6
$\begingroup$

I'm trying to understand the proof of the Generalized Equipartition Theorem in the Microcanonical Ensemble. The proof that appears in Wikipedia is the same that can be found in Kerson Huang's Statistical Mechanics (page 137 of the 2nd edition).

There is a step that I am not able to understand. After getting to the following expression

\begin{aligned}\int _{H<E}x_{m}{\frac {\partial (H-E)}{\partial x_{n}}}\,d\Gamma &=\int _{H<E}{\frac {\partial }{\partial x_{n}}}{\bigl (}x_{m}(H-E){\bigr )}\,d\Gamma -\int _{H<E}\delta _{mn}(H-E)d\Gamma \\&=\delta _{mn}\int _{H<E}(E-H)\,d\Gamma ,\end{aligned}

both Huang and Wikipedia state the following: "The first term on the right hand side of the first line is zero (it can be rewritten as an integral of $H − E$ on the hypersurface where $H = E\,$)".

I suppose that they are using Gauss' divergence theorem to rewrite the integral of the divergence of a vector field over a (hyper)volume as the integral of the vector field over the (hyper)surface that bounds the (hyper) volume. But I do not see how the integrand

\begin{aligned}\frac {\partial {x_{m}(H-E)}}{\partial x_{n}}\end{aligned}

can be seen as a divergence.

$\endgroup$

1 Answer 1

6
$\begingroup$

It's not strictly a divergence, but it can be rewritten as a boundary term which vanishes. Let $M$ denote the region for which $H \leq E$. It is known that $$ \int_M \vec{\nabla} \psi \, d^N V = \oint_{\partial M} \psi \, d^{N-1} \vec{S}, $$ or, looking at the $n$-component of this, $$ \int_M \frac{\partial \psi}{\partial x_n} \, d^N V = \oint_{\partial M} \psi \, d^{N-1} S_n. $$ Choosing $\psi = x_m (H - E)$ then lets us conclude that $$ \int_M \frac{\partial [x_m (H-E)]}{\partial x_n} \, d^N V = \oint_{\partial M} x_m (H - E) \, d^{N-1} S_n, $$ which (as you note) vanishes on the boundary of the region $M$ because $H = E$ there.


As an aside, the identity I started with above does follow from the divergence theorem as you know it. If $\vec{c}$ is any constant vector field, then we have \begin{align*} \vec{c} \cdot \left[ \int_{M} \vec{\nabla} \psi \, dV \right] &= \int_{M} \vec{c} \cdot \vec{\nabla} \psi \, dV \\ &= \int_{M} \vec{\nabla} \cdot (\psi \vec{c}) \, dV & \text{(since $\vec{\nabla} \cdot \vec{c} = 0$)} \\ &= \oint_{\partial M} \psi \vec{c} \cdot d\vec{S} & \text{(divergence theorem)} \\ &= \vec{c} \cdot \left[ \oint_{\partial M} \psi \, d\vec{S} \right]. \end{align*} Since this identity holds true for any constant vector field $\vec{c}$, then the vectors in square brackets at the start and at the end must be equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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