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For which of the invertible transformations [itex](\mathbf{q}, \mathbf{p}) \leftrightarrow(\mathbf{Q}, \mathbf{P})[/itex]

[itex]\mathbf{Q}(\mathbf {q}, \mathbf {p}, t)[/itex]

[itex]\mathbf{P}(\mathbf{q}, \mathbf {p}, t)[/itex]

is it so that for every Hamiltonian [itex]\mathcal{H}(\mathbf {q}, \mathbf {p}, t)[/itex] there is a [itex]\mathcal{K}[/itex] such that

[tex]\dot{Q}_i = \frac{\partial\mathcal{K}}{\partial P_i} \;\;\;\;\;\;\;\; \dot{P}_i = -\frac{\partial\mathcal{K}}{\partial Q_i}\; ?[/tex]

Stationary action should correspond, and that condition is met if

[itex]\sum p_i\dot{q}_i - \mathcal{H} = \sum P_i\dot{Q}_i - \mathcal{K} + \frac{dF}{dt}[/itex],

since integrating [itex]\frac{dF}{dt}[/itex] results in something only dependent of the endpoints.

Question

Consider this part of Goldstein'sClassical Mechanics.

rearranging 9.13 to make this clear:

[tex]\mathcal{K} = \mathcal{H} + \frac{\partial F_1}{\partial t} + \sum_i \dot{Q}_i\left(P_i - \frac{\partial F_1}{\partial Q_i} \right) + \sum_i \dot{q}_i\left(\frac{\partial F_1}{\partial q_i} - p_i\right)[/tex]

I guess I might like this explained a more. Why aren't we able to to have the coefficients of [itex]\dot{q}_i[/itex] or [itex]\dot{Q}_i[/itex] be non-zero, and have the difference absorbed into [itex]\mathcal{K}[/itex] ?

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# Canonical Transformations

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