Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 9E (Classical Dynamics)

Use the canonical Poisson bracket convention $\{H,G\}={\sum }_k(H_{q_k}{G}_{p_k}-H_{p_k}{G}_{q_k})$. The Hamiltonian flow generated by $G$ has ${\dot{q}}_k=G_{p_k}$ and ${\dot{p}}_k=-G_{q_k}$. The chain rule therefore givesThus $H$ is invariant under every such trajectory exactly when $\boxed{\{H,G\}=0}$ identically. Antisymmetry then gives $dG/dt=\{G,H\}=0$ along the flow of $H$, so $G$ is a constant of motion.

For $G=\alpha {\sum }_k{p}_k$, the generated equations are ${\dot{q}}_k=\alpha$, ${\dot{p}}_k=0$, hence $q_k(t)=q_k(0)+\alpha t$, $p_k(t)=p_k(0)$. Invariance under this simultaneous translation implies conservation of $\boxed{\sum _k{p}_k}$ when $\alpha \ne 0$; the zero generator alone conveys no additional conservation law.

Write $H=F(P,Q)$ with $P=\sum p_k^2$ and $Q=\sum q_k^2$. The angular momentum component $L_i={ϵ}_{ijk}{q}_j{p}_k$ has ${\partial }_{p_{\ell }}{L}_i={ϵ}_{ij\ell }{q}_j$ and ${\partial }_{q_{\ell }}{L}_i={ϵ}_{i\ell k}{p}_k$. Thereforesince a symmetric product contracts to zero with the antisymmetric Levi-Civita symbol. All three components of angular momentum are conserved by this rotationally invariant Hamiltonian.