In Hamill's "A Student's Guide to Lagrangians and Hamiltonians", section 5.2, the equations for a canonical transformation $(q,p) \to (Q,P)$, induced by the generating function $F(q,Q,t)$ are derived using the principle of stationary action and the fact that the equations of motion are indifferent to the addition of a complete time derivative $d{\Lambda}(q,t)/dt$ (not a function of $\dot{q}$!) to the Lagrangian; this indifference being because the small variations $\delta q$ to the extremal path vanish at the endpoints.

In particular, the author states that:

$\delta \displaystyle\int_{t_1}^{t_2} \left( P\dot{Q} -K(Q,P) \, + dF(q,Q,t)/dt \right)dt = \delta \int_{t_1}^{t_2} \left( P\dot{Q} -K(Q,P)\right)dt$

(where $K$ is the new Hamiltonian) because supposedly we have: $$\delta\int_{t_1}^{t_2} (dF(q,Q,t)/dt)dt=\delta \left(F\left(q(t_2),Q(t_2),t_2\right)-F\left(q(t_1),Q(t_1),t_1\right)\right)=0$$ for the reasons stated above for $\Lambda$.

I don't understand why this can be assumed. In general, $q$ may be a function of both $Q$ and $P$, and the small variations of $P$ at the endpoints need not necessarily vanish, and therefore $F$ may obtain different values at the point $t_1$ (and of course at $t_2$) for different variations of $Q$..

All this can also be stated in terms of $q$ and $\dot{q}$, in which case I would say that since $Q$ is a function of $q$ and $p$ (and $p$ is a function of $\dot{q}$), we get $F(q,\dot{q},t)$ which may vary at the endpoints since $\delta \dot{q}$ need not vanish at the endpoints.

This is an important step in the derivation of a canonical transformation using a generating function... What am I missing?


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