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.
John Marshall says:
January 21, 2012 at 2:34 am
John, I don’t “dismiss them” I do note that:
1. We understand the mysteries of our own climate only poorly, so our knowledge of the inner workings of e.g. Neptune’s climate must perforce be orders of magnitude worse. This makes it hard to draw any solid conclusions.
2. The atmospheres of the other planets are very, very, very different from the Earth. For example Venus, which is in some ways is the most similar, has temperatures and pressures at the surface that make the CO2 into a supercritical fluid:
Venusian atmosphere and spectroscopic properties of CO2
As can be seen from above link, partial pressure of CO2 in the Venusian ground atmosphere amounts to ~89 bar and ground temperature to 737 K on average. The “critical point” data of CO2: p(crit.) = 72.9 bar; T(crit.) = 304 K (handbook of physics). A substance beyond the critical point is neither gas nor fluid, though in general referred to as supercritical fluid. It surely is not a gas and very surely not a trace gas. SOURCE
John, perhaps you could explain to us all what an atmosphere where CO2 is not a trace gas, not a gas at all, not a liquid, but a supercritcal fluid can teach us about the Earth’s climate?
Yes, as you point out, planetary atmospheres all obey the same laws of physics as they apply to gases, but so does my grandma’s dying breath … so what?
w.
Never mind all that detail.
“The atmosphere ALWAYS reconfigures its energy distribution so that the Atmospheric Thermal Effect/Adiabatic Lapse Rate is maintained.”
Wilde’s Law.
On the WUWT threads many are doing all they can to ignore the implications of the Ideal Gas Law and N & Z’s ATE ( formerly widely accepted as the Adiabatic Lapse Rate or ALR).
Just look at Wikipedia here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas.”
and:
“the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature normal with increasing distance from the conductive source.”
which together show that we are dealing here with long established science although recent shenanigans have made it necessary for N & Z to try to redefine the adiabatic lapse rate (ALR) as ATE for the current uneducated generation.
So the issue isn’t whether the phenomenon is the ATE (Atmospheric Thermal Effect) or the ALR (Adiabatic Lapse Rate)
We can see that they are one and the same.
The issue is whether increasing GHGs can make a difference to the ATE/ALR by virtue of their radiative characteristics.
Well what Ned is doing is simply refining the data and the S-B calculations to show that the effect of more GHGs is zero as against the AGW theory that more GHGs can ADD to the ATE/ALR.
Where I propose a novel idea is in showing WHY the extra GHGs have zero or near zero effect on the ATE/ALR. Ned demonstrates the fact that there is a zero effect and I say WHY there is a zero effect.
The fact is that the entire atmosphere responds to ANY forcing that attempts to disrupt ATE/ALR by altering the tropopause height and that change in height then allows the surface pressure distribution to slide poleward or equatorward beneath the tropopause to negate the thermal effect of any change in tropopause height either at equator or poles.
The same process occurs on EVERY planet that has an atmosphere which tries to disrupt ATE/ALR. The atmosphere ALWAYS reconfigures so that ATE/ALR is maintained.
There is no other possible solution. IMHO.
To my mind the jigsaw is complete.
Stephen Wilde says:
January 21, 2012 at 11:11 am
Wilde’s Law.
With a little help from his friends. 😉
It may turn out to be consistent with Willis’ favourite the Constructal Law too….
Tallbloke says:
>So for a 1m^2 column we could generate how much power?
That is immaterial. ANY violation is a violation.
>According to Graeff’s experiments, the gradient was re-established within 36 hours
You don;t descibe the experiment in detail or provide a link, but only tiny imbalances are need to create a lapse rate of 10 K/km in the atmosphere (less that 1 mW/m^2 out of the top of the column. It would take VERY careful experimentation to get close enough to “perfect insulation” to see the effects we are discussing.
>Where is the energy going to come from to warm it up again?
From the ground in this case. The bigger question is where will the energy go? We have a single “hot reservoir”. You would take energy from this hot reservoir, create a machine (a pair of tall cylinders with a thermopile at the top), and turn thermal energy directly into usable work with no energy rejected to any “cold reservoir”.
How about we start with the definition of free energy as described in Chapter 21 (Systems Subject to a Gravitational Field) of Klotz/Rosenberg’s Chemical Thermodynamics 5th Ed:
G = f(T, P, n1, n2, …, ni, x) Eqn 21-6
The total differential is therefore:
dG = (dG/dT)*dT + (dG/dP)*dP + dG/dn1*dn1 + dG/dn2*dn2 + … + dG/dni*dni + dG/dx*dx Eqn 21-7
Now let:
dG/dx = mg Eqn 21-2
dG/dT = -S Eqn 11-3
dG/dP = V Eqn 11-4
with the assumption that the composition of the atmosphere consists of indistinguishable particles such that:
dG/ni = 0
Equation 21-7 becomes:
dG = -SdT + VdP + mgdx
or in terms of molar quantities:
dG = -Sm*dT + VmdP + Mgdx
The molar volume of an ideal gas is given by:
Vm = V/n = RT/P
The molar entropy of an ideal gas with a constant specific heat can be taken as:
Sm = CvlnT + RlnVm + Sm0 Eqn 6-111
After making the appropriate substitutions, we’ll get:
dG = -(CvlnT + Rln(RT/P) + Sm0)*dT + (RT/P)*dP + Mgdx
So, all one needs to do is solve this differential equation to answer the original question. At equilibrium, dG = 0 by definition, thus:
0 = -(CvlnT + Rln(RT/P) + Sm0)*dT + (RT/P)*dP + Mgdx
The solution is trivial if dT = 0, which is consistent with one formulation of the zeroth law of thermodynamics. It is of interest that Klotz & Rosenberg specify that the column is isothermal in their example.
Michael Larkin says:
January 21, 2012 at 7:38 am
Michael, the only stupid questions are the ones you don’t ask. And in this case, you’ve asked good ones that go to the heart of the discussion.
I don’t know what mechanism Jelbring thinks will restore the initial heat differential, but he definitely thinks it will be restored.
He does not say that, and in fact vigorously denies it. It is true nonetheless.
Saying as Jelbring does that gravity will cause an on-going, restorable temperature differential in an energetically isolated column of air is the same as claiming a perpetual motion machine. Here’s why.
IF gravity could cause such a temperature differential, we could use that differential to produce work. Producing work from heat has to cool the warm end and warm the cool end of the system. This is because you have removed heat from the warm end to do the work, and rejected the waste heat to the cool end of the system.
Jelbring claims that after such extraction of energy or any other disturbance, gravity will restore the prior temperature differential. This means that we could pull energy out of it forever, which is a perpetual motion machine.
And this means, of course, that gravity cannot create a temperature differential in an energetically isolated system such as Jelbring describes.
This is the part that all of the folks who are arguing about the mechanism are missing. The problem isn’t in the mechanism. It is in the outcome, which violates conservation of energy, so considerations about possible mechanisms by which gravity could/should/must cause a temperature differential are meaningless.
w.
On reading the discussion, I feel drawn to the conclusion that DALR defines a set of physical conditions where no useful work can be extracted from a compressible fluid using a heat engine.
Temperature difference alone is not decisive.
Consider a heat engine with a high altitude heat sink and low altitude heat source. Energy transfer is by physical contact only (conduction). For now, assume molecules which can transfer a discrete amount of energy on contact with the source or sink.
Can we get useful work from temperature difference at DALR ?
Firstly, consider a single molecule which has just gained additional energy at high altitude from the engine’s heat sink (high potential energy + increased temperature (kinetic energy)). The molecule then drops to the heat source where potential energy is lower but kinetic energy is high (boosted by the energy it had gained at the sink). The engine appears to have found a way to boost its own performance by transferring heat back from its own sink to its source.
This violates thermodynamic principles and the safest conclusion is that something is wrong with this modern-day version of the well-crafted pile of wood at the top of this thread.
I believe the mistake is to assume that only one physical property (temperature) is decisive in the potential to do work from conditions at the source and sink.
Second case: a heat engine with no capacity to store energy and therefore operates due to simultaneous events at the source and sink (single molecule doesn’t do anything here, we need to consider many molecules).
If a molecule is available to transfer-in energy at the source, the engine needs a molecule at the sink to transfer energy out. More generally, if there are ‘n’ molecules at the sink, the source can only accept the energy from ‘n’ molecules despite there being many more molecules available at the source.
It is not necessary to stick to the above constraints. My contention is that a DALR heat engine doesn’t work because the conditions at the sink limits the potential to receive energy at the source.
I would respectfully suggest is a mistake to postulate machines which assume only temperature difference is relevant to operation.
If number of molecules is equated to the density of a compressible fluid and density is a function of pressure, the above arguments suggest that pressure difference cancels temperature difference at DALR.
The capacity to do work is determined by total potential at different altitudes (pressure and temperature) for compressible fluid.
And this guides me to the view that there is nothing thermodynamically wrong with a sustained (permanent) temperature gradient in a gravity field.
OK guys – shoot me down
😉
Dr Brown says:
“Without the rapid cooling of the upper atmosphere that keeps it cold relative to the ground — cooling that is strictly radiation, because sooner or later the Earth has to lose the incoming heat from the sun — the adiabatic warming profile would not exist, and in parts of the atmosphere that profile inverts even as it is (for example, over the arctic in the long arctic night) as further proof that this isn’t an atmospheric compression effect, it is plain old convection.”
I would suggest that the arctic atmosphere does not “invert” in the sense of the action verb as a result of seeking an equilibrium but instead inverts as a consequence of differential cooling between the atmosphere and the surface. To claim that as evidence against Jelbring’s world one has to acknowledge the differential cooling, subtract that out and then demonstrate that whats left is something seeking an isothermic equilibrium. Jelbring’s world was defined in such a way that one cannot use inversion layers caused by differential surface cooling as evidence.
So I am blowing the whistle on that play. . . .out of bounds!!!!
Digging down a little further, just what influence does gravity have on creating heat further down into the earth’s crust? IF it does have a participatory influence, then why not the atmosphere where closer to the earth the greater the pressure?
A Physicist & DeWitt Payne: Thank you both for your efforts.
A Physicist, I appreciate your heroic effort to reach a level of concreteness that a layman like me can comprehend. Unfortunately, even that much concreteness left your explanation at a level of abstraction I cannot reliably navigate; in the corn and soybean fields of Indiana, where I live, we think “chemical potential” has something to do with batteries.
And, in any event, you addressed Loschmidt, who apparently argued for the adiabatic lapse rate at equilibrium, whereas my question was instead about Levasco et al., who agreed that isothermality prevails, but only in the limit: they found a small but finite lapse rate for finite numbers N of molecules, and their formula yields a value for N = 1 that I would have thought hard to argue with.
DeWitt Payne, you I could understand, and your explanation sounds admirably plausible. Although you, too, addressed Loschmidt (or, equivalently, Jelbring) instead of Velasco et al., I think I can adapt your argument to Velasco et al. well enough. You would begin by recognizing that, unlike in the adiabatic-lapse-rate configuration for which I understand Loschmidt argued, the hypothetical rising packet in a configuration where the atmosphere has Velasco et al.’s (less-than-adiabatic) lapse rate would absorb heat from its surroundings. But you would point out that the dQ/T difference prevailing in that heat transfer would not be as great as in the zero-lapse-rate, i.e., isothermal, situation. Quite understandably, therefore, you would conclude that the isothermal configuration is the higher-entropy one, and you would pronounce your proof successful. Q.E.D. Game over, thanks for playing. Exeunt omnes.
But here’s the problem. People who know this stuff tell me that, from a statistical-mechanics point of view, entropy is the log of the number of microstates. And, as I understand it, when Velasco et al count those up, they find that the configuration with the most microstates actually is one that has a slight but non-zero temperature lapse rate: it’s not isothermal–although their lapse rate does approach zero asymptotically as the number of molecules approaches infinity.
Or at least that’s how this layman understands what they say. And, if what they say is correct, the arguments you all are making for why Jelbring got it wrong have to be adjusted.
So I’m left to guess at who’s right. And I have to say that the approach Velasco et al. took appears from where I sit to be the more rigorous one. Couldn’t one of you smart guys out there actually engage with the Velasco et al. and Martin et al. papers tallbloke provides links to in his Loschmidt thread–and show us where they or I went wrong?
Ok, energy from the surface that would otherwise escape to space is returned to the surface by the action of GHG, where it warms the surface temperature over what it would otherwise be. We call this back-radiation.
By the same principle, energy from the surface that would otherwise escape to space is conducted back to the surface (eg: at night and towards the poles) by N2 and O2, where it warms the surface temperature over what it would otherwise be, We call this back-conduction.
If GHG can heat the planet through back-radiation, then non GHG must heat the planet as well through back-conduction.
http://tallbloke.wordpress.com/2012/01/20/greg-elliott-use-of-flow-diagrams-in-understanding-energy-balance/
DeWitt Payne said @ur momisugly January 20, 2012 at 8:54 pm
I stand corrected; it was but a single lecture. The Goodsteins’ book has considerable supplementary material and that made it seem to me much longer. I thoroughly enjoyed working through the reasoning, though his students were less enthralled. They would have preferred the “shorthand” of calculus. Of course Newton had to present his argument in the language of the day, rather than the language of calculus that he (and Liebnitz) were developing.
Also out of bounds on the former play is the contention the earth has to eventually lose the incoming heat from the sun. Thats not necessarily the case in Jelbring’s world as there is no incoming solar energy and both that and its losses are out of bounds.
We can all understand that jelbring’s world is impossible but Jelbring is in essence asserting that none of that is relevant to the question at hand as he asserts he was fair handed in drawing the boundaries of the issue by removing what is a balanced equation.
So we either have to criticize Jelbring’s fairness in defining the imaginary world we are dealing with or keep the play inside the bounds.
My apologies for being a self-annoited moderator but I find all this extremely thought provoking.
tallbloke says:
January 21, 2012 at 7:39 am
Thank you for the ideas and the comment, TB. Is there a book somewhere that contains that rule? Because it’s a curious rule if so. You claim that to falsify a perpetual motion machine I must first “demonstrate that it will produce work”????
In any case, let’s follow your logic. I have specified the machine, a tall insulated cylinder of air. Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.
As to your claim that I have failed to demonstrate that work can be produced out of a temperature differential … surely you are kidding. That’s how heat engines work. Are you claiming that a temperature differential cannot be used to produce work?
So as far as I can see, I have done what you have asked. I have specified the machine. Jelbring claims that it will produce a temperature differential, which means it will produce work.
Finally, if people do not respond to you, there are two possibilities you seem to have overlooked:
a) They didn’t notice the post in a thread that now has over 500 comments, or
w) They thought there was no point in answering, that the debate wasn’t worth it, either because the science expressed in the post was so colossally bad or because they reckoned your mind is made up and you don’t want to be bothered with further facts.
I fall into “Group W”, tallbloke. From my perspective, I’d done what you ask above. I’d specified the machine as a tall insulated cylinder of air, and Jelbring says gravity will thermally stratify it, cold at the top and warm at the bottom. I never even considered the possibility that you would think work could not be extracted from a temperature differential. It can. So Jelbring’s hypothesis is falsified.
So perhaps my silence “speaks volumes” … but it’s not saying what you think.
w.
PS—Please note that Robert Brown may be in “Group W” as well regarding your claim, viz:
He may not be all that impressed by some random guy on the internet claiming to find a violation of the zeroth law …
To me, the problem with your claim is the idea that there is a temperature gradient across A, so that each microscopically thin slice of A is a bit cooler than the previous slice … but there is no difference between the last slice in A and the first slice in B, you claim that they are the same temperature. That’s not a temperature gradient.
So you do not have a “thermal gradient” across them as you claim. You have a gradient everywhere but at the interface between A and B.
If that is the case and the last slice of A is at the same temperature as the first slice of B, then momentarily no heat will flow from A to B. As a result, heat will run down the thermal gradient from the next-to-last slice in A to the last slice in A, warming it slightly. In the same way, heat will flow down the thermal gradient from the first slice in B to the second slice, cooling the first slice slightly.
At that point you will have a true thermal gradient across the stack, where every slice is warmer than the previous one, including from the last slice in A to the first slice in B. And of course, as the zeroth law states, then heat will flow from A to B and on down the line.
However, the idea that Robert Brown is somehow afraid to confront your “proof” that the zeroth law of thermodynamics is wrong is … well … like I said elsewhere, I’m reforming, so let me just call it “less than probable”. Robert doesn’t have time to address every claim, and a claim that you’ve overthrown centuries of science and shown that one of the laws of thermodynamics is wrong, well, that is one claim that I can see he might not want to get involved in, that one is well out in the fruit-loops zone … if you are right, you’re up for a Nobel Prize, and Jimmy the Greek is not giving good odds in Las Vegas on that happening.
All the best,
w.
Willis and Dr Brown are completely right here of course.
The system proposed by Hans Jelbring would allow the construction of a PM machine.
Unless Jelbring is willing to state that the Thermodynamic laws are wrong or incomplete then his hypothesis must be rejected.
There is a problem with Gedanken experiments though, they can allow you to construct something that seems viable and yet breaches agreed physical laws.
I could invent a closed system that was initially composed of a diffuse cloud of particles in
thermodynamic equilibrium. Now that system would be near maximum entropy. However the addition of gravity starts to cause the particles to compress and voila I now have a system like the solar system and I have reduced the entropy of the system without the addition of energy or increasing the entropy of another system.
Perhaps someone could say that this proves gravity can reverse the flow of entropy in a closed system and the Thermodynamic laws need revising.
However, in the real universe we inhabit we cannot create such a system in such an inital state. Perhaps a supreme, all powerful being could but I am not holding my breath! The existence of such a being would invalidate all known physical laws in any event.
So we have to be careful with gedanken experiments. I tend to the view that if such a system is proposed, that is in breach of thermodynamic laws, that it is either in error or could not ever be created in the universe we inhabit.
Alan
“Jelbring claims that gravity will thermally stratify it, producing hotter air at the bottom and cooler air at the top.”
Only when energy is being added from an external source.
Gravity stratifies by mass not temperature.
Read up on the Ideal Gas Law and adiabatic lapse rate.
Just look at Wikipedia here:
http://en.wikipedia.org/wiki/Lapse_rate
“the concept can be extended to any gravitationally supported ball of gas.”
and:
“the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.”
You, Willis appear to be trying to rebut 150 years of established science in favour of some 20 years during which some bozos failed to appreciate that radiative physics is a minor player and readily offset by the atmospheric response.
All your recent threads have been corrupted by that misapprehension which is sad given the quality of your Thermostat Hypothesis (even though you really should have extended it globally whereupon you would have realised your current errors).
If you had extended your Thermostat Hypothesis to the entire planet you would have realised exactly why you are so wrong now.
davidmhoffer says:
January 21, 2012 at 8:29 am
Equation 8 does indeed contain two variables. One is So, the surface pressure. The other is NTE(Ps).
But NTE(Ps) is in turn a function with three variables, Ts, Ps, and Tgb. As a result, if we expand equation 8, it is obvious that there are FOUR FREE PARAMETERS in equation 8—So, Ts, Ps, and Tgb. You sure you understand this “algebra” deal? Let me go over it.
IF
a = b + c + d, and
final_answer = f + a
then we can expand the second equation by substituting b + c + d for “a” to give us
final_answer = f + b + c + d
In other words, although (as your claim states) there are only two variables listed in final_answer = f + a, in fact there are four variables that make up the final_answer. The exact same thing is true of equation 8.
Save your disgust, your snark, and your accusations of lying for when you are right, david. They don’t taste so good when you have to swallow your words.
w.
Thank you, Joe!
Chemical potential is a pretty dang tough concept to grasp, and arguably it’s best for everyone to find their own path to it. Because in thermodynamics, kinetic theory, and transport theory there is no “royal road”, and the struggle is a necessary part of the learning process.
For folks who like to see their physics come-to-life concrete numerical simulations (as I do), one good free-as-in-freedom reference is the article “Understanding the temperature and the chemical potential using computer simulations” (2004), which includes both links to free software and a very physical explanation of temperature, chemical potential, free energy, and other mysterious topics.
Joe Born says:
January 21, 2012 at 9:23 am
“Brown and Eschenbach” have been engaging you, tallbloke and a host of others on these very questions for several threads now. The idea that we are “failing to engage” fails the laugh test. Yes, at times we might not answer a post for some reason, mainly (in my case) raw incredulity about the claims made … sometimes I just shake my head and move along, I can’t answer every wacko theory about how the world works.
But indeed, the noble firm of “Brown and Eschenbach” and a host of other good folks here are doing all we can to fight the amazing ignorance of thermodynamics so ably demonstrated by you, tallbloke, Jelbring, and others …
w.
This a comment addressed to all that believe a non-zero temperature gradient develops in an atmosphere in a gravitational field at equilibrium. This builds on upon Dewitt’s two cylinders and Rasey’s two tubes comments above. We will build a two gas, double loop system that according to believers in a steady-state non-zero lapse rate, must work as a perpetual motion machine.
Construct a rectangular pipe loop, height H, width W. The pipes are thermally insulated. Whether they are optically transparent will not matter for now, but later we might want them to be so. Arbitrarily, we call upper right point P, and clockwise the corners are Q, R, S, the base being QR, verticals PQ and RS. On the length PQ, we drill holes and insert thermally conductive plugs (C) along its length, then wrap all pipes in insulation. Have a thermal conductive plug at Q. This is Loop 1. Copy it and make Loop 2. Arrange them in mirror images with PQ of Loop 1 nearest PQ of Loop 2. The Bases QR(i) are on an isothermal ground and horizontal. Final configuration Loop 1, clockwise from upper right, PQRS, Loop 2 CounterClockwise from upper LEFT, PQRS. QR(1) and QR(2) are at the same gravitational potential. PS(1) and P2(2) are also horizontal and at the same gravitational potential. There is an arbitrary separation between PQ(loop2) and PQ(loop 2), insulated, so no thermal exchange between them…. Yet.
Fill Loop 1 with Helium. Fill Loop 2 with Argon. Expose the plugs at Q on an isotheral ground. Let them come to thermal equilibrium and via the plugs at Q
Now, to be clear, I am in the isothermal camp. I believe (along with Willis, Dr. Brown, Dewitt) that the equilibrium state is an atmosphere on a dead planet is isothermal.
But let’s play by the theory of Jelbring that an DALR will be the equilibrium state. If so, at equibrium, temperature T at Q Loop 1, T(q1) = T(q2). Then T(r1)=T(r2). By Jelbring’s reasoning, T(p1) > T(p2) because the lapse rate in PQ(1) helium is lower (closer to zero) than it is in PQ(2) argon. T(sp1) is hotter than T(sp2). T(s1) = T(p1) > T(p2) = T(s2). PQ(1) has a different temperature. If it was isothermal T(q1)=T(p1)=T(p2)=T(q2).
Now, rip off the thermal insulation from PQ1 and PQ2, slap them together so the conductors are in contact and reapply the insulation, including at Q to decouple from the ground. The two loops that were in isolated equilibrium are no-longer in equilibrium because heat if flowing from P1 to P2. The Helium at P1 is cooling, the Argon at P2 is warming. Al along PQ1 the conductors are moving heat to the colder PQ2 except at point Q where they are the same temperature. As the Helium PQ1 cools, it becomes more dense. As the argon PQ2 heats, it becomes less dense. Both loops become unstable. Helium flows clockwise PQRS, Argon flows CLOCKWISE, PSRQ. We have started the system in motion and that should be disturbing, but it doesn’t prove perpetual motion yet. Sooner or later, the system will reestablish equibrium. If you believe as I, then the equibrium will be isothermal, all points in loops 1 and 2 at the same temperature. But if you believe as Jelbring, the system will have to find some equilibrium where T(p1) < T(q1) and T(p2) < T(q2), but T(P1) = T(P2) and T(q1) = T(q2) via the conductors.
But we face a great big problem. At equilibrium RS(1) lapse rate should approach the ALR for Helium and RS(2) should approach the ALR for Argon. But PQ(1) an PQ(2) must be at the same real lapse rate because of the shared conductors. That means the verticals of loop 1 have different lapse rates, which means different densities, so they are pneumatically out of balance and must flow with greater density toward the bottom. The same goes for Loop 2.
The only way PQ(1), RS(1), can stay in balance is if they have the same lapse rate. Likewise for P2(2), RS(2). And PQ(1) must have the same real lapse rate as PQ(2) since they are thermally conducted. The only solution is for the real lapse to be zero — Isothermal.
Jim D says:
January 21, 2012 at 10:18 am
Since it takes energy to do the mixing that you propose, please come back when you’ve considered what would happen if you start out isothermal and you don’t add mixing energy. Remember, we’re discussing the atmosphere on Jelbring’s thought experiment planet of Stygia (or equivalently the air in a tall thermally insulated column). In such a situation you can’t add work to do the mixing you describe.
Will the air move away from the isothermal state if there is no energy to mix it?
w.
gnomish says:
January 21, 2012 at 9:13 am
jelbring:
“If there is a temperature gradient between two parts of a system, net heat flows from the warmer part to the cooler part. If there is net heat flow within the system, it is not in equilibrium.”
this is a definition. it is not wrong. get literate – then you can get logical – then you can get scientific. you aren’t close yet.”
Agree. Thank You. My misstake reading the opposite and being in a hurry late after midnight.
DeWitt says:
If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top. The cylinders will then be isothermal. We disconnect the heat engine. If gravity then re-establishes the DALR,
I agree the helium column will cool, but disagree that the gas in the cylinders will become isothermal, because convection would occur which would maintain the adiabatic lapse rate, as the second law demands.
Once the helium column has cooled by 3C and the Xenon column has warmed by 3C, the heat engine stops.
What is going to warm the helium column up again so the heat engine can be run a second time?
This is not a perpetual motion machine. All you have done is exploit the Gibbs free energy between the two dissimilar gases until it’s gone.
How about we start with the definition of free energy as described in Chapter 21 (Systems Subject to a Gravitational Field) of Klotz/Rosenberg’s Chemical Thermodynamics 5th Ed:
Very clear, and pretty much inarguable. BUT, anybody that is silly enough to still believe that gravity is the long-sought Maxwell’s Demon, capable of spontaneously sorting out fast moving and slow moving molecules in thermal equilibrium after the discussion and law-of-thermodynamics-violating counterexamples presented so far will simply allow their eyes to glaze over and pretend that your textbook presentation is bad science, while Jelbring or N&Z — since they state things that support their fantasy that greenhouse gases have nothing to do with the average temperature of the Earth — are all “good science”.
I swear, I’m not a “warmist” or a “coolist” — I’m a physicist, and all I care about is objective truth as best we can make it out with good science — but it’s getting to where I can see where some of the “warmists” are coming from.
Look — the Earth’s climate isn’t driven by the things you want it to be driven by. That’s “confirmation bias”, visible in spades as people continue to try to defend the indefensible (because things that openly and obviously violate the laws of thermodynamics are indefensible). If you ever have spontaneous, stable separation of an adiabatic (thermally isolated) system in good thermal contact into hot and cold reservoirs, you have already violated the second law of thermodynamics. I don’t care what internal mechanism you try to invent to act as Maxwell’s Demon to accomplish it. One of the first things you learn about in stat mech is detailed balance — it is the basis, in a manner of speaking, of stat mech.
It is this sort of nonsense that permits the warmists to loudly ignore its challengers. The greenhouse effect is established science. Get used to it. Get over it. It’s experimentally verified by IR spectroscopy from the top of the atmosphere. Claiming that there is no such thing in the face of direct experimental evidence is just plain stupid. Claiming that the warming of the Earth is accomplished by a mechanism that creates a thermal gradient in an isolated atmosphere is just plain stupid. Claiming that “gravity” is in any sense whatsoever directly responsible for the warming is just stupid.
None of these observations of things that are stupid suggest that there isn’t a thermal gradient to the atmosphere, only that it exists because the atmosphere is differentially heated at the bottom and cooled primarily at the top. This creates several mechanisms that establish the gradient — convection, radiation, conduction — where convection described as adiabatic expansion of a parcel of dry air provides not a law or universal rule, but rather a useful heuristic baseline for understanding the bottom to top lapse rate.
The bottom to top cooling not only isn’t a law, it isn’t even universally maintained in the troposphere! There are plenty of times the air overhead is warmer than the air on the ground. That’s when things like freezing rain happens. It happens as standard practice over Antarctica during the long winter night — the ground layer of air is often colder — sometimes much colder — than most if not all of the rest of the troposphere. The atmosphere is no longer being heated at the bottom, you see.
I will point out that I can keep this up indefinitely. Thermodynamics isn’t the easiest physics in the world, but this particular issue isn’t subtle. It’s one of the first things you learn in any introductory class in thermodynamics — heat flows from hot to cold, never the other way around, no matter what kind of system you have on either end, as long as there is any interaction between the hot and cold ends that can carry energy. It can be conduction, convection, or radiation. Energy transfer can be mediated by electromagnetic forces, gravity, whatever you like. If you watch the system over times long compared to the thermalization interaction time, it will come into thermal equilibrium at the same temperature.
Will the air move away from the isothermal state if there is no energy to mix it?
The answer, in case you didn’t get it from the way he asked the question, is “no”. It will move towards it. Otherwise, you violate the second law and can trivially create any of a number of perpetual motion machines of the second kind between the top and the bottom of the air column.
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