Confirm my reasoning on a generating function proof

AI Thread Summary
The discussion centers on understanding equation (6.5) and its implications for variations in the variables q and Q. It emphasizes that variations must be zero at the endpoints for both action integrals to be valid, preventing non-trivial changes in momentum. The correctness of the reasoning is affirmed, noting that the term F must depend solely on q, Q, and potentially t, without involving canonical momenta. The conversation also highlights the distinction between Lagrangian and Hamiltonian formulations, with the latter allowing more general variations in phase space. Overall, the insights clarify the conditions under which the action integrals are identifiable.
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It's about equation (6.5) I'm not entirely getting the reasoning explained by the author so I came up with the following, can anyone confirm or refute. One way to look at equation (6.5) would be:

We create variations on the ##q## variables, in the form of ##\delta q(t)##. Since ##Q=Q(q,p,t)## the former variation induces a unique variation ##\delta Q(t)##. Since I want both of the action integrals to make sense, both ##\delta q(t) = \delta Q(t) ## have to be ##0## on the end points. This is not so subtle because technically a variation in ##q(t)## could change ##p(t)## in a non trivial way and ##\delta Q(t)## wouldn't be zero on the end points. However I have to make thsi a condition for both integrals over the Lagrangians to be identifiable as the action.

Under these conditions it is clear that the extra term with the ##F## vanishes.

Is this totally wrong, is this partially correct?
 
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It's correct. That's why ##F## must depend only on ##q##, ##Q## (and perhaps explicitly on ##t##). Note that there are no canonical momenta involved here since the proof deals with Hamilton's principle in the Lagrangian form. I'd also not call these transformations "canonical", because in the Lagrangian formalism you can only do point transformations, i.e., transformations between old and new configuration-space variables, since in the Lagrangian form the variation is with respect to curves in configuration space.

It's important to note that the Hamilton principle in the Hamiltonian form is more general than in the Lagrangian form, because it's about variations in phase space (unrestricted for the momenta and at fixed boundary values for the position variables). Then you have more freedom and thus also more symmetry transformations, namely the canonical ones of the Hamilton description of the dynamics (see the other thread about canonical transformations, last discussed yesterday).
 
Thanks so much for helping me out. About yesterdays thread, I wrote some kind of a summary at the end. I would appreciate it if you could confirm or refute
 
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