Past exam of the kinetic theory course of the University of Oxford / 2025 / 1

Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one and write $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$, $d{z}_i=d^3{x}_i{d}^3{v}_i$. The number-normalized one- and two-particle distribution functions areThus $\int fdz=N$ and $\int f_2(z,z_{\star })d{z}_{\star }=(N-1)f(z)$. These conventions absorb the otherwise explicit factor $N-1$ in the first BBGKY hierarchy equation.

Separate the Hamiltonian into kinetic energy, external potential energy, and pair potential terms. Their Hamilton's equations giveThe sign follows directly from the stated Poisson bracket: ${\dot{v}}_{i\alpha }=\{v_{i\alpha },H\}=-{\partial }_{x_{i\alpha }}H$. The Hamiltonian Liouville equation is consequentlyIntegrate over $z_2,\dots ,z_N$ and multiply by $N$. Assume sufficient decay in velocity space and either decay or periodic boundary conditions in position, so that integration by parts removes all transport terms with $i\ge 2$. The surviving interaction term iswhere indistinguishability makes all $N-1$ terms equal. Relabeling $(z_1,z_2)$ as $(z,z_{\star })$ yields the exact resultNo assumption of molecular chaos for the Boltzmann equation has entered: the dependence on the two-particle distribution function is still exact.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position $\mathbf{x}$.

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as $f{f}_{\star }$. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This givesHere $d\Omega =\sin \theta d\theta d\phi$; when the measure is written as $d\theta d\phi$, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in $B$. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time $\tau >0$:It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating $\tau$ as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose $f^{(0)}$ to match the collision invariants $1$, $\mathbf{v}$ and $|\mathbf{v}{|}^2$ of $f$. For unit particle mass, this meansHere $\rho$ is the mass density, $\mathbf{u}$ is the mean velocity, and $\Theta$ is the temperature in squared-velocity units, $\Theta =k_{B}T/m$ before setting $m=1$. In particular,Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

Use the BGK model from b. Multiply it by $v_i$ and integrate over velocity space. Matching the collision invariants makes the collision contribution vanish. For the external force term, integration by parts givesDefine the momentum flux and force density byThe resulting momentum conservation equation isThe pressure tensor is the second central velocity moment,The zeroth velocity moment also supplies the continuity equation ${\partial }_t\rho +{\partial }_j(\rho u_j)=0$, which will be needed in d.

Write $c_i=v_i-u_i$ and introduce the raw third velocity moment $M_{ijk}$ and the third central velocity moment $Q_{ijk}$:The expansion of their product givesThe heat flux is $q_i=Q_{ijj}/2$ for unit mass.

Multiply the BGK model by $v_i{v}_j$. The external force term isTherefore the raw second velocity moment satisfiesThe matching local Maxwellian has ${\Pi }_{ij}^{(0)}=\rho u_i{u}_j+\rho \Theta {\delta }_{ij}$, soTo remove the bulk motion, let $D_t={\partial }_t+u_k{\partial }_k$ be the material derivative. The continuity equation and the momentum conservation equation implyand henceSubtract this identity from the raw second velocity moment equation and insert the expansion of $M_{ijk}$. The terms $u_i{\partial }_k{P}_{jk}$ and $u_j{\partial }_k{P}_{ik}$ cancel by the product rule. The external force terms also cancel, since a common acceleration changes the local mean velocity without directly changing the relative velocities. What remains isThe trace of the relaxation term is zero because $P_{ii}=3\rho \Theta$. Thus the BGK model relaxes anisotropy of the pressure tensor while preserving the local random kinetic energy. The evolution is still a velocity-moment hierarchy, since $Q_{ijk}$ requires further information about $f$.