Guest post by Robert G. Brown
Duke University Physics Department
The Problem
In 2003 a paper was published in Energy & Environment by Hans Jelbring that asserted that a gravitationally bound, adiabatically isolated shell of ideal gas would exhibit a thermodynamically stable adiabatic lapse rate. No plausible explanation was offered for this state being thermodynamically stable – indeed, the explanation involved a moving air parcel:
An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field.
This argument was not unique to Jelbring (in spite of his assertion otherwise):
The theoretically deducible influence of gravity on GE has rarely been acknowledged by climate change scientists for unknown reasons.
The adiabatic lapse rate was and is a standard feature in nearly every textbook on physical climatology. It is equally well known there that it is a dynamical consequence of the atmosphere being an open system. Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.
Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work. As is usually the case, violation of the refrigeration statement allows heat engines to be constructed that do nothing but convert heat into work – violating the “no perfectly efficient heat engine” statement as well.
The proposed adiabatic thermal lapse rate in EEJ is:
where g is the gravitational acceleration (presumed approximately constant throughout the spherical shell) and cp is the heat capacity per kilogram of the particular “ideal” gas at constant pressure. The details of the arguments for an adiabatic lapse rate in open systems is unimportant, nor does it matter what cp is as long as it is not zero or infinity.
What matters is that EEJ asserts that 
in stable thermodynamic equilibrium.
The purpose of this short paper is to demonstrate that such a system is not, in fact, in thermal equilibrium and that the correct static equilibrium distribution of gas in the system is the usual isothermal distribution.
The Failure of Equilibrium
In figure 1 above, an adiabatically isolated column of an ideal gas is illustrated. According to EEJ, this gas spontaneously equilibrates into a state where the temperature at the bottom of the column Tb is strictly greater than the temperature Tt at the top of the column. The magnitude of the difference, and the mechanism proposed for this separation are irrelevant, save to note that the internal conductivity of the ideal gas is completely neglected. It is assumed that the only mechanism for achieving equilibrium is physical (adiabatic) mixing of the air, mixing that in some fundamental sense does not allow for the fact that even an ideal gas conducts heat.
Note well the implication of stability. If additional heat is added to or removed from this container, it will always distribute itself in such a way as to maintain the lapse rate, which is a constant independent of absolute temperature. If the distribution of energy in the container is changed, then gravity will cause a flow of heat that will return the distribution of energy to one with Tb > Tt . For an ideal gas in an adiabatic container in a gravitational field, one will always observe the gas in this state once equilibrium is established, and while the time required to achieve equilibrium is not given in EEJ, it is presumably commensurate with convective mixing times of ordinary gases within the container and hence not terribly long.
Now imagine that the bottom of the container and top of the container are connected with a solid conductive material, e.g. a silver wire (adiabatically insulated except where it is in good thermal contact with the gas at the top and bottom of the container) of length L . Such a wire admits the thermally driven conduction of heat according to Fourier’s Law:
where λ is the thermal conductivity of silver, A is the cross-sectional area of the wire, and ΔT=Tb–Tt . This is an empirical law, and in no way depends on whether or not the wire is oriented horizontally or vertically (although there is a small correction for the bends in the wire above if one actually solves the heat equation for the particular geometry – this correction is completely irrelevant to the argument, however).
As one can see in figure 2, there can be no question that heat will flow in this silver wire. Its two ends are maintained at different temperatures. It will therefore systematically transfer heat energy from the bottom of the air column to the top via thermal conduction through the silver as long as the temperature difference is maintained.
One now has a choice:
- If EEJ is correct, the heat added to the top will redistribute itself to maintain the adiabatic lapse rate. How rapidly it does so compared to the rate of heat flow through the silver is irrelevant. The inescapable point is that in order to do so, there has to be net heat transfer from the top of the gas column to the bottom whenever the temperature of the top and bottom deviate from the adiabatic lapse rate if it is indeed a thermal equilibrium state.
- Otherwise, heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.
It is hopefully clear that the first of these statements is impossible. Heat will flow in this system forever; it will never reach thermal equilibrium. Thermal equilibrium for the silver no longer means the same thing as thermal equilibrium for the gas – heat only fails to flow in the silver when it is isothermal, but heat only fails to flow in the gas when it exhibits an adiabatic lapse in temperature that leaves it explicitly not isothermal. The combined system can literally never reach thermal equilibrium.
Of course this is nonsense. Any such system would quickly reach thermal equilibrium – one where the top and bottom of the gas are at an equal temperature. Nor does one require a silver wire to accomplish this. The gas is perfectly capable of conducting heat from the bottom of the container to the top all by itself!
One is then left with an uncomfortable picture of the gas moving constantly – heat must be adiabatically convected downward to the bottom of the container in figure 1 in ongoing opposition to the upward directed flow of heat due to the fact that Fourier’s Law applies to the ideal gas in such a way that equilibrium is never reached!
Of course, this will not happen. The gas in the container will quickly reach equilibrium. What will that equilibrium look like? The answer is contained in almost any introductory physics textbook. Take an ideal gas in thermal equilibrium:
where N is the number of molecules in the volume V, k is Boltzmann’s constant, and T is the temperature in degrees Kelvin. n is the number of moles of gas in question and R is the ideal gas constant. If we assume a constant temperature in the adiabatically isolated container, one gets the following formula for the density of an ideal gas:
where M is the molar mass, the number of kilograms of the gas per mole.
The formula for that describes the static equilibrium of a fluid is unchanged by the compressibility (or lack thereof) of the fluid – for the fluid to be in force balance the variation of the pressure must be:
(so that the pressure decreases with height, assuming a non-negative density). If we multiply both sides by dz and integrate, now we get:
Exponentiating both sides of this expression, we get the usual exponential isothermal lapse in the pressure, and by extension the density:
where P0 is the pressure at z=0 (the bottom of the container).
This describes a gas that is manifestly:
- In static force equilibrium. There is no bulk transport of the gas as buoyancy and gravity are in perfect balance throughout.
- In thermal equilibrium. There is no thermal gradient in the gas to drive the conduction of heat.
If this system is perturbed away from equilibrium, it will quickly return to this combination of static and thermal equilibrium, as both are stable. Even in the case of a gas with an adiabatic lapse rate (e.g. the atmosphere) remarkably small deviations are observed from the predicted P(z) one gets treating the atmosphere as an ideal gas. An adiabatically isolated gas initially prepared in a state with an adiabatic lapse rate will thermally equilibrate due to the internal conduction of heat within the gas by all mechanisms and relax to precisely this state.
Conclusion
As we can see, it is an introductory physics textbook exercise to demonstrate that an adiabatically isolated column of gas in a gravitational field cannot have a thermal gradient maintained by gravity. The same can readily be demonstrated by correctly using thermodynamics at a higher level or by using statistical mechanics, but it is not really necessary. The elementary argument already suffices to show violation of both the zeroth and second laws of thermodynamics by the assertion itself.
In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down. Reverse that to a cooling, like those observed during the winter in the air above Antarctica, and the lapse rate readily inverts. Follow the air column up above the troposphere and the lapse rate fails to be observed in the stratosphere, precisely where vertical convection stops dominating heat transport. The EEJ assertion, that the dry adiabatic lapse rate alone explains the bulk of so-called “greenhouse warming” of the atmosphere as a stable feature of a bulk equilibrium gas, is incorrect.
“”””” Joe Born says:
January 24, 2012 at 8:21 am
That Dr. Brown has it wrong is readily demonstrated by a thought experiment nearly any layman can perform.
If an ideal monatomic gas subjected to gravity in a thermally isolated container consists of only a single molecule, its kinetic energy K–and thus the mean translational kinetic energy–at any altitude z is given by K = mg(z_max -z), where m is molecular mass, g is the acceleration of gravity, and mgz_max is the total (kinetic + potential) energy of the gas. “””””
Well Joe, I DON’T agree that Professor Brown has it wrong; but I DO agree that nearly any layman can perform your thought experiment; that’s how they earn the prestigious title of “Layman” rather than “Professor.”
So to your layman thought experiment. you state:- “”””” If an ideal monatomic gas subjected to gravity in a thermally isolated container consists of only a single molecule, “””””
There you just shot your self in the head. A gas, ideal or not, cannot consist of a single molecule.
when in a thought experiment you say “”””” any altitude “”””” that is taken to mean “”””” any altitude “””””, tha’ts the terrific advantage of a thought experiment; an infinifte sized container is easy to get by declaration; and your container must be infinite in size for your molecule to be able to be at any altitude.
Therefore your single molecule never hits the wall of the container, and there is nothing else present to hit either, so no collisions occur, so your molecule has no Temperature. It’s energy is indeterminate since according to Einstein there is no absolute frame of reference, and your molecule is simply in free flight.
it also clearly doesn’t have any Maxwell-Boltmann energy distribution either since only one molecule is present; further proof that it has no Temperature; or at least it has no Temperature different from zero Kelvins.
Maybe if you listen a bit more closely to Professor Brown, you could eventuall discard your layman’s mortarboard, and tassel.
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Dr Bown. With regards to your thought experiment with the silver wire, you say that:
“””Heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.”””””
Sorry, I am only an aviator and meteorologist, but are you not confusing ‘heat’ and ‘temperature’ here?
The individual molecules in the upper atmosphere can indeed be very hot (high kinetic energy), but there are so few of them, their total temperature on any thermometer is very low.
Thus the top of your silver wire can be as hot as it likes (say 35oc), and yet the individual molecules at 50,000 feet are not going to accept any of that heat because they are already (individually) quite hot themselves. (But because there are so few of them, the atmosphere at this level feels very cold).
Does not your thought experiment fail, because most of the molecules in the atmosphere are all at the same heat (kinetic energy), while the difference in temperature with altitude (on a thermometer) is simply an effect of the number of molecules you meet (pressure and density).
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I am not denying greenhouse effect here – any meteorologist will know this is a real effect. But it is surely also true that an atmosphere warmed at its base by conduction will transmit that heat throughout the atmospheric column, maintaining its temperature and lapse rate, yet with most of the molecules in that column having the same kinetic energy (the same heat, but not the same temperature).
Where Tallbloke et al seem to fall down, is believing that conduction-convection is the sole method of decreasing LW heatloss from a planet (thus deriving a warmer surface). It is not. The absorption and reemission of LW by clouds and certain gasses (the greenhouse effect) is a much more potent effect.
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Paul Bahlin says:
January 25, 2012 at 9:47 am
Nope. The premise is that all collisions are elastic, both with each other and with the container walls. See for example the Wikipedia article on Kinetic Theory:
The particles must also be small so that the total volume of the particles is much smaller than the volume of the container.
Robert Clemenzi says:
January 25, 2012 at 12:42 pm
Water vapor becomes optically thin far below the tropopause. Look at the IR emission spectrum of the atmosphere from space. The water vapor lines and bands are far more intense than the CO2 band. That’s because water vapor, which has a scale height about 1/4 that of the noncondensable gases (2km compared to 8km) becomes optically thin (optical density < 1) at lower altitudes where its warmer. At 17 km for a tropical atmosphere the amount of water vapor in a given volume is less than 1% as much as the amount of CO2.
Ralph says:
January 25, 2012 at 12:54 pm
This piece of misinformation seems to be even harder to kill than a gravity maintained adiabatic lapse rate. Temperature, as long as the pressure is high enough it can be defined, is a function of the average kinetic energy of the gas only. What happens at low pressure is that it takes longer for a thermometer to equilibrate with the gas. If the thermometer has a large thermal mass, on the order of the heat content of the volume of interest, it will perturb the actual temperature.
“On the influence of gravity on the thermal conductivity”
M. Tij (1), V. Garzó (2), A. Santos (2) ((1) Départment de Physique, Université Moulay Ismaïl, Meknès, Morocco; (2) Departamento de Física, Universidad de Extremadura, Badajoz, Spain)
(Submitted on 25 Feb 2000)
In this paper we evaluate the corrections to the Navier-Stokes constitutive equations induced by the action of a gravitational field in a gas subjected to a thermal gradient parallel to the field with no convection. The analysis is performed from an exact perturbation solution of the BGK kinetic model for Maxwell molecules through sixth order in the field. The reference state (zeroth order approximation) corresponds to the exact solution in the pure planar Fourier flow, which holds for arbitrary values of the thermal gradient. The results show that the pressure tensor becomes anisotropic, so that the momentum flux along the field direction is enhanced. In addition, the heat flux increases (decreases) with respect to its Navier-Stokes value when the gas is heated from above (below).
full text:
http://arxiv.org/pdf/cond-mat/0002397v1.pdf
These authors said the same thing I did in a lot more detail and precision numbers.
snip:
So there. A gravity field introduces a first order effect that causes coefficient of thermal conductivity to be asymetric in direction of flow.
Maxwell’s Demon in other words sorting out molecules by kinetic energy content. I’m afraid you boys are going to have to come to grips with this. There’s no violation of thermodyanmics. The Demon creates a thermal gradient at the expense of an equal and opposite potential energy gradient. You can’t make a perpetual motion machine out of this. Every attempt at finding a way for the Demon to get a free lunch has failed. This is no exception.
Gravity-Induced Electric Fields in Metals
Canadian Journal of Physics, 1971, 49:(22) 2754-2767, 10.1139/p71-334
Just to muddy the waters further. There’s an electric field on that wire due to gravity – what if the hypothetical gas can conduct electricity – won’t it warm up because of the current flowing between the ends of the wire?. (I have become confused – I thought I ‘got’ lapse rates.)
A good argument : http://www.youtube.com/watch?v=RDjCqjzbvJY
Let’s try that quote from the paper again. Somehow the last few words in the paragraph I clipped didn’t come through and they were important (my bold).
“The main results concerning the transport of momentum and energy are that
the external field induces (i) anisotropy in the pressure tensor, (Pzz−p)/p ≃
845 ǫ2g∗2, and (ii) deviations from the Fourier law, qz/q(0)z − 1 ≃ 58
5 ǫg∗. While the first effect is of second order, the correction to the heat flux is
of first order, so that it depends on the sign of the thermal gradient. As
a consequence, the heat transport is inhibited when the gas is heated from
below (ǫ 0).”
Again, the source:
On the Influence of Gravity on the Thermal
Conductivity
http://arxiv.org/pdf/cond-mat/0002397v1.pdf
Well, I still haven’t worked out what any of you are talking about.., but isn’t the opposite and equal energy gradient to make up for the gravitational energy gradient, pressure?
Gold star for Myrrh.
Some fun for everybody:
http://phet.colorado.edu/en/simulation/gas-properties
A final exam you all might want to look at, especially problem 1. Would you flunk?
http://www.physics.sc.edu/~yar/phys706_2011/…/final_solutions.pdf
I know, I know, those of you who are devoted to the idea of a lapse rate at true equilibrium to the extent that you ignore the fact that the solution openly violates the second law won’t be swayed by a little thing like the fully worked out solution — which is the one I have in the article at the top, but this exam goes ahead and computes various quantities of interest and shows that they do the right asymptotic things. If the debate continues, I have a lovely contribution from a list-lurker that does the textbook exercise of showing that isothermal is indeed the maximum entropy solution. And on the list today it was verified (not by me) that the full stat mech computation is isothermal. Finally, Caballero has it as a homework exercise (2.17).
Is this starting to look like a conspiracy of some sort? A nefarious plot by evil warmists, who’ve managed to corrupt every thermo and stat mech textbook in the country? Who are brutally grading students incorrectly on their final exams, all to keep them from questioning Greenhouse Warming by hiding the True Solution from them? If so, you might want to ask yourself — is it just barely possible that all of these Ph.D. physicists who teach and do research and all that stuff in the general field of thermal physics are all right, and it is me that is wrong?
It truly would be lovely if, with all of these guns brought to bear to back up what was already a perfectly adequate proof by contradiction above, we could all just agree that Jelbring’s EE paper is categorically incorrect, because the state with a DALR that he asserts as the stable thermal equilibrium of an isolated ideal gas violates the second law of thermodynamics, fails of detailed balance, is not (in fact) the maximum entropy solution, and doesn’t make sense as no air parcels are being moved around in an atmosphere in static equilibrium.
Just a thought. A foolish one, I know — a mad, mad, dream. But y’know, I’ve gotta lay it out there.
rgb
Darn it. It was still cut off.
While the first effect is of second order, the correction to the heat flux is of first order, so that it depends on the sign of the thermal gradient. As a consequence, the heat transport is inhibited when the gas is heated from below, while the opposite happens when the gas is heated from above.
I really needed that last bit where it is specified that heat transport is accelerated when heated from above in addition to being inhibited when heated from below.
I highly recommend reading the whole paper. This corrects both Brown AND Nikolov neither of whom have characterized the situation correctly either conceptually or with mathematical rigor.
DeWitt Payne says:
January 25, 2012 at 1:15 pm
Ralph says:
January 25, 2012 at 12:54 pm
“This piece of misinformation seems to be even harder to kill than a gravity maintained adiabatic lapse rate. Temperature, as long as the pressure is high enough it can be defined, is a function of the average kinetic energy of the gas only. What happens at low pressure is that it takes longer for a thermometer to equilibrate with the gas. If the thermometer has a large thermal mass, on the order of the heat content of the volume of interest, it will perturb the actual temperature.”
It’s harder to kill because it’s correct. All matter above absolute zero has a temperature. There is no constraint on some minimum amount of mass involved. Any amount will do. The problem you are struggling to understand is that there is a Bolztman distribution of kinetic energy such that you cannot obtain a reliable temperature for an ensemble by measuring one individual molecule’s temperature. It’s sort of like trying to figure out the average intelligence quotient here based on measuring one individual. It does not follow that only groups have an IQ. Every molecule has a temperature.
The way to think about the difference between the stratosphere and the troposphere is to use the slab gray atmosphere toy model ( see 7.3.2 here, for example. Petty goes into more depth).
Thanks, looks very useful. I do so hate the toy model with downwelling radiation, but the general idea looks good and I’ll see what I can make of it.
rgb
DeWitt:
You asked elsewhere about why a laser can cut metal even when the metal gets very hot. But Claes Johnson’s paper is only about spontaneous blackbody emission, not induced emission as in a laser. Induced emission does not have a frequency proportional to the absolute temperature of the source, as does spontaneous blackbody emission in accord with Wien’s Displacement Law. (When lasers are used to cut steel there is also a blowing process helping to cool the metal.)
You yourself quoted empirical proof of what Claes has proven computationally when you talked about a gas not absorbing radiation from a cooler emitter, but doing so when the emitter got warmer than the gas. I would very much appreciate the reference for this experiment if you could oblige.
Robert Brown says at 10:40am:
“…the point is that a stable thermal equilibrium of an isolated ideal gas with a lapse rate violates the second law of thermodynamics…the zeroth law clearly states that the two locations (with different temperatures) are not in thermal equilibrium.”
Well, the top post does tell us:
“Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.
Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics…”
Why is that last true? I agree, it should be so but it flies in the face of hydrostatic equilibrium actually being in “temp. equilibrium” in Caballero. I noted that the 1st time you sent me there last week. Seems so long ago now…ha.
I am truly curious. So far as I can see in the top post this is just announced, 2nd law violation is not irrefutably proven up there. In the gas ideal gas column of interest, no process is irreversible in there. The molecules bounce around elastically with everything, entropy does not go down. Couldn’t entropy just stay the same in this idealization & therefore pass the 2nd law test? In real world, of course entropy goes up due to real inelastic collisions.
But then , in sec. 2.3 Caballero writes being non-isothermal is acceptable for hydrostatic equilibrium. Stunning. Caballero even goes so far in Sec. 2.3.1 to write: “Overall, it can safely be stated that atmospheric motions with horizontal scales > 10 km are always in hydrostatic equilibrium.” Meaning non-isothermal. That is a stunningly tall column & long wire to insert but it works in reality, no perpetuum, no 2nd law violation. Why?
Maybe I am about to be directed to another text book deep dive but that would be ok & progress.
Robert, the teach-a-holic – Thank you for an enjoyable renewed science discourse.
DeWitt Payne said @ur momisugly January 25, 2012 at 1:15 pm
Ain’t that the truth?
More accurately, measuring temperature with a thermometer always changes the temperature. Most often that can be neglected because the change is trivial, but there’s many a researcher come unstuck from not realising that they are changing what they measure.[/nitpick]
George E. Smith; says:
January 25, 2012 at 12:52 pm
“There you just shot your self in the head. A gas, ideal or not, cannot consist of a single molecule.”
Ummm… I think it can’t exist as a liquid or solid because that requires proximal arrangement with neighboring molecules. A molecule is a gas generally when it is very isolated from neighbors so that it can flit about traversing a great number of molecular radii without hitting anything else. There’s probably a better definition but in general a gas is a gas because the individual molecules are a great distance from neighboring molecules compared to liquids and solids.
thepompousgit says:
January 25, 2012 at 10:55 am
Myrrh said @ur momisugly January 25, 2012 at 4:19 am
Brownian motion has nothing whatsoever to do with electrons; it’s the motion of small particles of matter (originally pollen grains) being jostled by the random movement of molecules in a fluid.
That was my point. Not only does it not have anything to do with electrons, it has eff all do mixing carbon dioxide thoroughly in the atmosphere…
…though idiotically given as proof is a typical non-experiment from the AGWSF department, by opening a bottle of scent in a classroom saying it proves the scent is spread by Brownian motion, that’s when it’s not being not being idiotically explained by using ideal gas properties of elastic collisions in empty space as if ideal gas, but more often than not, claiming both these processes happening at the same time – seemingly as unconcerned as Willis about context.
Photons are packets of energy being exchanged between atoms. They are not heat so they don’t care about the temperature of those atoms. Heat flows from hot to cold only. Physics is divided into Classical and Quantum descriptions of the world. They describe the same world in different ways. From your questions, and there’s nothing wrong with your questions, it’s clear that these two views are conflated in your mind. You definitely need to learn some basic physics. Mine came from Resnick, Halliday & Walker in 1969. While the papers being discussed are recent, the physics isn’t.
Not conflated in my mind, conflated in minds that can’t separate contexts one from another.
As I showed in the link to an exchange I had about ‘net heat flow including heating flowing from cold to hot’
Visible light is not a radio wave, for example, it has distinct properties in its own right and these properties act in distinct ways on meeting matter, to reduce this to some as yet unproven idea of photons and claim that all photons in transferring energy heat matter oblivious to other uses of energy while claiming to be discussing science of the physical world around us we can see and taste and hear and which we do understand empirically well how it impinges on us and we on it, is frankly pathetic coming from those claiming themselves educated in this. It’s not I who needs to learn some basic physics..
An ad hominem BTW is when you claim that what someone claims is false by virtue of who they are. Insulting someone is not an ad hominem. The first is a logical fallacy, the second is being rude and not a logical fallacy.
I know the difference. I meant the ad hom the like which you’ve repeated, “You definitely need to learn some basic physics” and “Bottom line is: get that basic physics under your belt” …
I suggest some return to the real physical world around us and stop blinding themselves with their own imagined quantum brilliance, which may or may not yet have arrived from the future carried by a non-existant photon..
A good insulator is one that conducts heat slowly. A bad insulator is one that conducts heat quickly. All matter conducts heat, but the rate at which it conducts depends on the state (solid/liquid/gas) and chemical composition. Air conducts heat very slowly, but perforce will reach an equilibrium temperature very slowly in the absence of convection. In any real atmosphere, convection will predominate over conduction.
Thank you, I hope that helps Professor Brown to better articulate what he meant.
Unfortunately Myrrh you have chosen the wrong classroom in which to learn the basics. There’s not just basic thermodynamics and quantum physics, there’s basic boundary layer climatology being discussed here. A lot of the discussion is frankly a display of ignorance. There’s nothing at all wrong about ignorance per se, but wilful ignorance is a different matter. This muddies the water and makes learning extraordinarily difficult for those who have yet to grasp the fundamentals.
You err, I have grasped the fundamentals well enough to see that many hide the fact they don’t know the basics by playing the ‘superior because so very well educated in science card’ and worse, refusing to engage and distracting from this by the use of the ad hom technique that those questioning them have nothing worth listening to because they don’t have science phd’s coming out of their arses.
I call bullshit on the lot of it.
To teach is to learn twice.
And what are you learning twice if teaching nonsense imaginary physics about a fictional world?
When some confuse photons with discrete packets of visible light and conclude that visible light must therefore be heating water when real physics says water is transparent to it? And when having given the properties of one thing to another by claiming visible and short wave heat land and oceans they junk the real great thermal energy coming to us direct from the Sun, because they can’t find a role for it in their so called ‘energy budget’..? From which they have also expunged the great cooling role of the greenhouse gas water vapour, simply by ignoring it and sticking their fingers in their ears whenever it’s mentioned?
I call bullshit on the lot of it.
Live long and prosper Myrrh. Your questions are not stupid. The answers you have received are not stupid, either. Confusing you, yes, but not stupid.
Thank you, I add be happy to you. The answers I have received as explanations for ‘greenhouse gases warm the Earth’ are most definitely stupid, I’m not the one confused here, as I’m still trying to point out..
Perhaps this will settle things, for anyone still listening that thinks that Jelbring’s paper is correct:
http://www.physics.sc.edu/~yar/phys706_2011/…/final_solutions.pdf
I have waiting in the wings a user-contributed textbook demonstration that isothermal is maximum entropy with gravity, as well. Then there is Caballero’s detailed balance assignment (2.17). There is Paul Birch’s analysis of Velasco, which seems to be a full stat mech computation that arrives at the same conclusion. And finally, there is the clear violation of the second law the article above makes very clear.
Are we done yet?
Tallbloke, from much earlier: Since Robert Brown is setting a refutation of Hans Jelbring’s 2003 paper, it would be a common courtesy to provide a link to that paper in the headline post.
This is my bad, Tallbloke. My only excuse is that I wrote the paper in latex and submitted it to Anthony to format the equations, and while I did put the reference in I forgot the link (and had no way to add it afterwards). I appreciate your doing so.
For everybody: We’re quickly getting to where there are three or four complete algebraic demonstrations that the equilibrium state of a thermally isolated ideal gas in a gravitational field is isothermal that have been linked, directly indicated in replies, or posted and discussed. Stat-Mech and Thermo textbooks are more or less unanimous about isothermal equilibrium, of course — it is derived in stat mech very early in the process, in a way that is more or less independent of the details of the system in question. In Thermo, it is the zeroth law and the second law — any time one has a proposed system with a non-isothermal equilibrium one can trivially violate Kelvin-Planck and Clausius with it, unless it is a very odd system indeed. A column of ideal gas isn’t odd, it is textbook.
It would be nice to lay this to rest soon. Sure, anyone who wishes to can claim to be smarter than all of the authors of all of the standard physics texts on thermo, but it looks pretty unanimous out there. There are more interesting and useful things to discuss than a paper that is really pretty obviously wrong, however well intentioned.
If anybody has a serious argument — one that can stand up to e.g. the exam question and solution above, for example — I’d still be happy to address it as soon as I have time again, but it might be a day or so.
rgb
Joe Born says:
January 25, 2012 at 10:24 am
“First to your contention that we are not talking about the microcanonical ensemble. Here a little context is in order. Recall that both Robert Brown in this thread and Willis Eschenbach in the previous one were addressing themselves to refuting the Jelbring paper. That paper began with a hypothetical ideal gas G disposed between concentric spherical surfaces A and S. Jelbring said that “A and S are thermally insulated preventing heat from entering into G and infrared radiation to reach space.” It is no doubt to parallel this condition that Robert Brown says of his thought experiment that it involves an “adiabatically isolated column of an ideal gas.” In short, heat can flow neither into nor out of the gas, so by definition it is indeed a microcanonical ensemble.”
They are isolated with respect to heat flow to and from the outside universe. Not from the walls of the container, or at any rate, the floor – the planetary surface – with which they are in thermal equilibrium. So, except in the ludicrously pedantic sense in which no ensemble is truly canonical unless it includes every single particle in the entire universe, the canonic limit applies. None of which actually matters, so long as you don’t insist on filling the volume with a hard vacuum orders of magnitude more rarefied than even intergalactic space! Jelbring explicitly states that his shell is no higher than the 100mbar level. Robert Brown has also indicated somewhere (I think in one of the previous threads) that he is not considering cases in which the gas is so thin that the concept of temperature goes pear-shaped.
Joe: “Next I consider your statement that “for any reasonable number of particles, it makes no difference” whether we’re dealing with the microcanonical ensemble or not. I agree with you that as a practical matter any lapse rate as small as I’m saying Equation 8 implies would be too small to measure.”
No, that’s not what I’m saying. It is a general principle of statistical mechanics that the microcanonical converges to the canonical, as Velasco et al themselves point out. Equation 8 does not imply any lapse rate at all. All it implies is that the details of statistical mechanics calculations are messy for small numbers.
Joe: “You make two further points that at base are not technical arguments so much as statements of your point of view. You say, “However, even in this extreme case, the temperature at equilibrium will still be the same throughout the entire height, in the crucial sense that no net work could be extracted from the gas by connecting different levels, by any means whatsoever.”
On the contrary, this is an utterly fundamental technical argument. The very definition of “same temperature” for connected regions is “no net flow of thermal energy”. Hence “no net work” can be done by the gas.
Joe: “You additionally observe that I am “taking the extreme and irrelevant sub-thermodynamic case of a minuscule total number of isolated particles – in which regime the macroscopic temperature is increasingly ill-defined and no longer simply proportional to the kinetic energy per particle.” Let’s set aside what exactly you may mean by imprecise terms like “sub-thermodynamic,” “macroscopic,” and “increasingly ill-defined.””
Let’s not. Let’s face up to it instead. If we have a large number N of monatomic particles we say that the thermodynamic temperature is T, where the total kinetic energy is 3/2 NkT. If there is only one isolated particle it doesn’t have a temperature at all! In a thermal system the motion of the particles is isotropic; there is no directional bias; and this is not possible for a single isolated particle. It’s kinetic energy, not thermal energy; we cannot put T=(mv**2)/3k. If anything, T=0. What about N=2 or 3 then? Somehow we have to cobble a fit across the middle between these two contradictory extremes. In statistical mechanics, this is possible, but messy, if you are very, very careful about your definitions, your scenario, your boundary conditions, and the limits on your sums and integrals. Even then you are quite likely to get it wrong. Whether Velasco et al have got it right I don’t know. What I do know is that if your results purport to show any non-zero temperature gradient in a state of thermal equilibrium, then somewhere or other you’ve made a boo-boo. That result cannot be correct. I am enormously more certain of that than I ever could be of any piece of statistical mechanics, and so would just about every other physicist I know.
Joe: “And let’s concede that the case I discussed was an extreme case. That case nonetheless remains relevant, because it illustrates by exaggeration something that remains true …”
Fine, but you’ve chosen the wrong extreme to illustrate this problem. It doesn’t illuminate the solution, it obscures it. It’s as if you were discussing the tendency of buses to come in threes, and spent all your time trying to analyse the behaviour of two buses setting off from the terminus 100Gyr apart.
separation science, especially that surounding the separation of uranium gases in centrifuges, give you all the evidence you need to that this is a load of crap.
blog comments dont overturn working engineering.
Steven Mosher,
“separation science, especially that surounding the separation of uranium gases in centrifuges, give you all the evidence you need to that this is a load of crap. ”
Yup separation science. that means centripetal force, separates things of different density I believe. Lemme see, what in our atmosphere is different temperature and where do they end up?? Oh yeah, gas particles are different temps and are generally separated, or stratified, by their density!!! What does it? Well, my barely HS education tells me GRAVITY!!!!
What were you saying again??
DeWitt Payne said @ur momisugly January 25, 2012 at 12:31 pm
Thank you your majesty 🙂
Robert Brown:
You invited “a serious argument” so I put one:
Empirical evidence using spectroscopy proves that a gas does not absorb spontaneous emission from a body which is significantly cooler than it, but it does absorb (and spectral lines thus appear) when the same body is made warmer than the gas.
Q.1: Why?
Q.2: Does this extend to the oceans and/or land surfaces and thus imply that a warmer solid or liquid surface does not absorb radiation from a cooler atmosphere?
Q.3: Can you refer me to any experiment proving empirically that backradiation from a cooler atmosphere can warm the surface and/or slow its rate of cooling, as claimed by the IPCC?
DeWitt Payne says: January 25, 2012 at 1:15 pm
This piece of misinformation seems to be even harder to kill than a gravity maintained adiabatic lapse rate. Temperature, as long as the pressure is high enough it can be defined, is a function of the average kinetic energy of the gas only. What happens at low pressure is that it takes longer for a thermometer to equilibrate with the gas. If the thermometer has a large thermal mass, on the order of the heat content of the volume of interest, it will perturb the actual temperature.
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But that is the whole point, isn’t it. just what is Dr Brown supposed to be measuring? He says ‘temperature’, which is a wooly term unworthy of a scientist.
Does he mean the individual the temperature of individual molecules (molecular kinetic energy), or does he mean the temperature of the total airmass? (See my earlier post, a few scrolls up.)
If he means the temperature of individual molecules, then thermal equilibrium in the ‘silver wire experiment’ has already been achieved (between the top and bottom of the air column). The individual molecules at altitude have about the same temperature (kinetic energy) as those at sea level. Thus the warm wire at the top of the collumn (at say +30 oc ) cannot ‘heat the air’ at high altitude, because the individual molecules at altitude are already at something like +30oc (in terms of their individual kinetic energy)..
If, on the other hand, he means the temperature of the total airmass, as measured by a standard thermometer, then he should know that you could never get the air at 50,000 ft to be +30oc. Just not possible. Never going to happen. Crazy suggestion. So Dr Brown cannot mean this.
So when DR Brown says:
“”””Heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.””””
He cannot be referring to the total airmass temperature, as measured by a thermometer, as you could never get the air at 50,000 ft to be +30oc. So he must be referring to individual molecular temperatures (molecular kinetic energy). But these molecules at altitude are already at something like +35oc (individually) already. So there will be no heat flow in the wire, because the molecules at altitude are already at the same temperature as those a sea level and the same temperature as the top of the wire.
In short, I fail to see what this thought experiment is supposed to prove !!
There could never be any flow of temperature in the silver wire. If there were, then there would already be a massive flow of energy through the atmosphere itself, through conduction and convection. And while there IS a certain amount of heat transfer through the atmosphere from surface to tropopause (as we know), this heat transfer has never made the upper atmosphere +35 oc — not even in several hundred million years of heat transfer between surface and upper atmosphere (with or without a silver wire).
Again, I must ask, what is this thought experiment supposed to prove?
Please see my previous posts at:
Ralph says: January 25, 2012 at 12:27 pm
and
Ralph says: January 25, 2012 at 12:54 pm
.
Nullius in Verba says:
January 25, 2012 at 12:05 pm
“… convection on a world with no GHGs …
I think it would work a bit like the convection pattern of the thermohaline circulation, inverted. … With GHG-free atmosphere, you have to swap warm for cold and up for down, and equator for pole. Thus, air is warmed at the equator, rises, and spreads polewards. Its potential temperature (adjusted for lapse rate) remains constant then, and most of the atmosphere is equatorially hot, even at the poles. The thin layer in contact with the ground there cools, and flows back to the equator over the surface. … ”
I agree with the overall scenario (and was actually thinking of posting something similar), except that I suspect that instabilities and the Earth’s rotation would cause the pattern to break up into relatively narrow latitudinal bands perhaps as little as a few tens of kilometres wide, with a sort of helical circulation in each (like a rope). The bands would probably meander a bit, depending on the local topography, and also wander up and down in latitude with the weather. The bands would sometimes break up further into strings of cyclones and anticyclones. We’d still have a troposphere and tropopause, and, as now, the temperature in the convective cells and at the top of the troposphere would fall fairly gradually from equator to pole. The lapse rate in the active parts of the cells would be more-or-less adiabatic; in between it would be smaller (with temperature inversions common at night).
Hey Myrrh you are not stupid. Since last year I’ve been chasing up your statement that visible light does not create heat and believe me the info has not been easy to find and even when I find some it always conflicts with the one I found before.
Anyway I’ve come to the conclusion that you are correct. IR radiation is what warns the atmosphere and not visible radiation. Thank you.