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Show that transformation is Canonical

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the transformation [tex]Q=qe^{\gamma t}[/tex] and [tex]P=pe^{-\gamma t}[/tex] with the Hamiltonean [tex]H=\frac{p^2e^{-2\gamma t}}{2m}+\frac{m\omega^2q^2e^{2\gamma t}}{2}[/tex] show that the transformation is Canonical


    2. Relevant equations

    I know that the condition for a transformation to be canonical is [tex]\sum_i p_i\dot{q_i}-P_i\dot{Q_i}=H-K+\frac{dF_2}{dt}[/tex] but don't know what to do with it...

    I've managed to prove from [tex]\frac{dF_2}{dq_i}=p_i[/tex] and [tex]\frac{dF_2}{dP_i}=Q_i[/tex] that [tex]F_2=qPe^{\gamma t}[/tex].
     
    Last edited: May 19, 2012
  2. jcsd
  3. May 19, 2012 #2
    I have another question related to this problem. Does having the generating function guarantee that the coordinate transformation associated with it is canonical? Or is it a necessary but not sufficient condition?
     
  4. May 20, 2012 #3

    fluidistic

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    Gold Member

    I know a way to check out whether the transformation is canonical but it doesn't involve the Hamiltonian.
    If you know Poisson's brackets, if you can show that [itex][Q,P]_{q,p}=1[/itex] then you'd be done.
    To answer your question "Does having the generating function guarantee that the coordinate transformation associated with it is canonical? Or is it a necessary but not sufficient condition?", I'm not 100% sure but I'd say yes (99% sure).
     
  5. May 20, 2012 #4
    I've been messing around with the coordinates and got this:

    If we have a general invertible transformation of the type [tex]Q=Q(q,p,t)\Leftrightarrow q=q(Q,P,t)[/tex] and [tex]P=P(q,p,t)\Leftrightarrow p=p(P,Q,t)[/tex] then the following is true:

    \begin{matrix}
    \dot{q}=\frac{\partial q}{\partial Q}\dot{Q}+\frac{\partial q}{\partial P}\dot{P}\\
    \dot{p}=\frac{\partial p}{\partial Q}\dot{Q}+\frac{\partial p}{\partial P}\dot{P}
    \end{matrix}

    From Hamilton's eqs we have (inserting the given coordinate transformations):
    \begin{matrix}
    \dot{q}=\frac{\partial H}{\partial p}=\frac{\partial H}{\partial P}e^{-\gamma t}=\frac{\partial q}{\partial Q}\dot{Q}\Rightarrow \frac{\partial H}{\partial P}=\dot{Q}\\
    \dot{p}=-\frac{\partial H}{\partial q}= -\frac{\partial H}{\partial Q}e^{\gamma t}=\frac{\partial p}{\partial P}\dot{P}\Rightarrow -\frac{\partial H}{\partial Q}=\dot{P}
    \end{matrix}

    The form of Hamilton's equations is the same but from what I've seen in Goldstein, when one makes a canonical transformation the form is preserved not for the initial Hamiltonian but for the "Kamiltonian" [tex]K=H+\frac{dF_2}{dt}[/tex] i.e.

    \begin{matrix}
    \frac{\partial K}{\partial P}=\dot{Q}\\
    -\frac{\partial K}{\partial Q}=\dot{P}
    \end{matrix}

    and because I've already proved that there exists a function [itex]F_2[/itex] I get the feeling that the proof is not correct. What am I doing wrong here?

    Thanks for the advice! Actually, calculating with the Poisson brackets is another part of this problem, but instead it is asked to evaluate [itex][Q,Q],[P,P],[Q,P],[q,H],[p,H][/itex] without explicit calculation.

    For the first two the answer is obviously zero because of the crossed derivatives, the third one, being this a suposedly canonical transformation is 1 and for the last two I'm having a little bit of a problem because if I use [itex]\frac{du}{dt}=[u,H]+\frac{\partial u}{\partial t}[/itex] on q and p I am not sure if it is [itex]\frac{dq}{dt}[/itex] or [itex]\frac{\partial q}{\partial t}[/itex] (for p too) that is zero. I know, this is a dumb question, but when a function depends only on t does the partial derivative looses its meaning?
     
  6. May 20, 2012 #5

    fluidistic

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    Gold Member

    Ok, since you found out [itex]F_2[/itex] then I think you are done. Let me give you a useful link: http://books.google.com.ar/books?id...a transformation is canonical poisson&f=false.
    About the Poisson's bracket question, you are right for the 3 first. As for the last 2 I must say I'm not sure; I'd consult Landau's book on classical mechanics first and then other books.
    About your question about partial derivatives; when you have a function of only 1 variable a partial derivative is the same as a total derivative.
    But I doubt q only depends on t.
     
  7. May 20, 2012 #6
    Thanks for the book! I tried to see if there was anything in Landau that could help but didn't find it yet. Anyhow, I found this http://solar.physics.montana.edu/dana/ph411/p_brack.pdf and I can't figure out why, in page 2 after equation (3) [itex]\frac{\partial q_1}{\partial t}=0[/itex]. Isn't [itex]q_1[/itex] also a function of t? Why isn't [itex]\frac{dq_1}{dt}[/itex] also zero?
     
  8. May 20, 2012 #7

    fluidistic

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    I'm in the same boat as you. I do not understand why it is so.
    For sure, [itex]q_1[/itex] isn't a constant of motion, otherwise we'd have [itex]\{ q_1 , H \}=0[/itex]. So there's indeed a dependency of [itex]q_1[/itex] with respect to time. Thus, I do not understand how it's possible that [itex]\frac{\partial q _1 }{\partial t } =0[/itex].
    I wish someone could enlighten us!
     
  9. May 20, 2012 #8
    I'm just posting this so that the thread doesn't seem closed.
     
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