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

Solution

Put

z_{i} = (x_{i}, v_{i})

and

d z_{i} = d^{3} x_{i} d^{3} v_{i}

. Normalize the symmetric

N

D_{N}

f (z_{1}) = N \int D_{N} d z_{2} \dots d z_{N}, f_{2} (z_{1}, z_{2}) = N (N - 1) \int D_{N} d z_{3} \dots d z_{N} .

Thus

\int f d z = 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.

H = H_{1} + H_{rest} + H_{int}

, where

H_{1} = \frac{1}{2} ∣ v_{1} ∣^{2}, H_{int} = \sum_{j = 2}^{N} ϕ (∣ x_{1} - x_{j} ∣),

and

H_{rest}

contains all terms involving only particles

2, \dots, N

\dot{x}_{i} = v_{i}, \dot{v}_{i} = - \sum_{j \neq = i} \nabla_{x_{i}} ϕ (∣ x_{i} - x_{j} ∣) .

\partial_{t} D_{N} + \sum_{i = 1}^{N} [v_{i} \cdot \nabla_{x_{i}} D_{N} - \sum_{j \neq = i} \nabla_{x_{i}} ϕ_{ij} \cdot \nabla_{v_{i}} D_{N}] = 0.

Assume 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_{rest}

terms by integration by parts. The terms in

H_{int}

differentiating an integrated velocity also vanish. Indistinguishability makes the remaining

N - 1

terms equal, so

\partial_{t} f + v_{1} \cdot \nabla_{x_{1}} f = N j = 2 \sum N \int \nabla_{x_{1}} ϕ_{1 j} \cdot \nabla_{v_{1}} D_{N} d z_{2} \dots d z_{N} = \int d z_{2} \nabla_{x_{1}} ϕ_{12} \cdot \nabla_{v_{1}} f_{2} .

Relabeling the two particles gives the first BBGKY hierarchy equation:

\partial_{t} f + v \cdot \nabla_{x} f = \int d^{3} x_{⋆} d^{3} v_{⋆} \nabla_{v} f_{2} \cdot \nabla_{x} ϕ (∣ x - x_{⋆} ∣) .

This is exact for a differentiable pair potential; no independent-particle closure has been made.

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