Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 8B (Classical Dynamics)

The Jacobi identity for the Poisson bracket isFor quantities with no explicit time dependence,If $F$ and $G$ are conserved, the first two terms of the Jacobi identity vanish. Henceor, by antisymmetry,Thus the poisson bracket of conserved quantities is itself conserved.

For $Q=\lambda q$ and $P=\lambda p$,The one-dimensional criterion for a canonical transformation therefore givesFor either value, the type-two generating function for a canonical transformationworks, because $p=S_q=\lambda P$ and $Q=S_P=\lambda q$; when ${\lambda }^2=1$, the first equation is equivalent to $P=\lambda p$. For every other $\lambda$, the Poisson bracket is not one and the map is not canonical.

For $Q=q$ and $P=\lambda p$,Thus the map is canonical only forAt that value it is the identity and is generated by

For $Q=p$ and $P=\lambda q$,The map is therefore canonical only forThen $Q=p$ and $P=-q$. The type-one generating function for a canonical transformationgivesas required.