5
$\begingroup$

In all derivations of the equipartition theorem I can find a thermodynamic equilibrium distribution is used to show it's validity.

But more vague sources (physics.stackexchange answer by Luboš Motl, wikipedia, Hyperphys) state that only equilibrated temperature with surroundings is required.

To further stress it:

From wikipedia:

"More generally, it can be applied to any classical system in thermal equilibrium"

"Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat."

"Systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true."

Thus, how can the equipartition theorem be shown without assuming thermodynamic equilibrium?

Maybe an explanation of why this is important:

In deriving the Stokes Einstein relation: $$\gamma D = k_B T $$ ($\gamma=$ friction coeff., $D=$ diffusion const.) In some derivations the thermodynamic equilibrium is used and in some just the equipartition theorem.

If equipartition is not a thermodynamic equilibrium phenomenon then the relation holds in a much wider variety of cases. Basically it could not be used to determine if thermodynamic equilibrium is at hand by measuring $D$, $\gamma$ and $T$ which I was told is an important applicaiton of the relation.

$\endgroup$
  • $\begingroup$ where have you seem a demonstration that requires thermodynamical equilibrium? (which is different than showing that the ET is valid in examples that also happen to be in thermodynamic equilibrium) $\endgroup$ – Wolphram jonny Jun 17 '16 at 20:25
  • $\begingroup$ well in every proof, i have seen, a thermodynamic equilibrium is required including the examples in the question. Basically what you are asking is my question. Is there a proof that does not require thermodynamic equilibrium? $\endgroup$ – pindakaas Jun 19 '16 at 16:59
  • $\begingroup$ I'll try to find one online, unfortunately do not have my books on statistical mechanics here $\endgroup$ – Wolphram jonny Jun 19 '16 at 17:01
  • $\begingroup$ Here is one: sbfisica.org.br/bjp/files/v30_176.pdf $\endgroup$ – Wolphram jonny Jun 19 '16 at 17:17
  • $\begingroup$ I am not sure this paper actually proves that: "Summing up, we have shown that Hamiltonian dynamical systems decribed by phase space distributions depending only on the Hamiltonian comply with the strictures of a generalized energy equipartition theorem. Homogeneous terms in the Hamiltonian exhibiting the same degree of homogeneity and depending on the same number of canonical variables give the same contribution to the total mean energy. In the case of Gibbs canonical ensemble our version of the theorem reproduces the standard well-known results." $\endgroup$ – pindakaas Jun 20 '16 at 7:19

Your Answer

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

Browse other questions tagged or ask your own question.