# The Classical Virial Theorem (microcanonical derivation)

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

Theorem: Classical Virial Theorem

The classical virial theorem states that

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

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

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

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
 $$\langle x_i \frac {\partial H}{ \partial x_j}\rangle$$ $$\frac {C}{\Omega(E)} \frac {\partial}{\partial E}\int dx x_i \frac {\partial H}{\partial x_j} \theta (E-H(x))$$ $$\frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x) However, writing $x_i \frac {\partial (H-E)}{\partial x_j} = \frac {\partial}{\partial x_j} \left [x_i(H-E)\right ] - \delta_{ij}(H-E)$ allows the average to be expressed as  \(\langle x_i \frac {\partial H}{ \partial x_j}\rangle$$ $$\frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x) 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:

$\langle x_i \frac {\partial H}{ \partial x_j}\rangle = \frac {C\delta_{ij}}{\Omega(E)} \frac {\partial}{\partial E}\int_{H(x)<E} dx (E-H(x))$

$= \dfrac {C\delta_{ij}}{\Omega(E)} \int_{H(x)<E} dx$

$=\dfrac {\delta_{ij}}{\Omega(E)} \Sigma (E)$

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

$\langle x_i \dfrac {\partial H}{ \partial x_j}\rangle= \delta_{ij} \dfrac {\Sigma(E)}{\frac {\partial \Sigma(E)}{\partial E}}$

$= \delta_{ij} \frac {1}{\dfrac {\partial \ln \Sigma(E)}{\partial E}}$

$= k\delta_{ij} \frac {1}{\dfrac {\partial \tilde{S}}{\partial E}}$

$= kT\delta_{ij}$

which proves the theorem.

Example
Example $$x_i = p_i$$: and $$i = j$$ The virial theorem says that
 $$\langle p_i \frac {\partial H}{ \partial p_j}\rangle$$ $$KT$$ $$\langle \frac {p_i^2}{m_i} \rangle \(kT$$ $$\langle \frac {p_i^2}{2m_i} \rangle$$ $$\frac {1}{2} kT$$

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

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