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.
Robert Brown says at 1/27 9:45am:
“Also bear in mind that speed is not temperature.”
Right, temperature is KE, a constant mass times speed squared so it varies as the speed varies. If the speed is constant with h, KE is constant top post Fig. 1, temperature is constant with increasing h in top post Fig. 1 & it is isothermal.
If speed, thus KE varies with h in Fig. 1, then T is non-isothermal as per Rodrigo Caballero on line text linked in Perpetuum Mobile thread:
“Temperature is just another name for the mean kinetic energy density of molecular motion.”
The Rodrigo Caballero on line text as of this post still says for top post Fig. 1:
“Mean velocities will be greater near the bottom of the box than near, the top…”
This text means KE thus temperature will be greater near the bottom of Fig. 1 than near the top, so non-isothermal. Though now have some input around here that this text may change in some way for the next edition.
Dr. Brown,
I apologize for attributing motive to your statements. You had fun calling me on that, and insulting my intelligence and education.
But my confusion persists. Could you please address my example or answer my questions, repeated here:
Dr Brown,
Expanding: I think that you are arguing that your heat conductor can convey energy (heat) from the ground to the sky without any net work being done. You have not shown how this is actually possible.
If you have a mechanism that worklessly exchanges a hot molecule at the bottom of your solid conductor with an identical but cold molecule at the top of your conductor it look like no gravitational work was done, right? The daemon lives! and we have free energy!
But if you do that exchange, the conductor still looks the same, but it isn’t really still exactly the same. The net height of the conductor doesn’t change but a larger (hotter) molecule from the bottom of the conductor has been replaced by a smaller (colder) molecule from the top of the conductor. The molecules of the conductor, save the top one, are slightly closer to earth. Gravity has taken won a little advantage in return for the heat you have conducted to the colder top of the conductor.
Please explain some mechanism that conducts heat to the top of you conductor without loosing work to gravity, I don’t see any such mechanism 😉
Robert Brown says:
January 26, 2012 at 10:12 pm
With respect to the Tij et.al. paper that seems to have had such an effect on Joules Verne, some numbers might be useful:
h=kb*T/(m*g) = 9.08km for N2 at 300K
λ = 68nm at 101325 Pa
To get λ ~0.01h where the deviation from the Fourier Law is ~10% requires a reduction of pressure of 11 orders of magnitude to ~ 1E-6Pa or an increase of g of a similar magnitude or some combination of both that increases λ and reduces h.
I’m not going to touch the temperature gradient as that’s even worse.
Paul Birch says:
January 27, 2012 at 7:34 am
LaTex is indeed a markup language, or more strictly speaking, a typesetting language, as it can be used in other than HTML formats. Search on “latex tutorial” and you get lots of hits. I learned a little once and promptly forgot it. The other unfortunate thing is that like HTML, the conventions may be slightly different from site to site. Some testing may be necessary to get started. Either use a page with a preview feature or use Firefox and CA Assistant, which also, unfortunately, doesn’t work everywhere.
[Moderator’s Note: WUWT also has a test page here where you can test a post for formatting before adding it to a thread. The usual policy rules will still apply, however. -REP]
One (hopefully) last comment about KE & PE in the atmosphere as related to thermodynamics.
KE energy of a particle is indeed constant –> when averaged over long periods of time.
PE energy of a particle is indeed constant –> when averaged over long periods of time.
Total energy of a particle is indeed constant –> when averaged over long periods of time.
For example, in a solid there are only three axes, but there is KE and PE associated with each axis, so there are SIX degrees of freedom and each will have an average energy 1/2 kT when averaged over sufficiently long periods of time.
In the atmosphere, this is slightly different, with an energy of kT associated with PE (see htttp://en.wikipedia.org/wiki/Equipartition_theorem#Sedimentation_of_particles). This means that in equilibrium, a molecule should have (on average) mgh = kT, or h = kT/mg. For an N2 molecule at T = 300 K, this works out to ~ h = 9 km. So in an equilibrium atmosphere with T=300 K, when averaged over long periods of time, a molecule of N2 in such an atmosphere will have an average altitude of 9 km. Not surprisingly, this is indeed about the “average” height of the earth’s atmosphere.
NOTE 1: Thermodynamic equilibrium deals with long-term averages, not with short-term motions. Looking at 1 flight of one molecule is not sufficient. So the fact that one particular molecule of N2 might happen to have an above average PE does not affect its chances of being above or below average in KE (ie in temperature), just like a high speed in the x direction has no bearing on its chance of being above or below average in the y speed.
NOTE 2: This also explains why molecules do not separate by mass, with the heaviest molecules all at the bottom. N2 will have an average height of 9 km independent of any other molecules present, and this will follow an exponential decline with altitude. Helium, with 1/7 the mass would have an average height of 7*9km = 63 km (which explains why it can drift ouf into space more easily. CO2, with about 1.5 times the mass, will have an average height of ~ 6 km. So the heavy gases are enriched at low altitudes, but do not completely displace the lighter molecules.
NOTE 3: Earth is NOT perfectly insulated and is NOT in equilibrium. There is a constant heat flow up thru the atmosphere due to cooling at the top by GHG’s. This makes a few adjustments for the actual situation on earth. For example, the temperature does drop with altitude, and convection mixes the gases in the troposphere. These do not negate the basic conclusions.
So explain how the silver wire has precisely the same conductivity to the gas at low pressure at the top of the tube as it does to the gas at high pressure at the base of the tube. It cannot as there are less gas molecules coming in contact with that wire at the top than there are at the bottom – that is the definition of pressure.
through the wire. Even you aren’t asserting that the conductivity isn’t zero, you are just asserting that it is rate limited by the area in contact with the gas at the top. This area in contact is a variable under my control, of course, but even exactly as drawn we seem to agree that the conductivity isn’t zero.
You will find that when you take the different conductivity across the silver-to-gas boundary into account that the high gas-to-silver conductivity at the base is in lapse rate proportion to the lower silver-to-gas conductivity at the top.
No, I won’t find any such thing because it isn’t true.
Besides, what is to stop me from adjusting the areas in contact with the gas in both venues using a heat sink?
Finally, the only way heat won’t flow is if the conductivity is zero, and all I need for a second law violation is any non-zero transfer of heat:
So I take it that, now that I’ve clearly shown that your “objection” in no way affects my argument, you’re ready to concede, right?
rgb
Expanding: I think that you are arguing that your heat conductor can convey energy (heat) from the ground to the sky without any net work being done. You have not shown how this is actually possible.
If you have a mechanism that worklessly exchanges a hot molecule at the bottom of your solid conductor with an identical but cold molecule at the top of your conductor it look like no gravitational work was done, right? The daemon lives! and we have free energy!
But if you do that exchange, the conductor still looks the same, but it isn’t really still exactly the same. The net height of the conductor doesn’t change but a larger (hotter) molecule from the bottom of the conductor has been replaced by a smaller (colder) molecule from the top of the conductor. The molecules of the conductor, save the top one, are slightly closer to earth. Gravity has taken won a little advantage in return for the heat you have conducted to the colder top of the conductor.
Please explain some mechanism that conducts heat to the top of you conductor without loosing work to gravity, I don’t see any such mechanism 😉
Sigh. Look, I’m really trying to keep up, but if you aren’t willing to actually study the answers to these questions-posed-as-statements, how are you going to learn? Do I have to hand-teach you all one at a time?
Imagine that you have a stack of pool balls a kilometer high. Whack the bottom one. What happens to the top one? If the pool balls are perfectly elastic, like the molecules of an ideal gas, the top one pops up a bit in reaction because of conservation of momentum and energy. You can do the same experiment with a line of balls in an ordinary pool table, or you can look at:
http://en.wikipedia.org/wiki/Newton%27s_cradle
with its running demo. The case of the wire is almost identical, although the wire allows a bit more variation in the transmission of energy along the way. As hot molecules strike the bottom, they successively warm higher and higher layers in the wire until heat input at the bottom is picked up by the colder air molecules at the top, cooling the bottom and warming the top.
There’s nothing magical about this. Heat flows from hot reservoirs to cold. Gravity doesn’t act on heat. Heat is not pulled towards the ground. The fallacy in your argument is simple — nothing in the wire changes its height while it conducts the heat! Gravity doesn’t do any work on the system. All that happens is that all of the silver molecules oscillate back and forth in place a little more violently, with exactly the same average gravitational potential energy, per molecule.
If you think that metals don’t conduct heat vertically, place a pan on a stove, put your hand on the top of the pan, and turn on the stove. Remove hand when convinced.
rgb
Robert Brown,
just gotta say that reading these threads is always an education on new physics that my ignorant mind has never heard of before.
” All that happens is that all of the silver molecules oscillate back and forth in place a little more violently,”
With this quick sentence you have destroyed millenia of observations that metals and most other materials expand when they are heated!! HAHAHAHAHAHAHAHAHA Wonderful, continue the teaching of the new IPCC Physics!!!
Tim Folkerts says:
January 27, 2012 at 1:06 pm
Not true. That would imply that the volumetric mixing ratio of CO2 would be higher at low altitude than at high altitude in the lower atmosphere. But this doesn’t happen because stratification isn’t allowed at LTE. If it were true then the partial pressure of CO2 at the surface would be different than the volumetric mixing ratio times the surface pressure. There’s about 6 kg CO2/m². If no other gas than CO2 were present, the surface pressure would be ~59Pa instead of the ~39.5Pa that it actually is. See a more complete discussion of the problem and solution here. The elevator speech version: The m in the scale height equation kbT/mg is the average molecular weight of the atmosphere, not the weight of the individual molecules. Stratification only occurs at very low pressure/very high altitude when LTE no longer applies.
Rodrigo Caballero says at 1/27 6:07:
“..see recent discussion by Verkley and Gerkema, J. Atmos. Sci. 2004”.
Geez, Verkely seems to find both non-isothermal AND isothermal T fields depending on constraints for top post Fig. 1.
Verkley et. al. paper linked by Guinganbresil at 1/27 9:38am.
Some issues interested posters might discuss in the Verkley (V) paper.
1) The air column in Verkley is different than the control volume of Fig. 1 top post. The paper says the column Verkley et. al. is analyzing is permitted to let work out above and below the control volume: “(V) Fig 1.: The column is assumed to exchange no net heat with its surroundings but may perform work on the air above and below the column. “
This at 1st look seems to be a change from what has been discussed here, where no work is allowed across the control volume above or below. Not sure what this means yet. Need help, maybe from my Boren text due in a few days or from other posters.
2) After Equation 10 the words “Since this should be true for all variations dT”. Why do they not say “must be true”? Equation 11 approaches infinity as T approaches zero. Is this why? The Velasco issue again: isothermal is wrong until can “ discern between the cases of a finite system and an infinite system.”?
3) In part b, Verkley et. al. finds a non-isothermal T field in eqn. 18 !! Don’t hit me!
4) In part c, they write: “The above calculations show that the result of the maximization process depends on the constraints that are used. “ I say: no kidding.
5) Verkley et. al. write in conclusion: “..this constraint choice is of course open for debate.”
So Robert Brown et. al. can say, view agrees with Verkley part a, just choose my constraints for isothermal.
So Trick et. al. can say, view agrees with Verkley part b, choose my constraints for non-isothermal.
Very unsatisfying. Might be a long thread, the open question for Velasco conclusion in 1996 seems still open in conclusion in Verkley et. al. 2004.
Infinity is a real demon. Go figure.
At the minimum, it seems for an isothermal T field conclusion, Fig. 1 in the top post must be modified to show at the top and bottom of the column: “…may perform work on the atmosphere air above and below the column. “
Consider, for example, a centrifugal compressor running continuously under steady conditions. The compressor draws air from a temperature and humidity controlled space.at 1 bar, and discharges it into a process at 2 bar. .The entering air is steadily and continuously heated from ambient (Ti) of 15°C to an exit temperature, (To),which will be 78°C or higher,. It is a function of the ratio of the outlet pressure (Po) to the inlet pressure, (Pi) and the efficiency of the compressor; which can be close to, but will be less than one. To = Ti * (Po/Pi)^((R m eff)/Cp mean), R is the universal gas constant, and m is the molecular weight of the gas. Is it thermodynamically stable?. Why is the discharge hotter?
You are doing work on the system, are you not? It is an open system with a continuous input of work and for that matter mass. It is not an isolated system. There are all sorts of ways one can construct things that move heat from cold to hot reservoirs or create temperature differentials in initially isothermal air. We call most of them “refrigerators”, and no, refrigerators are not “thermodynamically stable” in the usual meaning of the terms, which would be “in thermodynamic equilibrium”. They can be stable in the sense that they are predictable in their cyclic operation, so a steady pattern of input work, mass transport, and heat flow can be established.
You don’t have to work so hard to make up an example. Go into your house. Open the refrigerator door. Cold inside. Feel the coils in back or underneath. Hot. Pay the electric bill on the motor. Heat input. Sure, one can use work lots of different ways to make air colder or hotter, but what you cannot do is violate the second law of thermodynamics.
So why is this example at all relevant to either Jelbring’s paper or my rebuttal? In fact, it isn’t. It is a straw man argument, proposed to try to convince — I dunno, yourself? — that because “refrigerators exist” Jelbring’s second law violating one can exist, or something like that. If you have some other point, feel free to make it, but to me this seems like nothing but non sequitor.
As for insulting your education and so on, look. Open a completely random college level introductory physics textbook. Turn to the chapters on thermodynamics. Just read them, please, like a novel. Don’t stop, don’t put the book down, just read the words and try to understand them, look at the examples. If you feel inspired, do a few homework problems. Pay special attention to the zeroth, first and second laws, get to understand heat engines, refrigerators, limits on efficiency and coefficient of performance. Learn the Kelvin-Planck and Clausius statements of the second law. Learn the entropy statement. Then come back to the discussion. You’ll be better for it, more able to make useful contributions, and I suspect that you will no longer argue that the second law can be violated or that the atmosphere can somehow refrigerate or heat the Earth per se.
rgb
A physicist quoted Eddington:
I love this quote. The quote, and its context comprise a work of art, the beauty of which I can but poorly appreciate, and not relate at all.
QDaniels,
the Eddington quote is very interesting. I have been assured by well educated warmers that even conduction breaks the 2nd at very low levels at times. Whether they are claiming that this is measured or a result of solving statistical mechanics they did not explain. So, because it is a really small violation, do I apply the quote or not?? 8>)
The Rodrigo Caballero on line text as of this post still says for top post Fig. 1:
“Mean velocities will be greater near the bottom of the box than near, the top…”
This text means KE thus temperature will be greater near the bottom of Fig. 1 than near the top, so non-isothermal. Though now have some input around here that this text may change in some way for the next edition.
You did read Caballero’s actual post on the list earlier today, right? Where he directly stated that an isolated atmosphere in a gravitational field reaches isothermal equilibrium, as first demonstrated by Gibbs some 120 years ago?
I really don’t think you should be quoting Caballero to support these statements, because the point isn’t that adiabatic lapse can occur, it is that static, thermodynamic equilibrium is isothermal without a lapse.
Just a thought.
rgb
PS Prof Bob’s silver wire is too ductile to ring nicely, which is why I chose Aluminium.
Nice post, although by now I’m convinced that it is fruitless. Even Caballero weighing in today hasn’t cut off the nonsense.
At the moment, I think the score is this
100% of the Ph.D. physicists on list think that equilibrium is isothermal
80% of the engineers and other scientists do
maybe 60-70% of the lay people (with no real physics education) do
with a small, stubborn, 30% overall who just don’t seem to be able to grasp the ideas of thermal conductivity, thermal equilibrium, entropy, work, the first and second laws of thermodynamics, and so on, who just don’t want to give up the notion that temperature is what they want it to be instead of what it actually is: nothing but a measure of comparative thermal equilibrium.
This last group relies on heuristic arguments that seem to be attempts to do statistical mechanics in their head or by constructing a mental picture that suits them, drawing on a high school level understanding of mechanics. They neglect the fact that thermodynamics is the valid macroscopic heuristic for statistical mechanics, and that stat mech computations cannot be done on an heuristic basis at all; they are horrendously difficult, involving taking limits of nearly infinite sums in just the right way to get the relevant part of the answer and discard the parts that don’t scale up to relevance as one goes to large systems. I’m unaware of any general heuristic that permits one to skip doing the real work in this case other than thermodynamics.
I’m getting tired of replying. The issue is settled. Really. It was settled well over 100 years ago, and if you’ve been following along you have heard not one, but a half dozen arguments that are all openly endorsed by the list physicists, however enthusiastically they are challenged by people who do not, in fact, understand physics. I’m not trying to be an elitist, only pointing out that if you already know everything, what point is there in me trying to teach you? If you are certain that you know better than physics textbooks and people who teach and do research in the field, well then, the best of luck to you.
So I think that this will probably be my last post on this thread. We’re down to a core of people that will never be convinced, because they don’t want to be, people who have diverted the thread into side discussions (always interesting, but…) and people who understand the issues well but are trying to herd cats. On to other things!
rgb
Robert Brown
You have created a perpetual motion machine of your own if you totally ignore the gravitational effects of charges and atoms in your silver wire.
http://arxiv.org/ftp/arxiv/papers/0908/0908.2820.pdf
Tim Folkerts says:
>> So the heavy gases are enriched at low altitudes, but do not completely
>>displace the lighter molecules.
DeWitt Payne says:
>Not true. That would imply that the volumetric mixing ratio of CO2 would be higher
>at low altitude than at high altitude in the lower atmosphere.
I respectfully disagree. This is certainly not a key issue in this discussion, but equipartition is equipartition. In equilibrium, molecules of one mass have the same average kinetic energy as molecules of a different mass. Similarly, molecules of one mass have the same average potential energy as molecules of a different mass. This implies they must have different average altitudes, just like they have different average speeds.
It is certainly true that the ACTUAL atmosphere does not show this behavior in the troposphere, but that would be because it is not in equilibrium. Specifically the actual atmosphere is well mixed by convection — a process that would be completely missing in an equilibrium atmosphere. So the actual atmosphere has constant mixing ratios even though this would not be the equilibrium condition. At least, that’s my take on this side issue.
“Imagine that you have a stack of pool balls a kilometer high. Whack the bottom one.”
Takes a lot of work to add the mgh to move all the balls and displace the top ball That’s gonna be quite a big whack. That’s all JimZ said – the vertical wire doesnt look like a work-free transfer to altitude.
Could I venture another question as I read the thread. If we depart from perfect elastic collisions, is there a mechanism to suggest that greater frequency and more energetic collisions at higher pressure would be associated with higher pressure? The degree of this phenomenon might be dictated by the (ahem) “collision efficiency”?
Dr Brown says:
“There’s nothing magical about this. Heat flows from hot reservoirs to cold. Gravity doesn’t act on heat. Heat is not pulled towards the ground. …”
I didn’t say that gravity acts on heat. Heat is not real, it is a useful concept, and you’re correct, gravity doesn’t act on heat. Heat without matter is a pointless idea, and matter is subject to gravity.
Kinetic energy will not be conducted from low gravitational potential to high gravitational potential without work against gravity; there isn’t kinetic energy without mass. The atmosphere and the conductor will have the same temperature gradiant, in a energy equal partitioned steady state, and neither will be conducting heat from the warmer to the colder.
As you wrote, wacking a stack of billiard balls on the bottom is the exact situation we are arguing about. The billiard balls are elastic AND they are compressible, just like molecules, just like all solid matter. Electrical force is conservative, so there is no energy loss in the billiard ball column when the wack is propagated . But the balls are compressible and momentum is conserved. Wacking the bottom of the column upward does cause some displacement of the matter in the column, and the displacement works against gravity. So wacking the bottom of the column with a finite energy will not lift the ball on the top of the column, when the column is taller than some finite height, depending on the compressibility of the balls (same is true with wacking the bottom of a stack of molecules). True in theory and true in practice.
Your argument does work with elastic and in-compressible matter. The heat would be conducted as you say when your conductor is made of ideal matter that is elastic and not compressible.
Tim says;
quote
I respectfully disagree. This is certainly not a key issue in this discussion, but equipartition is equipartition. In equilibrium, molecules of one mass have the same average kinetic energy as molecules of a different mass. Similarly, molecules of one mass have the same average potential energy as molecules of a different mass. This implies they must have different average altitudes, just like they have different average speeds.
end quote
Molecules of one mass don’t have the same average kinetic energy as more massive or less massive molecules. It is equipartition of the total energy of a each of the molecules. It is conservation of energy AND momentum; heavy molecules move slowly and light molecules move fast. That’s how gas chromatography works, all molecules at the same temperature but heavy and light ones at different speeds. Molecules of different mass don’t have the same average potential energy.
PS, there isn’t any “equilibrium” in this argument.
This paper proves the Kelvin-Helmholtz mechanism in general relativity and also says that even objects in gravitational equilibrium will emit radiation (published in Physical Review D of the American Physical Society).
http://arxiv.org/pdf/gr-qc/0605066.pdf
A million Sols
If there was a million stars providing the same solar flux as our sun does and they surrounding earth in every direction, what temperature earth be?
I think it would be around 120 C.
I think the surface would be 120 C, and I think the air temperature would be 120 C.
Everyplace would 120 C or hotter- underground it would be hotter.
It seems if the earth had no atmosphere or ocean- or the moon would temperature of 120 C.
Though with Earth and it’s atmosphere the surface only get 1000 watts instead of 1300 watts, so the surface couldn’t be heated by the Suns to 120 C. And I guess “somehow” the atmosphere would make up the difference- and the atmosphere may heat the surface so it’s higher than 120 C. But also it seems to me that the atmosphere would have a lapse rate- 10 Km above the surface it would be cooler than the surface.
If we assume that earth would have a lapse rate [can’t see how it wouldn’t] what would the lapse rate be? would be around the current average lapse of 6.5 C per 1000 meter. Or would it be higher- like 9.8 C per 1000 meters. Or say 1 C per 1000 meter?
Next question is what would the ocean temperature be. Our sun gets the ocean as warm as 30 C.
If the entire ocean is 30 C, one could see all those sols warming it to 40 or maybe 50C. But how does it get much warmer than that?
So a question is what kind of world be coolest, if surrounded by a million sol?
What our water world be the coolest?
What if you have entire planet of water- something around same gravity as earth, but about twice the diameter.
Such a world would be quite strange. The world is say 99.9 % water, and without nitrogen sky, it would be black. The water pressure would exceed any pressure human have ever made, oh I guess not:
“typical pressures reached by large-volume presses are up to 30-40 GPa, pressures that can be generated inside diamond anvil cells are ~320 GPa, pressure in the center of the Earth is 364 GPa, and highest pressures ever achieved in shock waves are over 100,000 GPa. ”
3000 km [3 million meters] of water would be 300,000 atm- not even enough to make diamonds.
320 GPa is 3.1 million atm.
Anyways strange things would happen at + 300K atm. But need 2 gee water planet to make diamonds.
Question is would a planet comprised solely of H2O be cool or hot if surrounded by 1 million Sols?
Expanding on the wacking of a stack of billiard balls: Consider a slinky spring coil toy, if physics departments still have them. Consider a long one hung from some high point. Give it a small bump upwards to the bottom of the spring coil. The displacement that propagates upwards has a definite kinetic energy. You can see the kinetic energy (the displacement) move up the spring, then disappear (for reasons that don’t matter to this idea), and them see the displacement reappear moving down the spring at the same speed that it went up the spring. (If you bump the spring upwards too much, the displacement will reach the top of the spring, it will be reflected, and then it will be a different problem not applicable to this one). To abuse the terminology, you can see a quantity of kinetic energy be subject to gravity, just as when a ball is thrown upwards and then comes back down. The same is true of a very tall stack of billiard balls, except the displacement is too small, and moves too fast, and moves way too far up the stack of billiard balls to be easily watched. None the less, the it is the same.
If the spring coil toy (or the billiard balls) were infinitely stiff (incompressible), the kinetic energy (the displacement) would propagate up the coil (billiard column) infinitely far and infinitely fast. If the spring or the billiard balls were massless, there wouldn’t be any kinetic energy and gravity wouldn’t do anything, so that condition isn’t relevant.
So far as this ongoing question goes; Heat is an imaginary quantity, equilibrium is an imaginary condition, incompressible ideal matter is an imaginary idea. Gravity is real, mass is real, kinetic energy is real, momentum is real, and compressible matter is real. The question is how to formalize the problem of conducting either heat or kinetic energy upwards against gravity. Imaginary heat is not subject to gravity and will always be conducted from warm to cold when a pathway can be imagined. Real kinetic energy is subject to gravity, and needs real particles with real mass, to bridge the distance between warmth and cold.
steve mosher says;
“separation science, especially that surounding the separation of uranium gases in centrifuges, give you all the evidence you need to that this is a load of crap.
blog comments dont overturn working engineering.”
That is correct, but there is an issue of parametric scale, mass difference (force) vs statistical distribution (total energy). Separative efficiency of centrifuges is small, hence the large cascades. The separative efficiency of the Earth’s gravitational atmosphere is very good at one parametric scale, 100% efficient at separating helium form the rest of the atmospheric gases 😉
Ian W says:
January 27, 2012 at 5:34 am
Here’s the key, Ian. Any conduction is sufficient to disprove the theory. Doesn’t matter if the conduction is minimal, or is limited by density. The question is not how much heat will flow. It is if heat will flow at all.
w.
Willis Eschenbach,
“Here’s the key, Ian. Any conduction is sufficient to disprove the theory. Doesn’t matter if the conduction is minimal, or is limited by density. The question is not how much heat will flow. It is if heat will flow at all. w.”
In one of Climate Science’s favorite configurations you have two objects radiating against each other. When they finally equilibrate do they stop radiating??
Myrrh says:
January 27, 2012 at 5:51 am
Thanks, Myrrh. An underlying problem is that of turbulence in the vortex, which costs efficiency. I’ve always suspected that this could be reduced by using laminar air flow on the inlet, but never was able to test that. However, the ugly fact is that at the speeds and radii involved, turbulence will always be present and will cost electricity.
In addition, you need to factor in the inefficiencies involved in compressing and then expanding the air. If you are looking for heat, you could conceivably utilize the heat of compression, or run it with hot compressed air.
But if what you are looking for is heat from (presumably) an electric air compressor by way of a vortex tube … why not use electrical resistance heating? It’s basically 100% efficient energy transformation right where you want it.
There’s no free lunch in the vortex tube. It looks like magic because it splits a stream of ambient temperature compressed air into a hot flow and a cold flow. But there is huge work in compressing the air.
w.