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.
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.”
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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.
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“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.
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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