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quasar987

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Say I have a canonical transformation Q(q,p), P(q,p).

In the {q,p} canonical coordinates, the Hamiltonian is

[tex]H(q,p,t)=p\dot{q}-L(q,\dot{q},t)[/tex]

And the function [itex]K(Q,P,t)=H(q(Q,P),p(Q,P),t)[/itex] plays the role of hamiltonian for the canonical coordinates Q and P in the sense that

[tex]\dot{Q}=\frac{\partial K}{\partial P}, \ \ \ \ \ \ -\dot{P}=\frac{\partial K}{\partial Q}[/tex]

My question is this:

A priori, I would say that it is certainly not necessary and that the two ways are equivalent [that is, in the event that it is even

In the {q,p} canonical coordinates, the Hamiltonian is

[tex]H(q,p,t)=p\dot{q}-L(q,\dot{q},t)[/tex]

And the function [itex]K(Q,P,t)=H(q(Q,P),p(Q,P),t)[/itex] plays the role of hamiltonian for the canonical coordinates Q and P in the sense that

[tex]\dot{Q}=\frac{\partial K}{\partial P}, \ \ \ \ \ \ -\dot{P}=\frac{\partial K}{\partial Q}[/tex]

My question is this:

**must I first express all the [itex]\dot{q}[/itex] in there in terms of q and p before transforming my Hamiltonian into [itex]K(Q,P,t)=H(q(Q,P),p(Q,P),t)[/itex], or can I just compute [itex]\dot{q}(Q,\dot{Q},P,\dot{P})[/itex] and substitute?**A priori, I would say that it is certainly not necessary and that the two ways are equivalent [that is, in the event that it is even

*possible*to invert [itex]p=\partial L /\partial \dot{q}[/itex] !], but I have evidence that it's not and that the second way leads to equations of motion that are wrong.
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