Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 15D (Classical Dynamics)

The Poisson bracket is $\{f,g\}={\sum }_i(f_{q_i}{g}_{p_i}-f_{p_i}{g}_{q_i})$. Canonical variables satisfy $\{Q_i,Q_j\}=\{P_i,P_j\}=0$ and $\{Q_i,P_j\}={\delta }_{ij}$.

To first order the conditions are ${\partial }_{p_j}\delta q_i-{\partial }_{p_i}\delta q_j=0$, ${\partial }_{q_j}\delta p_i-{\partial }_{q_i}\delta p_j=0$, and ${\partial }_{q_j}\delta q_i+{\partial }_{p_i}\delta p_j=0$. The choices $\delta q_i=\epsilon F_{p_i}$ and $\delta p_i=-\epsilon F_{q_i}$ satisfy them by equality of mixed partials.

Taylor expansion gives $\delta H=H_q\cdot \delta q+H_p\cdot \delta p=-\epsilon \{F,H\}$. Hamilton's equation gives $\dot{F}=\{F,H\}$, so $F$ is conserved exactly when the generated infinitesimal canonical transformation leaves $H$ invariant.

For a constant translation vector $a$, take $F=a\cdot {\sum }_i{p}_i$. Then $\delta r_i=\epsilon a$ and $\delta p_i=0$. The potential depends only on position differences and the kinetic energy only on momenta, so this is a symmetry and total momentum is conserved.