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.

Mr. Brown:
I commend you on your continued posts here.
Thank you.
If sheer volume of words carried scientific weight, Robert and Willis would be winners by a mile.
However, Robert still hasn’t addressed my short proof that no heat has to flow in a gravity induced thermal gradient at energetic equilibrium.
Here it is again if he feels like getting round to it.
Robert says:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
“… all considerations about the mechanism go nowhere. Consider the outcome. If it’s true that gravity can separate molecules by temperature, then we can pull energy out of tall insulated cylinders of air …”
But whether the lapse is enabled by gravity or just thermal gradients, we do already have a lapse rate in our current atmosphere, so of these magical vertical lapse rate generators that you and Dr. Brown keep speaking of, why have these not already been built?? Sound so simple. I’ll take some of that forever free energy Willis no matter what forms the lapse, just show us the schematics…. PLEASE? ☺
Willis, I am pointing out why the isothermal state is not the maximum entropy state, which you have not disputed yet. An isothermal or any stable stratification with a linear profile is stable against diffusion. There are multiple stable states possible when only local diffusion is permitted, but the only one stable to eddy mixing is the isentropic one. If you start isothermal, it will stay isothermal. If you start isentropic it will stay isentropic, and if you start half way between these, it will stay like that too.
Robert Brown says at 1/21 10:01am
“..the molecules can share their kinetic energy around and come to thermal equilibrium without moving up or down and involving gravity.”
Certainly the delays in many conversations are irritating especially stopping in and out like me. But it is interesting and fun pastime to engage on this particular thermo topic.
To regroup: control volume is an adiabatic GHG-free tall air column in the presence of inexplicable gravity. I continue to be on the side of those thinking this column must stratify while at the same time it cannot generate energy forever. No energy must be harmed in this thought experiment. Or created.
Have I detected at least a little backing off by Robert Brown no-stratify position by above seeking to take out the n*g*h PE term by molecules exchanging KE limited to lateral movements only? Or is this somehow out of context?
Remember in the top post Willis’ quotes Robert Brown writing: “On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down…”
This last statement seems to be a violation of conservation of energy in a gravity field where each molecule PE + KE = constant since in my view: each molecule moving up in the presence of an inexplicable gravity field cannot have exactly the same velocity thus kinetic energy as those moving down in presence of gravity field PE (= n*g*h) since KE changes with h for total energy to be constant.
That stratifying seems to square with 0th,1st, 2nd Law and ideal gas law. I don’t think Maxwell-Boltzmann distribution can be invoked in a gravity field since it deals with statistics of molecular movements w/o external forces. Here gravity is an external force. M-B can only be invoked if gravity is turned off.
In this 1 system cv with gravity no heat can flow (there is no system 2), cannot reduce the KE + PE of any h > 0 upper molecule, total is always constant. Cannot increase the KE +PE of any lower h~0 molecule, total is always constant. Thus temperature has got to vary because KE varies as P*V varies with h (via PE = ngh).
My view this cv has to stratify by temp. consistent with all the laws and therefore by density thru PV=nRT. If it doesn’t, total energy of each molecule is not conserved. Same reason it can’t generate energy forever. There is no hot & cold source within the CV maintained forever if molecules always move with same constant energy.
Like Tallbloke said, build a Perpetuum Mobile if you can – let us know your design but energy conservation will always and everywhere stop it.
Oh wait, I see now Willis says at 1/21 12:34pm:
“I have specified the machine, a tall insulated cylinder of air. Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.
As to your claim that I have failed to demonstrate that work can be produced out of a temperature differential … surely you are kidding. That’s how heat engines work. Are you claiming that a temperature differential cannot be used to produce work?”
Willis’ should read the 2nd Law v-e-r-r-y slowly, there has to be two systems – one hot and one cold for heat to flow when they touch. In this cv there is only 1 system, heat cannot flow. There IS no temperature differential between two systems here, only one system. Or show me the second system. Otherwise, Willis’ specified Perpetuum Mobile machine is not demonstrated. One system cv: no heat flows or work produced. Can’t even be said the insulated cylinder of air system contains heat (does contain energy) if I read my Maxwell definition right (heat only flows).
Oh Willis, one more thing, if this is another 5km high wire gizmo, thermocouple or not, I’ve already tried that… no thanks… burned down the shed it was anchored to and I’m sure it was not from the lapse rate but more from the electro-static differential found in our atmoshere at times! ☺
The perpetuum mobile won’t happen because we talk about an hypothetical planet with no radiative transfer. Try to build a thermocouple with no emissivity and then we’ll talk. If your thermocouple emits heat, eventually the planet will cool.
The zero adiabatic lapse rate cannot happen without a Maxwell demon. Remember, the number of molecules traveling upward is equal to the number traveling downward if the atmosphere is in balance. Now, if you have the same temperature everywhere, it means that molecules at higher altitudes are traveling faster. So, at a certain height, the molecules traveling downward should be warmer than the molecules traveling upward. Except if you have a Maxwell demon blocking the faster molecules traveling downward.
The only way PQ(1), RS(1), can stay in balance is if they have the same lapse rate. Likewise for P2(2), RS(2). And PQ(1) must have the same real lapse rate as PQ(2) since they are thermally conducted. The only solution is for the real lapse to be zero — Isothermal.
Well done, but way too much work. The easiest PPM2 one can build is a simple thermocouple that lights a light bulb or turns an electrical motor. Place the upper contact in the supposedly cold air at the top of the static column. Place the lower contact in the warm air at the bottom. Insulate the heat transfer medium (e.g. silver) all the way down to the actual junction, so that heat is conducted from the hot bottom to the cold top, through the thermoelectric junction.
The motor turns, doing work as heat is conducted from the warm bottom to the cold top. But the top never gets any warmer, as the DALR is supposedly the stable equilibrium state of the gas. No matter how much heat we remove from the bottom and deliver to the top or middle, it somehow “falls” back to the bottom to re-establish the DALR.
Even simpler, don’t worry about the actual motor — simply connect the top and bottom with a silver rod insulated on the sides. Bottom is hot, top is cold, so heat flows along the rod — forever. As fast as you deliver heat to the top, it is sorted out by gravity and returned to the bottom to keep it from actually cooling. The system never reaches equilibrium! Which violates so much simple common sense that maybe — although I doubt it — even Jelbring and Tallbloke with concede that this cannot happen, and that the assertion that any spontaneous stable process can create a permanent thermal gradient in an adiabatic system is false (since this argument catches them all, which is why it is used to show the equivalence of the various forms of the second law in intro thermodynamics textbooks)..
rgb
Willis Eschenbach says:
January 21, 2012 at 12:34 pm
PS—Please note that Robert Brown may be in “Group W” as well regarding your claim, viz:
This goes back to the proof I [tallbloke] offered Duke.edu physicist Robert Brown up near the top of the thread. he said:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.”
And I [again tallbloke, not me —w.]] pointed out:
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B. QED.
He may not be all that impressed by some random guy on the internet claiming to find a violation of the zeroth law …
To me, the problem with your claim is the idea that there is a temperature gradient across A, so that each microscopically thin slice of A is a bit cooler than the previous slice … but there is no difference between the last slice in A and the first slice in B, you claim that they are the same temperature. That’s not a temperature gradient.
So you do not have a “thermal gradient” across them as you claim. You have a gradient everywhere but at the interface between A and B.
If that is the case and the last slice of A is at the same temperature as the first slice of B, then momentarily no heat will flow from A to B. As a result, heat will run down the thermal gradient from the next-to-last slice in A to the last slice in A, warming it slightly. In the same way, heat will flow down the thermal gradient from the first slice in B to the second slice, cooling the first slice slightly.
At that point you will have a true thermal gradient across the stack, where every slice is warmer than the previous one, including from the last slice in A to the first slice in B. And of course, as the zeroth law states, then heat will flow from A to B and on down the line.
I’ll be interested to see if Robert endorses this laughable response to my proof. Clue, no heat is going to run down the thermal gradient, because KE+PE at any altitude h is in equilibrium over the height of the column.
The rest of your longer than elevator length reply is more ad hominem attack. You seem to be all out of science, so I’ll leave you to it and go do my write up for the blog.
Cheers
TB.
Smokey said @ur momisugly January 20, 2012 at 6:32 pm
Actually, no. Magic means something is caused supernaturally. Quoting Feynman; he asked rhetorically “What is behind the Law?” and answered, “There is nothing behind the Law”. Many (most?) of us accept that. Some believe that God is behind the Law. I have no quibble with that belief (I just don’t share it), but that belief is magical (supernatural).
DeWitt Payne says:
January 21, 2012 at 8:45 am
“You can hand wave all you want about non-equilibrium thermodynamics…”
I am no longer handwaving. The question is moot. All the other side discussions are moot. All the rest of this blog thread is moot. I am now proclaiming Bart’s Law as settled:
Bart’s Law:
It follows that surface temperature sensitivity to the addition of radiative gases is negative: the addition of more radiative gases will tend to lower the surface temperature. This is a sensitivity – it does not exclude the possibility of feedback from other processes tending to resist the change.
Bart’s Law is justified by the following line of reasoning:
Convergence is initially superlinear, because the more you bite into the areas of the tails, the faster you increase the backradiation and hence surface temperature. Eventually, you reach a point where successive bites approach zero, and there you stabilize.
Willis Eschenbach:
Please take note of the above. Here is a physically viable alternative to the standard greenhouse theory.
Trick says:
January 21, 2012 at 1:42 pm
Excellent comment Trick.
See my attempted disproof of DeWitt Payne’s twin cyclinder version of Willis’ one lunger above at
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871872
Thoughts?
Willis wrote;
(In response to my expressed umbrage at allusions to ignorance amongst those of use that do not believe in the GHE)
“Dang, bro’, with all due respect, I didn’t say you were.”
Willis, my sincere apology, I believe I may have read another comment about those of us that do not believe in the GHE as being too ignorant to understand it. I then read your post about being deluged with ignorant people and mingled the two posts together in error.
But those that think that all people who do not accept the GHE are ignorant may learn otherwise in the next few years when we all finally figure this thing out.
Please keep trying to explain things, and eventually we all might reach a better understanding about what is going on.
Cheers, Kevin
Tallbloke, may I suggest that you (and other skeptics) simulate that precise experiment, using the methods and software of Understanding the temperature and the chemical potential using computer simulations (2004).
To handle the reservoir coupling, have the simulation’s “demon” exchange energy and atoms with *both* reservoirs A and B. Furthermore, have that demon exchange particles and energy only with the top 1/10 of reservoir A (the lower reservoir) and the lower 1/10 of reservoir B (the upper reservoir). And finally, don’t forget to assign the particles in reservoir B a higher gravitational potential energy than those in reservoir A.
Start the simulation in some arbitrary atomic configuration, and let it run for awhile. You will find that it approaches an equilibrium in which:
(1) Reservoirs A and B are at the same temperature
(as mediated by the exchange of energy)
(2) Reservoirs A and B are at the same chemical potential
(as mediated by the exchange of particles)
(3) Reservoir A (the lower reservoir) has a higher density and pressure than B
(in consequence of B’s higher gravitational potential)
Then after thinking some more, you will appreciate that microscopic atomic simulations will always respect the laws of thermodynamics, in the sense that their “demons” will always act so to equalize temperature and chemical potential, in accord with orthodox thermodynamical theory (naturally), and entirely at variance with “gravito-thermal theory.”
Speaking personally, I have mixed feelings about claims that the GHE is “established science.”
Skepticism about established science is good, to the extent that this skepticism motivates people to study these questions seriously and in-depth. And from the point of view that skepticism can be an excellent motivation for learning, this entire WUWT discussion has been terrific!
And yet what Robert Brown says is 100% correct too: “The greenhouse effect is established science. Get used to it. Get over it.” In particular, the “gravito-thermal” theorists have been making extraordinary claims, but they have not brought forth any of the extraordinary evidence — mathematical, theoretical, computational, experimental, or observational — that is needed to support their claims.
So perhaps the best balance for rational skepticism is this: “The greenhouse effect is established science. Get used to it. Get over it — after you have checked its key ideas for yourself.”
tallbloke said @ur momisugly January 21, 2012 at 9:31 am
I am truly gobsmacked! A definition is a tautology. A tautology is a proposition that is unconditionally true. But not unconditionally true according to TB. Quoting Aristotle (translated):
“one cannot say of something that it is and that it is not in the same respect and at the same time”.
Post-modernists do this all the time of course, thus excluding themselves from rational discourse.
Robert Brown says at 1/21 1:52pm
“The easiest PPM2 one can build is a simple thermocouple that lights a light bulb or turns an electrical motor. Place the upper contact in the supposedly cold air at the top of the static column.”
You are mixing metaphors. The cv of interest allows only one air system. This PPM2 is more than one system. And things are going across the cv. Try again.
However, Robert still hasn’t addressed my short proof that no heat has to flow in a gravity induced thermal gradient at energetic equilibrium.
Here it is again if he feels like getting round to it.
Robert says:
“if A and B are placed in thermal contact, they will be in mutual thermal equilibrium, specifically no net heat will flow from A to B or B to A.” That’s the zeroth law.
Assuming your A and B have at least some dimension, then a thermal gradient across them would mean that the top surface of A will be at the same temperature as the bottom surface of B where they contact. Therefore no heat will flow. Even so, the average temperature of the whole of body A will be higher than that of B.
Sure, and in return you can address the actual topic of the thread, which is how any system that spontaneously generates a thermal gradient violates the second law of thermodynamics and thereby enables the construction of PPM2K’s, including the trivial thermocouple example in the previous post.
First of all, what you offered is not a proof. You simple restate your conclusion. You postulate two systems in thermal contact, one at a higher average temperature than the other, and then say “no heat will flow” as long as the boundary in between them is at some (presumably intermediate) temperature.
The problem is that temperature is a meaningless concept for a two dimensional surface. What you’re saying, in a nutshell, is that you have two volumes neither of which is in thermal equilibrium (if they were, they’d have the same temperature throughout each volume.
But let’s see if I can come up with a simple, simple argument that will convince you.
Go into your column of air with its imagined lapse rate. Get perfectly insulating jar and fill it with air from the top. Get a second jar and fill it with air from the bottom. By hypothesis, these jars are in thermal equilibrium, but have different temperatures. Note that once you fill the jars, you can move them up or down all you like — the gravitational field doesn’t vary (by enough to matter). You’ve simply replaced the fluid that was supporting the air with the jar walls — nothing else changes. Which, incidentally, is why your entire argument is wrong — no gravitational potential energy is involved in heat exchange or energy balance. You move the jar slowly up, move it slowly down, nothing in the jar knows that it has been moved at all!
Now put the two jars in thermal contact. I don’t care how — move them together, use an insulated silver conductor of heat to couple their otherwise adiabatic walls without moving them up or down — it doesn’t matter, does it. What happens? Well, they are at different temperatures, so heat flows. If you change the temperature across any jar heat flows. There is nothing special about the heat flow. It will flow up or down or side to side to neutralize a thermal gradient no matter what you do to the otherwise unaltered jars of air.
The existence of the jar itself doesn’t matter. Put a side-insulated silver rod into the air column and heat will flow in it forever. Put a heat engine into the air column and you can run it off of that temperature difference forever.
The point is that your air column isn’t in equilibrium. Heat can easily be conducted through the air — the equilibrium distribution of the air (and its temperature) isn’t just determined by convection. Once it settles down into a density profile, heat will flow throughout that profile until the temperature is the same because the air at the top cannot tell any difference between heat that arrives through the intermediary air and heat that arrives through a silver rod. The point is that heat spontaneously flows from hot to cold, never the other way around, in the absence of something that does continuous work to push it the other way.
Nothing about the density of air in a jar tells you its temperature. Heat is perfectly happy being conducted uphill, downhill, or sideways. Warmth has no mass, and objects placed in gravitational fields do not spontaneously separate into warm side down. Indeed, in general it is rather the other way, at least at first! But equilibrium is always isothermal in a case like this with no meaningful constraints.
rgb
I’m getting frustrated by the doziness of some of the contributors here, including some Phds.
The column only becomes graded by temperature when energy is added from an external source.
Until there is an external source of energy the column gets graded by MASS not temperature.
Gravity works only on mass. It ensures that the greatest density is at the bottom of the column.
Apart from that gravity does no work and supplies no energy.
In the absence of an external energy source the column will become isothermal i.e. the same temperature from top to bottom.
But
there will be a higher energy content at the bottom because there will be more mass per unit volume at the bottom due to the effect of gravity.
Then when the whole column is irradiated by an external source of energysuch as the sun the temperature will rise most at the bottom because the incoming radiation has more molecules to react with where the density is greater.
This is well established physics of more that 150 years provenance as here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas”
and
“temperature decreases with altitude at a fairly uniform rate. Because the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.”
This is all well proven basic physics.
What is going on here and in other threads that seem to deny basic well demonstrated physical principles ?
DeWitt Payne says:
January 20, 2012 at 6:51 pm
If as you say the TOA temp is the same as at the surface, then there would be an abrupt stepdown to the near-zero K of space. That’s not a satisfactory explanation. I fear that something essential is being missed by positing thermal equilibrium, which preordains the isothermal conclusion. In other words, it becomes a tautology. Space cannot store energy, only transmit it radiatively. Molecules, on the other hand, can. After all, temperature is the manifestation of the kinetic energy of molecules,and depends upon their DENSITY in any given volume. Don’t have time to say anything more.
Robert Brown says:
January 21, 2012 at 1:52 pm
Even simpler, don’t worry about the actual motor — simply connect the top and bottom with a silver rod insulated on the sides. Bottom is hot, top is cold, so heat flows along the rod — forever. — even Jelbring and Tallbloke with concede that this cannot happen.
Excellent. another suggested experiment. My old dad is a fine qualified civil engineer with a wealth of fluid dynamics experience. He told me earlier today that he once did a set of calcs which showed that a water pipe of infinite length with infinite pressure at one end would have no flow at the other due to frictional loss. I suspect the heat flow in your 10km silver wire might have similar conduction issues due to atomic interstice losses. By all means give it a try though.
Robert Brown says:
January 21, 2012 at 2:22 pm
Robert,
Thank you. I will give your reply careful consideration before responding. I hope we can sort this out together.
Cheers
TB.
RE: Robert Brown says: 1:52 pm about Rasey’s 1:08 Helium-Argon Loops
Well done, but way too much work.
Thank you. You and Willis and others had already done the thermocouple PPM2 refutation. I decided to play in Jelbring’s ballpark and stay strictly in lapse rates and pneumatics. Let’s let him find non-zero real lapse rate in the argon and helium loops that work at equilibrium. I’m not even trying to extract work out of the system. What’s is the system’s equilibrium state in a gravitational field if not isothermal?
On the planet Stygian, you have one house on the (hot) beach, and your sister has another house on the (cold) mountaintop.
Fortunately, your family possesses two one-meter^3 blocks of copper. You keep one (enormously heavy!) copper block at the top of the mountain, the other at the bottom.
Every morning you and your sister exchange these two copper blocks (this requires zero work, as the two blocks weight the same—a simple pulley raises the beach-house block at the same rate that the mountain-house block descends).
You use the mountain-cooled copper block to air-condition your beach house all day and even (via a thermoelectric generator) to power your beach-house computer and beer cooler all night.
Similarly, your sister uses the beach-warmed copper block to heat her mountain house, and to power her computer and hot-chocolate-maker.
Next morning, exchange the now-warmed beach block for the now-cooled mountain block, and do it again. Which amounts to this: free heating, air-conditioning, and electrical power for you and your sister, forever, with zero work input! Thank you, thermo-gravitic theory!
Willis said
“I realized after posting this that some people might say “well, maybe the isothermal case is the high energy state”. There’s an easy way to settle that.
If an energetically isolated system is in its lowest energy state, it cannot perform work.
If the isolated atmosphere in Jelbring’s thought experiment is warm at the bottom and cold at the top, I can stick a thermocouple into it and use the temperature differential to generate electricity to perform work.
Therefore, the isothermal state (same temperature everywhere) is the lower of the two energy states, since I cannot use it to do work. ”
Actually this might not be a proof.
One at least has to reflect on the [miniscule] gravitational effect on the squillions of electrons in the thermocouple wires .Or to make it simpler consider the electrons in one wire subject to the Seebeck effect. Maybe, just maybe, the gravitational effect on the electrons in the wire exactly balances the gravitational effect on the air columns temperature gradient.
-In a thermocouple the gravitational effect in each wires material might be proportional to its Seebeck coefficent thus no work could be done 😉