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.
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 @ur momisugly 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
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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?
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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.