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Specific question on Goldstein section on Time-independent perturbation theory

  1. Jun 25, 2012 #1
    I apologize that this is rather specific, but hopefully enough people have used Goldstein. I have a basic grasp of action-angle variables, and I'm going through the time-independent perturbation theory section in Goldstein (12.4).

    In this section we seek a transformation from the unperturbed action-angle variables [itex](J_0, w_0)[/itex], to a new set [itex](J, w)[/itex] such that the [itex]J[/itex]'s are all constant, and therefore the [itex]w[/itex]'s are all linear functions of time.

    In the first equation of the section, 12.61 (all equation numbers from the 3rd edition), he writes [itex]q_k[/itex] as a multiple Fourier expansion of all the [itex]w_0[/itex]'s. Specifically, this assumes that [itex]q_k[/itex] is a periodic function of each of the [itex]w_0[/itex]'s. They make this point again just before equation 12.74, saying that in order for the q's and p's to be periodic in both [itex]w_0[/itex] and [itex]w[/itex] then ...

    What I don't understand is: Why should q be periodic in the unperturbed angle variables [itex]w_0[/itex]? I agree that they should be periodic in the perturbed angle variables [itex]w[/itex]. The argument in one dimension would run like this:

    If you move the system through one cycle of q, the change in [itex]w[/itex] is given by:

    [itex]\Delta w = \oint \frac{\partial w}{\partial q} dq[/itex]
    [itex] = \oint \frac{\partial^2 W}{\partial J \partial q} dq[/itex]
    [itex] = \frac{\partial}{\partial J} \oint \frac{\partial W}{\partial q} dq[/itex]

    where the partial with respect to J can only be taken outside the integral because it is constant!

    [itex] = \frac{\partial}{\partial J} \oint p dq = \frac{\partial}{\partial J} J = 1 [/itex]

    so every time w advances by 1, q returns to its same value, and is therefore periodic in w with period 1.

    But I don't see how this works for [itex]w_0[/itex]! The argument runs analogously until the third line:

    [itex]\Delta w_0 = \oint \frac{\partial w_0}{\partial q} dq[/itex]
    [itex] = \oint \frac{\partial^2 W_0}{\partial J_0 \partial q} dq[/itex]

    but now you shouldn't be able to take out the partial with respect [itex]J_0[/itex], since it will no longer be constant, right? [itex]J_0[/itex] was the constant action variable in the unperturbed problem, but once you add a perturbation, the perturbation Hamiltonian will in general depend on [itex]w_0[/itex], i.e. [itex] \Delta H = \Delta H(w_0, J_0, t)[/itex]. Therefore [itex]\dot{J_0} = -\frac{\partial \Delta H}{\partial w_0} \neq 0[/itex]

    I know it's pretty specific, but if anyone could help me I'd really appreciate it!

  2. jcsd
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