Show that a transformation is canonical

  1. fluidistic

    fluidistic 3,365
    Gold Member

    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. fluidistic

    fluidistic 3,365
    Gold Member

    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
    I)[itex][Q,P]_{q,p}=1[/itex]
    II)[itex][Q,Q]_{q,p}=0[/itex]
    III)[itex][P,P]_{q,p}=0[/itex].
    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
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?