Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 15B (Classical Dynamics)

For a regular Lagrangian, define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and invert this relation for $\dot{q}(q,p,t)$. Its Legendre transform is the HamiltonianSince $Ldt=p\cdot dq-Hdt$, the action isFor independent variations with $\delta q(t_1)=\delta q(t_2)=0$,Stationarity for arbitrary $\delta p$ and $\delta q$ gives Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$.

The defining relations for a type-2 generating function giveHenceThe exact differential changes the action only by an endpoint term. Thus the transformed variational principle has canonical one-form $P\cdot dQ-Kdt$ and yields Hamilton's equations in $(Q,P)$ with

For $H=H(p)$, Hamilton's equations giveThus $P=p$ and $Q=q-t{H}^{\prime }(p)$. The Poisson bracket is $\{Q,P\}=1$, so the transformation is canonical for every $t$. A type-2 generator isIndeed $p={\partial }_q{F}_2=P$, $Q={\partial }_P{F}_2=q-t{H}^{\prime }(P)$, and the transformed Hamiltonian is $K=0$, as expected because $Q,P$ are constants of motion.

For the harmonic oscillator Hamiltonian $H=(q^2+p^2)/2$,This rotation preserves the symplectic form $dq\wedge dp$, so it is a canonical transformation for every $t$. Where $\cos t\ne 0$, solving for the initial variables givesand hence the type-2 generatorIts derivatives give the displayed transformation and $H+{\partial }_t{F}_2=0$. At $\cos t=0$ the canonical rotation remains perfectly regular, but its projection to $(q,P)$ is singular, so no type-2 generating function in those variables exists there; another generating-function chart covers those isolated times.