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.

Robert Brown says:
January 21, 2012 at 1:52 pm
How do you know this? Heat must move from bottom to top mediated by silver atoms. Agitated silver atoms at the bottom are at times displaced upwardly, striking atoms higher up and so on transferring some of their energy, right? The trouble is that they must do this in the presence of gravity. A silver atom moving up loses a little energy to gravitational potential energy. This can’t be conducted. As the atom moves back down it accelerates and its kinetic energy increases. Lower kinetic energies mean lower temperatures. Higher kinetic energies mean higher temperatures. A temperature gradient must exist.
Tallbloke says 1/21 at 2:06:
“Thoughts?”
It was obvious DeWitt had two cv DALR (note says DALR – must think stratified) particular gas systems at different T which as Robert Brown says can always run a heat engine when touching from hot source, outputting work, to cold sink. Stops at 3C as all agree.
The connection of two cylinders is a different problem than Willis’ cv of interest in a gravity field, like connecting his to earth surface; it is not by adiabatic definition. After each cylinder is disconnected & made adiabatic in a gravity field, each is one system again & in my view has to be stratified to obey all thermo laws being discussed, harm or create no energy.
Robert Brown said @ur momisugly January 21, 2012 at 1:20 pm
Substitute “philosopher/historian” for “physicist” and it’s ditto for the rest. Further, no warmist has managed to budge my convictions by a single iota (so far), but many of the “sceptics” here may well achieve what the warmists have failed to do. It’s a strange world…
More insight into Bart’s Law:
The equilibrium temperature will be established by the local maximum of the most substantial peak of the radiative response of the atmosphere. If the peak is in the IR, it will be slightly above the peak emission for the IR gases alone, because the total includes emissions from higher energy radiative sources.
If you add more IR radiating gas, that should tend to pull the peak lower in wavenumber and/or broaden the distribution below, resulting in surface cooling. If you add more gases which radiate at higher frequency than the peak, that will tend to push the peak to a higher wavenumber and/or broaden the distribution above, and you should get surface heating.
The key to all of this is stability. In Bart’s Law, it is stated that the temperature will “rise” until equilibrium is reached, and the atmospheric emissions keep it from rising further. The greenhouse theory implicitly says it will fall until equilibrium is reached, and the atmospheric emissions will keep it from falling further.
But, the situation is not symmetric. More energy is emitted than retained. Therefore Bart’s paradigm meets increasing resistance going up, but the GHG paradigm meets decreasing resistance to going down. Therefore, Bart’s paradigm is stable and bounded, but the GHG paradigm is not.
sky says:
January 21, 2012 at 2:27 pm
“In other words, it becomes a tautology.”
Glad someone else recognizes that. But, I have found what I believe to be the solution to the conundrum which slices through the Gordian knot. In any real atmosphere, there are always emitters, and that makes it possible for temperatures to increase within standard SB limitations until such as time as balance is achieved.
Robert Brown says 1/21 2:22pm:
“…You move the jar slowly up, move it slowly down, nothing in the jar knows that it has been moved at all!..”
Robert please slow down along with Willis, I am sure you can think this thru way better s-l-o-w-l-y.
Here you now have three systems & F=ma; everything in those jars & even the jars themselves know they have been moved. Mixing metaphors again. Just slow down. Please stay with Willis’ original cv of one system GHG-free air in a tall column in the presence of inexplicable gravity field.
We can learn from you, no kidding. I’m glad you’re engaging. My & other’s view is gas in this cv has to stratify, M-B can’t be invoked due to the ngh term; your view it does not stratify & invoke M-B. Let’s engage on that cv – Willis’ premise. Please.
Robert continues:
“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.”
Yes, yes we get this Robert, you have two systems now, hot jar connected to cold jar works in two systems. Now get into Willis’ one system cv again. Put down the jars. Then do not cross the cv. Help me et.al.
“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, Tallbloke etc:
Off the shelf Thermal engines (and vortex tubes)
stirling engines running of ice/hand/coffee with videos.
http://gyroscope.com/default.asp
As Willis would put it, I’m just a “random guy.” And, frankly, I have almost no knowledge of thermodynamics. For that matter, I didn’t understand all of the paper by Velasco, Roman, and White, whom I’d never heard of. So, how can I have the chutzpah to credit that paper over the opinions of Robert Brown, a Duke professor, and Willis Eschenbach, who regularly demonstrates more physical-science insights in a month than I’ve probably had in my life–and over the opinions of many other undoubtedly well-credentialed people at this site who have patiently explained to me that I’m ignoring one of the most basic laws of physics?
Four reasons.
Three of the reasons are two monatomic molecules and the following equation:
= (3E/(5N-2))(1-mgz/E),
where is mean single-molecule kinetic energy, N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude.
Place the two monatomic molecules by themselves into the isolated thought-experiment column we’ve belabored. Have them share a total (potential + kinetic) energy of 2mgz_mid. Let them equilibrate. When we plug this little system’s values into the equation above, which is adapted from Velasco et al.’s paper, we get (3/4) mgz_mid(1 – z/z_mid) for the mean per-molecule kinetic energy in our gas column.
According to Velasco et al., that is, the mean per-molecule kinetic energy in our gas column is 0 at 2z_mid, (3/8) mgz_mid at z_mid, and (3/4) mgz_mid at the bottom of the column. On the other hand, Brown, Eschenbach, and the many physicists whom I hereby thank for having patiently given their time here to “fight the amazing ignorance of thermodynamics [that I’ve] so ably demonstrated” say that the mean kinetic energy at equilibrium is independent of altitude: it’s the same at z=z_mid as at z=0>
A thermodynamics innocent, I cannot understand how what Brown et al. say can be so. On the other hand, my simple mind can indeed believe the story the Velasco et al. equation tells.
So, yes, my ignorance of thermodynamics is indeed amazing. But until the experts can explain to me in terms I can understand just which of Velasco et al.’s equations is wrong and why, Ill have to believe Velasco et al.
Oh, I said there was a fourth reason why I credited the unknown Velasco et al. over the obviously brainy people who opine on this site. It’s that I spent my career having such folks as clients, and my experience is that things tended to go badly when I took their word on complicated technical matters instead of forcing them to make me understand.
“Tallbloke
January 21, 2012 at 2:27 pm
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 o f 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.”
Frankly, Tallbloke, I am dissapointed with your responses to the science posts on here.
Initially I thought there was a whiff of reflection about the science being presented to you. But now there is a positive stench of desperation in your responses.
Give it up man, we are all wrong at some point in our lives, no shame in that.
I am very very skeptical about the CAGW hypothesis. My skeptism does not involve trying to deny that changing CO2 concentrations in the atmosphere has no effect, I think it does, it is the magnitude of these effects, given all the other connected feedbacks, where my doubts lie. However I am open to evidence which will show that I am wrong.
You are giving the cause of CAGW skeptism a bad name by this ludicrous defense of a flawed hypothesis. An hypothesis so flawed it is clear to people who have not completed a tertiary education.
Alan
Scot Allen says at 1/21 2:49pm:
Fine silver atom thoughts. I figured to eventually go there. When Robert Brown drops an insulated silver rod into Willis’cv tall air column, remember these thoughts. To make progress in understanding Willis’ original top post premise.
My view the silver rod, insulator and ideal gas all equilibrate, no heat flows, no energy forever b/c even though we have 1 stratified gas system, and 2 solid bodies they are all in energy equilibrium & 1 cv system obeys all the laws we’ve discussed. M-B not applicable since have ngh PE term. No energy harmed or created.
Willis Eschenbach;
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.>>>
From N&Z:
Equation (7) allows us to derive a simple yet robust formula for predicting a planet’s mean surface temperature as a function of only two variables – TOA solar irradiance and mean atmospheric surface pressure, i.e
Ts=25.3966(So+0.0001325)^0.25*NtePs
Two variables.
Yes, you need multiple variables to calculate NtePs in the first place because mean surface pressure is not uniform. If you understood N&Z (which you state that you don’t) you would understand that a big part of what they have done is provide a mechanism to calculate NtePs. It could just as easily be measured in which case it would indeed represent a single variable. The problem is that measuring it is just as complex as measuring the temperature of the earth compared to the insolation because they don’t vary with each other. P varies with T^4. All N&Z are showing you in this part of their equation is how they got to the correct value in the first place, which is not straight forward. But if we could just stick a pressure meter into the earth and get that number, it would indeed be a single variable.
My accusation and snark stands.
Hi Robert, and thanks again for your reply. I’m going to split my response into two halves. the first will deal with your statements and claims about my position. The second half will deal with the argument you feels sufficiently addresses the matter in hand.
You say that:
“The problem is that temperature is a meaningless concept for a two dimensional surface. ”
Well, I was only trying to address your ‘slices’ argument from the original post, so I humbly submit that you are the one who started it with the surfaces malarkey.
and you say that I should:
“address the actual topic of the thread, which is how any system that spontaneously generates a thermal gradient violates the second law of thermodynamics”
Well you can’t define us out of existence. Total energy = KE+PE. The second law I’m working from (feel free to define the one you’re using) tells me that in a gravitational field, the higher up you are, the less Kinetic Energy you’ll have and the more Potential Energy you’ll have. Total energy must be conserved. Everyone is agreed on that one, so, less KE = less heat high up.
Now, I’m aware that there are some very sophisticated arguments about this stuff which use statistical mechanics, but as Joe Born will tell you, he’s been through the math in the Coombes and Laue paper, and the Valesco et al paper, and he has found the reason C&L come to the isothermal conclusion. Joe says it doesn’t work. They are playing with limit cases which don’t relate to the mechanical behaviours of actual atmospheric gases.
“and thereby enables the construction of PPM2K’s, including the trivial thermocouple example in the previous post.”
The only viably well specified machine I’ve seen in this thread fails the logic test quite quickly. keep trying though.
“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.”
Your problem here is that you keep going back to your own preconceptions instead of following our argument. KE+PE is constant, so as we go up, PE increases. Therefore KE (and temperature) falls. So an air packet is in energetic equilibrium when the temperature gradient across it is such that it matches the lapse rate defined by gravity and it’s specific heat at constant pressure. Loschmidt held it should be that way for solids too by the way, so your silver rod may not behave as you’d expect.
Now, your capturing of gases in jars argument seems hopelessly confusing to me. That may be my fault or yours or just that we’re seeing things differently. Try this explanation of why we have ended up with differing interpretations and see if it makes sense to you.
H/T Frank
“In a laboratory setting, we say that the pressure is the same in all directions and therefore that the kinetic energy associated with movement of molecules (thermal diffusion) is the same in all directions. On a planetary scale, we say that pressure change is associated with the weight of gases overhead, and forget to consider what this means about the behavior of molecules. If there is a pressure gradient with height, it can only be caused by molecules moving upward more slowly than they move downward. The obvious reason for this is that the molecules moving upward are being slowed down by gravity and converting kinetic energy to potential energy (and cooling when we consider a large enough group to treat mean kinetic energy as temperature). The opposite is true for molecules moving downward.”
Now from what I read in an earlier comment, someone claimed that you’ve said that the KE of molecules won’t be affected by gravity. If that’s so, please could you explain why they are exempt from the law of gravity. What goes up must lose velocity. If it’s mass stays the same, it ends up with less KE. QED
Thanks and I hope we can continue discussion.
TB.
Robert Brown says:
January 21, 2012 at 8:39 am
“get an intro physics textbook and read it…….”
=============
I’ve always wanted one, does Amazon sell a good one ?
Do you sell a good one ?
Don’t be bashful, just tell me which book to buy.
Thanks.
respect.
Alan Millar says:
January 21, 2012 at 3:32 pm
Frankly, Tallbloke, I am dissapointed with your responses to the science posts on here.
The feeling is mutual Alan. Except I’m disappointed, rather than dissapointed.
thepompousgit says:
January 21, 2012 at 2:17 pm
tallbloke said @ur momisugly January 21, 2012 at 9:31 am
It is a definition. One that is severely lacking in universality. It fails as a general law as soon as it is placed orthogonally to a gravitational field.
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.
Wipe your windows pompous. I didn’t criticise it as a definition, I pointed out it failed as a general law. And while your smug logico-philosophical point is trivially true, the problem we face is people running riot with limited definitions of the thermodynamics laws which are useful heuristics in limited settings and pushing them to the status of generally applicable ‘laws of nature’.
This is the reason for most of the confusion in this thread.
Sorry, folks, the web site ate part of the equation in my last post; it must not have liked the mean-indicating angle brackets
Here it is without the mean-indicating brackets:
K = (3E/(5N-2))(1-mgz/E),
where K is mean single-molecule kinetic energy, N is the number of molecules, E is total system energy, m is molecular mass, g is the acceleration of gravity, and z is altitude.
As I said before: Place the two monatomic molecules by themselves into the isolated thought-experiment column we’ve belabored. Have them share a total (potential + kinetic) energy of 2mgz_mid. Let them equilibrate. When we plug this little system’s values into the equation above, which is adapted from Velasco et al.’s paper, we get (3/4) mgz_mid(1 – z/z_mid) for the mean per-molecule kinetic energy in our gas column.
According to Velasco et al., that is, the mean per-molecule kinetic energy in our gas column is 0 at 2z_mid, (3/8) mgz_mid at z_mid, and (3/4) mgz_mid at the bottom of the column. On the other hand, Brown, Eschenbach, and the many physicists whom I hereby thank for having patiently given their time here to “fight the amazing ignorance of thermodynamics [that I’ve] so ably demonstrated” say that the mean kinetic energy at equilibrium is independent of altitude: it’s the same at z=z_mid as at z=0.
Tallbloke your reply to Robert is just another obfuscation.
You have been shown that Jelbring’s system allows heat to flow indefinitly. Robert’s silver rod shows that. You have not attempted to disprove that by the use of the current thermodynamic laws. No, you have attempted to disprove that by trying to say that whist the laws apply in general they do not apply in the particular.
You have said, in refutation, that ‘something’ in atomic structiures will prevent connected heat sources, at different temperatures, exchanging heat. That something is presumably due to the distance between the heat sources and/or the nature of the connection or perhaps it is something else.
You have to publish the details man, a Nobel prize awaits. Nobody could deny you it. You would go to Norway garlanded, as the man who has added to and improved, the laws of thermodynamics.
That is assuming you have some evidence or mathematics to back up this claim!!!
Do you?
Alan
Would an experiment resolve these issues?
For example, Take a sealed cone filled with non GWG and spin it about so that the pointed end is on the extremity and the flat end is nearest the centre. Measure the temperatures near the pointed end and near the flat end.
tallbloke says:
January 21, 2012 at 1:17 pm
You didn’t pay close attention to the specification of my device. The two cylinders are thermally coupled at the bottom. So the gas at the bottom of the column in contact with, say, a copper plate that is otherwise insulated must be at the same temperature. So the cooled helium that falls to the bottom of the cylinder causes heat transfer from the bottom of the xenon cylinder cooling the xenon at the bottom while the heat engine (or a silver rod) warms the top of the xenon and cools the helium. In the end, both cylinders are isothermal at the same temperature. So now we have two isothermal cylinders. But if Jelbring is correct and we thermally disconnect the top of the cylinders, the lapse rates will be re-established and more work can be done. Or, if we extract work slowly enough, the engine will run forever. A classic perpetual motion machine of the first kind. Or possibly it’s the second kind. I haven’t actually done the calculations to see if the heat content of the cylinders decreases over time.
tallbloke said @ur momisugly January 21, 2012 at 4:27 pm
Wipe your own windows TB. The Laws of Thermodynamics are called Laws just because they are universal, show no sign of being contradicted by any repeatable experiment, are unchanging etc. There is no chance of any of these Laws being repealed by those posting to this thread who view them as “limited heuristics in limited settings”. They have been accepted as Laws for over a century not just because they make sound theoretical sense, but also because they have withstood every test of them so far.
Please note that I am not saying that they will remain as Immutable Laws for Ever and Ever Amen. Paradigm shifts do occur. But every paradigm shift I have studied was enabled by people who clearly understood the paradigm they were overturning.
Most of the people on this thread exhibiting confusion are doing so because they do not understand basic physics.Reading Feynman’s The Character of Physical Law might help remedy this.
BTW Tautologies are far from trivial; they are the building blocks of the edifice we call mathematics, science and logic.
Alan Millar says:
January 21, 2012 at 4:43 pm
Robert’s silver rod does not show that. The atoms in the silver rod are affected by gravity in the same way molecules of air are affected. Hotter silver atoms at the bottom deliver their energy to those above, and those to more atoms above and so on during conduction. Each tiny temporary displacement carries kinetic energy, but some of that energy is converted to gravitational potential energy. This stored potential energy can’t be conducted. It is converted back to kinetic energy as the atoms displace downward. These differences in kinetic energies between layers must create a temperature gradient.
There are really only two possibilities.
1) The second law works and thus the temperature must be uniform.
2) KE + PE is constant, so the particles cool as they go up, and thus the temperature drops.
I can give the “elevator speech” for why #2 is suspect. For any given trip, the KE + PE is indeed constant, but for different trips, the value is different (ie the boltzman distribution). If you look near the top, the “low energy tail” never gets that high, so you are only looking at self-selected particles that started in the high energy tail. These originally-high-energy-particles have indeed lost some KE on the way up, but they started with extra on average. This can (and does) leave this SUBSET of particle with the right average energy to be at the same temperature as the WHOLE set was at the bottom.”
I certainly haven’t proven this rigorously in this non-mathematical paragraph. However, it is clear the “lose KE and lose temperature” argument has a huge hole in it. Those who want to pursue this line of reasoning IN THE FACE OF STRONG COUNTER-ARGUMENTS, would need to determine the distribution of energies of particles at any altitude, and show that those remaining particles are indeed lower KE then the whole set was at the bottom. I am sure they can’t do this because I am sure they are wrong.
But hey, that Nobel Prize is still waiting to be claimed….
sky says:
January 21, 2012 at 2:27 pm
But the gas density decreases to zero with altitude. For an isothermal atmosphere, the volumetric energy density decreases with altitude exponentially because the volumetric kinetic energy density decreases much faster than the volumetric gravitational potential energy increases. There is no kinetic energy temperature for space. The pressure is too low and the mean free path is too long for a Maxwell-Boltzmann kinetic energy distribution to form. Local thermodynamic equilibrium doesn’t exist, so a gas kinetic temperature can’t be defined. That’s not to say there isn’t an effective radiative temperature. There is, but it would have no effect on the kinetic energy of an individual molecule. Any molecule that might be excited would radiate long before it could transfer energy to another molecule and change the kinetic energy of either molecule. Similarly, collisions between molecules that resulted in transferring enough energy to excite one of the molecules so it could radiate and result in a loss of total kinetic energy would be extremely rare.
Also, at low density, the thermal diffusivity is extremely high, leading to a rapid decrease in any existing temperature gradient if there is no other source of energy, which there isn’t by the definition of the problem.
DeWitt Payne, where are you getting your lapse rates for helium and xenon? Are they empirical or calculated?
Alan Millar says:
January 21, 2012 at 4:43 pm
Tallbloke your reply to Robert is just another obfuscation.
You have been shown that Jelbring’s system allows heat to flow indefinitly.
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Isn’t the confusion here that between perpetual motion and recycling? The Water Cycle is recycling, no one thinks of it a perpetual motion..
Taking energy out of a system is going to have some effect on the system, how that plays out if taking energy from the gravity piston is what has to be calculated, (gravity doing the work of a piston by default as it were, the gases changing density by temp using gravitational potential by being in the different levels by weight/density, rising and falling, so energy extracted from anywhere along that is going to have to be replaced).
For example, extracting wind energy to power must be having some effect, altering wind patterns perhaps?