Specific question on Goldstein section on Time-independent perturbation theory

[Bosh]

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!

Thanks.

 

[Valerie Park]

The fact that q should be periodic in w_0 is actually not required for the perturbation theory. The idea is that you want to find a transformation from the unperturbed action-angle variables (J_0, w_0) to a new set (J, w) such that the J's are all constant, and therefore the w's are all linear functions of time. This transformation can be found by solving the equations of motion for the perturbed system, which will be of the form:\dot{J_0} = -\frac{\partial \Delta H}{\partial w_0} \dot{w_0} = \frac{\partial \Delta H}{\partial J_0} These equations can be solved for the transformation of the variables without needing to assume that q is periodic in the unperturbed angle variables w_0. The assumption of periodicity is used in the section to make the solution simpler, but it is not necessary.