I'm reviewing the Feynman Lectures and in them he states that the rotational kinetic energy of a monoatomic molecule in a gas is insignificant due to a small moment of inertia.

But wouldn't the equipartition theorem simply make it so the angular velocity would be large enough to make up for it? I.e. shouldn't the random collisions eventually dump an equal amount of energy into each degree of rotational motion as in each degree of translational motion?


I'm coming at this purely from a classical POV, so please avoid Quantum Mechanical answers if possible.

  • 1
    $\begingroup$ Please edit your question to give a specific reference to a section and page number. I don't think this can be explained properly without quantum mechanics. An atom in a monoatomic gas can't rotate due to symmetry, but Feynman may not say that if quantum mechanics hasn't been introduced yet at that point. $\endgroup$ – user4552 May 28 '19 at 12:13
  • 1
    $\begingroup$ Let's assume we are talking about macroscopic spheres of matter for the sake of argument. As a side note, you say monatomic gas's cant rotate due to symmetry - but don't alpha particles have internal structure and thus should be able to be distinguishable by orientation? $\endgroup$ – charlysotelo May 28 '19 at 12:52
  • 2
    $\begingroup$ Because the quantum energy levels are spaced far enough that the equipartition theorem (a fact of classical physics) doesn’t hold anymore. $\endgroup$ – knzhou May 28 '19 at 13:56
  • 2
    $\begingroup$ @knzhou: It's not just that the energy levels may be widely spaced, it's that they essentially don't exist for a monoatomic gas. Single-atom species don't have rotational bands. (In, e.g., monoatomic hydrogen you can generate angular momentum by electronic excitations, but that doesn't give a rotational band.) $\endgroup$ – user4552 May 28 '19 at 17:13

You are correct, the fact that some group of molecules has much smaller moment of inertia than the rest does not make the group except from the equipartition theorem.

But there is a physical reason for why sometimes certain regions of phase space or degrees of freedom are discarded as non-contributing to total energy. In other words, there is a reason for why equipartition theorem or equal treatment of all degrees of freedom is not accurate.

This reason is "extremely slow equilibration". Some degrees of freedom are in only a very weak interaction with the rest of the system. For example, collisions of spheres do transfer energy from translational to rotational energy, but only very slowly. If the system achieved temperature by quickly accepting energy from outside, some rotational degrees of freedom may not have enough time to acquire the same temperature as the other degrees of freedom. The more smooth the spheres, the longer the time to equilibrate. If this time is astronomical, for some regions of phase space it makes sense to just exclude them from calculations, because the probability they are realized is extremely small so they do not contribute much.

Another similar example: atmospheric gas in a closed vessel with microscopic hole in the wall through which gas molecules escaping to vacuum with only a slow rate. Because of the hole, the phase space available is infinite, and the equilibrium state is that density is close to zero everywhere. But because the hole is so small, the gas in the vessel achieves temporary quasi equilibrium with itself, with some high temperature and high density. To find properties of the gas, only volume of the vessel is important, not the vast space outside it. Again, a part of the phase space is discarded because it is not relevant in the time period considered.

There are other examples: such as oxygen and hydrogen gases in a cold vessel forming water after a long time, or escape of atmosphere molecules from Earth's gravity field. These processes are so slow they are usually ignored.

  • 3
    $\begingroup$ This seems to be answering a different question than OP’s. The answer to OP’s question involves quantum effects. $\endgroup$ – knzhou May 28 '19 at 13:57
  • 2
    $\begingroup$ What would the relaxation time be classically for OPs example? Because if it turns out to be seconds, then this doesn't answer the question, but if it's much longer (how much I'm less sure of) then it does. $\endgroup$ – jacob1729 May 28 '19 at 15:27
  • $\begingroup$ Very interesting... Is there a way to estimate this slow equilibration time? Can you refer me to some resources that discuss this? $\endgroup$ – charlysotelo May 28 '19 at 15:45
  • 2
    $\begingroup$ @DuncanHarris I don't see why that's an issue. The rest of the calculation doesn't rely on knowing the substructure accurately. $\endgroup$ – jacob1729 May 28 '19 at 16:07
  • 1
    $\begingroup$ @charlysotelo I found this paper, but cannot access it: researchgate.net/publication/… $\endgroup$ – Ján Lalinský May 28 '19 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.