 Solution

ID: past-exam-of-the-classical-mechanics-course-of-the-university-of-oxford/2023/3/c/i/solution

(q, P)

p = \frac{\partial F _{2}}{\partial q}, Q = \frac{\partial F _{2}}{\partial P} .

F_{2, qP} \neq = 0

so that

p = F_{2, q} (q, P)

can be solved for

P

. The differential identity

d F_{2} = p d q + Q d P

gives

d (F_{2} - PQ) = p d q - P d Q

, expressing preservation of the canonical symplectic form.

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