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

With Hamilton's equations ${\dot{q}}_j=H_{p_j}$ and ${\dot{p}}_j=-H_{q_j}$, the chain rule givesFor 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 $H=-(\kappa /2){\sum }_{i<j}\log D_{ij}$. Since $F_{q_j}=2q_j$, $F_{p_j}=2p_j$, its pairwise contribution to the Poisson bracket isEvery 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.