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Show that a transformation is canonical

  1. Feb 29, 2012 #1


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    1. The problem statement, all variables and given/known data
    Show that the following transformation is canonical:
    [itex]Q=\ln \left ( \frac{\sin p }{q} \right )[/itex], [itex]P=q \cot p[/itex].

    2. Relevant equations
    A transformation is canonical if [itex]\dot Q=\frac{\partial H'}{\partial P}[/itex] and [itex]\dot P =-\frac{\partial H'}{\partial Q}[/itex].
    H' is the Hamiltonian in function of Q and P.

    3. The attempt at a solution
    I've calculated [itex]\dot Q = \dot p \cot p - \frac{\dot q \sin p }{q}[/itex] and [itex]\dot P=\dot q \cot p - \frac{q \dot p}{\sin ^2 p}[/itex].
    Now I guess I must check out if the conditions about the partial derivatives of the Hamiltonian are satisfied but I do not know either H(q,p) nor do I know H'(Q,P). I'm stuck here.
  2. jcsd
  3. Mar 2, 2012 #2


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    Ok guys, discard my attempt in my last post.
    I've found in some class notes and in Landau&Lifgarbagez's book the necessary condition for a transformation to be canonical. If I'm not wrong, I must show that
    Where [itex][f,g]_{p,q}[/itex] denotes the Poisson brackets of f and g with respect to p and q. In other words, this is worth [itex]\frac{\partial f}{\partial p} \frac{\partial g}{\partial q}-\frac{\partial f}{\partial q} \frac{\partial g}{\partial p}[/itex].
    Now I have that [itex]Q=\ln \left ( \frac{\sin p }{q} \right )[/itex] and [itex]P=q \cot p[/itex].
    I tried to be careful in doing the partial derivatives. I found out that:
    (1)[itex]\frac{\partial Q}{\partial p}=q \cot p[/itex]
    (2)[itex]\frac{\partial P}{\partial q}= \cot p[/itex]
    (3)[itex]\frac{\partial Q}{\partial q}=-\frac{1}{q}[/itex]
    (4)[itex]\frac{\partial P}{\partial p}=q(-1- \cot ^2 p)[/itex]

    Using this, I found out I) to be worth [itex](1-q) \cot ^2 p-1[/itex]. Unfortunately this isn't worth 1. What am I doing wrong?

    Edit: I just found a mistake in my definition of Poisson's brackets. But now I find I) to be worth [itex]1+ \cot ^2 p-q \cot ^2 p[/itex] which is still wrong. I've rechecked the partial derivatives, I do not see any mistake, yet my result keeps being wrong.

    Edit 2: This can be rewritten as [itex]\frac{1}{\sin ^2 p}-q \cot ^2 p[/itex]. This is not worth 1 so that the transformation isn't canonical, which is absurd. I do not see any mistake.

    Edit 3 : I am nut guys! I found out the mistake in a derivative. I now reach all the results I should, problem solved!
    Last edited: Mar 2, 2012
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