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My professor mentioned: A simple way of testing whether a mapping $(q,p)$ to $(Q,P)$ is canonical is by examining:

$$P · dQ − p · dq$$

and if it equals to $dA$ (a differential) then it is canonical.

However, I'm wondering why is this the case, since the requirements for canonical map is that at first is $$P ·dQ − Kdt = p·dq − Hdt + dS$$ (so that the closed contour integral of $P ·dQ − Kdt$ to equal that of $p·dq − Hdt$. Then what about the $Kdt$ and $Hdt$?

David Z
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1 Answers1

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Be advised that there are different definitions of a canonical transformation (CT), cf. e.g. this Phys.SE post.

Your last definition of a CT agrees with the definition in e.g. Landau & Lifshitz and Goldstein, while your professor is listing a sufficient condition for a symplectomorphism, which is called a CT by e.g. Arnold.

Qmechanic
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