Past exam of the classical mechanics course of the University of Oxford / 2023 / 3 / c

A type-two generating function for a canonical transformation uses the mixed variables $(q,P)$ and definesFor a local invertible canonical transformation, require $F_{2,qP}\ne 0$ so that $p=F_{2,q}(q,P)$ can be solved for $P$. The differential identity $d{F}_2=pdq+QdP$ gives $d(F_2-PQ)=pdq-PdQ$, expressing preservation of the canonical symplectic form.

On the interior of the principal inverse-cosine branch, $0<q<\pi$, the given canonical transformation hasIntegrating $Q={\partial }_P{F}_2$ suggests $F_2=\cos q\sin P+C(q)$. Its other derivative is ${\partial }_q{F}_2=-\sin q\sin P+C^{\prime }(q)$; comparison with $p$ requires $C^{\prime }(q)=0$. Hence a suitable type-two generating function for a canonical transformation isBoth derivative equations reproduce the specified transformation exactly. Its mixed derivative is $F_{2,qP}=-\sin q\cos P$, so it defines a local canonical transformation where $0<q<\pi$ and $\cos P\ne 0$, equivalently where $|Q/\cos P|<1$. At $q=0,\pi$ or $\cos P=0$, the stated formulas fail to give a regular coordinate chart. These restrictions explain the square-root branch and the local meaning of canonicity. This is a trigonometric type-two canonical transformation.

For a time-dependent type-two generating function for a canonical transformation, the differential identitygives $K=H+{\partial }_t{F}_2$, with the partial derivative taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the Hamiltonian correction isAs always, adding a function of time alone to the new Hamiltonian leaves Hamilton's equations unchanged. Choosing a time-independent type-two generating function for a canonical transformation fixes that irrelevant freedom and gives the displayed $\Lambda =0$.