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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-4/15b/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 15B / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

With Hamilton's equations

\overset{q}{˙}_{j} = H_{p_{j}}

and

\overset{p}{˙}_{j} = - H_{q_{j}}

, the chain rule gives

\frac{d F}{d t} = j \sum (F_{q_{j}} H_{p_{j}} - F_{p_{j}} H_{q_{j}}) = [F, H] .

For the vortex problem, combine the two ordered terms for each unordered pair, writing

D_{ij} = (p_{i} - p_{j})^{2} + (q_{i} - q_{j})^{2}

and

H = - (κ /2) \sum_{i < j} lo g D_{ij}

. Since

F_{q_{j}} = 2 q_{j}

,

F_{p_{j}} = 2 p_{j}

, its pairwise contribution to the Poisson bracket is

- \frac{2 κ}{D _{ij}} [(q_{i} - q_{j}) (p_{i} - p_{j}) - (p_{i} - p_{j}) (q_{i} - q_{j})] = 0.

Every pair cancels, so

[F, H] = 0

and $\sum_{j} (q_{j}^{2} + p_{j}^{2})$ is conserved along any collision-free solution. The logarithmic Hamiltonian is undefined at coinciding vortices, so the derivation applies on its natural domain.

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