# 3.3: The Classical Virial Theorem (Microcanonical Derivation)

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Consider a system with Hamiltonian $$H (x)$$. Let $$x_i$$ and $$x_j$$ be specific components of the phase space vector.

##### Theorem $$\PageIndex{1}$$: Classical Virial Theorem

The classical virial theorem states that

$\langle x_i \dfrac {\partial H}{ \partial x_j}\rangle = kT\delta_{ij} \nonumber$

where the average is taken with respect to a microcanonical ensemble.

$\langle x_i \dfrac {\partial H}{\partial x_j}\rangle = \dfrac {C}{\Omega (E)} \int dx x_i \dfrac {\partial H}{\partial x_j} \delta(E-H(x)) \nonumber$

where the fact that $$\delta (x) = \delta (-x)$$ has been used. Also, the $$N$$ and $$V$$ dependence of the partition function have been suppressed. Note that the above average can be written as

\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\int dx x_i \frac {\partial H}{\partial x_j} \theta (E-H(x)) \\[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial H}{\partial x_j} \\[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial (H-E)}{\partial x_j} \end{align*}

However, writing

$x_i \frac {\partial (H-E)}{\partial x_j} = \frac {\partial}{\partial x_j} \left [x_i(H-E)\right ] - \delta_{ij}(H-E) \nonumber$

allows the average to be expressed as

\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx \left \{\frac {\partial}{\partial x_j} \left [ x_i(H-E)\right ] + \delta_{ij}(E-H(x))\right \} \\[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\left [ \oint_{H=E} x_i (H-E) dS_j + \delta_{ij} \int_{H<E}d{x} (E-H(x)) \right ] \end{align*}

The first integral in the brackets is obtained by integrating the total derivative with respect to $$x_j$$ over the phase space variable $$x_j$$. This leaves an integral that must be performed over all other variables at the boundary of phase space where $$H = E$$, as indicated by the surface element $$dS_j$$. But the integrand involves the factor $$H - E$$, so this integral will vanish. This leaves:

\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C\delta_{ij}}{\Omega(E)} \frac {\partial}{\partial E}\int_{H(x)<E} dx (E-H(x)) \\[4pt] &= \dfrac {C\delta_{ij}}{\Omega(E)} \int_{H(x)<E} dx \\[4pt] &=\dfrac {\delta_{ij}}{\Omega(E)} \Sigma (E) \end{align*}

where $$\Sigma (E)$$ is the partition function of the uniform ensemble. Recalling that $$\Omega(E) = \frac {\partial}{\partial E} \Sigma (E)$$ we have

\begin{align*} \langle x_i \dfrac {\partial H}{ \partial x_j}\rangle &= \delta_{ij} \dfrac {\Sigma(E)}{\frac {\partial \Sigma(E)}{\partial E}} \\[4pt] &= \delta_{ij} \frac {1}{\dfrac {\partial \ln \Sigma(E)}{\partial E}} \\[4pt] &= k\delta_{ij} \frac {1}{\dfrac {\partial \tilde{S}}{\partial E}} \\[4pt]&= kT\delta_{ij} \end{align*}

which proves the theorem.

##### Example $$\PageIndex{1}$$

$$x_i = p_i$$: and $$i = j$$ The virial theorem says that

$\langle p_i \frac {\partial H}{ \partial p_j}\rangle = kT \nonumber$

$\langle \frac {p_i^2}{m_i} \rangle = kT \nonumber$

$\langle \frac {p_i^2}{2m_i} \rangle = \frac {1}{2} kT \nonumber$

Thus, at equilibrium, the kinetic energy of each particle must be $$\frac {kT}{2}$$. By summing both sides over all the particles, we obtain a well know result

$\sum_{i=1}^{3N} \langle \frac {p_i^2}{2m_i} \rangle =\sum_{i=1}^{3N} \langle \frac {1}{2}m_i v_i^2 \rangle = \frac {3}{ 2}NkT \nonumber$

This page titled 3.3: The Classical Virial Theorem (Microcanonical Derivation) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark E. Tuckerman.