Past exam of the kinetic theory course of the University of Oxford / 2024 / 1 / d / Solution

Spatial homogeneity makes $\rho$ independent of position and makes the two-particle number density depend only on the separation. Isotropy removes its directional dependence. ConsequentlyHere $g_2$ is a radial distribution function, with the finite-$N$ normalization inherited from a; statistical independence is unnecessary. In particular, the translated pair in c has separation $\mathbf{R}$ for every $s$, so its $s$ integral is simply ${\rho }^2{g}_2(R,t)$.

Isotropy of the random velocity $\mathbf{c}=\mathbf{v}-\mathbf{u}$ makes its second velocity moment proportional to the identity. Taking its trace fixes the proportionality constant:Thus $\theta$ is the one-component random velocity variance; for unit mass in thermal equilibrium, $\theta =k_{B}T$. A Maxwell-Boltzmann velocity distribution is not required for the pressure tensor to be isotropic.

For the interaction contribution, the angular integral is $\int d\Omega {\hat{R}}_i{\hat{R}}_j=(4\pi /3){\delta }_{ij}$, by isotropy and its trace. With $d^{3}R=R^{2}dRd\Omega$, c becomesCombining the two pressure tensors gives the virial pressure from pair correlations:For a repulsive pair potential, the outward radial force is $-{\varphi }^{\prime }(R)\ge 0$. Since $g_2\ge 0$, the second term in $p$ is nonnegative, and positive when occupied separations feel nonzero repulsion. It expresses additional momentum transfer through interactions. The displayed integral assumes a differentiable pair potential and convergence; a hard core requires a contact formula or a limiting argument.