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I was wondering what are the sufficient conditions for the Equipartition Theorem. I know there is another question (For which systems is the equipartition theorem valid?) that somewhats answers this question, but it only addresses the need for the system to be in thermal equilibrium.

However, I was wondering if the system has to be ergodic as well. I've read this in several places before however I haven't been able to find a good explanation or derivation of the Equipartition Theorem that mentions the system must be ergodic. I was wondering if this actually is true and if so, could anyone direct me to an explanation of why or provide an original one. Thanks.

Community
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MonkSphere
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2 Answers2

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The generalized equipartition theorem states that if $x_i$ is a canonical variable (position or momentum variable), then

$$\left\langle x_i \frac{\partial \mathcal{H}}{\partial x_j}\right\rangle = \delta_{ij}\ k T$$

where the average $\langle \cdot \rangle$ is taken over an equilibrium probability density $\rho(p,q)$:

$$\langle f(p,q) \rangle = \int dp dq \ \rho(p,q) \ f(p,q)$$

This probability density can for example be the canonical probability density (for a system at fixed $N,V,T$)

$$\rho_c(p,q)_ = \frac{e^{-\beta \mathcal H(p,q)}}{N! h^{3N} Z}$$

or the microcanonical probability density (fixed $N,V,E$):

$$\rho_{mc}(p,q) = \delta(\mathcal H(p,q)-E)$$

The theorem is indeed not valid if the system is not ergodic. You can find a good explanation of why this is the case here.

valerio
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The equipartition theorem says that if the system is in contact with thermal reservoir of temperature $T$ and has Hamiltonian description with Hamiltonian being a sum of terms, one of which is quadratic function of some canonical variable $q_c$ ($aq_c^2$, where $a$ is constant), then the expected average value of $aq_c^2$ is $k_B T/2$.

This theorem follows from the above assumptions and the assumption that

$$ \rho(q,p) = \frac{e^{-H(q,p)/(k_BT)}}{Z} $$

is the probability density for the values of coordinates and momenta.

Whether the last assumption is right does not depend so much on the kind of system, but more on the situation the system is set up in. If the system interacted long time with much greater system that is macroscopically in equilibrium, then this function is the best assumption for most systems with short-range interaction.

For systems with long-range interaction (gravity), this function is not justifiable and consequently the equipartition theorem cannot be derived.

Ján Lalinský
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