# 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: Classical Virial Theorem |
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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. |

To prove the theorem, start with the definition of the average:

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

\(\langle x_i \frac {\partial H}{ \partial x_j}\rangle\) | \( \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\int dx x_i \frac {\partial H}{\partial x_j} \theta (E-H(x))\) | ||

\(\frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial H}{\partial x_j}\) | |||

\(\frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial (H-E)}{\partial x_j}\) |

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

\(\langle x_i \frac {\partial H}{ \partial x_j}\rangle\) | \(\frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx \left \{\frac {\partial}{\partial x_j} \left [ x_i(H-E)\right ] + \delta_{ij}(E-H(x))\right \}\) | ||

\( \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\left [ \oint_{H=E} x_i (H-E) dS_j + \delta_{ij} \int_{H<E}d{x} (E-H(x)) \right ]\) |

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 | ||||||||||||
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Example \(x_i = p_i \): and \(i = j\) The virial theorem says that
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 \[\sum_{i=1}^{3N} \langle \frac {p_i^2}{2m_i} \rangle =\sum_{i=1}^{3N} \langle \frac {1}{2}m_i v_i^2 \rangle = \frac {3}{ 2}NkT\] |