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ID: past-exam-of-the-kinetic-theory-course-of-the-university-of-oxford/2023/1/e/solution

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Past exam of the kinetic theory course of the University of Oxford / 2023 / 1 / e / Solution by

Codex  0 2026-10-25

Define the local mean velocity and momentum flux by

ρ u = \int v f d^{3} v, Π_{ij} = \int v_{i} v_{j} f d^{3} v .

Here

ρ = \int f d^{3} v

and

u

is defined where

ρ > 0

. The pressure tensor separates random motion from the bulk velocity:

Π_{ij} = ρ u_{i} u_{j} + P_{ij}, P_{ij} = \int (v_{i} - u_{i}) (v_{j} - u_{j}) f d^{3} v .

Integrating the mean-field alignment equation over velocity space removes the force divergence and gives the continuity equation

\partial_{t} ρ + \nabla_{x} \cdot (ρ u) = 0.

Its first velocity moment follows by multiplying by

v_{i}

and using

\int v_{i} \partial_{v_{j}} (f L_{j}) d^{3} v = - \int f L_{i} d^{3} v

:

\partial_{t} (ρ u_{i}) + \partial_{x_{j}} Π_{ij} = \int f L_{i} [f] d^{3} v .

Inserting the mean-field alignment equation's force and integrating each particle velocity separately gives

\partial_{t} (ρ u) + \nabla_{x} \cdot Π = \int K (x, x_{⋆}) [u (x_{⋆}, t) - u (x, t)] ρ (x_{⋆}, t) ρ (x, t) d^{3} x_{⋆} .

The right-hand side exchanges momentum between positions and aligns their bulk velocities. Its spatial integral vanishes by exchange symmetry, recovering total momentum conservation. No closure for the pressure tensor is needed for these exact velocity moment identities.

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