The generalized equipartition theorem (where variables need not be quadratic) 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 $\rho(p,q)$:

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

In the most general case this probability density is the canonical ensemble's. For the theorem to hold ergodicity is also required. However, I'm having trouble finding a rigorous prove where the assumptions are explicitly used in the derivation.

Could you provide such a prove, or a reference to a paper/book where it can be found?

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    $\begingroup$ The proof is essentially identical to the proof of the standard equipartition theorem. What about the proof on the wiki page are you not satisfied by? The assumption of equilibrium and erodicity are simply used to obtain the form of $\rho(p,q)$, which will be covered in book on Stat Mech which you find sufficiently rigorous. $\endgroup$ Commented Feb 5, 2017 at 16:36
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    $\begingroup$ Have you looked at the Wikipedia article and the links therein? en.wikipedia.org/wiki/Equipartition_theorem $\endgroup$
    – Farcher
    Commented Feb 5, 2017 at 16:38
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    $\begingroup$ I have read the proof on the wikipedia page, but I don't see clearly where ergodicity is assumed. I imagine this is related with the Gibbs distribution in the canonical ensemble; however, I'm not an expert in the field, and that's the reason I asked ;) $\endgroup$
    – nabla
    Commented Feb 5, 2017 at 20:32
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    $\begingroup$ If you take a canonical ensemble, the ergodicity is already implied. $\endgroup$
    – Roger V.
    Commented Oct 11, 2021 at 4:59
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    $\begingroup$ > "In the most general case this probability density is the canonical ensemble's. For the theorem to hold ergodicity is also required." -- These two claims seem to be wrong. Canonical ensemble probability is not "the most general", it is simply the appropriate distribution for the canonical ensemble constraints (fixed volume, temperature). Ergodicity is sometimes suggested as important when time averages are to be equal to ensemble averages, but relevance of ergodicity is controversial since that equality is not a necessary part of statistical physics. $\endgroup$ Commented Jun 13, 2022 at 14:32

2 Answers 2


The generalized equipartition theorem is derived in Section 6.4 of the famous Huang's Statistical Mechanics book (1987, 2nd edition).

In order to prove the "generalized equipartition theorem", Huang uses the microcanonical ensemble, which is the "standard" choice for systems that can be regarded as isolated (in the sense that the energy is a constant of the motion). This "standard" choice of the microcanonical is somehow justified if we believe that the system is ergodic, or if we assume the ergodic hypothesis. In other words, to prove equipartition, ergodicity may only be needed as a very first step to justify the use of the microcanonical ensemble in the first place, see e.g. this, this and this answers. Is ergodicity necessary?

Let me stress that Huang does not invoke the ergodic theorem to justify the microcanonical ensemble, even though it seems to be the justification we need: this is discussed in Section 4.5 of the same book. In fact, another way to look at the microcanonical ensemble is the principle of indifference, as proposed by Jaynes in "Information and Statistical Mechanics", see also this and this answers. Jaynes' argument provides a justification for the use of the microcanonical ensemble (not its "realism") that is alternative (or complementary) to the ergodic hypothesis.

In other words: if our isolated system is ergodic, then the microcanonical ensemble can (in principle) be realized, see e.g. the historical review by Gallavotti. If we are not sure about the ergodicity of the system, we can still use the microcanonical ensemble via Jaynes' argument, even though it may not be physically realized. There seems to be no other connection between the equipartition theorem and ergodicity other than this, quite indirect, link. I may be wrong, but it seems to me that there is the same (indirect) connection between "ergodicity" and any possible result pertaining to the equilibrium state, see e.g. this answer.


J. A. S. Lima and A. R. Plastino published the article On the classical energy equipartition theorem back in 1999. In their article they derive a generalized equipartition theorem. Their generalized approach is valid for systems with arbitrary distribution functions and for systems with non-quadratic terms in the Hamiltionian. A link to their article is https://doi.org/10.1590/S0103-97332000000100019. This article should answer most of your questions.

Their article can be summarized as follows:

Suppose there is a system with $f$ degrees of freedom and the Hamiltonian is \begin{equation} \mathcal{H} = g(x_1,...,x_L) + h. \end{equation} $(x_1,...,x_L)$ is a subset of the phase-space coordinates for positions and momentas. $g$ is homogeneous such that

\begin{equation} g\left(\lambda x_{1}, \ldots, \lambda x_{L}\right)=\lambda^{r} g\left(x_{1}, \ldots, x_{L}\right). \end{equation} For systems that are distributed according to a Boltzmann distribution you can derive the expression

\begin{equation} \langle g\rangle=\frac{L}{r} k_B T. \end{equation} Here $k_B$ is the Boltzmann constant and $T$ is the temperature. This is the generalized form of the equipartition theorem that you are looking for. For example, for a monovalent ideal gas you have $r=2$ and $L=3N$ such that you recover the famous result $U = \frac{3}{2}N k_B T$ where U is the internal energy of the gas.


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