# Specific question on Goldstein section on Time-independent perturbation theory

1. Jun 25, 2012

### 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 $(J_0, w_0)$, 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.

In the first equation of the section, 12.61 (all equation numbers from the 3rd edition), he writes $q_k$ as a multiple Fourier expansion of all the $w_0$'s. Specifically, this assumes that $q_k$ is a periodic function of each of the $w_0$'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 $w_0$ and $w$ then ...

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

If you move the system through one cycle of q, the change in $w$ is given by:

$\Delta w = \oint \frac{\partial w}{\partial q} dq$
$= \oint \frac{\partial^2 W}{\partial J \partial q} dq$
$= \frac{\partial}{\partial J} \oint \frac{\partial W}{\partial q} dq$

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

$= \frac{\partial}{\partial J} \oint p dq = \frac{\partial}{\partial J} J = 1$

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 $w_0$! The argument runs analogously until the third line:

$\Delta w_0 = \oint \frac{\partial w_0}{\partial q} dq$
$= \oint \frac{\partial^2 W_0}{\partial J_0 \partial q} dq$

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

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

Thanks.