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Question about canonical transformation

  1. Dec 4, 2013 #1
    I was going through my professor's notes about Canonical transformations. He states that a canonical transformation from (q, p) to (Q, P) is one that if which the original coordinates obey Hamilton's canonical equations than so do the transformed coordinates, albeit for a different Hamiltonian. He then considers, as an example the Hamiltonian

    [tex]H=\frac{1}{2}p^2,[/tex]

    with a transformation

    [tex]Q = q,[/tex]
    [tex]P = \sqrt{p} - \sqrt{q}.[/tex]

    The notes state that "this transformation is locally canonoid with respect to H," and that in the transformed coordinates the new Hamiltonian is

    [tex] K = \frac{1}{3} \left( P + \sqrt{Q} \right)^3.[/tex]

    I don't understand how we know that this is locally canonical, or what it really even means to be locally canonical. Also, where do we get K from. Since the inverse transformation would be

    [tex]q=Q[/tex]
    [tex]p=\left( P + \sqrt{Q} \right)^2[/tex]

    why isn't the new Hamiltonian

    [tex]K= \frac{1}{2} \left(P + \sqrt{Q} \right)^4,[/tex]

    where all I've done is plug the inverted transformation into the original Hamiltonian. I'm a bit confused by all this. Would appreciate any help.

    Thanks.
     
  2. jcsd
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