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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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    Hi delickcrow123, I rolled back your edit since it goes a little further than is really appropriate for an edit. The new formula (1), representing an entirely separate way of checking whether a mapping is canonical, is out of scope for this question. But you could potentially ask that as a separate question. Feel free to reapply the bounty if you would still like to have it on this version of the question, and you're also welcome to make minor edits to phrasing and such - just make sure not to change what the question is asking. – David Z May 01 '18 at 06:58

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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  • Is my professor's definition also a sufficient condition for my definition of the canonical transformation? He seems to use both definitions interchangeable. –  Apr 30 '18 at 11:23
  • No, for starters because your professor's definition does not specify $H$ and $K$. – Qmechanic Apr 30 '18 at 13:05
  • May I ask if you could be more precise what do you mean by "your professor's definition does not specify H and K"? Do you mean he removed H and K from the my definition of a canonical transformation, hence requiring only P·dQ=p·dq+dS? –  May 01 '18 at 11:26