2. An atmosphere, with or without greenhouse gases, can reduce the temperature drop from variations in temperature. It does this by reducing the amount of day / night ∆T and equator / polar ∆T.
3. A transparent GHG-free atmosphere cannot warm the surface above the isothermal state of the theoretical S-B temperature.
Therefore: at a minimum, the greenhouse effect of GHGs and clouds must be responsible for at least the thirty degrees C that the earth is above the theoretical S-B temperature.
In addition, since the greenhouse effect undoubtedly reduces the day/night temperature swings, some additional amount of the warming beyond the 33°C is also due to the greenhouse effect.
Finally, the rest of the warming is due to the atmosphere itself reducing things like the equator / polar temperature differential through the physical export of heat from the tropics to the poles. This is a separate and independent effect from the greenhouse effect.
w.
Joel Shore says:
January 22, 2012 at 11:02 am
@Joules Verne and tallbloke: The average surface temperature of the moon is the perfect debating point if your goal is to never resolve anything. You can debate this all day but unless you carefully define how the average is taken (e.g., are you looking at the first fraction of a mm or at the first meter of the surface), you’ll get all sorts of different answers!
And, this is precisely because there are lots of different temperature distributions having lots of different average temperatures that are all compatible with the moon being in radiative balance.
And the range is?
On another thread where I addressed your concern around this issue, you started waffling about average temperature a metre down in the regolith and asked where the surface is. On the surface was my reply. DIVINER carries an MSU instrument and it measures the radiation leaving THE SURFACE, not the temperature 5cm 50cm or 100cm below the surface.
I submit that you are the one wishing to make the debate endless, in order to delay the inevitable, preferably forever.
davidmhoffer says:
January 22, 2012 at 10:26 am
Tallbloke;
I’ll be providing a mathematical proof of the misapplication of S-B on my site soon. Watch this space.>>>
To limit my misuse of time on the internet I long ago made a deal with myself to participate in one an one only forum. Please provide that mathematical proof as soon as you can and I will break my three year old deal with myself on the spot.
David, thanks for all your excellent application of reason on this thread. If you just sign up to follow my blog, you can ignore the intervening posts you will be notified of via email, and you can be in early when the S-B law misapplication post comes up.
Cheers
TB.
Willis Eschenbach: “Joe, you claimed above that you misunderstood it until you read the equations that preceded it. I went back and read them. I could not see how they changed the conclusion in any way, much less in the way that you claimed.”
Those of you who, like me until today, tended to give Willis’s conclusions a good deal of weight, would do well to look at Velasco et al.’s Equation 8 and see for yourself whether it supports the conclusion that Velasco et al. found a non-zero lapse rate. You may revise your assessment of Willis’s ability to do the math.
I know I did.
I must also emphasize that this system (Jelbring) does not represent a perpetuum mobile – it is strictly a result of elementary thermodynamics. If this is a perpetuum mobile, then the rotation of the planets around the sun in the solar system is a perpetuum mobile – I don’t think Newton would approve! He understood gravity.
Of course it does, because it violates elementary thermodynamics.
At this point, the demonstration has been made many, many times. I will repeat it. Pay attention.
First, reference has been made to one textbook with the proper thermodynamic argument for the specific case of gravity given in the text itself.
Second, reference has been made to another textbook where proving that there is no lapse rate in the specific case of a gas in gravity is a student exercise.
Third, the example of Maxwell’s Demon has been given repeatedly — any time a natural force is asserted to cause the spontaneous separation of an isolated system into reservoirs at different temperatures, one can build a heat engine that does nothing but convert heat into work with no other effect, violating the second law.
Fourth, both Willis and I have undertaken — repeatedly — to show that this is the case with specific examples. Examples include:
* Two different columns of gas with different lapse rates. Place them in good thermal contact at the bottom, so that the bottoms remain at the same temperature. They must therefore be at different temperatures at the top. Run a heat engine between the two reservoirs at the top and it will run forever, because as fast as heat is transferred from one column to another, (warming the top) it warms the bottom of that column by an identical amount, causing heat to be transferred at the bottom to both cool the column back to its original temperature profile and re-warm the bottom of the other column. The heat simply circulates indefinitely, doing work as it does, until the gas in both columns approaches absolute zero in temperature, converting all of their mutual heat content into work.
* A single, isolated column of gas with a different temperature at the bottom and the top that is maintained by gravity as the stable thermal equilibrium. Since the temperatures at the bottom and top are different, one can run a heat engine between them. Examples of heat engines that could be run that have been given include a Carnot Cycle (ideal) engine, an electrical thermocouple, a precisely balanced “dippy bird”. All of these are normal thermodynamic heat engines that take heat from any reservoir at high temperature and deliver some of that heat to a cold reservoir at a lower temperature, converting the rest into work.
If the heat engine is located completely inside the insulated column of gas, it will do work that is ultimately transformed back into heat inside the column. Since the stable lapse rate is given as g/C_p (where I have no idea why you are using C_p in your example above, since P is varying along the gas column, but it doesn’t really matter) no matter how much heat you deliver to the top of the column from the bottom in this way, no matter how much heat is transformed into work (and ultimately back into heat) the energy sorts itself out again so that the stable lapse rate is maintained. The heat engine (of any sort) will run forever, literally converting heat into work over and over and over again.
Or, one can run a heat engine that removes the work done from the reservoir and releases it into the outside environment regardless of its temperature. In this case the heat content of the gas column will be gradually transformed into work until it reaches absolute zero in temperature. Congratulations! You’ve now violated the refrigerator statement of the second law as well!
For some reason, nobody in this thread wants to confront this head on. They continue to write absurdity after absurdity, basically stating that the gas columns they describe are in thermal equilibrium because they satisfy some equation or another that they pretty much made up and asserted as the definition of thermal equilibrium. They do not bother to take ordinary and elementary precautions and determine whether or not this assertion is consistent with the laws of thermodynamics and actual definition of equilibrium in terms of heat flow. They ignore well over 100 years of physics and laws that every physicist knows cannot be broken, by asserting a system that is a textbook example of violation of the second law. Any bright undergraduate, given this as a homework problem without being told what mechanism causes the “miraculous” spontaneous separation of a gas column into hotter and colder gas in equilibrium should be able to demonstrate that it violates the second law!
As I said, IMO Jelbring shouldn’t waste his intellect on trying to explain the warming of planets. He should go ahead and patent “Jelbring’s 100% efficient heat engine”. All he needs is a tower full of, say, Xenon, a km high. Then there are any number of designs for a heat engine that take up heat at the bottom and deliver it to the top as long as there is a temperature difference between the bottom and top, and of course there will always be such a temperature difference because as fast as heat is delivered to the top, gravity will sort it back out so it falls to the bottom to maintain the thermal gradient.
That patent can join the vast list of similar patents for PMMs, at least as soon as one builds a working model. The patent office no longer accepts applications for PMMs without a working model.
BTW, it is absolutely trivial to test the Jelbring hypothesis, and for that matter the N&Z variant (which I’m still working through, as N&Z’s applies to an open system where Jelbring’s is closed by assertion and hence easier to instantly contradict, thanks Anthony for the new post of N&Z with more detail and annotation). It is asserted that air will have an approximate static lapse rate of C_p/g in thermal equilibrium. The actual atmosphere is never, of course, in anything like thermal equilibrium — heat is always flowing from the ground out to space over 80-90% of the Earth’s surface at any given time, and where it isn’t it is only because there is a thermal inversion that is “impossible” according to both N&Z and Jelbring, e.g. over Antarctica during its winter.
However, it is a tabletop experiment to equip a simple centrifuge with recording thermistor buttons glued into the “top” and the “bottom” of a Dewar cylinder containing ordinary air and leave it long enough to reach thermal equilibrium inside the cylinder. We’re talking undergraduate science here, something you could very likely manage with items purchased at Wal Mart. Spin the radially aligned Dewar up to 100 to 1000 g and you should be able to get an easily measureable permanent thermal gradient in the pseudogravitation of the centrifuge.
You won’t, of course. If you could, once again such a system could be used to build a PMM — build large toroidal disk at a Lagrange point, spin it up to 1000 g’s, run a series of thermocouples between the inner and outer thermal reservoirs, and it will generate electricity until it cools the gas in the cylinder to zero without energy input (or if the thermocouple runs inside of the adiabatically isolated torus, it will drive a motor forever).
Looking forward, therefore, to easily reproducible tabletop verification of Jelbring’s hypothesis, followed by any number of PMM patents.
rgb
jae says:
January 22, 2012 at 7:09 pm
Willis, don’t you not have anything to say about the empirical evidence? Why do other planetoids with atmospheres show about the same amount of “warming, DESPITE the amounts of GHGs present? “Why is it not getting warmer these days, when CO2 emissions are increasing drastically? The “bottom line” is always empirical evidence, but you just seem to shake it off!
I keep wondering just WHY you will not address this issue. You have no problem addressing other issues. Come on!
Could you have a case of confirmation bias?
jae, I think you’re being a little hard on Willis. Think of the wrongly named GHE this way. The atmosphere is like an electrical wire. The energy absorbed by the surface travels through the atmosphere to the effective radiating altitude where it is emitted to space. The warming of the atmosphere is like the resistance in the wire.
Now, what allows the energy to flow are the GHGs. They allow the atmosphere to conduct the energy. Without them there would be no way to the energy to flow. Hence, GHGs are required to close the circuit if you will. It would do no good for Willis to deny that GHGs are required.
I think this analogy may be more realistic than many think. If the effective radiation altitude is set by the relationship found by N&Z (i.e. the turbulence of gases in a gravitational field) then you have the empirical evidence that increases in CO2 would do little to change the temperature and yet still require GHGs like CO2 to be present to enable the flow of energy.
tallbloke says:
Frankly, that is just a very silly statement. I have been the one who has consistently said that the debate over the temperature of the moon is a distraction from the discussion…from the very first time I commented on it over two weeks ago: http://wattsupwiththat.com/2011/12/29/unified-climate-theory-may-confuse-cause-and-effect/#comment-857787 And, I have very clearly explained why this is the case.
I have also predicted that a lot of time would be wasted arguing about irrelevancies such as the average surface temperature of the moon: http://wattsupwiththat.com/2011/12/29/unified-climate-theory-may-confuse-cause-and-effect/#comment-860185
Joel Shore says to Tallbloke
“Frankly, that is just a very silly statement. I have been the one who has consistently said that the debate over the temperature of the moon is a distraction from the discussion”
Well in the Halpern et al paper that Joel signed up to section 3.3 starts
“The Moon is a good example to contrast with the Earth”
Perhaps a little consistency is required here!
Or maybe like everyone else you realise that the Halpern et al paper is riddled with mistakes.
gbaikie says at 1/22 8:30pm:
“I don’t see why anyone would think temperature is related to potential energy.”
If temperature is not related to molecule’s PE then can construct a Perpetuum Mobile!
Consult Caballero text top post link for ideal gas p. 12: ” The number of molecules can be expressed as a density, and, as we will see, the mean speed can be expressed as a temperature….Temperature is just another name for the mean kinetic
energy density of molecular motion.”
Consider Willis’ premise of a GHG-free atmosphere air in adiabatic (insulated) tall cylinder in earth’s gravity field g at room avg. temperature. Zoom in on one air molecule speeding around in that cylinder (an ideal control volume allowing no energy in or out).
This molecule has a mass m and speed v. As gbaikie writes, the molecule’s kinetic energy KE is 1/2mv^2.
This molecule is of mass m & at any height h has potential energy PE of m*g*h.
The molecule’s total energy is KE + PE.
The thermo 1st law tells us total energy cannot be created or destroyed but it can be changed from one form to another (thermal to potential to chemical and back – if this process is irreversible 2nd law deduces entropy increases).
The molecule’s total energy = KE + PE which therefore must = constant always since energy can’t be created or destroyed.
Hence as the molecule speeds around in the cylinder & colliding w/o energy loss (an ideal gas!) 1st Law tells us molecule must conserve total energy changing PE into KE and back & forth w/no entropy increase ideally since their sum always = constant for each molecule. The molecule’s thermal energy must change with PE introduced by changing height in the gravity field. Why?
Consider molecule happens to move up against the gravity field or h increases. Compute molecule’s PE = m*g*h so it increases.
The molecule’s speed decreases as it moves up against gravity field so its KE decreases as 1/2mv^2 decreases.
This is consistent with 1st Law: PE up, KE down so KE + PE = 1/2mv^2 + mgh = constant & unchanged or conserved. Focus on KE decreasing here.
Robert Brown & Caballero tell us the thermal energy of the molecule is its KE and ideal gas law tells us PV = nRT. In the cylinder: V,n,R are bulk constants. So P ~ constant * T in the cylinder. If KE decreases with height thermal energy decreases with height.
Air is compressible gas & since P is higher at bottom of cylinder due to weight of molecules above in the gravity field, T must be higher because P proportional constant * T.
Robert Brown (and thus Willis’) ignores the upward moving molecule’s decreasing thermal energy thus decreasing T (from decreasing KE and increasing PE) and thus they ignore 1st Law of thermo. This enables Robert and Willis’ to construct a Perpetuum Mobile by inserting a silver rod in the gas cylinder.
Again, that is why it is important that temperature is related to potential energy of the molecules w/gravity.
If you don’t do the KE + PE = total energy math right, a Perpetuum Mobile looks possible to construct. It is not possible; Robert and Willis’ haven’t done the total energy math right & thus they don’t apply thermo physical laws correctly in the presence of gravity field.
In real life, I can really construct a highly (but not perfectly) insulated cylinder of air. I can really drop in an (insulated or not, superconducting or not) silver wire. No Perpetuum Mobile results. Not even close.
Why? 2nd thermo Law tells us “differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system”, no heat energy flows in Willis’ premise one cylinder in equilibrium. Why is this so hard?
Since apparently even the past thermo grand masters & their non-ideal real experiments could not agree whether the T and P fields are ideally stratified or isothermal, that can & will be debated all you want. Have at it.
NB: if Willis’ premise air column really is isothermal as Robert Brown claims, buoyancy disappears from Willis’ premise – a story for another day.
Ideally stratified or isothermal answer really doesn’t matter much in real life since the atmosphere is agreed to really have a lapse rate. Real air molecule KE & thus air thermal energy or temperature generally decrease with height as the air molecules move up against gravity, generally decreasing KE and increasing PE keeping their total energy constant.
(I write generally atmosphere T decreases with height – lapses – but that is not always true b/c the real atmosphere has very complex heat flowing & V is not constant unlike in Willis’ premise.)
For my view, entropy inside Willis’ ideal air column is constant & there is a stratified P & T field since ideally pressure and temperature decrease w/increasing height. Makes it easier to think thru at least.
This became important to consider recently because of the discussion of the new N&Z paper(s) and how much of the real atmosphere lapse rate is due to P&T field gradients and how much lapse rate is due to earth’s atmosphere radiating to very cold space.
No matter what you drop into Willis’ control volume, thermocouple, silver wire, drinking bird, movable open or closed jars positioned anywhere, heat or chemical energy cannot be made to flow in this one reservoir at equilibrium due to 2nd Law. There is no possible Perpetuum Mobile in Willis’ premise.
Willis’ air column is only one block of energy! Put two of Willis’ premise “cylinders” together and then heat will flow if Tavg. is different. As Robert points out repeatedly.
Apologies if the following has already been pointed out, but if I wait to reply until I’ve read the whole thread I’ll forget what I wanted to say! I’ve been following the discussion on previous threads but only now on my new computer am I again able to post comments.
First, I’m delighted that someone has at last realised that an atmosphere in thermal equilibrium must be isothermal. “Same temperature” and “thermal equilibrium” are just different ways of saying the same thing.
Jelbring erroneously “thinks total energy evenly distributed per molecule is the final state of energetic equilibrium”; however, Willis is also mistaken in saying “the equilibrium state is when the energy is evenly distributed per volume and not per molecule”. There is no such requirment at equilibrium (in an isothermal atmosphere there is a lot more energy per unit volume at low altitudes, where the air is denser).
Willis is also mistaken in claiming that Jelbring’s model violates energy conservation. It does not violate the First Law of Thermodynamics (permitting the operation of a Perpetual Motion Machine of the First Kind), but the Second Law (a Perpetual Motion Machine of the Second Kind, which destroys entropy). The power from Willis’s thermopile, doing work within the model environment, will endlessly dump its waste heat back into that atmosphere. Still impossible, of course.
Jelbring’s hypothesis founders on that basic error, but, like Willis, I am unable to make sense of the Nikolov & Zeller theory. It reads like gibberish. So I’m not sure if they’re making the same main mistake, or a different one.
Caveat 1: A planet can continuously radiate more power than it receives from the sun due to the release of gravitational potential energy in its atmosphere. Jupiter is an example. It has been doing so for billions of years, gradually shrinking in size (since it has no real “surface” we can consider it all atmosphere, for present purposes). Such gravitational collapse can even heat the core of the very largest planets to the point at which thermonuclear fusion ignites (we call such planets “stars”!). Energy can also be released by tidal friction (Io is a good example). Or, as in the case of the Earth, by radioactive decay (fission is negligible), which contributes some four orders of magnitude less power than the sun, and is therefore considered climatalogically insignificant.
Caveat 2: In General Relativity, an equilibrium state is possible in which the temperature is higher at the bottom of a gravitational well than at the top. Consider a large gravitating body, surrounded at a distance by shell at uniform temperature T1. Let the red shift climbing out of the well from the surface to the shell be Z. Then the temperature of the surface, at equilibrium, will be T1.Z. The surface radiates Z**4 times as much power as the shell.
So how is this equilibrium? First, every outgoing photon drops in energy by the same factor Z. Second, time runs Z times more slowly at the surface. Third and fourth, only some of the photons manage to reach the shell, the rest following trajectories that curve back down to the surface, a factor of 1/Z in each of two orthogonal directions (other ways of looking at this are to note that the aberration of light compresses the sky into a cone, or that the horizon curves up into bowl). And that makes up the Z**4.
It is not possible to use this gravitational temperature “difference” to drive a perpetual motion machine because any connection between surface and sphere has to deal with this same fundamental time differential or change of reference frame. Convert the proper temperature from one frame to the other (by means of the factor Z), and there’s no temperature difference any more. [This is in sharp contrast to the nonrelativistic Jelbring model, in which the top and bottom of his atmosphere could be connected in parallel by two columns of differing scale height and lapse rate (eg, air and helium).]
This is my closing comment. I have already shown explain my view about a static atmosphere here: Marc77 says: January 22, 2012 at 8:59 am
Said shortly, back-radiation of IR photons is a back-radiation of momentum and kinetic energy. The phenomenon is similar to how gravity “back-radiates” momentum and kinetic energy of gas molecules. Gravity also cool photons as they get out of a gravity field based on general relativity. The gradient of cooling as a particle goes up in a gravity field is proportional to the ratio of momentum to kinetic energy. So heavier the gas molecule will lead to a higher lapse rate. This effectively leads to the possibility to generate energy from the differential of temperature at different heights. This means that a static atmosphere under gravity is highly unstable. So gravity alone is enough to generate convection cells.
That’s it. My model of a static atmosphere is right in predicting an exploitable differential of temperature. It means that convection cells spontaneously appear on nearly every planets, even if the surface is uniformly heated and the atmosphere is perfectly transparent. These convection cells are necessary to come close to an isothermal equilibrium. But don’t forget that a planet with a temperature is not in thermal equilibrium with space, so it is normal to never have an isothermal equilibrium.
Paul Birch said:
“The surface radiates Z**4 times as much power as the shell.”
Sorry, I should have said that the surface radiates Z**4 times as much power as the shell per unit area
Joe Born says:
January 19, 2012 at 6:57 pm
“If you reduce the number of molecules to a fairly small number, you can see that Willis’s argument, beguiling as it is, is wrong. Consider a single-molecule system, for example. Does anyone doubt that the molecule’s translational kinetic energy is greater when it is lower in the gravitational field than when it is higher?”
Oddly enough, on average it isn’t. It just goes higher less often. The mean square velocity at any height is the same (given a Boltzmann distribution of velocities as the molecule leaves the surface).
Willis Eschenbach;
Actually the facts we have do allow us to draw some conclusions. Let me define the “theoretical S-B temperature” as the temperature of a blackbody with a uniform surface temperature corresponding to the amount of incoming radiation.>>>
And there lies your mistake Willis. If the earth had a uniform temperature and was exposed to uniform insolation, your position would have merit (though I’d still have a couple of quibbles but they are minor by comparison).
One the of (many) major points that N&Z are making is that earth is NOT a uniform temperature and is NOT exposed to uniform insolation. When one takes these factors into account, one arrives at completely different numbers and completely different conclusions. Of these, I class three as being most important:
1. Taking the lack of uniformity into account, we are seaking a mechanism by which the surface temperature of earth is raised, not a mere 33K above theoretical blackbody, but 133K. Unless we know what disparity we are trying to explain, we can’t theorize a mechanism to explain it.
2. Given that we are seeking to explain a disparity of 133K, can we, as you have done in your response, simply subtract the lower number from the higher number and attribute it to GHE? We cannot. That could only give us the value for GHE is all other factors are zero, and as you will see in my next point, they are not.
3. In a system in which neither surface temperature nor insolation are uniform, the relationship of P to T^4 biases the use of averaging from a static data set in multiple ways, but for the purposes of this discussion, I shall include only one. For lack of a better term, I shall call it “currency”.
In terms of temperature, an increase of one degree in the tropics requires considerably more watts/m2 than does an increase of one degree in the arctic. If we simply average temperatures (and more importantly, CHANGES to temperature) between the tropics and the arctic, what we are in effect doing is averaging two different currencies.
You live in the United States where you have a currency called the “dollar”. I live in Canada where we also have a currency called the dollar. For sake of illustration, let’s say that the exchange rate is that one Canadian dollar is worth $0.90 American dollars.
Now, suppose we each have $100 dollars. Who has more money? You or me? You do, because your dollars are worth more than my dollars. If you sent me your $100 and I converted it to Canadian dollars, how much money would I have? I would have $211! I just “invented” $11! If we ignore the currency conversion, we started out with $200 and an average of $100 per person. By sending your $100 to me in Canada, and performing the currency conversion, we now have an average of $105.50 each. But in terms of absolute buying power, the amount of money hasn’t actually changed. All that has changed is how we measure it.
Let’s bring this back to “temperature”. The tropics absorb more heat than they radiate. The extra heat is moved from the tropics to the temperate and arctic zones by various mechanisms. The tropics are as a result a number of degrees cooler than they otherwise would be, and the arctic zones are a number of degrees warmer. But we cannot average these numbers, even after adjusting for area, because the currency is different. If the tropics cool by 10 w/m2 and the arctic warms by 10 w/m2, and we average T, we will get a higher average T than we started with. The currencies are different. Watts/degree in the tropics is a different value than watts/degree in the arctic. The energy balance (buying power) is exactly the same, but the average of T (dollars per person) is higher due to the “currency conversion”.
Now, let’s wrap all of that up and return to calculating how much of the observed difference between temperature and theoretical blackbody we can attribute to GHE. 100%? Not a chance!
We first have to account for the amount of “degrees” that the tropics “sent” to the temperate and arctic regions and account for the watts (currency) conversion. I think this is where people get lost in understanding N&Z.
How does the currency get moved? It gets moved by a lot of different processes including conduction, convection, oceanic currents and yes, back radiation from GHG’s. Each one “moves” a certain number of degrees from one place to another, and we have to do the currency conversion (in watts) to understand how much is due to what. But this brings us to the seminal point of N&Z.
At what point does the earth stabilize in terms of how many degrees are sent from tropics to arctic? Answer: It stabilizes when thermal equilibrium is reached. Aha!
Since stability occurrs when thermal equilibrium is reached, or explained another way, when the temperate and arctic zones warm sufficiently that they are radiating at a net loss equal to the net gain in the tropics, it matters not in the least what the mechanism for re-distributing the “degrees” is. If the only possible mechanism is conduction, then earth will stabilize when the net absorption in the tropics is balanced by the net loss in the temperate and arctic zones. If it is possible to re-dsistribute the “degrees” by convection, conduction, back radiation, oceanic currents, guys in black suits with sunglasses and brief cases filled with bundles of new unmarked “degrees”, then the planet will reach thermal equilibrium when…the temperate and arctic zones warm enough to radiate a net loss that exactly equals the net gain in the tropics.
So mechanism for moving the “degrees” doesn’t matter. The only thing that matters is that there is at least one. Beyond that, one or twenty, thermal equilibrium can only be achieved when the arctic and temperate zones warm enough to reach a radiance that is a net loss that balances the net gain in the tropics. The “currency” conversion gives us the illusion of a higher temperature (more money because I moved the dollars to Canada) but there actually is no change in energy balance.
If I increase massively the number of degrees that are moved by GHG’s, or decrease it for that matter, I don’t change the fundamental requirement for the temperate and arctic zones to warm to the point where their net loss equals the net gain of the tropics. The only think that I change is which processes do how much work. The end result, thermal equilibrium, is the same.
Which is why GHG’s do not matter.
KevinK says:
January 19, 2012 at 7:42 pm
“I ask myself, if the Greenhouse … Effect is REAL … WHY is it that NOBODY has figured out how to apply it to any practical problem…?”
They have. Keeping plants warm. They’re called greenhouses.
People keep saying that the greenhouse effect is badly named. It isn’t. The analogy is actually pretty good.
Greenhouses keep things warm by
1) Blocking direct convection to the cold outside (though indirect convection on either side of the glass still occurs).
2) Making it harder for the thermal radiation to get out than for sunlight to get in (most glass has quite a high long-wave absorptivity/emissivity and even glazing that was long-wave transparent would still have long-wave absorbant water condensing on it at night).
The atmospheric “greenhouse” keeps things warm by
1) Blocking direct convection to space (because the density of the atmosphere drops off to near vacuum above ~100km) (though indirect convection to other parts of the globe still occurs).
2) Making it harder for the thermal radiation to get out than for sunlight to get in (because the main greenhouse gases like water vapour and CO2 are strong absorbers of long wave radiation, but transparent to visible light).
Paul Birch thinkstheresa greenhouse effect!
Here is an interesting paper especially as it comes from a source with no “spin” on the AGW debate.
The way I read the paper is it gives massive support for the conclusions of the famous Woods experiment.
Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.
Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.
The addition of IR absorbers to the plastic made it equivalent to “glass”
The results of the study show that( Page2 )
…”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf
Tallbloke
If you just sign up to follow my blog, you can ignore the intervening posts you will be notified of via email, and you can be in early when the S-B law misapplication post comes up.>>>
Been there, done that, looking forward to it.
Will you lift Joel Shore’s ban for that thread? He’s already stipulated to the accuracy of N&Z’s math in another thread, it should be interesting to see if he continues to try and dispute the matter in some other way.
Somebody needs to put the substance of this argument in context. Logically the effect of unwinding the lapse rate is going to equal the effect of the lapse rate.
So since we will never observe the unwinding of the lapse rate its a bit like trying to count angels dancing on the head of a pin. If everybody is in agreement the lapse rate has a warming temperature effect on the surface the focus needs to be on the quantity of that.
I forgot something in my last comment. I’ve shown that gravity does generate a lapse rate by itself. Now, a greenhouse gas molecule cannot heat the surface without a lapse rate. If the molecule receives as much energy from above than from below, it will also radiate equally in all directions and no heating will occur. So greenhouse gases amplify the lapse rate instead of creating it.
In conclusion, we have:
1- Gravity prevent the heat of the planet from going out too easily and creates a high lapse rate.
2- Greenhouse gases amplify this lapse rate.
3- The differential of temperatures drives convective cells and brings back a lower lapse rate.
In the end, if all radiation to space come from the ground, it will have the black body temperature predicted by S-B. The average height of emission to space determines the height of the black-body temperature.
I take a wild guesstimate on that. From eyeballing and averaging AMSU channels 5 and 6 to get a 500mb figure I come up with 246K, then subtract the effect of 102watts convection instant heat being transferred to that layer and get 207K. Then I am going to add the 102 watts back to the 288K average surface temperature and I get 306K. So I will propose the lapse rate creates 99K warming of the surface.
I made an error. Both gravity and greenhouse gases can create a lapse rate by themselves. The planet is warmer than space and greenhouse gases can turn this differential of temperature into a lapse rate within the atmosphere.
Finally(hopefully), if the model that I have expressed proved to be false. It could be that a gas is not really a bunch of discrete particles moving independently. Thanks to quantum mechanics, heat and the distribution of momentum could be a function of the whole gas. Previously, I have shown that the gravitational gradient of temperature is a function of the ratio of momentum to kinetic energy of the mediator that moves the kinetic energy around. I don’t know quantum mechanics very well, so I wonder if a group of gas particles could be able to exchange heat through virtual photons even if those gas particles are unable to emit those photons to space. If it was the case, the model that I have expressed above would not hold anymore. I have to say that the concept of discrete particles moving independently often seems to be in contradiction with the inability to transfer latent heat into work. If you could build a small ratchet, and plunged it into a gas of heavy particles at low density and high momentum, it seems the ratchet would constantly turn in the same direction. Also, how can hot air pack together and start rising together to form a convection cell if molecules really move independently? This is all out of my leagues, so I will leave with this question:
If you calculate the gravitational gradient of temperature in a gas, what is the average mass of the particle that moves heat around?
See my model above to understand what I mean:
Marc77 says: January 22, 2012 at 8:59 am
Robert Brown,
You say that the Jelbring hypothesis violates ”elementary thermodynamics”. But you have ignored the very elementary thermodynamics that I have outlined here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872145
and the post you are referencing here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872766
I also posted a similar reply to you on the Tallbloke blog, which you also ignored.
If you would address the thermodynamics that I have outlined, I would be very grateful. I have presented basic equations, not a “thought experiment”. If there is something wrong, I would very much like to know what it is. Point out the “absurdities” please.
But I have a bad feeling in my stomach when I see you say things like:
”Since the stable lapse rate is given as g/C_p (where I have no idea why you are using C_p in your example above, since P is varying along the gas column….)”
You also throw the word “equilibrium” and “thermal equilibrium” around very loosely and I would appreciate your thoughts on my comments with respect to that.
Please sir, address the thermodynamics I have outlined. Then we can discuss why gravity can drive a rotating solar system but cannot drive the mass transfer in the troposphere.
I would also appreciate your thoughts about the significance of an isothermal troposphere eliminating the “green house effect” completely.
Bill
davidmhoffer says:
January 23, 2012 at 9:26 am
Willis Eschenbach;
Actually the facts we have do allow us to draw some conclusions. Let me define the “theoretical S-B temperature” as the temperature of a blackbody with a uniform surface temperature corresponding to the amount of incoming radiation.>>>
And there lies your mistake Willis. If the earth had a uniform temperature and was exposed to uniform insolation, your position would have merit (though I’d still have a couple of quibbles but they are minor by comparison).
One the of (many) major points that N&Z are making is that earth is NOT a uniform temperature and is NOT exposed to uniform insolation. When one takes these factors into account, one arrives at completely different numbers and completely different conclusions. Of these, I class three as being most important:
The conventional calculation does not require uniform illumination.
The major reason they get ‘completely different numbers’ is their assumption of the darkside of any planet being always at ~3K, in other words negligible heat capacity. This doesn’t even come close to being true even for the moon. It would be a lower bound to the mean temperature but not a very useful one just like the conventional assumption is an upper bound (which is not a bad one for the Earth).
Hans Jelbring says:
January 20, 2012 at 4:55 pm
“…tell where I am wrong in my paper which is common rutin in science or skip the thing.”
You’ve been told, not only by Willis but by qualified physicists too. Wherever you have a temperature gradient you don’t have thermal equilibrium, by definition. This is basic thermodynamics. Your assumption to the contrary – that equilibrium equalises the total energy per molecule (including gravitational Potential Energy) – is plain wrong, contrary to the known principles of physics. It was an easy mistake to make, but now’s the time to bite the bullet and ‘fess up. Sorry.
William Gilbert says:
January 23, 2012 at 11:37 am
Apparently you missed my comment above. I’ll repeat in more detail.
Your thermodynamic argument is circular and assumes the conclusion. It’s the logical fallacy of begging the question. Requiring that U remain constant with altitude is exactly equivalent to requiring adiabatic expansion. Thus it is no surprise that the lapse rate under those conditions turns out to be the adiabatic rate. An isothermal column does not violate the First Law because the needed energy to increase U with altitude is supplied by conduction from the surface. Heat conduction will always happen when there is a temperature gradient. At low pressure, the temperature gradient will be reduced to zero faster because thermal diffusivity increases as the inverse of the density. As a result, the surface will be cooler and radiate less energy than it receives until the atmosphere above it is isothermal. Note that this does not happen unless the planet surface is isothermal because otherwise the temperature differential between the equator and the poles and between the night and day side will drive winds that will generate turbulent diffusion and convective heat transfer that will overwhelm conduction by orders of magnitude and force the lapse rate away from zero towards the adiabatic rate.
Thoughts? Of course there is no greenhouse effect if the troposphere is isothermal. Who ever said there would be? But there is also no greenhouse effect if the atmosphere is transparent regardless of the temperature profile.
Bart says:
January 20, 2012 at 6:04 pm
“Stefan-Boltzmann is an equilibrium relationship.”
Sorry, but that’s not true. The black body radiation from a surface (or volume) depends only on its temperature and emissivity. Whether or not it is at equilibrium from the point of view of radiation, conduction or convection is irrelevant. Oh, I dare say one could invent some clever material whose emissivity changes with heat flow (cf. photochromic glass), or somehow use the non-equilibrium energy flow to drive a laser, but that has nothing to do with the everyday thermal radiation from the Earth, for which the emissivity is almost always and almost everywhere very close to unity.
Paul Birch: “‘Does anyone doubt that the molecule’s translational kinetic energy is greater when it is lower in the gravitational field than when it is higher?’
“Oddly enough, on average it isn’t. It just goes higher less often. The mean square velocity at any height is the same (given a Boltzmann distribution of velocities as the molecule leaves the surface).”
Thank you for actually engaging; if I didn’t know better, I would have thought everyone here is afraid of math.
But perhaps you can help me out a little further with my own math. To keep things at a level I can wrap my mind around, let’s consider just a one-molecule system whose total energy E = mgz_max, where m is molecular mass, g is the acceleration of gravity, and z_max is some height above the bottom of the column.
Now, the only places I know for the energy in our hypothetical monatomic-ideal-gas system to go are (1) potential energy and (2) translational kinetic energy. If the potential energy at altitude z is mgz–and people who claim to know this stuff tell me that it is–then the kinetic energy K must be the total energy minus that potential energy: K = mg(z_max – z) = E(1 – mgz/E). No matter how frequently or infrequently the molecule visits altitude z, its kinetic energy will have that same, altitude-dependent value, no? So how would its average, i.e., the mean translational kinetic energy for that altitude–be anything other than that value?
And that value, which is exactly what Velasco et al’s Equation 8 gives for f = 3 and N = 1, decreases with altitude: there’s a temperature lapse rate.
But, then, I’m not a physicist. Maybe there’s something in that mean-square-velocity stuff you mentioned that makes the kinetic energy something different?
Anyway, any guidance you could give would be greatly appreciated.
Willis Eschenbach, just out of interest – have you downloaded and read “Fourier 1824” yet?
After all he is supposed to be the “father” of the GH theory.
Paul Birch says:
January 23, 2012 at 9:35 am
Yes!
I’ve been trying to make this argument to people who should know better without much success. Expect people to throw the 1909 Wood experiment back at you. But several people have demonstrated, starting in 1767 with Swiss physicist Horace de Saussure, whom Fourier mentioned in his 1824 paper as posted by O H Dhalsveen above, that IR opaque glazing does increase the temperature in an absorbing box compared to IR transparent glazing. Wood himself was rebutted at the time by Abbot, who also pointed out that Wood’s measured temperature increases were much smaller than he should have observed. Recent confirmation of de Saussure’s results and failure to confirm Wood include Professor Vaughan Pratt. A recent article claiming to have replicated Wood’s results can easily be shown to be incorrect and, in fact, proves de Saussure, Abbot and Pratt to be correct.
It’s an easy experiment to do and would be, in my opinion, make a great Middle School Science Fair project.
davidmhoffer says:
January 21, 2012 at 8:29 am
“Equation 8 in N&Z’s article has two variables, and anyone who can read can see that. All their results are calculated using THAT equation. Trying to represent it as having four variables by pointing to an intermediate step is an outright lie.”
Their equation has two variables and four free parameters. Do you not understand the difference between a variable and a free parameter? N&Z could also have chosen the form of the equation differently (for example, with three or four power terms instead of two); indeed, we have no way of knowing how many other forms they tried before coming across one that seemed to work. Given that there are only about six planets being fitted (excluding airless bodies) information theory tells us that such an empirical fit is wholly unconvincing. They do give what I suppose is intended as a physics justification for their model, but this I find incomprehensible – probably because it’s complete nonsense.
Joe Born says:
January 23, 2012 at 1:31 pm
“Paul Birch: “Oddly enough, on average it isn’t. It just goes higher less often. The mean square velocity at any height is the same (given a Boltzmann distribution of velocities as the molecule leaves the surface).”
But perhaps you can help me out a little further with my own math. To keep things at a level I can wrap my mind around, let’s consider just a one-molecule system whose total energy E = mgz_max, where m is molecular mass, g is the acceleration of gravity, and z_max is some height above the bottom of the column.”
But that’s the flaw. Even your one-molecule system doesn’t have the same energy all the time. Every time it hits the surface it bounces back with a different energy, sometimes higher, sometimes lower. That energy has a Boltzmann distribution (which you can calculate by summing the squares of a gaussian distribution of velocities along the three orthogonal axes, x,y and z). The molecule bounces in all different directions too. So the height it reaches is different each time. The greater the height the less often it reaches there, in fact, the probability of reaching a given height drops off exponentially. Only the most energetic bounces pass though the higher altitudes. So, surprisingly, the overall effect is that the mean square velocity turns out to be independent of height.
DeWitt Payne said @ January 23, 2012 at 1:39 pm
I respectfully disagree; I made a greenhouse so that I could continue to grow tomatoes, eggplants, peppers and cucumbers. I used IR transparent polythene for the cover as the IR opaque sort costs more. It seems to work more than well enough. The top is covered with 50% shadecloth to restrict heat gain. Most days during summer, the vents are wide open to restrict heat gain. It seems to work by restricting convection, not IR radiation.
Heat loss at night is an issue, but that is restricted by a second film of IR transparent polythene some 15 mm away from the outer film. This appears to work by reducing heat conduction, rather than IR radiation. While marginally more expensive than a single layer IR opaque film, this arrangement provides superior restriction of heat loss.
It seems an odd sort of analogy that equates reduced convection and heat conduction with radiative transfer of heat. Or are my aging brains missing something?
davidmhoffer says:
January 22, 2012 at 1:03 pm
Nick Stokes says:
January 22, 2012 at 12:12 pm
davidmhoffer says: January 22, 2012 at 11:24 am
“You mean other than attributing 33K to the GHE is totally and completely wrong?”
People sometimes say that Earth with the same atmosphere free of GHG would be 33K cooler. What you and N&Z claim is that Earth with no atmosphere at all would be some different temperature. This is a completely different proposition and, in my view, of very little interest.
REPLY: Nick and I find ourselves in rare agreement – Anthony
***********************
Then help me out guys. If we don’t know what the temperature of the earth is with no atmosphere, and we don’t adjust the temperature profile that results for heat capacity and latency (time constant), then on what basis do we quantify atmopsheric effects?
=====
Hmm, the problem here is “if we don’t know” doesn’t equal “is not known”.
If y’all would put the Water Cycle back into the ‘energy budget’, you could begin discussing the real world and not that imagined in KT97 and variations on the theme…
The standard figures for the cooling role of water in the water cycle is:
Earth with atmosphere as we have it now: 15°C
Earth without any atmosphere at all: -18°C
Earth with atmosphere but with no water: 67°C
As I’m sure you spotted immediately on reading this, the water cycle reduces the temperature of the Earth with our atmosphere by 52°C, to bring it down to the 15°C.
Without the Water Cycle, the Earth would be 67°C, water vapour is the main greenhouse gas, it cools the Earth, think deserts. Therefore:
greenhouse gases cool the earth
‘Greenhouse gas warming’ is a misnomer.
Time to overturn the AGWSF paradigm.
Paul Birch: “Every time it hits the surface it bounces back with a different energy, sometimes higher, sometimes lower.”
Thanks for your answer. I wonder if I might impose upon you to clarify a couple points in the above passage.
What “surface” did you mean? Actually, let’s just concentrate on the surface off which the molecule bounces at the bottom of the column. At that surface its potential energy is zero, right? So the “different energy”.you must be talking about is all kinetic energy? If the potential energy is zero, and the kinetic energy is “sometimes higher, sometimes lower,” then the total energy is changing? But those other guys say energy can’t be created or destroyed. Who’s right?
Joe Born says
“Thanks for your answer. I wonder if I might impose upon you to clarify a couple points in the above passage. What “surface” did you mean?”
In the derivation of the ideal gas laws the molecules are said to make perfectly elastic collisions with the container walls.
In reality the walls are ‘sticky’.
DeWitt Payne,
A few thoughts:
First, your link by Abbott is only Abbott’s argument. He apparently did not conduct Wood’s experiment in an effort to replicate it, so we can discard that link. It is replicating the experiment that is important. [Very reasonable sounding arguments were formulated over hundreds of years to explain the retrograde motion of planets, pre-Kepler. We know how those arguments turned out.]
Next, Prof Pratt does not replicate Wood’s experiment. Wood explains:
“The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed.”
Pratt’s boxes are completely empty, as can be seen. Wood packed his enclosures with cotton to eliminate convection; his experiment was based on the idea that the “greenhouse effect” was due to convection.
Finally, others have claimed to replicate Wood’s experiment exactly and with the same result. I agree that this would be an interesting high school experiment. But all variables must be accounted for, the thermometers must be switched and the experiment repeated, etc.
I agree that this seems an easy experiment to conduct. Unfortunately, Pratt eliminated a major factor, and so he cannot claim to have replicated Wood’s experiment.
Paul Birch says at 1/23 3:25pm:
“Even your one-molecule system doesn’t have the same energy all the time.”
It does if you mean total energy. It doesn’t if there is gravity field AND you mean thermal energy or mean kinetic energy. Check the 1st Law.
Have to be careful in that temperature (thermal energy) is really defined as a mean energy of system of molecules but thinking about the one molecule reveals some fundamentals easier. In the ideal GHG-free gas w/gravity in ideal container thought experiment, the one-molecule does have the same total energy all the time, TE = KE + PE (1st Law energy conservation). As the molecule zooms up in the gravity field, reduce KE (thermal energy) and increase PE for same TE. The molecule does have the same total energy all the time.
At bottom of control volume or cylinder, it is convenient to set h=o and at that point TE = KE +0 so we can know TE at all other h if we know the molecule speed and mass at h=0. I imagine, but would have to do the math, that there is an h at top of column where the molecule speed just hits 0 and TE = 0 + mgh, then falls back down under gravity. At top then the molecule reaches 0K thermal energy or absolute zero.
“That energy has a Boltzmann distribution….So, surprisingly, the overall effect is that the mean square velocity turns out to be independent of height.”
This should not be surprising, the Maxwell-Boltzmann distribution is for the special case of there being no gravity field operating. So, yeah, the weightless M-B particle speed distribution is independent of height, by its assumption of the molecules moving freely.
Bryan says:
Bryan:
You seem to have a problem understanding that context is important for a discussion. For the discussion we are having here, about the errors that render the N&Z paper nonsense, the debate over the temperature of the moon is a distraction.
Also, you will note that when we discussed the temperature of the moon in Halpern et al., we noted that the moon has wide temperature swings and for the most part avoided trying to define a single average temperature for a celestial object with such a wide temperature variation, noting only that a ballpark “average” that you get from taking the mid point of the mean day and night temperatures is somewhere in the range of what you expect, i.e., between the extreme of what would be predicted if the temperature just followed the local insolation and the other extreme of what would be predicted for the case of a uniform temperature.
We did not spend a lot of time arguing about exactly what that average temperature was. Hence, this actually shows a strong consistency in our position regarding getting bogged down in gory details of exactly what the “average temperature” is for a celestial object that has an extremely broad temperature distribution.
Paul Birch;
we have no way of knowing how many other forms they tried before coming across one that seemed to work. Given that there are only about six planets being fitted (excluding airless bodies) information theory tells us that such an empirical fit is wholly unconvincing. They do give what I suppose is intended as a physics justification for their model, but this I find incomprehensible – probably because it’s complete nonsense.>>>
I agree. Anything that I don’t understand is clearly nonense. For example, I don’t understand how gravity works, so it must be nonsense.
You’re essentially accusing them of curve fitting and noting that there are “only” six planets being fitted. “Only” six? They found ONE curve and it fits SIX planets.
Seems like a pretty good curve to me.
O H Dahlsveen says:
January 23, 2012 at 1:35 pm
Clueless. Got a link?
w.
davidmhoffer says:
January 23, 2012 at 6:07 pm
David … I truly don’t know what to say about this comment.
1. Paul didn’t say he didn’t believe equation 8 because he didn’t understand it. In fact, he doesn’t believe it because he does understand it. As a result, your attempt at satire is meaningless.
2. I’m not “essentially” accusing them of curve fitting, I am saying flat out that they fitted the curve … as they freely admit.
3. I’m sure it does seem like a “pretty good curve” to you, David … I’m sure it does.
w.
Smokey says:
January 23, 2012 at 4:41 pm
[my emphasis]
That would be the boxes were insulated with cotton around the outside. If the boxes had been filled with cotton that would have been packed with cotton. Also the boxes had been constructed with the window attached before cotton was mentioned. Also, there wouldn’t be much point in painting the inside of the box black and then filling it with white cotton. Are greenhouses filled with cotton? Not hardly. Vaughan Pratt’s boxes were insulated with 0.5″ thick foamed polystyrene art board (polystyrene foam with paper covering on both sides). My boxes (work in progress and the weather has been terrible here lately) are insulated with R-30 fiberglass batting inside a box made of 0.75″ foamed polystyrene board. I get an increase over ambient of on the order of 100C for a box with a polyethylene film window and 20-30C higher than that for a box with a glass window. The bottom surface of a triple layer IR opaque windowed box reached 170C. Wood only got up to 60C with a higher ambient temperature.
Why would Abbot replicate identically an obviously flawed experiment? Wood only achieved a temperature of 65C. Abbot’s experiment reached 118C, 102C above ambient. Wood didn’t even mention the ambient temperature. Wood’s results disagreed with both previous results and with theory. It was dismissed as irrelevant at the time. It wasn’t resurrected until 1990 by people who apparently didn’t bother to look at the complete story.
I think I know what Wood’s mistake was, but, as I said, the weather has been uncooperative so I haven’t been able to prove it. I’ve only had about 2 hours of clear sky in the last two weeks.
thepompousgit says:
January 23, 2012 at 3:40 pm
You’re not trying for maximum temperature. It would kill the plants. Any device that is intended to operate at a temperature different from ambient must be insulated first, like an oven or a refrigerator or your house or like a planet is insulated by the vacuum of space. Most people use polyethylene film for greenhouses because it’s cheap and easily replaced if it gets damaged by the weather and it’s good enough.
But some applications need better performance which can be provided by low-E glass ( http://www.florianproducts.com/low-e.html ). If Wood were correct, then low-E glass that is now standard in high quality double glazed windows would be a colossal scam. But he was wrong and it isn’t.
thepompousgit says:
January 23, 2012 at 3:40 pm
Most people use polyethylene film for greenhouses because it’s cheap and good enough. If you’re trying to demonstrate the greenhouse effect, though, you have to try harder.
There is a market for greenhouses constructed of high performance low emissivity glass like PPG Solarban80. If Wood was correct the entire low emissivity glass industry is a colossal scam. But he wasn’t and it isn’t.
thepompousgit says:
January 23, 2012 at 3:40 pm
My replies keep getting lost. Perhaps they’re getting bounced because I mention the trade name of a commercial product.
Anyway. Most people use polyethylene film for greenhouses because it’s cheap and good enough. But Wood was trying to demonstrate that even under what he thought were ideal circumstances, the IR transmission characteristics of the window made no difference. He was wrong and it’s not difficult to prove it.
Willis Eschenbach;
3. I’m sure it does seem like a “pretty good curve” to you, David … I’m sure it does.>>>
I once sat in a calculus class where the professor proposed a problem involving a case of beer at some unkown temperature being immersed in a lake of known temperature and the temperature of the case of beer being taken at a couple of points in time after the beer had been immersed. It was a long time ago, there may have been other factors, but that was the essence of the problem. The professer asked for equations to determine the temperature of the case of beer at various times in the future. I got the answer via three pages of calculus, as did much of the rest of the class.
The guy beside me however pulled out a sheet of natural log paper, plotted what little data we’d been given, struck a line through it, and got almost exactly the same answer.
That’s a pretty good curve.
Now if you can set your sarcasm aside for a moment, please help me understand this. There is in fact a curve, is there not? Yes, N&Z may have erred in how they determined the curve. I haven’t gone back through the math to make any judgment. There are those far more versed than me in both the calculus and the physics, both of which were a very long time ago for me. And, even if they erred, I refuse to throw the babay out with the bath water. They are correct on a number of points, including the fact that straightforward averaging of P and T arrives at conclusions regarding GHE and other issues that are unsound. The effective blackbody T of earth is not 255, it is less, a lot less. By moving energy from the tropics to higher latitudes, the energy balance can be completely undisturbed, yet yield a higher average temperature, without GHG effects to do it. They got a lot more right than they got wrong, and they got some things wrong for the right reasons. I think they erred in useing surface pressure for example, I think they should have focused on density. The higher the density, the more efficient processes such as conductance and convection are at moving energy from low to high latitudes, which in turn results in a higher average temperature though not a higher average radiance, and this relievs their theory of needing to account for higher temperatures derives from compression of gas. I maintain that the driving factor is the efficiency with which energy ismoved from low to high altitdues, and the higher that efficiency, the more uniform the temperature of the planet, and from that, the higher the average T though effective T via SB Law remains unchanged and no violation of laws of thermodynamics occurs.
But at days end, either there is a curve, or there isn’t. If I was intent on refuting N&Z, I’d start with the empircal data. What are the surface pressures of those planets and what are there temperatures? Does that yield a curve suggestive of a mathematical relationship or not? If the curve plot zig zags across the page, not. If it follows a smoothly curved line of some sort, then I have only one question.
What defines the curve?
If N&Z arrived at the right curve via the wrong means, there still remains a curve to be explained.
DeWitt Payne said @ January 23, 2012 at 8:18 pm
Lost posts: Sometimes they get stuck in the spam filter & get released by different mods. I find it helps to send a message to the mods to delete the “extras”. Sometimes the message really doesn’t get through. As a matter of course these days, I copy the post to the clipboard, or Notepad just in case.
Greenhouse cladding: I looked up the difference between IR treated polythene and ordinary. The treated gives between up to 1.5 C above ambient at night. On really cold nights it is next to useless so not much use in a greenhouse. I suspect that it might pay for itself in a hothouse when you are paying for fuel. I am familiar with LoE glass as we used double-glazed LoE when building The House of Steel. Gits like their comfort 🙂
I find most people haven’t read the historical research papers they so freely “quote”. Historians & philosophers read them; they are usually very instructive 🙂 I might have a look at those early experiments and have a play; I certainly have enough scrap materials lying around. First I have to build The Finished Greenhouse. The experimental one works, but the FG will be equipped with automated venting (thermal wax operated) and will be productive over a slightly longer season. SWMBO tends to frown if I play when there’s “real” work to be done.
I’d appreciate sharing what you come up with in regards to your experiment. My email is jonathan at sturmsoft dot com. Cheers…
I fear that lay readers visiting this site are going to conclude that a large portion of the physics community have taken leave of their senses.
Whatever the convoluted quibbles about the N & Z workings everyone knows that air above heated ground gets warmer than it would above unheated ground.
That means ALL the air not just any GHGs within it. And that the degree of warmth acquired will be density related
The primary mechanism is conduction and convection. The presence of GHGs might help the conduction and convection process by spreading upward longwave around more efficiently between adjoining non GHG molecules but equally it helps the conversion of conducted and convected energy back to outward longwave so the effect is pretty much neutral.
Arguing against the general principle is frankly an embarrassment to all.
Then all that N & Z did was to look at improved data for planetary surface temperaure, adjust for an apparent error in the previous application of the S – B equations and point out that in the real world the simple relationship betwen pressure and solar input comes remarkably close to matching actual surface temperatures without considering atmospheric composition at all.
In the process they find that the extra kinetic energy needed to explain surface warmth beneath an atmosphere cannot possibly be accounted for by the presence of GHGs alone.
So, chaps, instead of bleating that it isn’t possible and inventing all sorts of imaginative ‘flaws’ in the N & Z work (that do not address the essential point) I think a period of quiet reflection is called for on the part of the most assertive contributors here.
Willis Eschenbach says:
January 23, 2012 at 12:38 am
“Actually the facts we have do allow us to draw some conclusions. Let me define the “theoretical S-B temperature” as the temperature of a blackbody with a uniform surface temperature corresponding to the amount of incoming radiation. Here’s what we know.
1. For a given incoming radiation, any variation in surface temperature from the isothermal state will lower the mean surface temperature.
2. An atmosphere, with or without greenhouse gases, can reduce the temperature drop from variations in temperature. It does this by reducing the amount of day / night ∆T and equator / polar ∆T.
3. A transparent GHG-free atmosphere cannot warm the surface above the isothermal state of the theoretical S-B temperature.
Therefore: at a minimum, the greenhouse effect of GHGs and clouds must be responsible for at least the thirty degrees C that the earth is above the theoretical S-B temperature.
In addition, since the greenhouse effect undoubtedly reduces the day/night temperature swings, some additional amount of the warming beyond the 33°C is also due to the greenhouse effect.”
I strongly disagree. Fact is If insolation was uniform the passive solar water heating system could not even make it to the average local ambient temperature for the same reason I disagree with your conclusions.
Thus your conclusion that it would have to be 30 degrees warmer is no doubt wrong. There is not the slightest doubt in my mind that heat storage in the atmosphere is part of the greenhouse effect, if not all of it.
The only reason passive solar water heating technology works is because of the diurnal cycle of insolation. Otherwise the system would produce exactly what underground pipes produce now without a system namely water at your local average climate temperature which also happens to be about the temperature of the soil that the pipes run through.
The fact is simple physics ensures that at least a major portion of the “GHE” is due to atmospheric warming by the diurnal cycle.
Air is an excellent insulator and the GHE is really more like 16K because probably the correct way of measuring it is by taking its “average” throughout the layers of the atmosphere from the surface to that notion of TOA where 240 watts is emitted to the sky and since the lapse rate is linear the average should be half what is measured at the surface.
I have built a few water systems and would not be surprised that 16K is an average you can achieve.
Probably the only reason to not today rely exclusively on them is the fact that you can have extended periods of very heavy clouds that limit their potential on certain days in certain locations so you have to a back up system to deal with that.
But that is something even the greenhouse effect has no answer for.either!
Bryan:”In the derivation of the ideal gas laws the molecules are said to make perfectly elastic collisions with the container walls.
“In reality the walls are ‘sticky’.”
However that may be, the question before the house is what the temperature distribution would be in the ideal case.
A brief review of the bidding:
Jelbring says that at equilibrium a thermally isolated ideal-monatomic-gas column in a gravitational field will assume a configuration that exhibits the dry adiabatic lapse rate, and I believe he’s saying this is the maximum-entropy state. I won’t attempt to characterize his rationale.
Robert Brown disagrees. He says that the maximum-entropy configuration is isothermal. Unfortunately, his analysis is appallingly superficial for a physics professor. For all that is apparent in his responses, he remembers nothing more than the result that heat flows from hot to cold and apparently doesn’t recall the assumptions by which that result was reached. This appears to be the general level of understanding of the other physicists on this site, too. I’m embarrassed for Duke and, since it is one of the most well-regarded universities in the United States, for my country’s educational establishment.
To me it appears that neither Jelbring nor Brown is correct. Based on the result stated in the Velasco et al. paper, my impression is that the maximum-entropy configuration does have a non-zero temperature lapse rate, but one that is less than the dry adiabatic lapse rate for any significant number of molecules. But there are steps in Velasco et al. paper I was struggling with, and I had hoped that there was enough intellectual curiosity at this site that I would get help with that from the physicists here.
Unfortunately, none of the physicists engaged meaningfully, and the only two participants here who have engaged have been big disappointments. Willis Eschenbach claimed actually to have read Velasco et al.’s equations and found that they didn’t say what I contend they do. Since Equation 8 so clearly does yield a non-zero lapse rate, I briefly harbored bad thoughts about Willis’s veracity. After seeing him flail about with the math on another thread, though, I’m willing to accept that he actually had read the equations but just was not equal to the math. Be that as it may, it’s clear he’ll be no help.
Finally, Paul Birch weighed in on the single-molecule thought experiment to which I applied Equation 8, but he, too, appears able only to parrot rules he’s heard. I’ve been reduced to trying Socratically to lead him to the correct answer, but it’s tough sledding. And, in any event, it doesn’t seem likely that he’ll be equal to helping derive Equation 8 from Equations 5 and 6.
So my question to you is, Can you perform that derivation?
I take your kilometer’s tall cylinder of atmosphere in thermal equilibrium and flip it over, like flipping an hour glass. It involved no input of work as its height did not change, so the column’s energy remains constant. But now the gas at temp T that was at the top has been wildly compressed, making it much, much hotter, while the gas at temp T that was at the bottom has been expanded, making it much, much colder. If you let it get anywhere close to thermal equilibrium I’m going to flip it over again, and since I’m not putting in any work, I can do this all day, continuously forcing your air column back to the dry adiabatic lapse rate as I sip a margarita.
I love doing jobs that don’t involve an input of work and constantly overturn an idealized thermal equilibrium, because it’s just that easy.
You must be very strong, because the density of air in the column in thermal equilibrium is much greater at the bottom than the top. In fact, you did a rather lot of work, if the cylinder in question is very high. All the air in your cylinder will then fall back to the bottom and inelastically collide with it. Because the column is adiabatic (well insulated) and you’ve input work, the air will have more energy and will be at a higher temperature in equilibrium. If you keep doing this, it will get very hot indeed, and I suspect that you will get tired. If you’re planning on restoring your lost metabolic energy with margaritas, plan on getting a really, really big pitcher.
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Joe Born says:
January 23, 2012 at 4:17 pm
Paul Birch: “Every time it hits the surface it bounces back with a different energy, sometimes higher, sometimes lower.”
Joe: “What “surface” did you mean? Actually, let’s just concentrate on the surface off which the molecule bounces at the bottom of the column.”
OK. (In Jelbring and Willis’s set-ups there’s also the lid and side surfaces, but by all means let’s ignore them, since in the absence of collisions it’s obvious that the horizontal components of the molecule’s momentum keep the same “temperature”).
“At that surface its potential energy is zero, right?”
Zero or an arbitrary constant, yes.
“So the “different energy”.you must be talking about is all kinetic energy? If the potential energy is zero, and the kinetic energy is “sometimes higher, sometimes lower,” then the total energy is changing?”
Correct. The molecule is in thermal equilibrium with the surface, so at that surface its vertical velocity at each bounce is drawn from the Maxwell Distribution, which goes as exp(-p**2/kT), positive velocity on its way up, negative half on its way down. Energy is being swapped between surface and gas molecule on each bounce, but there is no net transfer over time. So the height the molecule can reach is different each bounce; thus, averaged over time, there is an probability density falling off exponentially with height. When you see the molecule high up, this is inevitably a highly selected outlier from the distribution – a bounce with an unusually high upwards velocity when it left the surface far below.
In my opinion, this is one case in which taking the extreme Knudsen case (single molecule, or gas too rarefied to have collisions) doesn’t actually make it any easier to understand what is going on – particularly since the gas is in thermal equilibrium and has a corresponding distribution of velocities.
Trick says:
January 23, 2012 at 4:50 pm
Paul Birch says: “Even your one-molecule system doesn’t have the same energy all the time.”
Trick: “It does if you mean total energy”
Not when it is in thermal equilibrium with a surface. It has an energy drawn randomly from the underlying Maxwell distribution. See my reply to Joe.
Joe Born you say;
“Based on the result stated in the Velasco et al. paper, my impression is that the maximum-entropy configuration does have a non-zero temperature lapse rate, but one that is less than the dry adiabatic lapse rate for any significant number of molecules. ”
If you post a link to the paper I will read it.
davidmhoffer says:
January 23, 2012 at 9:27 pm
“… a case of beer at some unkown temperature being immersed in a lake of known temperature and the temperature of the case of beer being taken at a couple of points in time after the beer had been immersed….The guy beside me however pulled out a sheet of natural log paper, plotted what little data we’d been given, struck a line through it, and got almost exactly the same answer.”
That is because the physics is trivial and he only needed the time constant. One free parameter and a known equation. So far as I can see, N&Z have no sound physics behind their equation and use at least four free parameters to fit no more than six inaccurate data points. That’s not a “good curve”. It’s worthless rubbish. I don’t even need to rework the fit, as Joel has done, to know it’s worthless. Common sense, my scientific intuition, and information theory alike all tell me so. Note that this doesn’t imply that N&Z’s hypothesis is necessarily wrong; what it says is that their purported evidence is meaningless; it doesn’t prove a thing (except perhaps that N&Z don’t seem to have a clue how physics works).
Robert Brown disagrees. He says that the maximum-entropy configuration is isothermal. Unfortunately, his analysis is appallingly superficial for a physics professor. For all that is apparent in his responses, he remembers nothing more than the result that heat flows from hot to cold and apparently doesn’t recall the assumptions by which that result was reached. This appears to be the general level of understanding of the other physicists on this site, too. I’m embarrassed for Duke and, since it is one of the most well-regarded universities in the United States, for my country’s educational establishment.
I’m sorry you find my analysis superficial. Perhaps you’ll find my top post on the issue (pending) more convincing. Also bear in mind that the only question I’m addressing is whether or not the adiabatic lapse rate is static thermal equilibrium as asserted by Jelbring. The adiabatic lapse rate completely ignores the mechanism of conduction and relies entirely on convection (the basis of adiabatic lapse).
I’m perfectly happy to look at an actual statistical mechanical computation that purports to examine the issue, as long as it is done in accordance with the axioms and methods of stat mech, that is to say, use a partition function or examine the density of states properly. Jelbring does none of that and his argument is absurd and wrong.
It is also certainly the case that — as my next top post will demonstrate pretty clearly — that including any sort of channel for heat conduction will either reduce any observed lapse rate to zero or cause a most peculiar flow of energy. I’d be interested in hearing your response to it.
The nice thing about the laws of thermodynamics is that they pretty much always hold. It requires very special circumstances for them not to hold — very small systems, open systems, strongly constrained systems. Analyzing a large systems in thermal equilibrium (not open) leaves only strong constraints, and those require very careful arguments and experimental support for experts to accept — the knee-jerk assumption is that equilibrium is isothermal and the first and second laws hold, built right in to the axioms of stat mech, detailed balance, the partition function. In the meantime, Caballero leaves it as a textbook exercise in detailed balance to show that there is no thermal lapse in equilibrium in the vertical direction in a uniform gravitational field.
I’ve already offered a verbal version of the argument (in another thread, IIRC) — the only way to achieve detailed balance across a vertical surface between neighboring differential slices of gas is if the distribution of velocities above and below the slices are the same, because they have to exchange equal numbers of particles going in the opposite directions. If the MB distributions are the same above and below, the two slices have the same temperature. If they have different temperatures, the peaks are shifted and detailed balance is impossible.
The point is that detailed balance is perfectly satisfied if — and I would argue only if, but I’m willing to listen to a counter argument — the MB distribution of molecular velocities is the same throughout the fluid. That goes without question, as a simple and obvious identity. It is trivial to balance forces with an isothermal distribution. So whether or not there is another solution, as you assert, there is absolutely no doubt that a stationary solution to both the static force balance and the requirement for detailed balance in energy flow is an isothermal column of fluid with the textbook expression for the pressure as a function of height. At any instant in such a fluid, the density will not change and the velocity distribution will not change, because equal numbers of particles will move up and down across any vertical surface and will have the same velocity as they do. Note that arguing that “gravity” speeds them up or slows them down ignores the fact that gravity does no net work in this system, and that the mean free path is much smaller than the distance for secular changes in the velocity due to gravity. Gravity heats things only when there are inelastic processes going on — dropping a big rock onto the Earth as a one time event. It is a bit difficult for me to see how gravity is going to move a system from detailed balance isothermal pressures to a distribution with a lapse and then maintain the lapse against the inexorable process of heat conduction the other way.
Perhaps there is an order of taking limits that is physically reasonable that permits this to happen, but surely the onus of proof is on the person making the assertion that it does to do a good job of proving it, as it is a most unusual and unexpected result. Jelbring’s E&E paper contains nothing of the sort, and it is what we are discussing, although feel free to top post about some other paper that proposes a lapse once we finish with Jelbring (later today) and N&Z (where I “finished” their empirical curve fitting T_s for the planets earlier this morning, demonstrating that the curve is completely meaningless, an artifact of the functional forms they used to fit with truly absurd physical numbers). That isn’t to say all of N&Z’s work is useless — I think their approach to computing a better approximate initial (pre-greenhouse) warming is fine and have proposed it myself on other threads, although their description and computation are not necessarily done “right”, yet. They are at least reasonable. I can’t assess their remarks on thermal lapse rates because they haven’t really explained them, or connected them in any way with the numbers in their “miracle” fit with its 54 Kbar and 202 bar reference pressures.
But when we finish with those two papers (and all of the attention they have attracted from confirmation biased skeptics who appear literally desperate to find some physics that shows that the GHE itself doesn’t exist, which is frankly absurd given top-of-atmosphere IR spectroscopy but go figure) I’d be happy to look at a paper that purports to derive a nonzero lapse rate with an open mind, especially if that lapse rate is almost vanishingly small, small enough to be a small correction to isothermal but nowhere near the observed adiabatic lapse rate that is clearly nowhere near a static thermal equilibrium.
In addition, I personally would be a whole lot happier for it to be experimentally verified. I already described a simple tabletop experiment to test the hypothesis — set up a horizontal dewar filled with e.g. Xenon as a centrifuge, place recording thermometers at both ends, and spin up the tube to 100-1000g, recording a long time series of the end temperatures. I’d expect the outer end to heat rapidly as you spin it up, and then relax to a uniform temperature throughout the tube, but I’d be happy to be experimentally shown to be wrong under controlled circumstances as long as the experimental run time exceeds the time required for conduction to thermalize the system and as long as the gas near the top/inside of the dewar never gets too thin for thermodynamics to work to describe it (in part because the mean free path gets so large that it is not in thermal equilibrium in the classical sense). I’d say spinning it up so that the top pressure is around what it is at the top of the troposphere (and the bottom is whatever, starting from 1 bar throughout) would be enough to see if there is a measurable lapse in equilibrium.
Perhaps that will be enough to restore your love and respect, to make my remarks and efforts to bring some small level of compliance with the actual laws of physics reflect well on Duke once again and stop embarrassing not only Duke, but all the physics departments in the United States. But somehow, I doubt it.
Sigh.
rgb
Correct. The molecule is in thermal equilibrium with the surface, so at that surface its vertical velocity at each bounce is drawn from the Maxwell Distribution, which goes as exp(-p**2/kT), positive velocity on its way up, negative half on its way down. Energy is being swapped between surface and gas molecule on each bounce, but there is no net transfer over time. So the height the molecule can reach is different each bounce; thus, averaged over time, there is an probability density falling off exponentially with height. When you see the molecule high up, this is inevitably a highly selected outlier from the distribution – a bounce with an unusually high upwards velocity when it left the surface far below.
In my opinion, this is one case in which taking the extreme Knudsen case (single molecule, or gas too rarefied to have collisions) doesn’t actually make it any easier to understand what is going on – particularly since the gas is in thermal equilibrium and has a corresponding distribution of velocities.
And I completely agree. Detailed balance. People are conflating increasing particle density and hence energy density with temperature, which is a measure of average energy per particle. Each particle on average doesn’t rise or fall in equilibrium, so gravity is completely irrelevant to its temperature. Where gravity matters is in establishing the pressure profile of the gas. Where the two get tied together is — only — in the density of an ideal gas as a function of its temperature.
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Joe Born
One thing I did notice recently when looking at the graphs of density against altitude for air is that in the real atmosphere(adiabatic) the density does not decrease as fast as a proposed isothermic solution.
For 10Km the USA standard atmosphere gives 0.41Kg/m3.
A formula used for an isothermal atmosphere at the same altitude gives 0.34Kg/m3.
This means that if a column of air was suddenly made thermally isolated the centre of mass would gradually drop.
This means that gravitational potential energy would be lost and must therefore be turned into kinetic energy.
In other words as well as the gas becoming isothermal it also increases its average temperature.
So it seems that the isothermal option has increased its entropy.
Joe Born says:
January 24, 2012 at 2:09 am
“Finally, Paul Birch weighed in on the single-molecule thought experiment to which I applied Equation 8, but he, too, appears able only to parrot rules he’s heard. I’ve been reduced to trying Socratically to lead him to the correct answer, but it’s tough sledding.”
Have a bit of patience! I don’t sit at this computer 24 hours a day, and you’re no Socrates! As an admitted non-physicist you seem to be of the opinion that this Velasco et al (whoever they may be) have proved that there is a non-zero lapse rate at equilibrium. You will pardon me if I suspect that you are simply misunderstanding what they say (unless they’re not physicists either, in which case they’re probably just wrong – or even if they are physicists, it’s a lot more likely that they’ve made a slip-up than that they’ve overthrown the principles of thermodynamics). So if you want physicists here to check out the paper, post a (non-paywalled) link where we can read it.
Paul Birch: “So if you want physicists here to check out the paper, post a (non-paywalled) link where we can read it.”
Links to the Velasco et al. and related papers are provide at http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/.
Bryan: “So it seems that the isothermal option has increased its entropy.”
Yes, it does seem so, and I believe it would. I’m actually making only a very modest claim. I’m sure that going from a significant lapse rate s large as the dry adiabatic lapse rate to isothermality would increase entropy. But going from such a lapse rate to the (very small) lapse rate Velasco et al prescribe would increase it more if Velasco et al. are right.
Robert Brown: “Perhaps that will be enough to restore your love and respect, to make my remarks and efforts to bring some small level of compliance with the actual laws of physics reflect well on Duke once again and stop embarrassing not only Duke, but all the physics departments in the United States. But somehow, I doubt it.”
I apologize that in a moment of exasperation I went over the top, although I must say I have indeed been taken aback by how unwilling the disputants here have been to consider the bases for their beliefs.
I would indeed welcome a chance to have people who do this stuff for a living vet Velasco et al. and Roman et al. But I’m guessing experimental proof will be hard to come by in view of the low lapse rate Velasco et al. dictate for any reasonable amount of gas.
By the way, I’m getting to this only after having moved my argument to your own thread, to which I suggest we all transfer this discussion.
Bryan: “If you post a link to the paper I will read it.”
Talllbloke has links to all the relevant papers in his Loschmidt thread here: http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/
Paul Birch says at 1/24 6:00am:
“Not when it is in thermal equilibrium with a surface.”
The ideal one-molecule construct for learning & discussing is in an adiabatic control volume, so it doesn’t come to any sort of thermal equilibrium with the surface b/c for discussion purposes the surface is adiabatic or ideally perfectly insulated so as to let no heat flow in/out for simplification – we are in an ideal world here. The one-molecule just elastically bounces off the adiabatic wall with no change in its total energy or thermal energy or potential energy.
It is true in the real world that the one-molecule will have a change in thermal energy & thus total energy as it contacts a real wall (if the molecule has enough contact time thermal equilibrium happens) because there are no real, perfect wall insulators or we could make a Perpetuum Mobile machine.
“It has an energy drawn randomly from the underlying Maxwell distribution. See my reply to Joe. “
Again, Maxwell-Boltzmann distribution is a special case where the particles are free to move. The particles are not free to move in an electrostatic field or a gravity field. Since the Perpetuum Mobile top post included gravity, M-B distribution is not applicable to the one-molecule in a gravity field by M-B definition.
Turn off gravity (and any other force field on the one-molecule ideal gas particle) and you can invoke M-B just fine.
Paul Birch:
Despite my suggestion that we all adjourn to Robert Brown’s thread, it seems appropriate to use this one to clear up a misconception.
What we’re (and Velasco et al.) are talking about is what I understand they refer to as a “microcanonical ensemble.” Whether I got the buzzword right or not, the point is that the total energy is assumed to remain constant, so none is exchanged with the walls. Hence my use of the word “isolated.”
Robert Brown says:
January 24, 2012 at 6:33 am
In addition, I personally would be a whole lot happier for it to be experimentally verified. I already described a simple tabletop experiment to test the hypothesis — set up a horizontal dewar filled with e.g. Xenon as a centrifuge, place recording thermometers at both ends, and spin up the tube to 100-1000g, recording a long time series of the end temperatures. I’d expect the outer end to heat rapidly as you spin it up, and then relax to a uniform temperature throughout the tube, but I’d be happy to be experimentally shown to be wrong under controlled circumstances as long as the experimental run time exceeds the time required for conduction to thermalize the system and as long as the gas near the top/inside of the dewar never gets too thin for thermodynamics to work to describe it (in part because the mean free path gets so large that it is not in thermal equilibrium in the classical sense). I’d say spinning it up so that the top pressure is around what it is at the top of the troposphere (and the bottom is whatever, starting from 1 bar throughout) would be enough to see if there is a measurable lapse in equilibrium.
This would be a nice experiment Robert, there are a few complications in practice. The centrifuges I use spin in a sealed chamber full of air which gets hot quite rapidly so we refrigerate the chamber to prevent overheating.
I would expect as you do that the gas in the bottom of the tube will rapidly compress and heat up (if you spun it up slowly perhaps no heating).
Joe Born says:
January 24, 2012 at 9:36 am
“What we’re (and Velasco et al.) are talking about is what I understand they refer to as a “microcanonical ensemble.” Whether I got the buzzword right or not, the point is that the total energy is assumed to remain constant, so none is exchanged with the walls. Hence my use of the word “isolated.””
Now I see what your misconception is. The models being discussed in this thread are not microcanonical ensembles, because they are in thermal equilibrium with a planetary surface, or in thermal contact with other surfaces such as thermocouples. They are canonical ensembles.
Your single molecule example only functions as an ensemble if its behaviour is averaged over multiple bounces and it exchanges energy with the surface. Otherwise it’s not a thermal system at all, merely a trivial example of frictionless ballistics, and thus irrelevant to this topic.
However, if you drop the single molecule condition, then even a completely isolated gas in a fixed volume in a gravitational field will have the same temperature throughout (bearing in mind that defining temperature rigorously for small numbers of molecules in an isolated system is quite a difficult and subtle problem).
Paul Birch says at 1/24 11:11am:
“However, if you drop the single molecule condition, then even a completely isolated gas in a fixed volume in a gravitational field will have the same temperature throughout.”
Not with the gravity field turned on, the speed of the molecules reduces as it moves up against the g field, so their KE reduces.
The gas molecule(s) will have the same total energy throughout by 1st Law. There is no law AFAIK that says the molecule(s) mean kinetic energy or temperature (as defined by Caballero in 2.1 top post link) has to be the same thoughout one thermodynamic system in a gravity field. The mean kinetic energy (temperature) of the molecule(s) varies with their potential energy which arises in a gravity field.
Just turn off the gravity field and then temperature will not vary (isothermal) because the mean kinetic energy IS the molecule(s)’ total energy which can’t vary by 1st Law.
Paul Birch:
I’m not sure we’ve yet converged about what the hypothetical is. Maybe it’s best for you to enter your remaining responses on Robert Brown’s “Refutation” thread, where he describes the gas at issue as “an adiabatically isolated column of an ideal gas.” I’ve stated my position over there.
Willis Eschenbach says on January 23, 2012 at 7:22 pm:
“O H Dahlsveen says:
January 23, 2012 at 1:35 pm
—-– have you downloaded and read “Fourier 1824″ yet?
Clueless. Got a link?
w.”
======
Yes, cut and paste: Burgess’ (1837) Translation of Fourier (1824)
Or simply click on: http://geologist-1011.mobi.
OHD.
Trick says:
January 24, 2012 at 12:52 pm
Paul Birch says at 1/24 11:11am: “However, if you drop the single molecule condition, then even a completely isolated gas in a fixed volume in a gravitational field will have the same temperature throughout.”
Trick: “The gas molecule(s) will have the same total energy throughout by 1st Law.”
No, they won’t. Molecules of higher total energy go higher; this selection effect has been pointed out repeatedly.
Trick: “There is no law AFAIK that says the molecule(s) mean kinetic energy or temperature (as defined by Caballero in 2.1 top post link) has to be the same thoughout one thermodynamic system in a gravity field.”
Yes, there is. The zeroth law that says that in thermal equilibrium the temperature is everywhere the same, and the second law that says you can’t extract net work from a system in thermal equilibrium, but can from any macroscopic temperature difference. Except for a negligible relativistic correction (of order 1E-12 for the Earth), gravity is irrelevant to this.
Fourier (1824, p. 140)
The heat of the sun, coming in the form of light, possesses the property of penetrating transparent solids or liquids, and loses this property entirely, when by communication with terrestrial bodies, it is turned into heat radiating without light.
This distinction of luminous and non-luminous heat, explains the elevation of temperature caused by transparent bodies.
Fourier (1824, p. 153)
It is difficult to know how far the atmosphere influences the mean temperature of the globe; and in this examination we are no longer guided by a regular mathematical theory. It is to the celebrated traveller, M. de Saussure, that we are indebted for a capital experiment, which appears to throw some light on this question. The experiment consists in exposing to the rays of the sun, a vessel covered with one or more plates of glass, very transparent, and placed at some distance one above the other. The interior of the vessel is furnished with a thick covering of black cork, proper for receiving and preserving heat. The heated air is contained in all parts, both in the interior of the vessel and in the spaces between the plates. Thermometers placed in the vessel itself and in the intervals above, mark the degree of heat in each space. This instrument was placed in the sun about noon, and the thermometer in the vessel was seen to rise to 70°, 80°, 100°, 110°, (Reaumur,) and upwards. The thermometers placed in the intervals between the glass plates indicated much lower degrees of heat, and the heat decreased from the bottom of the vessel to the highest interval.
The effect of solar heat upon air confined within transparent coverings, has long since been observed. The object of the apparatus we have just described, is to carry the acquired heat to its maximum; and especially to compare the effect of the solar ray upon very high mountains, with what is observed in plains below. This experiment is chiefly worthy of remark on account of the just and extensive inferences drawn
Fourier (1824, p. 154)
from it by the inventor. It has been repeated several times at Paris and Edinburgh, and with analogous results.
The theory of the instrument is easily understood. It is sufficient to remark, 1st, that the acquired heat is concentrated, because it is not dissipated immediately by renewing the air; 2nd, that the heat of the sun, has properties different from those of heat without light. The rays of that body are transmitted in considerable quantity through the glass plates into all the intervals, even to the bottom of the vessel. They heat the air and the partitions which contain it. Their heat thus communicated ceases to be luminous, and preserves only the properties of non-luminous radiating heat. In this state it cannot pass through the plates of glass covering the vessel. It is accumulated more and more in the interval which is surrounded by substances of small conducting power, and the temperature rises till the heat flowing in, shall exactly equal that which is dissipated. This explanation might be verified, and the results made more apparent, by varying the conditions and employing colored or blackened glasses, and exhausting the air from the intervals which contain the thermometers. When this effect is examined by the calculus, results are obtained in exact accordance with those of observation. It is necessary to consider attentively this order of facts, and the results of the calculus when we would ascertain the influence of the atmosphere and waters upon the thermometrical state of our globe.
If air that moves from high altitudes to low ones do not warm as they descends then why do so many people of different nationalities, in different places on this planet, talk about the dry winds which according to Wikipedia: – “regionally, these winds are known by many different names. These include:
• Föhn in Austria, southern Germany, and German-speaking regions of Switzerland, Liechtenstein and Northern Italy (even non-German-speaking regions)
• Bergwind in South Africa
• Chinook winds east of the Cascade Range in the United States and Canada, and north, east and west of the Chugach Mountains of Alaska, United States
• Favonio in Ticino and Italy
• Fogony in the Catalan Pyrenees
• Föhn in Wollongong and South Coast, NSW, Australia. Often associated with heavy orographic lifiting on the windward side of the escarpment
• Garmoosh, Garmesh, Garmbaad (Warm Wind): (Persian: گرمباد, Gilaki: گرموش) in Gilan region, in the south of Caspian Sea in Iran
• Halny in the Carpathian Mountains, Central Europe
• The Helm wind, on the Pennines in the Eden Valley, Cumbria, England
• Hnúkaþeyr in Icelandic
• Lyvas wind in Elefsina and Athens in Greece
• The Nor’wester in Hawkes Bay, Canterbury, and Otago, New Zealand
• Puelche wind in Chile
• Terral in Málaga (southern Spain)
• Vântul Mare in the Carpathian Mountains, Romania
• Viento del Sur (Southern Wind) in the Cantabrian region (northern Spain)
• Zonda winds in Argentina
The Santa Ana winds of southern California, including the Sundowner winds of Santa Barbara, are in some ways similar to the Föhn, but originate in dry deserts as a katabatic wind.”
Read up about it and then ask yourselves: Why is it sometimes so darn hot east of the Rockies?
O H Dahlsveen says:
Strawman.
Nobody is saying that air does warm as it descends or cool as it rises. That is a simple consequence of basic thermodynamics for an approximately adiabatic process. However, this fact does not mean that
(1) There is a lapse rate in ***THERMAL EQUILIBRIUM***.
(2) One can violate conservation of energy by having a planet continually emitting from its surface more power than it is receiving from the sun (without some real source of this energy such as gravitational collapse).
Joel Shore says onJanuary 24, 2012 at 4:24 pm :
O H Dahlsveen says: “———————“
Joel says:
“Strawman. —————–. ——- Nobody is saying that air does warm as it descends or cool as it rises. That is a simple consequence of basic thermodynamics for an approximately adiabatic process.”
=============
Pardon? – Nobody is saying that air does warm ——?
Anyway Joel as you seem to be shouting ***THERMAL EQUILIBRIUM*** at me Let’s, before I go to bed, see what Wikipedia says about thermal equilibrium:
“Thermal equilibrium is a theoretical physical concept, used especially in theoretical texts, that means that all temperatures of interest are unchanging in time and uniform in space. When the temperatures of interest are just those in the different parts of one body, the concept also requires that any flow of heat by thermal conduction or by thermal radiation into or out of one part of the body be balanced by a flow of heat in the opposite sense into or out of another part of the body.”
So then Joel, you think a rate, on this occasion The Lapse Rate (TLR) has got a “Thermal Equilibrium”?
Anyway, I think I know what you mean, so – I see it as follows; if a “blanket of still air” (say one meter thick) is at rest at ground level (in contact with the surface) has, say a temperature of 15 deg. C (°C) – and that for some reason, say one square meter (m²) of that surface absorbs more sunlight than it’s surroundings – the temperature of ± one cubic meter (1 m³) of air above it will form a pocket that warms more than adjacent air and it will expand, become lighter and eventually separate from the rest of the air blanket that surrounds it. This pocket has no choice other than to rise (let’s not now quibble about how high it rises).
Everybody, including me seem to agree that no heat is lost from the said air pocket by radiation or conduction, (well, I do say that some heat is lost by conduction, but never mind that for now). This is in spite of the fact that the air pocket may be saturated air (i.e. it contains lots of Water Vapor (WV), which is said to be a powerful GHG. So, even if you subscribe to the idea that N, O, O2 and Ar do not emit radiation then;
1) What happened to the law which says that any object (and molecules are objects) that has a temperature higher than zero Kelvin (0 K) must emit thermal radiation?
2) Why should an air pocket (which has no known radiation shield around it, any more than does the atmosphere itself have a lid on it) refuse to radiate away heat-energy?
Everybody keeps spouting the radiation law when it suits them. Yet no one can imagine that, in their many “Thought Experiments” they are mixing Infra Red (IR) radiation from the Sun up with thermal radiation from the ground which real “Field Experiments” more than 100 years ago (experiments which can easily be reproduced today) proved “dark radiation” from the ground was incapable of penetrating transparent solids and liquids.
Anyway, if radiations CAN NOT escape from neither an ascending nor from a descending air pocket then – just extend the “Air Pocket Concept” (APC) to include warm or cold fronts. – Which are nothing but very large air pockets – as far as I can make out.
And then what happens to the GHE?
It is important to note that this is only so because the two regions are parts of the same gaseous body. For example, consider an isolated region of gas: Go way up high in the real earth’s atmosphere with a bottle and trap some of that cold, sparse gas. (Put a stopper on it.) Now bring the bottle back down to earth. The gas in the bottle will still be cold, still sparse, and definitely not in thermal equilibrium with the ground air temperature. (The bottle, trapped gas included, will heat up.) But the real earth isn’t the body under consideration.
The “average energy per volume” condition you mention holds in thermal equilibrium because PV = nRT. For two regions of gas at the same T, PV/n is constant. That is, P times (V per n) is constant: or for each region containing n molecules, PV is constant. And that is a perfectly attainable steady state: regions of high P (low down) will have low V, and regions of low P (high up) will have high V.
The reason I labour this point is that there is nothing magic about average energy per volume: that is an accidental truism arising from the ideal gas law, as I have explained. If you reasoned from that as the initial premise, I believe you got the right answer by accident. It isn’t true of solids or liquids, for example, because they don’t obey this law. Two solids of different density in thermal equilibrium will most likely have different energy per volume, and yet have the same T.
O H Dahlsveen says:
January 24, 2012 at 9:07 pm
1) Let’s take a molecule in deep space. The concept of temperature may not apply to an individual molecule, but it can have kinetic energy and one can use that kinetic energy to calculate a non-zero ‘temperature’ of the molecule. Will that molecule radiate away its kinetic energy? Nope. Unless it collides with another molecule, that kinetic energy will stay kinetic energy and not vibrational or rotational energy that might be above the ground state and allow emission of radiation. Even when there are sufficient molecules to have a Boltzmann kinetic energy distribution, a symmetric molecule like nitrogen can only emit radiation at specific wavelengths plus some very weak continuum radiation. A monatomic gas like argon will emit even less.
2) Because the thought experiment specifies that it is perfect transparent so it can’t emit. Yes, no such animal exists in the real world. But we’re not talking about the real world but an abstraction.
The same applies to lapse rate. For a real planet there will be surface temperature differences leading to circulation and the atmosphere won’t be perfectly transparent which will prevent the atmosphere from being isothermal. But again, the thought experiment postulates that the surface is isothermal and the atmosphere is perfectly transparent. There will be no circulation or radiation and when that happens, conduction becomes the only form of heat transfer and the atmosphere will become isothermal.
Your dark radiation is in fact long wavelength infrared radiation (LW IR) or thermal IR, usually considered to be > 5μm wavelength. The near IR radiation in sunlight has a wavelength range from 0.8-5μm. There is, of course, solar radiation at longer wavelengths but it represents less than 1% of the total. There are substances that are transparent or nearly transparent to LW IR. Crystalline sodium chloride is one of them. The old dispersive optics IR spectrophotometers use an NaCl or CaF2 prism to separate those wavelengths. Polyethylene film and other polyolefin plastics like polypropylene and polymethylpentene (TPX) are also nearly transparent in the thermal IR.
The Fourier quote is interesting, but science has made a lot of progress since then. The de Saussure experiment reference is a good counter to those who believe that Wood’s very poorly documented 1909 experiment is somehow the gold standard.
DeWitt Payne says:
“Why would Abbot replicate identically an obviously flawed experiment?”
Because that’s how the Scientific Method works. Trust, but verify. [Actually, don’t trust; verify.] Otherwise, someone could claim an experiment proved something and it wouldn’t be verified. OTOH, if it’s a different experiment, then it’s a different experiment. Which is OK, but replicating the original experiment is critical. [See Fleischmann & Pons’ cold fusion experiment. No one was able to replicate their experiment. See why replication is so important?]
I agree that Wood didn’t convey his experiment very clearly. But he explained that it was a casual experiment intended to test his assumptions.
Also, I recall reading a comment posted here maybe a year ago by someone who conducted an experiment using flat mylar balloons filled with various gases: pure CO2, and air, and various other gases like nitrogen and argon. As I recall, he verified Wood’s results within a degree or two. And IIRC, the pure CO2 balloon didn’t heat any more than the partial CO2 balloon. He also provided pictures. Now I’m sorry I didn’t save the link.
DeWitt Payne says on January 25, 2012 at 8:41 am:
“1) Let’s take a molecule in deep space. ————–
2) Because the thought experiment specifies that it is perfect transparent so it can’t emit. Yes, no such animal exists in the real world. But we’re not talking about the real world but an abstraction.”
==================
Dear DeWitt Payne; – The thing about thought experiments is that they need not contain any real data, objects or situations and as soon as one realizes this the creator of such a “thought experiment” can think or say, just as you did: “Yes, no such animal exists in the real world. But we’re not talking about the real world but an abstraction.”
– What does that prove?
It proves you are talking about an abstraction called fantasy which has no place, at all, in science – apart from maybe as entertainment during the coffee-breaks. – You are of course correct in saying that an animal that does not exist does not radiate IR energy, or light. But who needs to take that into consideration?
Even lapse rates get a thought experiment which postulates that the surface is isothermal and the atmosphere is perfectly transparent. – Is there no end to this nonsense?
Then you say: “The Fourier quote is interesting, but science has made a lot of progress since then. The de Saussure experiment reference is a good counter to those who believe that Wood’s very poorly documented 1909 experiment is somehow the gold standard.”
=============
I know science has made a lot of progress since then but not this particular branch of it. – Science does not even know what his work was any more, which is a shame, because now they cannot falsify it. – Can today’s science falsify this sentence from 1824: “It is difficult to know how far the atmosphere influences the mean temperature of the globe.”?
– And by the way, have you read the full transcript of Wood’s 1909 experiment?
O H Dahlsveen says:
January 25, 2012 at 4:11 pm
Yes. And I’ve read a photocopy of Abbot’s reply to Wood. I’ve done similar experiments myself that agree with de Saussure and disagree with Wood.
Wood wrote another note with more details on the theory that I haven’t read, not having the requisite access privileges, but that was also rebutted because Wood failed to distinguish coherent from incoherent superposition of light.
See Vaughan Pratt’s page on Wood:
Yet somehow in spite of all the above this “casually performed and documented experiment” has become the modern gold standard. Confirmation bias in all its glory. Abbot, by the way, was the first to determine the value of the solar constant with reasonable accuracy for the time at 1346 W/m² (1.93 cal/cm²/min in units common at the time). So he was far more experienced in the field of accurate and precise measurement of incident solar energy than Wood.
Smokey says:
January 25, 2012 at 1:50 pm
You do know that polyester (Mylar®) is relatively opaque to thermal IR. A film thickness of 0.001 inches absorbs about 50% of incident thermal IR radiation. Balloons all by themselves in no way replicate anything like Wood’s experiment. All those gases are nearly completely transparent to solar radiation, especially over a path length as short as a balloon. You need well insulated boxes with a highly absorbing and emitting interior surface exposed to direct normal sunlight on a clear day. You need something like these boxes, in fact. I’ve since covered the front face of the insulating boxes with aluminum foil to reduce emission and absorption by the insulation.
DeWitt Payne,
Now all you need is some sunshine.☺
Seriously though, you need to replicate Wood’s experiment exactly. If you get significantly different results, then the next question is why Wood got the results he reported.
I don’t know much about emissivity of different materials. My carreer was in weather recording instruments. But I do know you need to replicate experiments. Experiments have to be reproducible, with at least close to the same results.
Smokey says:
January 25, 2012 at 5:51 pm
Yeah, right. I’m going to go out and buy some mercury thermometers (alcohol doesn’t go high enough) and a bunch of cotton and some clear rock salt plates. Not hardly. You can buy NaCl windows for IR spectrophotometry, but they’re very small, no more than a few centimeters in diameter, and they aren’t cheap. Wood doesn’t give the dimensions of his boxes or how thick the cotton insulation was or what type of glass he used. It’s literally impossible to reproduce Wood’s experiment exactly without a time machine.
Horace Greeley Abbot was the acknowledged master of measuring incident solar radiation flux at the time. He dismissed Wood’s results as flawed. The main reason being that the temperature Wood achieved was tens of degrees lower than he should have observed. I don’t actually need to know why Wood’s experiment was flawed, although I have a good idea or two, when I can prove to myself (and anybody else with an open mind) beyond a shadow of a doubt both theoretically and experimentally that if he had done the experiment correctly, he would have seen a significant temperature difference between the boxes.
DeWitt,
1. Mercury thermometers aren’t all that expensive. And they can be re-used indefinitely. I have several myself [or did, until I gave a couple away. Ebay and Craigslist are good sources for used merc thermometers, the only tricky part is finding two that are scaled to the same temperature range].
2. Since you’re apparently knowlegeable about materials, something other than rock salt with the same characteristics would be fine. So the cost of a rock salt window is irrelevant.
3. You say you don’t actually need to know why Wood’s experiment was flawed, so it sounds like your mind is made up already. You told a commenter upthread that he assumes the conclusion. Isn’t that what you’re doing? That sort of assumption can lead to unintentional bias.
I’ll be interested in the results of your experiment. Please post them here.
It is my understanding that Wood was experimenting regarding the effects of both convection and radiation. Despite his ambiguous wording, it seems that he packed the boxes with cotton to eliminate convection, since he argued that the greenhouse effect was due to the presence or absence of convection. If so, the experiment as I understand it would still show the difference between IR being present and being absent. Wood wrote:
It sounds to me like he packed the boxes with cotton. If so, there would be no convection, and the only variable would be the type of radiation passing through the windows.
But it’s your experiment. Let us know how it turns out. For myself, my view is strictly limited to questioning whether a doubling of CO2 would have any significant effect. [1°C is not what I would consider significant, and would be a net benefit, as would increased CO2.]
Paul Birch says at 1/24 2:46pm:
“Trick: “The gas molecule(s) will have the same total energy throughout by 1st Law.”
No, they won’t. Molecules of higher total energy go higher; this selection effect has been pointed out repeatedly. “
Paul Birch misinterprets or misunderstands Caballero. 1st Law says total energy (TE) of each molecule must remain constant because energy cannot be created nor destroyed. It is possible to compute that total energy constant from the velocity and height = 1/2mv(h)^2 + mgh. Just set h=o at a datum or bottom and measure the temp. at h=0 to get the velocity thus can solve for TE.
It makes no sense that some molecules never go higher than others, they pretty much all hit the top of the control volume (hmax) and the bottom (hmin) at some point in time, from random elastic collisions.
————-
“Trick: “There is no law AFAIK that says the molecule(s) mean kinetic energy or temperature (as defined by Caballero in 2.1 top post link) has to be the same thoughout one thermodynamic system in a gravity field.”
Yes, there is. The zeroth law that says that in thermal equilibrium the temperature is everywhere the same, and the second law that says you can’t extract net work from a system in thermal equilibrium, but can from any macroscopic temperature difference. Except for a negligible relativistic correction (of order 1E-12 for the Earth), gravity is irrelevant to this.”
Here Paul can’t extract work since the entropy is constant due to the process being reversible, the molecules gain just as much kinetic energy falling to the bottom as they lose rising to the top & vice versa: voila reversible. TE is constant by law in Willis’ premise gas column.
The 0th Law is happy and not violated once no heat flows in hydrostatic equilibrium, the limit dQ/dz goes to zero at each height, dT/dx = 0 as gravity isn’t acting horizontally.
That’s why Caballero states the molecular velocities are higher at the bottom than the top in what he says is a non- isothermal column, thus temperature is in equilibrium too. Just go read or re-read slowly his section 2.1 and 2.3 for yourself. You can rely on that and eventually come to grips with hydrostatic equilibrium situation. It is pretty basic thermo stuff.
Smokey says:
January 25, 2012 at 7:27 pm
Umm. If the boxes were packed with cotton, how was the light supposed to be absorbed by the walls of the box. Cotton isn’t exactly to most transparent of substances. There wouldn’t have been much point in painting the walls of the box black.
The obvious meaning of “the whole packed in cotton” is that the entire box with thermometer and window was surrounded with cotton for insulation leaving only the window exposed. He refers to constructing the boxes with windows attached before mentioning the cotton packing. Insulation of the walls is important if you want to see the effect. If there’s no insulation, there’s no greenhouse at all. A thin wall is rated at R1 in US units for insulation. That’s caused by the stagnant air film on each side of the wall and has very little to do with the thermal conductivity of the wall material. But R1 isn’t very much. If you don’t insulate the walls of the box, you only get a temperature increase of ~30C above ambient when the box is exposed to sunlight. A multi-layer windowed box I built reached 170C, about 150C above ambient.
I have thought about this a lot and done my homework, unlike the vast majority of people who rely on Wood. I’m using type K thermocouples instead of mercury thermometers. I have an 8 channel data logger so I can make continuous measurements. I’m using fiberglass batting instead of cotton. I’m using thin polyethylene film instead of rock salt. I have an IR thermometer as well as the thermocouples. The funny thing is that people who think Wood is the gold standard think I need to do all sorts of quality control like switching thermocouples between boxes and carefully calibrating them and such. Wood did none of that. Yet his results are now accepted by many as unquestionable. If Wood had really believed in his results, he would have replied to his critics by defending and extending his experimental results. He published two articles, both of which were promptly rebutted, and then dropped the subject. Does that sound like someone who believed his results were correct?
Bryan is fond of saying of Professor Vaughan Pratt, who also has found that a box with an LW IR transparent window is cooler than a box with an IR opaque window, that he isn’t defending his results. Besides being untrue, it applies to Wood in spades. Most so-called skeptics believe in Wood because it confirms what they want to believe, not because his results actually make sense. Because they don’t, except in bizzaro world.
DeWitt Payne says
Bryan is fond of saying of Professor Vaughan Pratt, who also has found that a box with an LW IR transparent window is cooler than a box with an IR opaque window, that he isn’t defending his results.
Professor Vaughan Pratts experiment has been criticised by many and twice on Climate Etc he refused to defend it saying he was looking for ways to improve it.
Everyone wishes you well with your experiment but its getting hard to see why you don’t publish full details so that others can attempt to replicate your results.
All these endless prequels will make folk lose interest.
Over on Dr. Browns latest thread there is another post which should be of interest.
……………………..
Bill Hunter says:
January 25, 2012 at 7:48 pm
The problem outlined by Dr. Brown is interesting but largely irrelevant to the issue of whether the surface would be warmer in the absence of greenhouse gases.
Here is my reasoning on that.
It’s the analogy to the passive solar water heating system. First lets get a few things clear on that.
A good passive solar system does not need a greenhouse for the collectors to operate well. Often they use just plain black piping. It works nearly as well as pipes in a greenhouse because the 1,000 plus watts of solar radiation far exceeds radiation losses without the greenhouse. The pipes do not ever get warm enough due to the convection occurring in the water system bringing constant cooler water to the inside of the pipes.
Greenhousing the pipes only adds a few degrees to the system and is often not done as that’s the most expensive part of the system. (convection also ensures the surface does not equilibriate to the average daily temperature reading above the surface. Error is greatest at night.)
The heat you get in a passive system is far in excess of the daily average temperature.
Its important to understand this as well. We are looking at incremental warming not warming from the greenhouse effect itself which is already incorporated in the ambient local temperatures.
Finally, we should note that a favorite tactic of folks arguing this point is they want to simplify it by applying uniform radiation. These are analogies like Willis was using to argue against the gravitational effect. But uniform radiation would cause our passive water heating system to fail. All we would have would be water at the ambient local temperature.
The passive system uses gravity and convection but it entirely depends upon a diurnal cycle to obtain a higher average and in no
Trick says:
January 25, 2012 at 8:08 pm
Circular logic or begging the question. The condition that the sum of potential and kinetic energy is constant is the definition of the adiabatic lapse rate. You have assumed your conclusion. The adiabatic lapse rate happens because the air packet is allowed to expand and do work against the surrounding pressure but is not allowed to reach thermodynamic equilibrium by heat conduction from the surrounding air or any other means for energy to enter or leave the parcel. There is nothing in thermodynamics or meteorology that says an air column can only have the adiabatic lapse rate. The adiabatic lapse rate is the upper limit for a stable atmosphere. If the lapse rate were greater than the adiabatic rate, convection would occur. But lapse rates less than the adiabatic rate are observed all the time. In fact, temperature increases with altitude in the stratosphere. Temperature increasing with altitude near the surface happens all the time at night. That’s how dew and frost form. In the Arctic over the sea ice or in the Antarctic during winter, the temperature inversions reach high into the atmosphere. That’s because the ground cools by radiation faster than the air can conduct energy to the ground. The important word in that sentence is conduct, which you seem to think never occurs.
DeWitt Payne,
Good luck with your experiment. You didn’t respond to the issues I raised [except for the cotton insulation, which I explained]. As I made clear, my view is strictly limited to questioning whether a doubling of CO2 would have any significant effect. If we are in agreement, there’s nothing to discuss, because as I said, I’m not doing the Wood experiment. Planet Earth is doing a fine experiment, and showing that CO2 is at best a bit player. And my oft-repeated hypothesis has never been falsified:
At current and projected levels, CO2 is harmless, and beneficial to the biosphere.
If we are in agreement with that, fine. If not, you need to explain why the rise in CO2 has no apparent effect on global temperatures. As you can see, it’s still not clear to me where you’re coming from.
Trick says:
January 25, 2012 at 8:08 pm
Trick: “The gas molecule(s) will have the same total energy throughout by 1st Law.”
Paul: “No, they won’t. Molecules of higher total energy go higher; this selection effect has been pointed out repeatedly. “
Trick: “Paul Birch misinterprets or misunderstands Caballero.”
I did not refer to Caballero. The selection effect is really easy to understand. Forget molecules. Think tennis balls, fired straight up by a row of spring loaded catapults. Some of the catapults are stronger, so some of the tennis balls shoot out faster. The faster they are (that is, the higher their total energy KE+PE) the higher they go. If you stand on a tall ladder, the only balls you will see passing you are the fast ones (the ones that had the highest muzzle velocity, the ones with the highest total energy). The slow ones never get that high. Your position up the ladder selects tennis balls of higher total energy.
————-
“Trick: “There is no law AFAIK that says the molecule(s) mean kinetic energy or temperature (as defined by Caballero in 2.1 top post link) has to be the same thoughout one thermodynamic system in a gravity field.”
Paul: “Yes, there is. The zeroth law that says that in thermal equilibrium the temperature is everywhere the same, and the second law that says you can’t extract net work from a system in thermal equilibrium, but can from any macroscopic temperature difference. Except for a negligible relativistic correction (of order 1E-12 for the Earth), gravity is irrelevant to this.”
Trick: “Here Paul can’t extract work since the entropy is constant due to the process being reversible, the molecules gain just as much kinetic energy falling to the bottom as they lose in rising to the top & vice versa: voila reversible. ”
The second law does not say that you can’t extract work from reversible processes; on the contrary, adiabatic expansion is the archetypical way to extract work from a gas reversibly (ie. isentropically). In this case the random motion of the particles in the gas is “reversible”, but cannot be used to extract work, because the gas is in thermal equilibrium, which is to say, all at the same temperature, as required by the zeroth law.
DeWitt Payne says at 1/26 7:03am:
“Circular logic..”
DeWitt – this experiment can really be performed in the non- ideal gas, near-adiabatic gravity infested real world. Given apparently no one has yet cited a ref. to it having been done appropriately & thus dismissing the demon(s), then I take that to mean experiment is difficult but logic is not circular.
Proof: insert the thermometers ideally in thought experiment – take a reading. They read the same or different. Circle ends.
Extending to atmosphere makes a difficult thought experiment compound the difficulty. Makes me ask, if we cannot yet dismiss the demons from the simpler thought experiment (a gas column in gravity field), then on what basis do we move on to the more difficult atmosphere?
I think the answer lies in – you need to operate with a view on the thermometer readings in ideal experiment, that enough physical laws (0th, 1st, 2nd) are reliable enough to allow moving on to the real atmosphere to draw law-consistent conclusions in agreement with stuff experimentally observed in the real atmosphere.
Among things we’ve learned can’t do so far to prove any conclusion physical: divide by 0, construct a Perpetuum Mobile esp. one that allows energy extraction, and decrease entropy.
Smokey says:
January 26, 2012 at 7:03 am
Obviously, if I believe that Wood’s results were wrong then I must believe that, all other things being equal, doubling the CO2 partial pressure will increase the average surface temperature. But, of course, all other things are never equal in a non-linear possibly chaotic system like the Earth and its atmosphere. I believe I fall into the lukewarmer camp because I think the overall climate sensitivity is at the low end of the IPCC range, on the order of 1.5 C/doubling. As far as policy, I think it’s pointless and possibly immoral to spend vast amounts of money to prevent a small increase in risk from things like extreme weather events when there are people at risk now from the extreme weather events that happen without an increase in temperature.
Skeptics should be concentrating their efforts not on the fundamental science, much of which is correct, and concentrate on the really flaky stuff like the IPCC Working Groups 2 and 3 reports on effects of warming and costs of mitigation. The first seem to be overestimated and the second wildly underestimated, not to mention unlikely to happen.
Smokey says:
January 26, 2012 at 7:03 am
Smokey, the corona (atmosphere if you will) of the sun is aprox 1-3 million K. Does the corona of the sun heat the sun?
DeWitt,
Thanks, looks like we’re pretty much in agreement: there is no looming disaster.
• • •
mkelly,
I’m not sure I understand your comment. Hotter radiating objects heat cooler objects, don’t they?
Bryan,
It’s winter in case you haven’t noticed. I need at least four hours of cloud free sunshine between 9:30AM and 3:30PM (otherwise the sun is too close to the horizon and I have a large hill on the western horizon as well) with little wind to get good results. I’m lucky to get that happening one day/week.
Vaughan Pratt refusing to get involved in a nit-picking session at Climate, Etc. is not the same as not defending his results. He hasn’t pulled the web page, and as someone once said about Wood, the results still stand. Look at the level of criticism here. I’ve been told by two different people that I need to put cotton inside the box to replicate Wood.
Right now I’m modifying one of my boxes so I can measure the air temperature profile from the exposed rear surface to the window. I’ll also switch the boxes between the different insulating boxes. But even if I can convince you that Wood was wrong, I’m betting that somehow you’ll manage to convince yourself that it doesn’t matter.
You also keep bringing up the Penn State experiment as if it somehow proves Wood was correct. It doesn’t. The important finding with regard to Wood from PSU is that the surface temperature at night with the IR transparent film can be lower than for the IR absorbing film. The reverse is never true. But this effect is small because the net radiation from ground to space on a clear, cool night with low humidity (which also doesn’t happen every night) is only ~100 W/m² compared to ~1,000 W/m² from direct normal sunlight on a clear day at local noon. Also, if there’s dew or frost, that makes both films IR absorbing. The geometry is such that cold air that forms near the IR absorbing film falls to the surface causing convective mixing while there is a temperature inversion in the IR transparent tunnels with no convection. The end result was that the IR absorbing film was not cost effective, not that it didn’t work.
DeWitt Payne says
“But even if I can convince you that Wood was wrong, I’m betting that somehow you’ll manage to convince yourself that it doesn’t matter. ”
On the other hand I have no doubt that you will carry out a careful and honest experiment.
After scrutinising your work I will try to replicate your results.
But I need to see a finished report before starting.
I have no hidden agenda.
The experiment tops any theory.
Roll on better weather
Smokey the answer is that it does not. It is not dense enough compared to the sun. If something that hot has no effect then how exactly does a few CO2 molecules.
mkelly,
It seems to me that if the Sun’s corona is at 3 million degrees, and the Sun’s surface is less than one one-hundreth of that, that the corona would heat the chromosphere. It just wouldn’t heat it very much, because the corona has so few molecules.
Paul Birch says at 1/26 7:08pm:
“The second law does not say that you can’t extract work from reversible processes…”
Paul ! – Yes it does. Double check. 2nd law forbids entropy to go down. You extract the work from a reversible process, drive entropy down. Can’t. Then it is irreversible, some energy left town.
Paul Birch continues:
“…the higher their total energy KE+PE the higher they go.”
Yes. But here none of the catapults are stronger by definition KE + PE = constant. 1st law. If the molecules all have the same constant total energy in the closed system, they all go just as high. Just like when your tennis balls are launched with the same total energy. Same height. 1st law.
Trick,
The First Law only requires that the total energy of a closed system remain constant. It does not require that the total energy at every location within, say, an insulated column of air, be constant. You can’t change the gravitational potential energy distribution without changing g, but the temperature profile can be anything within the constraints that the total energy be constant. It can increase or decrease with altitude. However, the system is only stable, i.e. in thermodynamic equilibrium, when it’s isothermal.
Trick says:
January 26, 2012 at 7:23 pm
Paul Birch: “The second law does not say that you can’t extract work from reversible processes…”
Trick: “Paul ! – Yes it does. Double check. 2nd law forbids entropy to go down. You extract the work from a reversible process, drive entropy down. Can’t. Then it is irreversible, some energy left town.”
Oh dear. Reversible processes are isentropic. Adiabatic expansion is one of them. The entropy in the gas stays the same, even though its enthalpy goes down.
Paul Birch: “…the higher their total energy KE+PE the higher they go.”
Trick: “Yes. But here none of the catapults are stronger by definition KE + PE = constant. 1st law. If the molecules all have the same constant total energy in the closed system, they all go just as high. Just like when your tennis balls are launched with the same total energy. Same height. 1st law.”
The molecules do not all have the same energy. They have a wide range of energies (or momenta) randomly drawn from a Maxwell distribution, from zero all the way up. All of the catapults are different.
It’s all about the energy. When a molecule collides with something (a photon, another molecule) there is an exchange of energy. The molecule can go faster or slower after the exchange.
If the energy is random kinetic, it is called heat. Temperature is the RMS average random kinetic energy. Note not all kinetic energy is random. The non-random part can do work. So can a difference in temperature, but not a uniform temperature.
The part of the energy that is not heat includes latent energy of evaporation, energy of position, and energy of motion – aka wind. These correspond to various degrees of freedom. In a real atmosphere, all of these are going on at the same time. Because of gravity, part of the motion (the non-random up-and-down part) gets converted from speed to energy of position. You can get an approximation of this by calculating the ratio of density to pressure. If this ratio stays the same, temperatures approximate isothermal..
But when they don’t, you get a change of temperature with altitude. Since in the real atmosphere we do get a lapse rate, which is less in a wet atmosphere due to part of the energy being in latent form. The details are very heavy math, which is why it’s easy to overlook some part.
Ocean heat: IR goes to evaporation until the air is saturated with water. Visible light goes to depth. In the North Atlantic, this leads to the Gulf Stream, and the heat radiates to the atmosphere and space over thousands of miles. In the Equatorial Pacific, the surface current is mostly East/West, so it gets warmer over time, until El Nino dumps the excess heat.
Ice Ages: should be caused by an increase in albedo from 30% to about 39%. This could happen from volcanoes, or from the heat engine slowing down. If the Polar areas remain frozen, while the tropics and temperate areas have more clouds, especially high level clouds, temperatures slowly drop. Or, in the case of the Younger Dryads, rapidly drop. We know that at the end of Ice Ages,, temperatures rapidly increase. The mechanism is so far unknown, but should have positive feedbacks until it gets close to current temperatures. So the rapid increase in temperatures from 1850 to 2000 is probably not CO2 related, and may have just ended around 1998. My guess is ENSO, and that could be a 1000 year cyclical pattern.
Repeat: temperature is the average random kinetic energy per molecule. If you double the number of molecules and double the random kinetic energy, the temperature remains the same. The lapse rate is due to the fact that a greater proportion of the total energy at altitude is in the form of positional and non-random kinetic energy. Jet streams at 8 km are often hundreds of km per hour, and since this is non-random, results in lower temperature. Winds are lowest at the surface, and this is part of the reason surface temperatures are higher. Energy, changes in the form of energy, and energy flows – including but not limited to radiation – explain the weather, and over time, the climate. Emissions are random, absorption is non-random due to the lapse rate, so H2O and CO2 concentrations play a role, but the IR path is only one of several paths that determine energy flows.
This is a nomination for a new top level post. It is a follow up to the comment above from Stephen Rasey above on Jan 21, 2012.
Two Sides of the Albedo; The Problem with 240 W/m^2.
by Stephen Rasey
240 Watts/m^2 is an important number in climate science[4]. People use it to calculate the black body temperature of the Earth ( Tbb=255 K = -15 C) and compare it to some measured average temperature, (Tm = 285K = +15C) which converts back to a blackbody heat flow of 390 W/m^2. The difference between 240 W/m^2 and 390 W/m^2 is explained as the Green House Effect.
There is a big problem here. Along the pathway from the sun to the surface of the Earth in this algebra, we have violated the laws of Thermodynamics at least twice.
Let me be clear at the beginning. I believe in a Greenhouse Effect (GHE). I am questioning the physical model and the arithmetic used to calculate the size of the GHE. In the end, I hope to convince you that a more useful estimate for solar heating of the Earth’s surface is 341 W/m^2, equivalent to about 279 K, not the 240 W/m^2 and 255 deg K generally assumed.
How people get to 240 W/m^2 is that take a solar constant (So = 1364 +/- 3 W/m^2) as the energy received full on at the distance of the earth. They then divide by 4 to spread that energy evenly over a sphere representing the earth’s top of atmosphere, yielding 341 W/m2.[4] THIS is the ORIGINAL SIN! Dividing by 4 leads to a dead planet; it is an isothermal planet without net heat flow, anywhere. Next, this 341 W/m^2 is passed through an albedo mechanism, reflecting 30% and transmitting a net 240 W/m^2 to the ground. At this point, it is equivalent to a Blackbody at 255K and thus the second sin against thermodynamics is committed.
In the past month on WUWT, there have been several highly enlightening posts by Brown, and Eschenbach [3] that have made clear that when any non-greenhouse gas[1] is in a gravity field of a planet, and the surface of the planet is isothermal, then the atmosphere must be isothermal at the temperature of the ground. The purpose of those posts was to refute the theory that a gravitational field can cause the lower part of the atmosphere to be warmer than the upper part. Until the temperature lapse rate goes to zero, work can be extracted from the temperature difference. Therefore, the atmosphere at equilibrium must be isothermal if the heating of the ground is uniform. There is no net heat flow. There is no convection, no net conduction, no latent heat exchange, no heat of vaporization or of fusion because there is no temperature differences to drive the heat flow. It’s a dead planet.
For this next argument, I ask you to imagine 4 concentric spherical shells, from inner to outer, call them G, B, T, S. G is the Earth Ground, a solid sphere with a gravitational field.. B is the Albedo layer. T is the Top of the Atmosphere with an ideal gas between T and G. S is the surface of solar irradiance. These shells are all spherically symmetrical with a common center. Why do we have S as a sphere? Because we divided solar insolation by 4 !! The only way to uniformly heat G is to have a spherically uniform radiator. When we do that, using the inverse square laws, we see that in order to irradiate G with X W/m2, surface S must also radiate at X W/m^2. S and G must be at equal Black Body temperatures for any X! We have not specified radius of S or G or T, except by “concentric” Rg .LT. Rb .LT. Rt .LT. Rs. Therefore, the temperature at T must also be equal to S. The Atmosphere between G and T is isothermal with everything at the same temperature. The Atmosphere has a pressure gradient, but is without a temperature lapse rate.
Wait a minute! What about the Albedo layer B? Don’t we have to multiply the Energy flow from S through T to G by the albedo fraction “A”, call it 30%. So, to plug in some numbers, S is radiating by 341 W/m2 at 279 K. T is at the same temperature. 341 W/m2 encounters the Albedo B, and only (1-A) or 240 W/m2 get through to the Ground. Therefore Ground must be bathed in 240 W/m2 that implies a black-body temperature of 255 K. Oops! Suddenly, we are confronted with a non-isothermal system, with the ground colder than the top of the atmosphere. Thermodynamic alarm bells should be going off in your heads.
We can plug Dr. Brown’s silver wire into the atmosphere near T and the other end into the Ground G and extract work from the temperature difference, warming Ground G until it warms back to near temp of T. We then unplug the wire. Ground G is again only receiving 240 W/m2 again. But who here believes that Ground G will drop in temperature again back to 255 deg K? How can it get colder than its surroundings? Of course, it will not.
At thermal equilibrium the Temp of G must be the same as T and S. Otherwise we could use the silver wire to extract work. The Albedo fraction (A) of Albedo B must make no difference in the temperature of G! Remember, this is a DEAD PLANET we are considering here. We made the Original Sin of dividing by 4 long ago.
Something is missing. There is an albedo layer. It is blocking about 100 W/m2 of downward solar irradiance heat flow. But, the albedo has two sides. It also blocks 100 W/m2 of heat flow upwards. It is serving as an insulator, a device that reduces the transmission of heat in both directions, but cannot by itself change the ultimate temperature on its two sides. An albedo is not a Maxwell’s Demon.
Don’t believe me? Try this on your next campout or hike. In the morning, fill your water bottle. Wrap it in a “space blanket” with an albedo at least 90%. So little of the sun’s heat will get to the bottle that the temperature of the water should plummet. By noon, you should have a welcome block of ice to cool you down. Read the directions of the space blanket carefully! If you put it on wrong side out, you’ll heat the water to boiling instead of freezing it. — No. Space Blankets don’t work that way. They have two equal sides. They cannot refrigerate their interiors; they can only retard the heat flow.
Albedo’s have two sides, too. They reflect heat away. They reflect heat back to the ground. They insulate. They retard the heat flow, not change the temperature of the end state. They do this without GHG’s (2). But an Albedo, whatever it reflectivity, cannot change the temperature of the ground —- ON A DEAD PLANET.
Put an Albedo on a rotating planet, illuminated by a near-point-source sun, possessing a night and a day, equator and poles, and make the albedo time varying by the minute and hour – it will make all the difference in the world! Solar insolation is anywhere from 1365 W/m2 to zero and all manner of heat capacity mechanisms come into play. But on that world, you don’t divide the solar irradiance by 4.
Stephen Rasey
Notes:
(1) The same is true if the atmosphere contains GHGs for the same reasoning, but I don’t want to go there, yet.
(2) But Albedo is from clouds and ice, and H2O is a powerful GHG! I’ll concede the point, but maintain we are talking about two separate properties of the same compound. Either way, a GHG driven albedo cannot act as a Maxwell’s Demon and keep the concentric spheres at different temperatures.
(3) Other relevant WUWT posts:
Brown, Jan 24, 2012: http://wattsupwiththat.com/2012/01/24/refutation-of-stable-thermal-equilibrium-lapse-rates/
Eschenback: Jan 19, 2012: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/
Eschenback: Jan 13, 2012: http://wattsupwiththat.com/2012/01/13/a-matter-of-some-gravity/
Brown, Jan 12, 2012: http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/
Rasey, Jan 21, 2012: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872239
[4] Just a small sample of references:
http://data.giss.nasa.gov/gistemp/2007/
Earth’s Climate History, Antón Uriarte
Image: Uriarte: Appendix 1 [Note Thermals and transpiration on an isothermal uniformly-insolated dead planet.]
http://www.soest.hawaii.edu/GG/FACULTY/POPP/Lecture2.ppt [same dynamic weather systems on a 342 W/m2 isothermal dead planet.]
Alley, Earth: The Operator’s Manual Page 352 #9
Anderson, Decadal Climate variability: dynamics and predictability. Pg. 296-7. The Figure 2 on 297 with “Surface temperature without Greenhouse effect” – Venus = (minus 46 deg C)
Wendisch-2012, “Theory of Atmospheric Radiative Transfer: A Comprehensive Introduction. Fig. 1.2 after Trenberth et al 2009. Same diagram as Uriarte above.
Stephen Rasey,
You seem to be laboring under two misconceptions.
1) “AVERAGING IS A SIN”
If an actuary say the average life-span in the US is 78.1 years, no one should interpret this to mean that everyone will always live exactly 78.1 years.
If an economist says that the US average income is $39,945, no one should interpret this to mean that everyone will always earn exactly $39,945.
If a scientist says the average incoming solar radiation is 340 Wm^2, no one should interpret this to mean that every square meter always will receive 340 W/m^2.
Any scientist who reports an average is well aware that averages do not tell you the value at every time for every location. The whole point in finding an average is to have a simple number that best summarizes the situation. Now, if a climate model actually assumed that the insolation was and even 340 W/m^2 everywhere, that would be Bad Science™. Fortunately, no climate model I know of makes such a stupid assumption.
2) “340 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM”
You explicitly use this assumption when you say “The only way to uniformly heat G is to have a spherically uniform radiator…. S and G must be at equal Black Body temperatures for any X!”
I could easily divide the outer shell (which you called “S”) into 1 m^2 regions, and then have 1/2 of each square meter surface radiating 2 * 340 W/m^2 and the other 1/2 radiating 0 W/m^2. Or have 1/10 radiating 10 * 340 W/m^2 and have 9/10 radiating 0 W/m^2. Or have 1/200,000 of the surface radiating at 200,000 * 340 W/m^2 and 199,999/200,000 radiating 0 W/m^2 (which would ~ correspond to the temperature of the sun & the faction of the sky covered by the sun). Or put another way, I could place a small, hot 340 W light bulb above every square meter of earth (with all of the light focused downward, of course) and produce a uniform 340 W/m^2 “insolation” at the TOA.
IF you HAD to make a model with globally uniform insolation, having1/200,000 of the shell “S” radiate at 200,000 * 340 W/m^2 would be the best option (for a clearly bad model that ignores night/day and poles/tropics). (By the way, this model WOULD show effects of albedo and of GHGs, and the surface could be above 255 K.)
Other than these two points, I agree with the rest of your analysis and conclusion. Surrounding the entire earth with a uniform 255 K shell (with any albedo) shining light onto the ground (with any albedo) with an intermediate layer (with any albedo) and any atmosphere (with or without GHGs) will cause the surface to be 255 K. (Actually, we would have to further assume no geothermal energy).
This would indeed be a very dead world, with no visiblelight for photosynthesis.
Tim, First your point 2:
240 W/m^2 is what is equated to a 255 K blackbody.
340 W/m^2 equates to a 279 K blackbody.
That difference is key to my central point that 279 K should be the ground temperature in an average insolation model. So the greenhouse effect is the increase from a base tempature of 279 K to some agreed upon average 288K.
Therefore the Greenhouse effect on Earth does not
add 33 deg K to a base like of 255 K,
but only 9 deg K to a base line of 279 K.
As to your point 1:
We are agreed that it is a stupid assumption. But lots of people, use 240 W/m^2. Each of the references I noted in [4] above made use of 240 W/m^2.
Not counting my own uses, there are 49 uses of it in this one thread.
I cannot speak to what GCMs use. But a lot of literature is littered with 240 W/m^2 and 255 K as the energy and temperature of the Earth without the greenhouse effect.
Google “240 W/m albedo” and look at the first couple of pages. Or Goggle ‘albedo “255 K”
Or better yet, Google ‘ “240 W/m” “255 K” ‘ and look at the list of hits. Look at the ones ending in “.edu”
Pull up the WUWT post: Earth’s baseline black-body, A damn hard problem. Do a find on “240”. A lot of people have fallen into the conceptual trap that the albedo itself, in a uniformly insolated planet, will change the temperature of the ground. Using Dr. Brown’s silver wire, you could pull power from the temperature difference across the albedo. It cannot be so.
@Tim Folkerts 9:38am,
First, the temperature would be 279 K, not 255 K. That is my essential point. I’m not giving ground on that.
Secondly, Divide solar insolation by 4 to average the insolation — EQUALLY — over the planet Earth. I didn’t make that up. A lot of climate scientists do that every day as a “Toy” model. It’s in the books. Its in the lecture notes. Equally over the planet means it is Spherically symmetrical. Therefore, the source of illumination must be spherically symmetrical. That is a direct consequence of dividing by 4.
Thirdly, by your 1/200,000 of the shell in radiating at 200,000 * 340/W ^2 and the rest at 0K, you basically make my point. Double the size of the shell. Each of those radiating areas have to spread their radiation out by the inverse square law so each only provides 1/4 of the energy per square meter from S to G. But each radiating point has expanded by a factor of 4. So four times the surface area providing only 1/4 the energy per meter of S to the Ground G equals no net change in radiation to G. G, or at least the top of the albedo layer B, receives 340 W/m^2, regardless of the size of S.
Fourth, because the ground is uniformly insolated, no point different than another, there would be an isothermal atmosphere, even in a gravitational field (See Brown: Refutation…), and there can be no temperature difference across any albedo layer, elsewise a silver wire across the layer would be able to extract work.
Albedos can change the temperature of the ground only when they are time-varying and/or there is a night and a day – i.e. non-uniform insolation.
That is why dividing by four is the Original Sin in climate science. Dividing by 4 creates an isothermal planet model. Any illustration (such as Uriarte) that shows 341 +/- 1 W/m^2 as incoming solar energy, and also shows almost 100 W/m^2 in “Thermals”, “Convection”, “Evapo-Transpiration” heat flow is inconsistent in its core assumptions.
Stephen, I thought I had responded earlier, but it looks like it got lost. So let me repeat the main points.
1) Yes, I should have said 2) “240 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM” (not 340 W/m^2). I got ahead of myself and wasn’t paying enough attention to the specific number. But this is really a secondary point.
The main point had to do with the SPECTRUM of the incoming light and this idea of yours: “A lot of people have fallen into the conceptual trap that the albedo itself, in a uniformly insolated planet, will change the temperature of the ground. ”
* If the incoming light is from a shell with a uniform temperature completely surrounding the earth, then everything within that shell will become that temperature independent of albedo, emissivity, atmospheric pressure, or GHGs. In this case you are right that worrying about albedo is “a trap”. The “sunlight” and the “earthlight” have the same spectrum and everything will be symmetric, so changing the interior cannot affect the balance of energy. You cannot have something on this world like clouds that reflect sunlight and absorb earthlight, because sunlight and earthlight are the same.
* If the incoming light is from “a million tiny stars”, then the other factors DO matter. There is no longer symmetry between the incoming sunlight (mostly 0.1-4 um) and the outgoing “earthlight” (mostly 4-100 um). Now albedo can and will matter, cooling the earth by reflecting away the incoming light. Now GHGs can and do warm the earth by letting sunlight in quite well, but not letting the earthling out as easily. Now, ignoring albedo is wrong, and your thinking would be “a trap”.
Since “a million tiny stars” is a closer analogy for uniform sunlight, then the effects of albedo & GHGs will matter, even in a uniformly illuminated world. Albedo WILL lower the effective radiating temperature from 279 K (340 W/m^2) to 255 K (240 W/m^2). GHGs WILL raise the surface temperature by preventing earthlight from escaping directly to space (which apparently amounts to ~ 33 K).
Stephen,
I did type too quickly earlier and got the wrong number. I should have typed
2) “240 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM”,
not 340 W/m^2, so we are in agreement there.
But that is not my key point. The key point is what the incoming “uniform radiation” is like. There are two main possibilities to consider.
1) If the incoming “sunlight” is from a shell with uniform shell with a constant temperature completely surrounding the earth (as you postulate), then the temperature anywhere inside will be the same temperature as the shell independent of albedo, emissivity, atmospheric pressure, or GHGs. The “earthlight” and “sunlight” will be identical, and anything inside (like clouds or space blankets or GHGs) will have an identical effect on energy flows up and down, and cannot change the equilibrium temperature.
If the shell is 340 W/m^2, the earth will be ~ 279 K.
If the shell is 240 W/m^2, the earth will be ~ 255 K.
Worrying about albedo would indeed be a “trap”, since it won’t affect this world with this “sunlight”.
2) If the incoming light is from “a million tiny stars” then the conditions of the earth can and will matter. The “sunlight” will be mostly 0.1-4 um, while the “earthlight” will be mostly 4-100 um. This means that clouds or GHGs can have a different effect on upward and downward light. Clouds can and will reflect away “sunlight” and absorb “earthlight” (lowering the effective blackbody temperature of the the earth from 279K to 255K). GHGs can and will absorb “earthlight” better than “sunlight”, which has the effect of warming the surface (apparently from ~ 255K to ~ 288 K based on actual observations).
The surface will be a uniform temperature, but the atmosphere will be cooler as you go up (due to IR radiation emitted from the atmosphere to space).
Worrying about albedo is now critical, and your thinking has now become the “trap”.
Both of these worlds are “uniformly illuminated”, but the second is a much better approximation of a uniformly illuminated earth with 340 W/m^2 of “sunlight”. The second has the correct spectrum of “sunlight” to match real sunlight. The second will have effects of GHGs and albedo, even on a uniformly illuminated world.
Tim, you bring up an excellent point that the spectrum of the energy received is significantly different than the energy emitted. Thank you.
I thought about this, too in a comment from Jan 12. (quote shortened, full comment here)
It is the value of “a=0” people use to get to 255 K if they assume A=30%
In the next comment, I showed that the total energy leaking through the albedo is the same (1-A) regardless of the value of “a”.
You discuss in Point 2 the issue of spectrum conversion very well. It is a serious issue, but I don’t think you can just assume the 255 K value is correct.
On the one hand, you argue that an Albedo of clouds lowers the blackbody temperature of the Earth, but then GHG’s ride to the rescue to raise its temperature. Wouldn’t make more sense to net out the change? “a = A” might not be right, but “a = 0” is quite likely wrong. For one thing, even at night, we can look up and see the bottom of the clouds from scattered light from the ground. There has to be some return.
Four thoughts:
1. There still exists the argument of the Isothermal atmosphere in a gravitational field and Dr. Brown silver wire which we bridge across the albedo layer. If “a” does not equal “A”, then we have a perpetual temperature difference across the albedo with which the silver wire can extract work. That is a good reason to say “a” must equal “A”. But it is not proof. Well, maybe it is, and I just don’t grok why, yet.
2. In your million star suns, the total energy may average 340 W/m^2, but spectrum will not be a 340 W/m^2 = 279 K blackbody, but a black body of much higher temperature and heat flux attenuated by distance (or spacing of the stars). So now we have 340 W/m^2 hitting the TOA, but not blackbody spectrum. How does that change arithmetic or Brown’s isothermal atmosphere? Does the albedo become a Maxwell’s Demon?
3. The literature doesn’t seem to spend much time on this spectrum conversion. Yes, solar energy is converted to heat. But at the same time, they talk of “thermals”, “latent heat”, “convection” in a model that needs to have an isothermal ground. At equilibrium there must be no net energy flow. There is no day and night, everywhere it is the same grey illumination. No wind, no weather. A significant piece of the literature is inconsistent in this regard.
4. The whole “divide by 4” average insolation is a bad idea to try to explain climate mechanisms. Might as well divide by zero. It is better to scrap it rather than fix it. Climate is a dynamic equilibrium, not a static one. Day and night cannot be averaged. Eppur si muove.
Two errata to my post from 9:59 pm above:
On the one hand, you argue that an Albedo of clouds [ made of GHGs ] lowers the blackbody temperature of the Earth, but then GHG’s ride to the rescue to [eliminate the drop and further] raise its temperature. Is the lowering of the blackbody temperature a real thing or only an accounting gimick?
3. The literature doesn’t seem to spend much time on this spectrum conversion. Obviously incorrect. Spectrum conversion is a big part of the literature. It is just not addressed much brief path from sun to ground. (1364 / 4 = 340) * 70% = 240 W/ m^2 implying 255 K black body.
Stephen,
This is getting a bit off-topic from the original post, but I will make a couple quick comments (or perhaps not so quick — we will see).
>Why do we account for albedo only on the direct-from-sun energy,
>but everyone seems to ignore it on surface-to-sky path?
For your analysis, there are not just two parameters for reflection (albedo for sunlight and earthlight). First, you need to consider reflection, transmission and absorption –> R+T+A = 1 (all light will either get reflected by, transmitted through, or absorbed by an object it interacts with) (albedo = 1 – reflectivity). You need to consider how these relate to both sunlight and earthlight (actually, you should consider it for each individual wavelength, but since sunlight (0.1-4 um) and earthlight (4-100 um) basically don’t overlap, you can pretty well treat them as two distinct ranges with two separate values) . You need to consider how these differ for the ground and the atmosphere.
So you would have a matrix of parameters like R_s_g (reflectivity of sunlight by the the ground) or T_e_a (transmission of earthlight through the atmosphere).
You say “a fraction “a” is reflected back downward”, so this would be R_e_a in my notation. But the atmosphere does not reflect much IR at all, so “a” would be very close to zero. And the ground does not reflect much earthlight (or “atmospherelight”, but “atmospherelight” would be the same basic wavelengths as earthlight).
Also, it is wrong to say “and (1-a) continues into space” since you did not include the energy absorbed by the atmosphere. With a cloud cover, nearly all of the earthlight is absorbed by the cliouds. This means “a” is very close to zero, but the fraction that escapes to space is ALSO very close to zero.
>If “a” does not equal “A”, then we have a perpetual temperature difference
>across the albedo with which the silver wire can extract work.
>That is a good reason to say “a” must equal “A”
The only problem comes if you try to make “a” and “A” different for the same wavelength. This would violate fundamental rules of physics, but you are not doing that There is no problem with perpetual temperature differences and perpetual extraction of work if there are “connections” to objects of different temperature (ie the sun and space). If you remove those connections (eg by putting an isothermal shell around the earth or enclosing the system in insulated walls), then perpetual temperature differences and perpetual extraction of work would be a problem.
>At equilibrium there must be no net energy flow.
It is important to distinguish between “steady-state” and “equilibrium” (which is much more restrictive than steady-state). A bathtub with water running in at the same rate it drains is a steady-state situation, but it is not an equilibrium situation. When the earth is gaining 240 W/m^2 of sunlight and losing 240 W/m^2 of earthlight, that is a steady-state situation, not an equilibrium situation. (This assumes an average over large areas and long times. For shorter times or areas, steady-state is not a good approximation.)
The earth is NOT in equilibrium — not even close. The earth IS, however, very close to steady-state (as far as net energy is concerned). Deviations from steady-state over periods of decades are pretty much synonymous with “climate change”.
4. The whole “divide by 4” average insolation is a bad idea
>to try to explain climate mechanisms.
Again, I don’t think anyone is using an average this to “explain climate mechanisms”. Averaging the incoming sunlight (or averaging evaporation or averaging outgoing thermal IR or … ) is simply a handy way to summarize a very complex situation. It is a way to present the basics to a wide audience. Actual analysis of details must (and does) worry about details. An audience like you or me or Willis or Dr Brown or Anthony or Dr Mann enjoys pondering the details, but IMHO that does not make the summary “a bad idea” for a different audience.
Tim,
Let me refocus our discussion to one point very much on topic to this thread of Perpentuum Mobile.
Back on Feb. 7, my long post was for the explicit purpose to show that climate scientists take average solar insolation (1367 / 4 ) W/m^2, then multiply by (1 – 30% albedo) to achieve a 240 W/m^2 and they equate that to a 255 K black body. I showed via the 4 concentric shells, that there would have to be a temperature of 279 K above the albedo, and 255 K below the albedo at equilibrium. This creates a Perpentuum Mobile, a violation of Thermodynamics. Therefore the model used to defend the 255 K black body temperature of the Earth is fatally flawed.
In your most recent post of Feb. 11, you said, I don’t think anyone is using an average this to “explain climate mechanisms”. “Anyone” is a pretty universal word. I only have to show a someone using it to invalidate the statement. My references in [4] of the Feb. 7 are sufficient for that. But I’m a sport. Show me ONE reference that derives a 255 K Earth ground blackbody average temperature that DOES NOT arrive at it via “average solar insolation”,
Your arguments concerning spectrum conversion and how the albedo is a function of wavelenght are very astute. They cannot be easily dismissed. But neither can Thermodynamic arguments be dismissed so as to create a Perpentuum Mobile.
The Earth is a system in dynamic equilibrium. No argument. But that is not the system under discussion.
“Average Solar Insolation” describes a planet with a uniform temperature. It is a “Toy” to “summarize a complex situation”, but summarize it so badly it leads to erroneous conclusions. Like the “Surface Temperature without Greenhouse Effect:”
Earth = -15 deg C, Venus = -46 deg C.
(From page 297, Anderson, Decadal Climate Variability: Dynamics and Predictability.)
Stephen, you are missing an important point, I think.
The average EMITTED radiation is often described as 255K blackbody radiation.
The average ABSORBED radiation is never described as 255K blackbody radiation.
They are equal in the sense that emitted and absorbed radiation are both ~ 240 W/m^2, but they are most certainly not the same spectrum, and never will be. This means they can (and do) have two very different effects on earth. It is kind of like saying a $1 bill and 100 pennies are the same. Yes, they have the same official value, but they don’t act the same in my pocket. I can’t put the pennies in a vending machine.
Giving me 100 pennies is distinct from giving me a $1 bill, even though in some sense they are the same.
Giving the earth 240 W/m^2 of sunlight is distinct from giving the earth 240 W/m^2 of thermal IR, even though in some sense they are the same.
PS. The outgoing radiation is, of course, modified by GHGs, so even it is not strictly 255 K BB radiation if GHGs are present.
PPS You say “I showed via the 4 concentric shells, that there would have to be a temperature of 279 K above the albedo, and 255 K below the albedo at equilibrium. ” but this is incorrect in your scenario. If the earth was surrounded by 279 K BB radiation, then the temperature everywhere inside would indeed be 279 everywhere, including below the “albedo layer”.
Instead of saying “The average ABSORBED radiation is never described as 255K blackbody radiation.”
I should have added” … except in thought experiments like yours or Jelbring’s.” Certainly we can postulate such a system, but it is so far from reality that it should not be equated in any way with statements about earth’s energy balance. A uniform shell around the earth is a useful alternate reality for sharpening our understanding, but it is definitely NOT what people are generally talking about.
Tim, we are talking past each other and If I am not clear, then I am sorry.
When I say 279 K above the albedo and 255 K at the ground, I am using the model that climate scientists use to achieve the figure of 255 K. I am showing that that model violates Thermodynamics and the 255 K must be wrong. You agree that 279 K must be the temperature below the albedo. Yes, it is a thought experiment. It is a simplied model to help determine the strength of the Greenhouse effect (255 K to 288K) and it is a Thermodynamic blunder. It is not my model; it is in many books and papers.
Average Insolation is a steady-state, static model. Jelbring is wrong. The atmosphere is isothermal with the ground to the TOA and the albedo cannot change its temperature. When we are in a steady state condition, there can be no net absorbtion. Your spectrum shifted example is one worth exploring, but since we cannot have any net absorption by the albedo layer and we must not commit the error of the albedo shell making a Perpetuum Mobile, we have to explore more what happens. Because whatever happens, it cannot result in 255K.
Stepehen,
I think we can agree that:
* it is NOT a violation of thermodynamics to run a machine perpetually using the hot sun (5770 K) and cold outer space (3 K) as the two thermal reservoirs.
* it IS a violation of thermodynamics to run a machine perpetually using the warm shell (279 K) and warm surface (279 K) as the two thermal reservoirs.
The “million tiny suns” earth is using the hot sun (albeit inefficiently) and cold outer space (albeit inefficiently) as the two thermal reservoirs. The various thermal connections cause the ground to be ~ 288 K (primarily warming from the sparse sunlight coming in, but also IR coming in from the atmosphere and IR going out to space). The various thermal connections cause the atmosphere to be ~ 220 K near the tropopause (continuously cooled by radiation to outer space). Two of the absolutely key factors are albedo (limiting the warming of the ground somewhat) and GHGs (allowing cooling of the atmosphere). Given these two temperatures (which are maintained by the hot sun and the cold space), we can indeed run a heat engine perpetually without any violations of thermodynamics.
You say “Your spectrum shifted example is one worth exploring, but since we cannot have any net absorption by the albedo layer”. But we can have a perpetual net absorption of energy from the hot reservoir (at least as long as the sun shines), and a perpetual net emission of energy to a cold reservoir, which is indeed exactly what is needed to run a heat engine perpetually.
PS Stephen, you might want to read up on Kirchhoff’s law of thermal radiation. http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation
Specifically note that “a corollary of Kirchhoff’s law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”
So the “albedo layer” must emit radiation just as well as it absorbs that same radiation. If this was NOT true, then your violation of thermodynamic laws could indeed happen. In a sense, you have rediscovered the reasoning that lead Kirchhoff to this law.
Very well argued, Tim. I’ve learned something. Thank you.
I concede that with the million starry sun model it is possible to drive an average insolated planet into a non-isothermal state if it has an albedo that varies with wavelength. The efficiency, as you say, is low, however the temperature difference across the albedo is likely small, but can be non-zero.
It provides a good explanation why in the formula: Normalized Energy to the ground = (1-A)/(1-a), that “a” does not have to equal “A” where “A” is top of albedo reflection and “a” is bottom of albedo reflection (or now modified to “absorption –> emission downward”). I’m still a long way from accepting “a = 0”, for as you note: the “albedo layer” must emit radiation just as well as it absorbs that same radiation. “ For that reason, I’m a long way from believing in the 255 K Earth “Surface temperature without Greenhouse Gas”. I’ll admit, it doesn’t have to be 279 K; might be a little more, or a little less, but I don’t see 255 K as reasonable since the albedo has two sides.
Here’s an interesting visual about that starry-sun-sky.
Since there are 41253 square degrees in a sphere and the sun takes up about 0.20 square degrees, there is room for about 200,000 suns in the celestial sphere. Let’s cut the sun up into 200,000 stars of 6000 deg K. Each will then be 12.6 square arc seconds or less than 4 arc seconds across. Jupiter, viewed from earth is 32 arc seconds or over 64 times bigger in area (though much dimmer!) than one of our 200,000 starry suns. One brilliant 4 arc-second spec for every spot in the sky the size of the sun. …. Hmmm…. too bad we cannot marvel at the spectacle because that uniform albedo is in the way. Just as well, it would hurt the eyes.