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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Roger, Your site is linked on the right column of this page, as always.
For context, Jelbring (2003) is cited by:
Gravitation and Gas Laws: An Alternative Approach to Climatology W Brune – Energy & Environment, 2009 – Multi-Science
and
The thermodynamic relationship between surface temperature and water vapour concentration in the troposphere WC Gilbert – Energy & Environment, 2010 – Multi-Science
Dear Mr. Brown,
I’m sorry but your conclusion about figure 2 is completely wrong as you expect the air reservoir to conduct heat in gravitational field and the silver rod to conduct heat in zero gravity.
I’m sorry to say that this assumption is incorrect and the example you are building your conclusions on is wrong. Even atoms of silver have the property of gaining potential energy and losing kinetic energy (therefore losing temperature) when traveling “up” in a gravitational field while transferring their temperature in that direction. The result is, the rod would demonstrate exactly the same temperature gradient the air container does. Your thermal engine would not work not because there is no temperature gradient but because it would not transfer any heat.
kdk33 says
“if the atmosphere was heated from the top there would be no convection, hence no lapse rate.
The lapse rate doesn’t apply to the ocean because water is incompressible. Hot water doesn’t expand as it rises, hence does not do work on the surroundings, hence does not change temperaure, hence no lapse rate.”
Do you not remember the school experiment with a beaker of water and a single copper sulphate crystal showing fairly rapid liquid convection?
How do you think an electric kettle heats water?
Bryan says:
January 24, 2012 at 6:59 am
For some of us, a thoughtful examination of the flame suffices. We don’t have to actually do the experiment and burn our fingers to know that it is hot.
If you wish to do the experiment, however, be my guest. I’ll just have a quick beer while I wait for your experimental refutation of a couple centuries of scientific thought …
w.
Joules Verne says:
January 23, 2012 at 4:25 pm
Gravity maintains TWO energy gradients. One kinetic and one potential. The kinetic gradient decreases with altitude and the potential gradient increases with altitude. The two opposing gradients cancel out and the column is isogenergetic. This is how you can have a perpetual temperature gradient yet not be able to extract any work from it for a perpetual motion machine – a temperature gradient can be nullified by an equal but opposite gradient of energy in a different form. You can’t connect the cold and hot sides of the atmosphere without climbing up in a gravity well and the useful energy represented by the change in temperature is exactly used up by the energy required to climb uphill against gravity. The books thus balance and conservation of energy is once again safe from the abuses of junk science.
Neat comment Joules. I can’t wait for the mad inventors to put their money where their mouth is and build one of the erroneously designed machines they propose. Problem is, when it fails to work they’ll come to the equally erroneous conclusion that it failed because the atmosphere is isothermal.
Hope no-one catches a nasty chill while bolting thermopiles together at the top of their 10km high rig. While they’re up there, they might notice how much thinner the air is too. That might give a bit of pause for thought about density and its effect on the ability of air packets at different altitudes to retain the heat of the Sun.
Look the adiabatic lapse rate exists this is basic physics equal partition of energy.
it is evident in the atmosphere but is upset by clouds at low level and UV absorption at higher altitudes.
If we cannot get even the basics correct we might as well give up.
For those who say there is no back radiation why is the sky temperature at night not 4K
For those who do not believe an warmer but still cold atmospheric layer cannot cause the surface to warm clearly do not understand the basic physics of radiative heat transfer. Why are layers of mylar used to thermally insulate space craft?
Unless we get the basic physics correct then the skeptical community will be undermined.
A gas giant planet like Saturn radiates more than twice the amount of radiation than it recieves from the Sun.
How do these planets ever die if gravity is constantly maintaining a heat gradient in the atmosphere? The Sun will die and become a cold white dwarf.
What is going to kill these gas giants? If gravity is constantly maintaining hotter gases at the bottom then convection will move gases around and therefore we seem to have an everlasting living planet.
What kills it and when, if Jelbring is correct?
Alan
The last few days have certainly been interesting regarding the debate of lapse rate in a gravity field. I think the point that is getting missed, but I see as intuitively obvious, is that in a real planet with a gas atmosphere, it is not and never can be in a state of thermal equilibrium.
The atmospheric column is always in direct contact with a massive heat sink(source) in the form of the planetary body, and the top of the column (at its effective radiating altitude) is a nearly ?? infinite heat sink at the temperature of interstellar space. Until or unless the planetary body is at the same temperature as deep space there will always be energy input at the bottom of the atmospheric column (and a temperature gradient) and there will always be heat loss by radiation (or some other means like boiling off of the atmosphere) at the top of the column.
In that sense, all these discussions regarding physically impossible constraints on the model are merely debating how many angels can stand on the head of a pin.
Lets look at real planets which are always warmer than interstellar space, and real atmospheres where at the base (troposphere) heat transport by convection dominates all other mechanisms by a huge margin. Under those conditions, there will always be a lapse rate in the atmosphere (in the troposphere) as long as it can find some means to lose energy to interstellar space by radiation or loss of mass.
The only discussion, is are there any realistic conditions where an atmospheric shell could not radiate heat to interstellar space?
If and only if that condition can be achieved do the discussions of a system in equilibrium have any meaning. In all other cases (with or without radiant heating by a near by star) the planet will always be losing heat to space, and will always have a troposphere layer (if it has an atmosphere of non-condensing gases) where heat transport is dominated by convection.
In all cases the presence of green house gases can only improve heat loss to interstellar space at the effective radiating height over the theoretical condition of no ability to radiate energy in the electromagnetic spectrum (note I did not limit energy loss to IR band width).
It is my view that the atmosphere will always find a way to lose energy to deep space by some means at or near its top, and will thus always have a troposphere lapse rate which implies that the planetary body will always be warmer with an atmosphere (regardless of gas mix) than it would be without one.
One other note that no one has mentioned. As I understand it, the average effective radiating altitude here on earth is around 4.5 km (14763 ft), that means that here where I live in Colorado at the summit of Pikes peak or Mt. Elbert I could just about throw a rock to the mean altitude where heat is lost to space by radiation. In short high altitude areas like the Himalayas, Andes, and the Rockies and a few volcanic peaks are “short circuits” in the troposphere, where energy absorbed by the ground from solar heating can be almost immediately radiated back to deep space directly from the ground. It would be interesting to see high resolution IR measurements of the heat loss from these areas directly to outer space.
Larry
I have a question along this topic which will likely show my ignorance. If gravity alone cannot induce a thermal gradient in a gas, how then are stars formed from gases where there is only gravity as an external force?
Let’s do some equation work.
Let’s say we take one mole of N2. That’s 0.028 kg. Now we take that mole at 15 C and move it up from the sea surface to the average top of the troposphere which is 17,000 meters high. Gravity constant will be 9.8 m/s^2. The total joules needed to do this is around 4660 J (E = mgh).
Now the heat capacity of N2 is 29.124 J·mol^−1·K^−1, so if we transformed 4660 J of heat kinetic energy, that would reduce the temp by -160 degrees, or so, or to -145 C. The real temp at the top of the troposhere is about -55 C, so we’re off by almost a factor of 3. Obviously, the atmosphere is also taking in energy from the sun (radiatively, and solar wind), there’s more than just N2, and the conversion of kinetic temperature energy to potential is probably more complex, to make up for this difference. If there was no energy from the sun, the atmosphere would shrink as energy was lost to space, and along with it potential energy, dropping molecules down till they settled on the surface of the planet and solidified (e.g. Pluto).
This is what I see. And again, I may be completely off base. But conservation of energy tells me we must be changing forms as we move up the atmosphere. So for the temperature to stay steady (which it doesn’t in real life), we’ve have to have even more energy at the higher altitudes in the air column than is at the surface. And would not this energy fall back down to the surface to equilibrate, driving the surface temperature even higher? The way I see it, there will always be a -temperature- difference at such large gravity distances if energy is in equilibrium.
Again, I could have done this all wrong.
John Marshall says:
January 24, 2012 at 7:26 am
It’s the disproof of Jelbring’s theory, which lately has attracted many supporters. As such it is an interesting and very relevant exercise.
w.
School physics.
The atmosphere stability condition is ds/dz>0, where s is the entropy density and z is the hight.
Atmosphere with ds/dz 0 are stable (including the isothermal atmosphere!)
What is all the ridiculous discussion about???
Dr. Brown, I’ve been reading this thread and the earlier thread with a great deal of interest and I understand the point you’ve raised. However, I note that it relies on the Ideal Gas Law being applicable.
My question is: Is the Ideal Gas Law applicable?
I’ve just done some quick research and it seems that the derivation of the Ideal Gas Law assumes that any gravitational field in question will have a negligible impact on the behaviour of the gas. That’s true in almost all reasonable cases, but it’s not true in the example we’re discussing. I would therefore ask for your comments on the underlying assumptions in the derivation of the Ideal Gas Law in this respect and if they are still true.
In particular, looking at the Empirical Derivation from the Wikipedia article:
http://en.wikipedia.org/wiki/Ideal_gas_law#Empirical
I noticed that there is a constant C that is directly proportional to the amount of gas. But in a gravitationally stratified column, the amount of gas is not constant throughout the column, and hence the “constant” is not really constant.
Similarly the derivation from statistical mechanics in that article assumes that there is no gravitational force involved. For small columns, that’s a reasonable assumption,Is it a reasonable assumption for large columns?
I don’t understand the theoretical derivation well enough, but again it seems to assume that there is no conversion of kinetic energy into gravitational potential energy – something that is core to this issue.
Again, thank you for your posts. I have enjoyed reading and learning from them.
Robert Brown says at 1/24 8:14am:
“…in an ideal gas the temperature is not determined by the total energy. That’s an absurd idea, given that one can perform a gauge transformation — change the zero of the total energy — without changing any of the physics.”
Ahhh..this is the big deal.
Consider this is equivalent given Caballero’s ref. in the Perpetuum Mobile thread w/gravity where in 2.1 find “Temperature is just another name for the mean kinetic energy density of molecular motion.” to Trick saying “in an ideal gas the mean kinetic energy density is not determined by the total energy”. If Trick said this, Trick would be going against 1st Law & Trick has not written any such thing. So in Trick’s view, what Robert Brown says at 8:14am is inconsistent with 1st Law. I am pretty sure but not perfectly sure that Trick’s view is consistent with all the thermo grand master Laws.
Robert Brown’s 8:14am violation of the 1st Law above, Robert Brown doesn’t grok yet. He will eventually! (Trick predicts the life of the universe is longer) b/c Robert’s smart and the thermo grand master’s are right. This is all like the slow progress of thermal conduction thru a near perfect insulator.
I admit this PE setting with h is actually a difficult preconceived notion to get over. It took me many years of practice to really come to grips with it. Notice that every time you see total energy equation written in thermo theory – it looks something like this: total energy = TE = PE + KE = constant.
That’s all the 1st Law really can do, it doesn’t tell us what that constant equals – it can’t and be general theory. 1st Law tells us (go look!) energy can be neither created or destroyed, but can be changed from one form to another, we have to do the deducing what that means.
The constant specific value is not set by nature; it is not a natural constant. TE is set in the specific experiment. Whatever the total energy in the white space top post is, it has to be constant to obey the 1stnd Law.
In Robert Brown’s top post we actually can just (out of thin air so to speak!) announce that our h=0 to be at the bottom of the white cylinder of gas. Here we know at h=0 that TE=PE +KE. Always. Cite the 1st Law!
At the bottom of the cylinder we just announce PE is 0 there. This is the tough nut to crack to mix my metaphors. PE is a potential energy – as such nature and the thermo grand masters allow this construct.
So we can move on. At Trick’s announced h=0: TE of the molecule(s) will be constant = TE = PE + KE = mgh + 1/2mv^2 = m*g*0 + 1/2mv(0)^2. Voila we have figured the TE constant.
In white area TE = constant = m*g*0 + 1/2mv(0)^2 = 0 + 1/2mv(0)^2 = 1/2mv(0)^2 where v is the velocity and m the mass of the molecule(s) at h=0. The TE constant is ½*m(0)*v(0)^2. KE varies with PE(h) in a gravity field… (can you say non-isothermal?) (Say it again: Deduce from 1st Law: molecule(s) KE or temperature must vary in a gravity field with PE, for total energy to be constant, OMG!)
But we are ok, what do you know, we have a formula consistent with 1st Law that found the total energy constant of the white area*. The constant = 1/2mv(0)^2 comes from finding the v of the molecule at h=0. For that, all we need do is find the KE or temperature (= kinetic energy density) at h=0 and we will know the temperature anywhere else in the one reservoir or one thermodynamic ideal system of the white area (i.e. at any h) from bottom to top thru TE = constant = ½*m(0)* v(0) = (mgh + 1/2mv^2). A thermometer placed at h=0 will work just great for this experiment.
Note 0 location IS arbitrary, I can pick ANY h and put my calibrated thermometer there and find m(h) and v(h) and get the same TE constant, it is just easier to think thru at h=0.
Robert Brown needs to not read FAST but s-l-o-w-l-y to grok this; to really come to terms with it, it will take some practice. The thermo grand masters did a fine job, Robert Brown can do one too.
Another big deal: Robert Brown cannot use, in Trick’s view, 1/24 8:14 statement to argue against Hans Jelbring’s 2003 paper. There may be other arguments but this Robert Brown one doesn’t work with the 1st Law.
Robert Brown continues:
“Is there something miraculously interesting in the thermal contact between silver and air that keeps heat from being conducted from hot to cold — in just this one special circumstance? I’m all ears.”
Nope there isn’t any Trick violation of 0thLaw equilibrium or 2nd Law entropy being allowed ideally constant, either. Robert Brown’s top post just needs to get rid of 1) the second thermodynamic system of the top post – that 2nd system is not in Willis’ Perpetuum Mobile original premise and 2) the perfect ideal insulator for compliance to all three Laws. Then Robert can do the easy math & will grok the big deal. Eventually Trick & Robert Brown will be in (Tallbloke’s) violent agreement.
Trick’s head cold is receding, I won’t be hanging here forever. Robert Brown should try to make this happen in a few days. If not, I predict violent equilibrium will eventually happen, that’s the 0th Law, LOL.
*barring typo’s and consistent with mass conservation, the molecule(s) mass does not change thus can assume E to mc^2 in total energy is justifiably being ignored here by all posters on the original Willis’ premise GHG-free air column.
“Now I can construct a heat engine which extracts useful work based on the temperature gradient and gravity will continue to organize the air column forever and my heat engine will never run out of “fuel”? Really??”
Nope! Impossible. As you took energy out, the molecules would not be able to move up as high in the gravity well (the pressure of the gas would drop), as there wouldn’t be enough energy to go into potential energy, and eventually the gas would condense and settle at the bottom of the container. It would cease being gas.
In short, the temperature at the surface level would steadily decrease as you took out energy, as potential energy and kinetic energy are -interchangable-, but temperature is -only- kinetic energy. And heat can -only- be transferred by kinetic collisions (if we ignore radiation). So your engine would take out the kinetic portion, leaving less to be turned into potential, making the air column decrease in height until what I said above happened. Same as what happens on Pluto.
Wayne2 says:
January 24, 2012 at 7:39 am
Excellent insight, Wayne. That is exactly what happens. In an isothermal column of air, individual molecules at high altitude have more energy because of gravity. But for exactly that same reason, there are fewer molecules at high altitude. As a result, and as we would expect, in the isothermal condition the energy is spread out evenly through space (equal energy per volume) rather than equal energy per molecule as Hans Jelbring and Mr. Verne assert.
w.
Willis Eschenbach says
“If you wish to do the experiment, however, be my guest. I’ll just have a quick beer while I wait for your experimental refutation of a couple centuries of scientific thought ”
Its always better to do the experiment!
If all the qualified scientists here were honest they would admit that this is an example of an experiment where prejudgement was entirely wrong.
http://en.wikipedia.org/wiki/Erasto_Mpemba
I think it is also important to note that Jelbring does not assert thermal equilibrium, they assert “energetic equilibrium” considering both heat energy and gravitational potential energy.
Larry
The iceman cometh says:
January 24, 2012 at 7:58 am
It is the formal disproof of Jelbring’s theory. It is idealized by its very nature.
w.
Thank you Dr. Brown.
A physicist says:
January 24, 2012 at 8:48 am
Although that may be true, I don’t see why, I don’t see Robert Brown making the claim, and I
Wait, wait, you claim to be a physicist, answer the question. Does heat flow forever in the silver wire or not? His ability to answer your random question about possible implications is immaterial to whether his proof is correct.
Does heat flow forever or not?
w.
Bryan/Willis:
I could be missing the whole point here but isn’t the Graeff paper (ref. http://tallbloke.files.wordpress.com/2012/01/graeff1.pdf) an example experiment?
h/t Lucy Skywalker
Robert, you’re making an inappropriate distinction here. When that hypothetical molecule hits another, and they both bounce off in random directions, how is that gravitationally induced motion distinguishable from thermal motion? Answer: they’re the same thing. Gravitationally induced motion within a gas is heat energy.
Robert G. Brown, thank you again for a good presentation and determinedly addressing the criticisms.