
#1
Jul713, 01:23 AM

P: 541

Background
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's Classical 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 nonzero, and have the difference absorbed into [itex]\mathcal{K}[/itex] ? 



#2
Jul713, 06:05 PM

P: 205

it's because K should not depend on the derivates of Q and q




#3
Jul813, 02:42 AM

Sci Advisor
Thanks
P: 2,153

I would argue as follows. The Hamiltonian version of the action is
[tex]S[x,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q}^k p_k  H(t,q,p)].[/tex] The trajectory in phase space is determined as the stationary point of this functional with the boundary values [itex]q(t_1)[/itex] and [itex]q(t_2)[/itex] fixed. Now if you want to determine new phasespace coordinates [itex](Q,P)[/itex] that describe the same system in the new coordinates by the variational principle, i.e., such that the phasespace trajectories are described by the Hamilton canonical equations, you must have [tex]\mathrm{d} q^k p_k  \mathrm{d} Q^k P_k  \mathrm{d} t (HK)=\mathrm{d} f.[/tex] From this it is clear that the "natural" independent variables for [itex]f[/itex] are [itex]q[/itex], [itex]Q[/itex], and [itex]t[/itex]. Then comparing the differential on each side leads to [tex]p_k=\frac{\partial f}{\partial q^k}, \quad P_k=\frac{\partial f}{\partial Q^k}, \quad K=H+\frac{\partial f}{\partial t}.[/tex] Then you can go over to other pairs of old and new independent phasespace variables in the "generator" using appropriate Legendre transformations. 


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