# Is Feynman talking about the zeroth law of thermodynamics?

In Volume 1 Chapter 39 of the Feynman Lectures on Physics, Feynman derives the ideal gas law from Newton's laws of motion. But then on page 41-1, he puts a caveat to the derivation he has just completed (italics in original):

Incidentally, when we say that the mean kinetic energy of the particle is $\frac{3}{2}kT$, we claim to have derived this from Newton's laws.... and it is most interesting that we can apparently get so much from so little ... How do we get so much out? The answer is that we have been perpetually making a certain important assumption, which is that if a given system is in thermal equilibrium at some temperature, it will also be in thermal equilibrium with anything else at the same temperature. For instance, if we wanted to see how a particle would move if it was really colliding with water, we could imagine that there was a gas present, composed of another kind of particle, little fine pellets that (we suppose) do not interact with water, but only hit the particle with "hard" collisons. Suppose that the particle has a prong sticking out of it; all our pellets have to do is hit the prong. We know all about this imaginary gas of pellets at temperature $T$ - it is an ideal gas. Water is complicated, but an ideal gas is simple. Now, our particle has to be in equilibrium with the gas of pellets. Therefore, the mean motion of the particle must be what we get for gaseous collisions, because if it were not moving at the right speed relative to the water but, say, was moving faster, that would mean that the pellets would pick up energy from it and get hotter than the water. But we had started them at the same temperature, and we assume that if a thing is once in equilibrium it stays in equilibrium - parts of it do not get hotter and other parts colder, spontaneously. This proposition is true and can be proved from the laws of mechanics, but the proof is very complicated and can be established only by using advanced mechanics. It is much easier to prove in quantum mechanics than it is in classical mechanics. It was first proved by Boltzmann, but now we simply take it to be true, and then we can argue that our particle has to have $\frac{3}{2}kT$ of energy if it is hit with artificial pellets, so it also must have $\frac{3}{2}kT$ when it is being hit with water at the same temperature and we take away the pellets; so it is $\frac{3}{2}kT$. It is a strange line of argument, but perfectly valid.

My question is, what is the proposition by Boltzmann that Feynman is referring to? I can think of three possibilities:

1. It could refer to the Equipartition Theorem, which was independently proven by Maxwell and Boltzmann, since the title of the section is "The Equipartition of Energy".

2. It could refer to the Zeroth Law of Thermodynamics, because "it will also be in thermal equilibrium with anything else" (especially when he emphasizes the "anything else") sounds like the fact that thermal equilibrium is an equivalence relation.

3. It could refer to Boltzmann's H-theorem, which was the derivation of the second law of thermodynamics from Newton's laws, because "if a thing is once in equilibrium it stays in equilibrium - parts of it do not get hotter and other parts colder, spontaneously" sounds like some formulations of the second law.

So which of the three is it, if it's any of the three, and how would you prove it from Newton's laws?

• I did not read Feynman, but based on the paragraph you quoted it looks like H-theorem, it was proved using classical mechanics and an indispensable assumption containing irreversibililty,i.e. the assumption of molecular chaos. In the quantum case the irreversibility comes from random phase assumption or decoherence. Either case H-theorem can't be derived solely form the dynamical equations. – Jia Yiyang Dec 27 '13 at 4:05
• Since you linked the wiki page, I just want to point out the quantum mechanical derivation given by the wiki page is at least misleading, if not wrong. Using Fermi golden rule as an approximation is neither the key of the derivation nor the source of irreversibility, you can check out Weinberg's QFT vol 1 for a derivation using only the unitarity of time evolution, while in fact the decoherence is assumed implicitly. – Jia Yiyang Dec 27 '13 at 4:20
• @JiaYiyang But doesn't "if a given system is in thermal equilibrium at some temperature, it will also be in thermal equilibrium with anything else at the same temperature" sound more like the zeroth law? You can look at the context here, by the way: www.feynmanlectures.caltech.edu/I_41.html – Keshav Srinivasan Dec 27 '13 at 7:00
• @Jia Yiyang: The golden rule does bring irreversibility into the derivation. Why do you think otherwise? Could you please give detailed reference ? (it is hard to wade through the whole book). – Ján Lalinský Dec 27 '13 at 10:09
• I do not know Boltzmann very well, however Gibbs' Statistical Mechanics does a pretty good job of proving thermal equilibrium as a generic thing (Zeroth law): page 36 (starting "The modulus Θ has properties analogous to those of temperature in thermodynamics.". – Nanite Feb 21 '14 at 22:32