Feynman explains how gravitational potential energy and kinetic energy convert to create the gravito-thermal greenhouse effect, without greenhouse gases

Excerpts from The Feynman Lectures, Chapter 40, The Principles of Statistical Mechanics, which prove the Maxwell/Clausius/Carnot gravito-thermal greenhouse effect is correct, and would occur in a pure atmosphere containing only the non-greenhouse gases N2 & O2 (99.94% of our atmosphere)(bolding and [additions] added):

If the temperature is the same at all heights, the problem is to discover by what law the atmosphere becomes tenuous as we go up. If N is the total number of molecules in a volume V of gas at pressure P, then we know PV=NkT, or P=nkT, where n=N/V is the number of molecules per unit volume. In other words, if we know the number of molecules per unit volume, we know the pressure, and vice versa: they are proportional to each other, since the temperature is constant in this problem. But the pressure is not constant, it must increase as the altitude is reduced, because it has to hold, so to speak, the weight of all the gas above it. That is the clue by which we may determine how the pressure changes with height. If we take a unit area at height h, then the vertical force from below, on this unit area, is the pressure P. The vertical force per unit area pushing down at a height h+dh would be the same, in the absence of gravity, but here it is not, because the force from below must exceed the force from above by the weight of gas in the section between h and h+dh. Now mg is the force of gravity on each molecule, where gis the acceleration due to gravity, and ndh is the total number of molecules in the unit section. So this gives us the differential equation Ph+dh−Ph= dP= −mgndh. Since P=nkT, and T is constant, we can eliminate either P or n, say P, and get

dndh=−mgkTn

for the differential equation, which tells us how the density goes down as we go up in energy.

We thus have an equation for the particle density n, which varies with height, but which has a derivative which is proportional to itself. Now a function which has a derivative proportional to itself is an exponential, and the solution of this differential equation is

n=n0e−mgh/kT.(40.1)

Here the constant of integration, n0, is obviously the density at h=0 (which can be chosen anywhere), and the density goes down exponentially with height.

Fig. 40–2.The normalized density as a function of height in the earth’s gravitational field for oxygen and for hydrogen, at constant temperature.

Note that if we have different kinds of molecules with different masses, they go down with different exponentials. The ones which were heavier would decrease with altitude faster than the light ones. Therefore we would expect that because oxygen is heavier than nitrogen, as we go higher and higher in an atmosphere with nitrogen and oxygen the proportion of nitrogen would increase. This does not really happen in our own atmosphere, at least at reasonable heights, because there is so much agitation which mixes the gases back together again. It is not an isothermal atmosphere. Nevertheless, there is a tendency for lighter materials, like hydrogen, to dominate at very great heights in the atmosphere, because the lowest masses continue to exist, while the other exponentials have all died out (Fig. 40–2).

40–2The Boltzmann law

Here we note the interesting fact that the numerator in the exponent of Eq. (40.1) is the [gravitational] potential energy of an atom. Therefore we can also state this particular law as: the density at any point is proportional to

e−the potential energy of each atom/kT.

That may be an accident, i.e., may be true only for this particular case of a uniform gravitational field. However, we can show that it is a more general proposition. Suppose that there were some kind of force other than gravity acting on the molecules in a gas. For example, the molecules may be charged electrically, and may be acted on by an electric field or another charge that attracts them. Or, because of the mutual attractions of the atoms for each other, or for the wall, or for a solid, or something, there is some force of attraction which varies with position and which acts on all the molecules. Now suppose, for simplicity, that the molecules are all the same, and that the force acts on each individual one, so that the total force on a piece of gas would be simply the number of molecules times the force on each one. To avoid unnecessary complication, let us choose a coordinate system with the x-axis in the direction of the force, F.

In the same manner as before, if we take two parallel planes in the gas, separated by a distance dx, then the force on each atom, times the n atoms per cm³ (the generalization of the previous nmg), times dx, must be balanced by the pressure change: Fndx=dP=kTdn. Or, to put this law in a form which will be useful to us later,

F=kTddx(lnn).(40.2)

For the present, observe that −Fdx is the work we would do in taking a molecule from x to x+dx, and if F comes from a potential, i.e., if the work done can be represented by a [gravitational] potential energy at all, then this would also be the difference in the [gravitational] potential energy (P.E.). The negative differential of [gravitational] potential energy is the work done, Fdx, and we find that d(lnn)=−d(P.E.)/kT, or, after integrating,

n=(constant)e−P.E./kT.(40.3)

Therefore what we noticed in a special case turns out to be true in general. (What if F does not come from a potential? Then (40.2) has no solution at all. Energy can be generated, or lost by the atoms running around in cyclic paths for which the work done is not zero, and no equilibrium can be maintained at all. Thermal equilibrium cannot exist if the external forces on the atoms are not conservative.) Equation (40.3), known as Boltzmann’s law, is another of the principles of statistical mechanics: that the probability of finding molecules in a given spatial arrangement varies exponentially with the negative of the potential energy of that arrangement, divided by kT.

This, then, could tell us the distribution of molecules: Suppose that we had a positive ion in a liquid, attracting negative ions around it, how many of them would be at different distances? If the potential energy is known as a function of distance, then the proportion of them at different distances is given by this law, and so on, through many applications...

40–4The distribution of molecular speeds

Now we go on to discuss the distribution of velocities, because sometimes it is interesting or useful to know how many of them are moving at different speeds. In order to do that, we may make use of the facts which we discovered with regard to the gas in the atmosphere. We take it to be a perfect gas, as we have already assumed in writing the potential energy, disregarding the energy of mutual attraction of the atoms. The only potential energy that we included in our first example was gravity. We would, of course, have something more complicated if there were forces between the atoms. Thus we assume that there are no forces between the atoms and, for a moment, disregard collisions also, returning later to the justification of this. Now we saw that there are fewer molecules at the height h than there are at the height 0; according to formula (40.1), they decrease exponentially with height. How can there be fewer at greater heights? After all, do not all the molecules which are moving up at height 0 arrive at h? No!, because some of those which are moving up at 0 are going too slowly, and cannot climb the potential hill to h. With that clue, we can calculate how many must be moving at various speeds, because from (40.1) we know how many are moving with less than enough speed to climb a given distance h. Those are just the ones that account for the fact that the density at h is lower than at 0...Since velocity and momentum are proportional, we may say that the distribution of momenta is also proportional to 

e−K.E./kT per unit momentum range. It turns out that this theorem is true in relativity too, if it is in terms of momentum, while if it is in velocity it is not, so it is best to learn it in momentum instead of in velocity:

f(p)dp=Ce−K.E./kTdp.(40.8)

So we find that the probabilities of different conditions of energy, kinetic and potential, are both given by e−energy/kT, a very easy thing to remember and a rather beautiful proposition.

So far we have, of course, only the distribution of the velocities “vertically.” We might want to ask, what is the probability that a molecule is moving in another direction? Of course these distributions are connected, and one can obtain the complete distribution from the one we have, because the complete distribution depends only on the square of the magnitude of the velocity, not upon the z-component. It must be something that is independent of direction, and there is only one function involved, the probability of different magnitudes. We have the distribution of the z-component, and therefore we can get the distribution of the other components from it. The result is that the probability is still proportional to e−K.E./kT, but now the kinetic energy involves three parts, mv2x/2, mv2y/2, and mv2z/2, summed in the exponent. Or we can write it as a product:

f(vx,vy,vz)dvxdvydvz∝e−mv2x/2kT⋅e−mv2y/2kT⋅e−mv2z/2kTdvxdvydvz.(40.9)

You can see that this formula must be right because, first, it is a function only of v2, as required, and second, the probabilities of various values of vz obtained by integrating over all vx and vy is just (40.7). But this one function (40.9) can do both those things!

40–5The specific heats of gases

Now we shall look at some ways to test the theory, and to see how successful is the classical theory of gases. We saw earlier that if U is the internal energy of N molecules, then PV= NkT= (γ−1)U holds, sometimes, for some gases, maybe. If it is a monatomic gas, we know this is also equal to 

23 of the kinetic energy of the center-of-mass motion of the atoms. If it is a monatomic gas, then the kinetic energy is equal to the internal energy, and therefore γ−1=23. But suppose it is, say, a more complicated molecule, that can spin and vibrate, and let us suppose (it turns out to be true according to classical mechanics) that the energies of the internal motions are also proportional to kT. Then at a given temperature, in addition to kinetic energy 32kT, it has internal vibrational and rotational energies. So the total U includes not just the kinetic energy, but also the rotational and vibrational energies, and we get a different value of γ. Technically, the best way to measure γ is by measuring the specific heat, which is the change in energy with temperature. We will return to that approach later. For our present purposes, we may suppose γ is found experimentally from the PVγ curve for adiabatic compression...

40–6The failure of classical physics

So, all in all, we might say that we have some difficulty. We might try some force law other than a spring, but it turns out that anything else will only make γ higher. If we include more forms of energy, γ approaches unity more closely, contradicting the facts. All the classical theoretical things that one can think of will only make it worse. The fact is that there are electrons in each atom, and we know from their spectra that there are internal motions; each of the electrons should have at least 12kT of kinetic energy, and something for the potential energy, so when these are added in, γ gets still smaller. It is ridiculous. It is wrong.

The first great paper on the dynamical theory of gases was by Maxwell in 1859. On the basis of ideas we have been discussing, he was able accurately to explain a great many known relations, such as Boyle’s law, the diffusion theory, the viscosity of gases, and things we shall talk about in the next chapter. He listed all these great successes in a final summary, and at the end he said, “Finally, by establishing a necessary relation between the motions of translation and rotation (he is talking about the 12kT theorem) of all particles not spherical, we proved that a system of such particles could not possibly satisfy the known relation between the two specific heats.” He is referring to γ (which we shall see later is related to two ways of measuring specific heat), and he says we know we cannot get the right answer...

Also see Feynman's lecture 42 in which he states,

"It is like the atmosphere in equilibrium under gravity, where the gas at the bottom is denser than that at the top because of the work mghneeded to lift the gas molecules to the height h."

."

n=(constant)e−P.E./kT