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

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:Thus $\int fdz=N$ and $\int f_2(z_1,z_2)d{z}_2=(N-1)f(z_1)$. For unit particle mass, $f$ also carries mass density. These conventions explain why no extra factor $N-1$ occurs in the result.

Separate the Hamiltonian as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, whereand $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\dots ,N$. The stated Poisson bracket gives Hamilton's equationsConservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equationAssume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over $z_2,\dots ,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by integration by parts. The terms in $H_{\mathrm{int}}$ differentiating an integrated velocity also vanish. Indistinguishability makes the remaining $N-1$ terms equal, soRelabeling the two particles gives the first BBGKY hierarchy equation:This is exact for a differentiable pair potential; no independent-particle closure has been made.

Take the zeroth velocity moment of a. The right side is a velocity space divergence with vanishing boundary contribution, giving the continuity equationFor the first velocity moment, use integration by parts componentwise:Write $\mathbf{c}=\mathbf{v}-\mathbf{u}$; then $\int \mathbf{c}f{d}^{3}v=0$, and the momentum flux splits into bulk transport and the pressure tensor:Thereforewhere repeated Cartesian indices are summed and $n_2=\int f_2{d}^{3}v{d}^3{v}_{\star }$. The negative sign is the attractive or repulsive force $-\nabla \varphi$, rather than an additional sign convention for the pressure tensor.

For fixed $\mathbf{R}$ and $t$, define $h(\mathbf{x},s)=n_2(\mathbf{a},\mathbf{b},t)$ with $\mathbf{a}=\mathbf{x}+(s-1)\mathbf{R}$ and $\mathbf{b}=\mathbf{x}+s\mathbf{R}$. The chain rule givesso $\boxed{{\partial }_{s}h=\mathbf{R}\cdot {\nabla }_{\mathbf{x}}h}$. The derivative with respect to $\mathbf{x}$ here translates both arguments of the two-particle number density; it is not a derivative of its first argument alone.

Take the spatial divergence of the proposed interaction pressure tensor. Its pair potential factor is independent of $\mathbf{x}$, so the preceding identity givesExchange symmetry of the two-particle distribution function implies $n_2(\mathbf{x}-\mathbf{R},\mathbf{x})=n_2(\mathbf{x},\mathbf{x}-\mathbf{R})$. In the second term, substitute $\mathbf{R}\mapsto -\mathbf{R}$. Since $R_i{\varphi }^{\prime }(R)/R$ is odd, both terms become equal:On the other hand, at fixed ${\mathbf{x}}_{\star }=\mathbf{x}+\mathbf{R}$, the first-argument derivative of the pair potential is ${\nabla }_{\mathbf{x}}\varphi =-\mathbf{R}{\varphi }^{\prime }(R)/R$. Thus the interaction force density in b is $-{\partial }_{x_j}{P}_{ij}^{\varphi }$. Moving it to the left yields local conservation of momentum:The integral over $s$ distributes each pair's momentum flux along the straight segment joining the particles. The factor $1/2$ removes the double counting of ordered pairs. Central forces also make $P_{ij}^{\varphi }=P_{ji}^{\varphi }$.

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.