Guest post by Robert G. Brown
Duke University Physics Department
The Problem
In 2003 a paper was published in Energy & Environment by Hans Jelbring that asserted that a gravitationally bound, adiabatically isolated shell of ideal gas would exhibit a thermodynamically stable adiabatic lapse rate. No plausible explanation was offered for this state being thermodynamically stable – indeed, the explanation involved a moving air parcel:
An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field.
This argument was not unique to Jelbring (in spite of his assertion otherwise):
The theoretically deducible influence of gravity on GE has rarely been acknowledged by climate change scientists for unknown reasons.
The adiabatic lapse rate was and is a standard feature in nearly every textbook on physical climatology. It is equally well known there that it is a dynamical consequence of the atmosphere being an open system. Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.
Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work. As is usually the case, violation of the refrigeration statement allows heat engines to be constructed that do nothing but convert heat into work – violating the “no perfectly efficient heat engine” statement as well.
The proposed adiabatic thermal lapse rate in EEJ is:
where g is the gravitational acceleration (presumed approximately constant throughout the spherical shell) and cp is the heat capacity per kilogram of the particular “ideal” gas at constant pressure. The details of the arguments for an adiabatic lapse rate in open systems is unimportant, nor does it matter what cp is as long as it is not zero or infinity.
What matters is that EEJ asserts that 
in stable thermodynamic equilibrium.
The purpose of this short paper is to demonstrate that such a system is not, in fact, in thermal equilibrium and that the correct static equilibrium distribution of gas in the system is the usual isothermal distribution.
The Failure of Equilibrium
In figure 1 above, an adiabatically isolated column of an ideal gas is illustrated. According to EEJ, this gas spontaneously equilibrates into a state where the temperature at the bottom of the column Tb is strictly greater than the temperature Tt at the top of the column. The magnitude of the difference, and the mechanism proposed for this separation are irrelevant, save to note that the internal conductivity of the ideal gas is completely neglected. It is assumed that the only mechanism for achieving equilibrium is physical (adiabatic) mixing of the air, mixing that in some fundamental sense does not allow for the fact that even an ideal gas conducts heat.
Note well the implication of stability. If additional heat is added to or removed from this container, it will always distribute itself in such a way as to maintain the lapse rate, which is a constant independent of absolute temperature. If the distribution of energy in the container is changed, then gravity will cause a flow of heat that will return the distribution of energy to one with Tb > Tt . For an ideal gas in an adiabatic container in a gravitational field, one will always observe the gas in this state once equilibrium is established, and while the time required to achieve equilibrium is not given in EEJ, it is presumably commensurate with convective mixing times of ordinary gases within the container and hence not terribly long.
Now imagine that the bottom of the container and top of the container are connected with a solid conductive material, e.g. a silver wire (adiabatically insulated except where it is in good thermal contact with the gas at the top and bottom of the container) of length L . Such a wire admits the thermally driven conduction of heat according to Fourier’s Law:
where λ is the thermal conductivity of silver, A is the cross-sectional area of the wire, and ΔT=Tb–Tt . This is an empirical law, and in no way depends on whether or not the wire is oriented horizontally or vertically (although there is a small correction for the bends in the wire above if one actually solves the heat equation for the particular geometry – this correction is completely irrelevant to the argument, however).
As one can see in figure 2, there can be no question that heat will flow in this silver wire. Its two ends are maintained at different temperatures. It will therefore systematically transfer heat energy from the bottom of the air column to the top via thermal conduction through the silver as long as the temperature difference is maintained.
One now has a choice:
- If EEJ is correct, the heat added to the top will redistribute itself to maintain the adiabatic lapse rate. How rapidly it does so compared to the rate of heat flow through the silver is irrelevant. The inescapable point is that in order to do so, there has to be net heat transfer from the top of the gas column to the bottom whenever the temperature of the top and bottom deviate from the adiabatic lapse rate if it is indeed a thermal equilibrium state.
- Otherwise, heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.
It is hopefully clear that the first of these statements is impossible. Heat will flow in this system forever; it will never reach thermal equilibrium. Thermal equilibrium for the silver no longer means the same thing as thermal equilibrium for the gas – heat only fails to flow in the silver when it is isothermal, but heat only fails to flow in the gas when it exhibits an adiabatic lapse in temperature that leaves it explicitly not isothermal. The combined system can literally never reach thermal equilibrium.
Of course this is nonsense. Any such system would quickly reach thermal equilibrium – one where the top and bottom of the gas are at an equal temperature. Nor does one require a silver wire to accomplish this. The gas is perfectly capable of conducting heat from the bottom of the container to the top all by itself!
One is then left with an uncomfortable picture of the gas moving constantly – heat must be adiabatically convected downward to the bottom of the container in figure 1 in ongoing opposition to the upward directed flow of heat due to the fact that Fourier’s Law applies to the ideal gas in such a way that equilibrium is never reached!
Of course, this will not happen. The gas in the container will quickly reach equilibrium. What will that equilibrium look like? The answer is contained in almost any introductory physics textbook. Take an ideal gas in thermal equilibrium:
where N is the number of molecules in the volume V, k is Boltzmann’s constant, and T is the temperature in degrees Kelvin. n is the number of moles of gas in question and R is the ideal gas constant. If we assume a constant temperature in the adiabatically isolated container, one gets the following formula for the density of an ideal gas:
where M is the molar mass, the number of kilograms of the gas per mole.
The formula for that describes the static equilibrium of a fluid is unchanged by the compressibility (or lack thereof) of the fluid – for the fluid to be in force balance the variation of the pressure must be:
(so that the pressure decreases with height, assuming a non-negative density). If we multiply both sides by dz and integrate, now we get:
Exponentiating both sides of this expression, we get the usual exponential isothermal lapse in the pressure, and by extension the density:
where P0 is the pressure at z=0 (the bottom of the container).
This describes a gas that is manifestly:
- In static force equilibrium. There is no bulk transport of the gas as buoyancy and gravity are in perfect balance throughout.
- In thermal equilibrium. There is no thermal gradient in the gas to drive the conduction of heat.
If this system is perturbed away from equilibrium, it will quickly return to this combination of static and thermal equilibrium, as both are stable. Even in the case of a gas with an adiabatic lapse rate (e.g. the atmosphere) remarkably small deviations are observed from the predicted P(z) one gets treating the atmosphere as an ideal gas. An adiabatically isolated gas initially prepared in a state with an adiabatic lapse rate will thermally equilibrate due to the internal conduction of heat within the gas by all mechanisms and relax to precisely this state.
Conclusion
As we can see, it is an introductory physics textbook exercise to demonstrate that an adiabatically isolated column of gas in a gravitational field cannot have a thermal gradient maintained by gravity. The same can readily be demonstrated by correctly using thermodynamics at a higher level or by using statistical mechanics, but it is not really necessary. The elementary argument already suffices to show violation of both the zeroth and second laws of thermodynamics by the assertion itself.
In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down. Reverse that to a cooling, like those observed during the winter in the air above Antarctica, and the lapse rate readily inverts. Follow the air column up above the troposphere and the lapse rate fails to be observed in the stratosphere, precisely where vertical convection stops dominating heat transport. The EEJ assertion, that the dry adiabatic lapse rate alone explains the bulk of so-called “greenhouse warming” of the atmosphere as a stable feature of a bulk equilibrium gas, is incorrect.
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One more nail in the coffin …
Velasco’s paper does NOT support the adiabatic lapse rate — not even close.
I actually took the time to read the paper (posted here if anyone is interested http://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf ). Joe Born said earlier in this discussion that “as a practical matter, this result differs only negligibly from the isothermality” — which is a drastic understatement as we will see.
It would take a top post to give details, but basically there are two problems with using this paper here.
1) It assumes a completely adiabatic container. This would mean that the surface temperature would not matter because the atmosphere and the surface are perfectly insulated from each other and cannot interact in any way. If the ground instead acted like a thermal reservoir (a more realistic assumption) then the temperature profile would indeed be exactly isothermal.
2) The paper is dealing with small numbers of particles. Ignore objection #1 for now, and assume we really do have a completely adiabatic container . Even then, if the number of particles is large, the atmosphere tends toward isothermal. I did the calculations in his paper (and could share the spreadsheet) . If the surface temperature of the gas is 300 K, then at 1 km altitude in the container, the temperature would be
* 80 K for 1 particle in the container
* 286 for 10 particles
* 299.87 for 1000 particles
* 299.9999 for 1,000,000 particles
So we are left with
* if the atmosphere does not interact in any way with the sun
AND
* if the atmosphere does not interact in any way with the surface
AND
* if the atmosphere contains no more than 1 mole of gas total
THEN
there will be a theoretical “thermo-gravimetric enhancement” of perhaps 0.00000000000000000000001 K from the top to the bottom.
“It seems to me you would cool the entire atmosphere.</i
Why? The wire and gas form a closed system in my picture above."
Yes. Without the wire you have:
"Thermal equilibrium is the state of maximum entropy. It is also the state where there is no free energy available to do work within the system. "
With the wire you have enter a new element into the system- and it isn't in thermal equilibrium.
A problem is the model says heat isn't conducted by the tube- and no details of how this done.
If the tube is conductive of heat it will act as the silver wire- which will diminish the lapse rate.
100 meter steel pipe vertical in the air which is insulated will not change very much over the distance of 100 meter. With in contact with interior of pipe, it means air will also not change much in the 100 meter elevation. Solids or liquids do not have lapse rates. 3000 meter of ocean depth does not increase the temperature of water at bottom. Remove the water, replace with atmospheric air and insulate the air from cold ocean and the air will warmer at the bottom than at the top. If it's not insulated, then air would be a cold as the ocean water.
Tim Folkerts: Two comments.
First, in asserting a non-zero lapse rate, your comment, which, as my comments weeks ago on Tallbloke’s Talkshop indicate, I largely agree with, confirms the point I have been making: since Dr. Brown’s explanation is based instead on the absence of any non-zero lapse rate whatsoever, no matter how small, it does not merit the non-critical gushing with which this thread is redolent.
Second, by stating that a temperature difference between the top and the bottom, no matter how small, you have expressed a position different from the one upon which after being faced with Velasco et al. Dr Brown has now seized, i.e., that no heat flow means no temperature difference.
I might add that my purpose in bringing up the Velasco et al. paper was not to contradict the conclusion that Jelbring’s theory is unsound but rather, since I had not fully comprehended all of Velasco et al.’s steps, to invite a thorough discussion of its reasoning and thereby arrive at a more-informed assessment of its validity. I was disappointed not only in that but also at the general shallowness that most comments directed to that paper exhibited.
Robert Brown says:
January 30, 2012 at 7:07 am
myrrh said some things about “You haven’t even read his explanation for writing it the way he did as your diabribe against him showed”
That’s because I don’t care why he wrote it badly. You show me one single original contribution in this article! It isn’t even a valid review article. I reiterate: He says “I’m going to prove that there is a stable thermal lapse in thermodynamic equilibrium”. He quotes a textbook that derives the DALR in a section on climate dynamics, atmospheric flow. He states “this is thermal equilibrium”. He concludes “I’ve proven that the DALR is thermal equilibrium and will heat an absolutely static, stable, isolated atmosphere with a fixed total energy content differentially after all thermalizing processes are finished.”
Where does he say that?
No he hasn’t, no it doesn’t. He hasn’t proven anything at all — he just begs the question by restating his assertion as his conclusion with a “QED”, and the conclusion he asserts violates the second law of thermodynamics.
I haven’t the faintest idea what you’re arguing about. I can’t find any of that in his paper.
Myrrh said: “I think you should begin again, Take II, and stick to his thought experiment and not your strawman silver wire deflection”.
By all means, let’s. That way we won’t be able to trivially prove that his assertion that the system is in equilibrium is false.
He set out what he meant by equilibrium here:
The silver wire is hardly a “straw man”, of course. It is just a proxy for heat conduction, something that he seems to have left out of consideration when he listed the agents responsible for establishing thermal equilibrium in his “thought experiment”. Unless you can show that no irreversible transport of heat energy is possible — even in fluctuations — within this gas, which is prima facie absurd, it is trivial to show that moving heat from the bottom to the top increases the entropy of the closed system. I’ve done so several times above.
You introduced “thermal equilibrium”.
You went off on a completely different tangent, and began arguing against something he hadn’t said and which has nothing to do with the poser he sets.
If you want to publish a paper for lay people to help them understand something, it helps to explain how the result doesn’t violate the second law of thermodynamics when obviously it does. Even for lay people. I mean, all it takes is a single course in introductory physics to see how his assertion fails. You don’t need a physics Ph.D, you just need to stop burying your head firmly in the sand because you don’t want to face the fact that a thermodynamically stable, DALR undriven by a thermal differential maintained by other means is bullshit magic.
Well, I’m a lay peeps, and I think I understand what he’s saying (I’ve explained myself in a post above), you however, don’t make any sense at all.
You’re quoting stuff I can’t find, you’ve ditched his scenario and put your own in its place, and changing energetic equilibrium for thermal equilbrium you then spend your whole time arguing that he can’t get an adiabatic lapse rate from a thermal equilibrium which he never said he could, and, you bring in conduction, the wire, when he is explicitly talking about gases. What the has solids to do in this? This is about our fluid gaseous atmosphere.
Now, if you can sort that out, where you’re getting his quotes, maybe I’ve missed them, why you’ve totally ignored his scenario, why you’re arguing against a straw man of thermal equilibrium of your own invention and brought in the totally irrelevant conduction through solids and all the arguments about the 2nd law with respect to that, maybe you could write something worth reading about his paper. But at the moment, all you’re arguing with everyone about is your own imagined paper.
Guest post by Robert G. Brown
Duke University Physics Department
The Problem
In 2003 a paper was published in Energy & Environment by Hans Jelbring
You still haven’t fixed the link.
that asserted that a gravitationally bound, adiabatically isolated shell of ideal gas would exhibit a thermodynamically stable adiabatic lapse rate. No plausible explanation was offered for this state being thermodynamically stable
Yes he did. Newton’s gravity. And note, thermodynamically stable adiabatic lapse rate, not your ‘thermodynamic equilibrium’.
Jelbring said: “The more atmospheric mass per unit planetary area, the greater GE has to develop. Otherwise Newton’s basic gravity model has to be dismissed.”
– indeed, the explanation involved a moving air parcel:
“An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field.”
This is standard meteorological background. For example:
“Thus, rising air is said to cool or warm adiabatically when its temperature changes are due entirely to pressure changes. In reality, some degree of energy exchange will always take place, but these are generally small on short timescales.”
http://www.st-andrews.ac.uk/~dib2/climate/lapserates.html
This argument was not unique to Jelbring (in spite of his assertion otherwise):
“The theoretically deducible influence of gravity on GE has rarely been acknowledged by climate change scientists for unknown reasons.”
He’s not saying it’s unique to him, he’s saying that this “has rarely been acknowledged by climate change scientists for unknown reasons”
Real scientists, such as meteorologists, know this. “Climate change” scientists use a lot of fictional fisics, besides such obvious boo boos like missing out the Water Cycle.
The adiabatic lapse rate was and is a standard feature in nearly every textbook on physical climatology.
That the adiabatic lapse rate is standard in “climatology” textbooks is irrelevant, he’s referring to real science and the point he’s making is that it is pressure related in the atmosphere and that there is no energy exchange in the process.
It is equally well known there that it is a dynamical consequence of the atmosphere being an open system.
Yeah, well, what do you mean by “well known”? Like, ‘well known that carbon dioxide is well-mixed and accumulates in the atmosphere for hundreds and thousands of years’? Like, ‘well known that shortwave visible heats oceans’?
But anyway, as I said in my post it’s good enough to approximate a closed system, and I’ve only just found the page I quoted from here, so my imagined cap to make it a closed system wasn’t quite accurately imagined, there is a cap on the dynamic system, it’s the inversion layer below the stratosphere.
There’s no reason to deviate from his scenario.
Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.
The scenario which you have put in place of Jelbring’s, and then continued to argue, and argue, and argue that he got wrong, when you imagined it yourself. That is straw man.
I suggest, again, that you go back to the beginning and make some effort to understand what he is saying first, before you think to refute him. At the moment, you’re simply arguing against yourself.
The link to his paper should be:
http://ruby.fgcu.edu/courses/twimberley/EnviroPhilo/FunctionOfMass.pdf
For some reason it’s put this discussion url in front which messes it up.
.
[I’ve been having some problem with word wrap, hope I’ve fixed it, or most of it..]
@Joe Kirklin Born
You have me thinking again, just after I had given up for the day. Are you suggesting that if there is a lapse rate, that the depth of the atmosphere makes a difference, in the sense of an effect emerging in a deep atmosphere of average gravity, might not be detectable in a shallow atmosphere with strong gravity? How about without a lapse rate?
Dr Brown says that an isothermal condition is the maximum entropy, even if there is an altitude difference in a gravitational field because ‘no energy transfer can take place’ which I take to mean ‘even if molecules move around’. This seems to be slightly circular because if an energy transfer can take place by some means other than hot-to-cold, he rules it out. If, prior to the isothermal state a mechanism causes stratification, either warmer at the bottom maintained by gravity and a dry adiabatic lapse rate or warmer at the top maintained by buoyancy and poor thermal conduction downwards, it is a pretty convenient argument to rule out energy transfer because it is isothermal.
Further, if ‘molecules can move around’ and the argument is that the lapse rate maintains the temperature constant (things already being isothermal) there is a problem if the isothermal condition was arrived at ‘because there is no lapse rate’. No lapse rate to achieve isothermal, maintained as isothermal by a lapse rate (exactly equal cooling of rising and warming of descending molecules).
To me, by definition of a lapse rate is cooling when rising. Because of the compression/expansion energy thing there obviously is a lapse rate prior to establishing an isothermal state, how is conduction going to overcome it? Or perhaps, which effect will dominate? Heat ‘attempting to flow’ up to colder gas by conduction opposed by some other mechanism may, at some total Delta T, constitute a net zero energy change without the temperature equalising. Is that the maximum entropy because no net energy transfer is taking place?
An earlier question no one picked up on is ‘is the planet rotating? Most planets rotate, at least a bit. The moon rotates slowly – it just appears not to from our perspective. Thus there will be rotation in cells away from the poles and a lot of vertical mixing and stirring. This would easily overcome tiny strange forces that might maintain a tiny temperature difference. To allow them to come into play we would again have to restrict the conditions to include a statement that it is a non-rotating planet. Do you hear the sound of time, wasting?
Joe,
I recognize your concerns. I don’t think they invalidate any of the comments made by Dr Brown.
1) thermodynamics is always about large numbers of particles, and in the limit of large numbers of particles, this solution becomes isothermal.
2) the completely adiabatic situation means no energy is exchanged with anything, so the atmosphere could not possibly be the cause of the warming of anything.
3) the introduction of any “thermal reservoir” in contact with the gas would cause the equilibrium condition of the gas to be isothermal, and any potential conflict with the 2nd Law disappears.
It is fascinating that people are this passionate about thermal physics, but this is a topic beyond what could be easily addressed in this sort of forum, and would more rightly belong in a classroom setting or as a potential journal article.
I agree, but those cans dissipate or equilibrate heat energy to the dominant system (free air or in the freezer). Convection excepted, the Earth’s atmosphere is closed, it IS the dominant system.
My experiment would involve a sealed conductive container placed at the top of an atmospheric column containing a temperature sensor but with an atmospheric pressure the same as that experienced by the bottom sensor. My guess ? yes, the top sensor would drop in temperature but not as much as an exposed sensor.
Atmospheric heat is contained heat, it cannot escape,except through convection.What if the air molecules at the bottom have not fallen, where do they get their heat?, borrow from the fallen ones? Your analogy does not come close to explaining atmospheric complexity, but I thank you, Robert, for taking the time and effort to discuss it.
I do not think that this topic is closed.
I suggest, again, that you go back to the beginning and make some effort to understand what he is saying first, before you think to refute him. At the moment, you’re simply arguing against yourself.
The link to his paper should be:
http://ruby.fgcu.edu/courses/twimberley/EnviroPhilo/FunctionOfMass.pdf
I find little to argue about in that paper. Quote:
“The main conclusion, derived from the model atmosphere of this paper, is the fact that
there has to exist a substantial greenhouse effect (GE) which is mass dependent and
which will develop independently of the amount of greenhouse gases in any real
planetary atmosphere.”
There is 10 tons of atmosphere per square meter- this has a high thermal capacity and during the night this air mass cools. It cools due to some degree due the surface radiating energy.
The paper mention that considerable amount energy is lost in higher atmospheric elevation- this must have to mostly be regarding “greenhouse gases” as non-greenhouse do not radiate a significant amount of energy, so I question this assertion. Air temperature cools significantly during the night, what portion of this cooling is losses to the ground vs some mechanism of air losing it’s kinetic energy to radiating gases?
I have no argument with that quote, I do have problem with
. It would depend upon the gas, water vapour has no problem ‘floating’ against gravity to rise above above the heavier gasses.
As cloud cover significantly reduces night cooling, I guess that radiative loss is significant.
“Tim Folkerts says:
January 30, 2012 at 4:18 pm
One more nail in the coffin …
Velasco’s paper does NOT support the adiabatic lapse rate — not even close.
I actually took the time to read the paper (posted here if anyone is interested http://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf )”
From the velasco.pdf:
“Coombes and Laue, there are two conflicting answers
to the above question:
(1) The temperature is the same throughout because the
system is in equilibrium.
(2) The temperature decreases with the height because
of the following two reasons.
(a) Energy conservation implies that every
molecule loses kinetic energy as it travels
upward, so that the average kinetic energy of
all molecules decreases with height.
(b) Temperature is proportional to the average
molecular kinetic energy.
Coombes and Laue concluded that answer (1) is the
correct one and answer (2) is wrong.”
Thereafter much mathematically exciting stuff.
But I would agree that: “The temperature is the same throughout because the
system is in equilibrium.”
Or I would say we aren’t dealing with one molecule of gas.
I would agree with 2 if there was only one molecule of gas.
But no doubt the math has more use.
Now, perhaps in the math proofs, but I didn’t see the part about:
“Velasco’s paper does NOT support the adiabatic lapse rate — not even close.”
Rather I thought paper was saying there was not paradox, and answer was (1)
which seems to fit my idea of what the adiabatic lapse rate is.
Anyhow, moving on:
“It would take a top post to give details, but basically there are two problems with using this paper here.
1) It assumes a completely adiabatic container. This would mean that the surface temperature would not matter because the atmosphere and the surface are perfectly insulated from each other and cannot interact in any way. If the ground instead acted like a thermal reservoir (a more realistic assumption) then the temperature profile would indeed be exactly isothermal.”
Ok.
“2) The paper is dealing with small numbers of particles. Ignore objection #1 for now, and assume we really do have a completely adiabatic container . Even then, if the number of particles is large, the atmosphere tends toward isothermal. I did the calculations in his paper (and could share the spreadsheet) . If the surface temperature of the gas is 300 K, then at 1 km altitude in the container, the temperature would be
* 80 K for 1 particle in the container
* 286 for 10 particles
* 299.87 for 1000 particles
* 299.9999 for 1,000,000 particles
So we are left with
* if the atmosphere does not interact in any way with the sun
AND
* if the atmosphere does not interact in any way with the surface
AND
* if the atmosphere contains no more than 1 mole of gas total
THEN
there will be a theoretical “thermo-gravimetric enhancement” of perhaps 0.00000000000000000000001 K from the top to the bottom.”
Ok so one mole of gas seems small. That is: “22.4 liters of gas is the volume of 1 mole of gas at STP (Standard Temperature and Pressure = 0 degrees C at 1 ATM)- http://www.exo.net/~emuller/activities/A%20Mole%20of%20Gas.pdf
What volume is the one mole of gas in? Planetary scale, it’s better vaccum then on the Moon- on in space itself. If per meter of surface it is a fair amount.
It has seemed to me that the “thermo-gravimetric enhancement” must be fairly tiny.
0.00000000000000000000001 K seems smaller than I thought, but I don’t how to relate to 1 mole of gas.
Tim Folkerts:
I agree with your numbered points and that in essence they support Dr. Brown’s conclusion. For reasons it seems pointless further to belabor, though, I believe there’s a gap in the reasoning by which Dr. Brown reached that conclusion. As I said, this in itself is a minor point, but I was trying, unsuccessfully, as it turned out, to use it as a can opener into a discussion of whether Velasco et al.’s conclusion is correct. That paper–or, more accurately, the Román et al. paper on which it relies–is the only one of which this layman is aware that treats the issue with any rigor, so I was interested in finding out whether it was accurate.
Crispin in Waterloo:
First, let me emphasize that I’m a layman whom a bunch of PhDs in this thread appear to have dismissed as a crank, so you may want to take that into account. Second, I confess to not quite following all that you’ve said. Third, although I’ve been guilty in the past of attempting qualitative explanations of the type you may be making, my bias now is not to trust them, because explanations that are completely contradictory can both seem plausible to me. That’s why I felt queasy at Dr. Brown’s explanation on which Willis Eschenbach based his Latin-titled post.
That being said, here’s my take. For Velasco et al.’s purposes, the state of a molecule is totally defined by its location and momentum, and the state of an ensemble of molecules is the combination of the individual molecules’ states: for an ideal-gas ensemble consisting of N monatomic molecules, each of which is characterized by x, y, and z components both of position and of momentum, the ensemble’s state can be represented by a point in 6N space, in which a surface I think of as a hyperparaboloid represents the states that exhibit a given total (potential + kinetic) energy. The infinitesimal volume between the hyperparaboloids for E and E + dE is taken as containing the states the ensemble of of fixed total energy E can assume. The likelihood that the ensemble’s state falls within any subregion of that infinitesimal volume is taken as the ratio of the subregion’s (infinitesimal) volume to the total infinitesimal volume between the hyperparaboloids.
Based on those assumption, Velasco et al.’s calculations came to the conclusion that the number of states that exhibit the kinetic-energy gradient their Equation 8 specifies is greater than the number that exhibit a gradient of exactly zero. And the size of the Equation-8-implied gradient is proportional to the ratio of the force of gravity on a single molecule to the ensemble’s total energy. So–and this may be the answer to the question you were posing–it will be higher for greater gravity and tend to be higher for fewer molecules.
What I haven’t figured out is by what proportion the number of states exhibiting a gradient within a suitably small neighborhood centered on the one implied by Velasco et al.’s Equation 8 exceeds the number that exhibit a gradient within an equally small neighborhood centered on zero. My understanding of the Second Law is that it’s not a law so much as, given the numbers involved, just a phenomenally safe bet: in circumstances normally considered, the number of states exhibiting isothermality is so astronomically many times the number that exhibit temperature segregation that the odds favor the house so much that you may as well take it as a law that the house always wins, i.e., that heat will invariably flow from hot to cold. But what some disputants above are, quite reasonably, saying is that the number-of-states ratio is not as great in cases where the Equation-8-implied gradient is significant as it is in situations to which we are entitled to apply thermodynamics.
Joe,
The key point in the Velasco paper is that the energy of the system never changes (known formally as a microcanonical ensemble). If there is one particle in the container, it will ALWAYS have an energy exactly “e”. It never exchanges energy with the surroundings. It always bounces perfectly with the walls, neither gaining nor losing energy. It does NOT have a boltzmann distribution of energies. The temperature of the walls have absolutely no effect on the particle. In such a case, you really can’t define “being at equilibrium” because by definition there is no exchange with anything.
In a more realistic setting, the particle would exchange energy with the walls, and the particle would have a boltzmann distribution based on the temperature of the walls of the container.
If there are 2 particles, the two have a total energy of exactly 2*e. So if one particle has energy “x”, the other will have “2e – x”. Again this is not a boltzmann distribution for either, but at least a particle can have a range or energies.
As you get more and more particles, each particle gets closer and closer to a boltzman distribution, and the lack of interaction with the walls becomes less and less important.
Tim Folkerts says:
January 30, 2012 at 4:18 pm
That’s not the way I read the paper. My understanding is that the temperature is always 300K regardless of the number of molecules. For small numbers of molecules, temperature is no longer strictly proportional to the average kinetic energy. You only get a lapse rate if you assume that temperature is always strictly proportional to average kinetic energy. But Velasco, et.al. specifically says this isn’t true.
[my emphasis]
Statement (2b): Temperature is proportional to the average kinetic energy.
DeWitt Payne:
I don’t want to speak for Tim Folkerts, but I think he and I are looking at the equations as being more reliable, whereas you’re going by the verbal description, which can only with difficulty be reconciled to what the equations say. But one thing you may want to consider is whether they are actually talking about temperature as a local quantity that does not vary with altitude or instead as a global quantity that has no relationship to altitude. I believe their definition of temperature is that it’s a quantity inversely proportional to the first partial derivative of the log of phase volume with respect to total energy.
Tim Folkerts:
I understand, and have from the first understood, everything you just wrote, with the exception of the thing about equilibrium not meaning anything in the microcanonical ensemble. As to equilibrium, I’ve been using that term to describe a configuration in which entropy is maximized, not as one that necessarily is characterized by a Boltzmann distribution, although an equilibrium microcanonical ensemble approaches such a distribution in the limit.
Joe,
Then it looks like we are pretty much on the same page. This stat mech stuff is tough, so it is difficult to estimate who is getting it and who is not, so I was trying to re-iterate my main points.
As to the comment about equilibrium, I was mostly talking about the system with a single particle. The particle cannot share energy with the walls; the particle cannot share energy with any other object within the system. So it cannot “equilibrate” with anything. For multiple particles in the system, I don’t have a problems talking about equilibration within the system.
DeWitt,
I should be more careful about the word “temperature”. What I was more specifically listing in the table was (2/3 KE/k), which should be the temperature for an equilibrium situation with a large number of particles.
So for one particle, it has some total energy at the bottom, “e”, which will entirely be KE and not PE. At the bottom we could define a temperature based on this KE using KE = 3/2 kT. As that single particle goes up, it loses KE, so we could say that its “effective temperature” drops to zero as it rise toward an altitude where PE = mgh = 3/2 kT = e and KE = 0.
For two particles, KE does not need to drop to zero when mgh = e, because it would be possible for one particle to have 2e of total energy, and have KE = PE = e at that altitude. So the “effective temperature” based on KE does not drop to zero at this altitude h, but rather at an altitude 2h.
But this is getting WAY to involved for this discussion. Lets just say that for large n, the energy distribution of the particles does indeed very well approximate a boltzmann distribution with a constant temperature at all altitudes.
With the wire you have enter a new element into the system- and it isn’t in thermal equilibrium.
. Kinda the thermodynamic definition of the change of entropy of a reservoir, isn’t it?
Sure it is. I put it in the system and left it for a very long time!
That’s what Jelbring is ignoring. The fact that air in fact conducts heat just like the silver, and that if you wait for equilibrium — whether or not it actually takes a very long time on a human scale to get there — the equilibrium reached will be isothermal or violate the second law, in particular by manifestly not being the maximum entropy state of the system. The cold air at the top gains more entropy than the warmer air at the bottom loses, quite independent of the density of the air, when heat is transferred by any irreversible means you like from the bottom to the top. There are always irreversible means available — in particular heat conduction through the air.
Elementary thermodynamics. And I do mean elementary:
rgb
A point that requires clarification.
Is ‘adiabatically isolated’ exactly the same as ‘thermally isolated’?
Tim Folkerts says
“In a more realistic setting, the particle would exchange energy with the walls, and the particle would have a boltzmann distribution based on the temperature of the walls of the container.”
This would mean that the walls need to match the temperature change in the gas for an adiabatic solution.
Previously I have worked on the assumption that there is no thermal interaction between the walls and the ideal gas.
The constant speed moving parcel of the ideal gas is the same solution as still air for hydrostatic equilibrium.
Robert Brown says
“an adiabatically isolated column of an ideal gas is illustrated.”
ILlustrated above is a rectangular or perhaps cylindrical tube.
Fixed diameter or base means the only variation in the tube is in the vertical
For adiabatic expansion in the atmosphere the rising parcel needs to expand.
The only volume expansion in this case is by changing vertically
So the formula – g/Cp = -9.8K/km for dry air would need amending.
Yes he did. Newton’s gravity. And note, thermodynamically stable adiabatic lapse rate, not your ‘thermodynamic equilibrium’.
The fact that a thermodynamically stable adiabatic lapse rate is inconsistent with physics, violating the second law, is the only point of my article. And it is obviously true.
You’ve now proven my point. All he does is assert an impossibility at the beginning, and use his assertion to prove that his assertion is true later. But his assertion that an adiabatic lapse rate is thermodynamically stable does not make it so. It is not.
rgb
Tim Folkerts says:
January 31, 2012 at 9:26 am
OK. I think I would make the single particle have the 300K KE at the mid-point of the column rather than the bottom, but it only makes a difference for very small numbers. The main point is still that Velasco, et.al. specifically deny that there is a vertical temperature gradient in the column at equilibrium for any number of particles. If the particles do not obey MB statistics, and they probably won’t for small numbers, the justification for converting average kinetic energy to temperature using MB statistics (2/3 KE/k) no longer exists.
[With the wire you have enter a new element into the system- and it isn’t in thermal equilibrium.]
“Sure it is. I put it in the system and left it for a very long time!
That’s what Jelbring is ignoring. The fact that air in fact conducts heat just like the silver, and that if you wait for equilibrium — whether or not it actually takes a very long time on a human scale to get there — the equilibrium reached will be isothermal or violate the second law, in particular by manifestly not being the maximum entropy state of the system.”
But silver [or solids and liquids] do not conduct heat just like an ideal gas. The temperature of an ideal gas is solely/mostly it’s velocity of it’s molecules. Liquids and solid molecules do not have velocity- they vibrate if above 0 K. Molecules of ideal gas can also vibrate, but the vibration of gas molecules does not cause a gas to have a temperature- temperature of gas is it’s density and velocity of the gas molecules.
Gases can be all about the vibration of gas molecule- rather than it’s velocity. One can “excite” gas molecules, but in terms of the conditions in earth’s atmosphere the energy of it’s ideal gases in terms what temperature they are, relates to velocity of gas molecules. A counter example is Northern lights- nothing to do with velocity of gas molecules and everything to do exciting the molecules of gas so they emit photons. Excited gas molecules do not remain excited, they will emit their photon. Whether gas has been excited are not has little to do with the temperature of the gas.
So in terms of liquids and solids their vibration is their temperature. And their molecules going nowhere, except in sense that you toss a brick and the molecules brick are moving. With ideal gases it’s not the vibration or rotation of gases which indicate how warm they are. It is a factor which could be included but is not a significant element of why 20 C room temperature air is 20 C.
But silver [or solids and liquids] do not conduct heat just like an ideal gas. The temperature of an ideal gas is solely/mostly it’s velocity of it’s molecules. Liquids and solid molecules do not have velocity- they vibrate if above 0 K. Molecules of ideal gas can also vibrate, but the vibration of gas molecules does not cause a gas to have a temperature- temperature of gas is it’s density and velocity of the gas molecules.
Gases can be all about the vibration of gas molecule- rather than it’s velocity. One can “excite” gas molecules, but in terms of the conditions in earth’s atmosphere the energy of it’s ideal gases in terms what temperature they are, relates to velocity of gas molecules. A counter example is Northern lights- nothing to do with velocity of gas molecules and everything to do exciting the molecules of gas so they emit photons. Excited gas molecules do not remain excited, they will emit their photon. Whether gas has been excited are not has little to do with the temperature of the gas.
So in terms of liquids and solids their vibration is their temperature. And their molecules going nowhere, except in sense that you toss a brick and the molecules brick are moving. With ideal gases it’s not the vibration or rotation of gases which indicate how warm they are. It is a factor which could be included but is not a significant element of why 20 C room temperature air is 20 C.
So many errors, so little time. Let’s start with the last one — with ideal monatomic gases, vibration or rotation of the gas molecules isn’t an important factor. With ideal diatomic gases they are, and worse, the dependency itself depends on the temperature of the gas and the energy required to excite rotations vs vibrations. Look it up, it’s in any intro physics textbook that covers thermodynamics at all (since they almost invariably cover the thermodynamics of ideal gases).
Second, it doesn’t make any difference “how” solids or liquids conduct heat versus an ideal gas. What matters is that heat conduction is a thermodynamically irreversible process because it invariably increases the entropy of the Universe as it occurs (instead of breaking even, the characteristic of reversible processes). If the gas permits irreversible heat transport at all by any means, including by mere thermal radiation (also irreversible) then the entropy of the system will increase as it becomes isothermal (assuming isolation and no external or internal sources of continuous work).
This is all stuff you should work to understand, since it is a major factor in the entire theory of thermodynamics.
rgb
Robert Brown says:
January 31, 2012 at 10:04 am
Yes he did. Newton’s gravity. And note, thermodynamically stable adiabatic lapse rate, not your ‘thermodynamic equilibrium’.
The fact that a thermodynamically stable adiabatic lapse rate is inconsistent with physics, violating the second law, is the only point of my article. And it is obviously true.
Sorry, I’ve only just notice it should be “your thermal equilibrium”, which you’ve put in place of what he’s said, your straw man. And, you’ve shown neither.
You’ve now proven my point. All he does is assert an impossibility at the beginning, and use his assertion to prove that his assertion is true later. But his assertion that an adiabatic lapse rate is thermodynamically stable does not make it so. It is not.
Hmm, the impossibility he asserts is bog standard physical science as observed and well known in meteorology – one has to understand it to understand his point. And, as before, you’ve proved nothing of the kind since you’ve been arguing your straw man thermal equilibrium. Read the page I quoted from, this is bog standard real physics of Newtonian gravity which is understood by countless weatherpeeps and used in their daily work calculations, thoroughly understood empirically for generations.
What is it with ‘climate’ scientists? Why haven’t you noticed that you’ve missed out all these real world physics from your thinking? This, in my humble opinion, is what I think is happening here, you just don’t know that it exists, as Jelbring says: “has rarely been acknowledged by climate change scientists for unknown reasons” So your tangent is naturally then to something that is in your ken.
But the one that really gets me is – how have y’all come to miss out the Water Cycle from your energy budget, KT97 and clones? How?! How have none of you noticed it’s missing?!
It’s bog standard industry figures that the Water Cycle brings down the temps 52°C to 15°C. Calculated on understanding the nature of our heavy gaseous ocean above us, weighing down on us a ton on our shoulders, in a thermodynamically stable adiabatic lapse rate.. ‘climate’ scientists teach this is empty space!
When one puts together all that you, ‘climate’ scientists, have missed out, and this current example just gets added to the list I’ve gathered so far, you’ve got a really weird world you’re describing. Alice through the looking glass impossible thinking before breakfast; visible light heating water and land, heavier than air carbon dioxide defying gravity and accumulating for hundreds and thousands of years, the heat from the Sun not reaching the Earth’s surface, cold objects heating warmer.., someone’s really been messing with your heads.
You’ve no ‘feel’ for the world around you through real physics, the wind might as well be a big wooden spoon stirring the atmosphere around, or the gods at the four corners in a blowing contest. It’s something I concluded a while back, that only someone who knows this physics really well could do the subtle tweaks across the range of disciplines involved.
But the effects have been dramatic. Reducing the argument to radiation as if it’s ‘all the same energy’ by stripping it of its individual properties and processes, has enabled the swapsies of properties and out of context use of laws to be made in the descriptions of energies, gases and processes, and, so ground into thinking through repetition in the education system that even the absence of the Water Cycle goes unacknowledged as you all busy yourselves arguing about the nuances of your fictional fisics with real world applied scientists who know better.
There’s still some around..
http://johnosullivan.livejournal.com/43659.html
..who know who’s really violating the second law.