- #1
etotheipi
We have some potential that depends on slowly varying parameters ##\lambda_a##. Using the angle-action variables ##(I, \theta)##, the claim is that we can define a two-form$$W_{ab} = \left\langle \frac{\partial \theta}{\partial \lambda_a} \frac{\partial I}{\partial \lambda_b} - \frac{\partial \theta}{\partial \lambda_b} \frac{\partial I}{\partial \lambda_a} \right\rangle$$such that the Hannay angle can be written$$\Delta \theta = \frac{d}{dI} \int_S W_{ab} dS_{ab}$$given that we already worked out that$$\Delta \theta = \int_S \left( \frac{\partial}{\partial \lambda_a} \left\langle \frac{\partial \theta}{\partial \lambda_b} \right\rangle - \frac{\partial}{\partial \lambda_b} \left\langle \frac{\partial \theta}{\partial \lambda_a} \right\rangle\right) dS_{ab}$$The author begins the proof by noting that the average of some quantity ##A## satisfies$$\langle A \rangle = \oint A(I, \theta) d\theta = \int A(q', p') \delta(I - I') \frac{dq' dp'}{2\pi}$$How has the author performed the change of variables here? The presence of the ##\delta(I - I')## means the integral only picks out when the action variable equals to ##I##. I tried to think about the Stokes' theorem, i.e. taking $$\begin{align*}
\langle A \rangle = \oint \omega = \oint A(I,\theta) \mathrm{d}\theta = \int \mathrm{d} \omega &= \int \mathrm{d} A(I, \theta) \wedge \mathrm{d}\theta \\
&= \int \left( \frac{\partial A}{\partial I} \mathrm{d}I + \frac{\partial A}{\partial \theta} \mathrm{d}\theta \right) \wedge \mathrm{d}\theta \\
&= \int \frac{\partial A}{\partial I} \mathrm{d}I \wedge \mathrm{d}\theta
\end{align*}$$Meanwhile we have that$$\mathrm{d}p \wedge \mathrm{d}q = \text{det}(\mathcal{J}) \mathrm{d}I \wedge \mathrm{d}\theta= \left(\frac{\partial q}{\partial \theta}\frac{\partial p}{\partial I} - \frac{\partial q}{\partial I} \frac{\partial p}{\partial \theta} \right) \mathrm{d}I \wedge \mathrm{d}\theta = \mathrm{d}I \wedge \mathrm{d}\theta$$and then$$\langle A \rangle = \int \frac{\partial A}{\partial I} dq' dp'$$Once in the required form I understand how to prove the result, but I can't figure out how to do this first step. What am I missing? Thanks
\langle A \rangle = \oint \omega = \oint A(I,\theta) \mathrm{d}\theta = \int \mathrm{d} \omega &= \int \mathrm{d} A(I, \theta) \wedge \mathrm{d}\theta \\
&= \int \left( \frac{\partial A}{\partial I} \mathrm{d}I + \frac{\partial A}{\partial \theta} \mathrm{d}\theta \right) \wedge \mathrm{d}\theta \\
&= \int \frac{\partial A}{\partial I} \mathrm{d}I \wedge \mathrm{d}\theta
\end{align*}$$Meanwhile we have that$$\mathrm{d}p \wedge \mathrm{d}q = \text{det}(\mathcal{J}) \mathrm{d}I \wedge \mathrm{d}\theta= \left(\frac{\partial q}{\partial \theta}\frac{\partial p}{\partial I} - \frac{\partial q}{\partial I} \frac{\partial p}{\partial \theta} \right) \mathrm{d}I \wedge \mathrm{d}\theta = \mathrm{d}I \wedge \mathrm{d}\theta$$and then$$\langle A \rangle = \int \frac{\partial A}{\partial I} dq' dp'$$Once in the required form I understand how to prove the result, but I can't figure out how to do this first step. What am I missing? Thanks
. Do you think it is still a mostly correct argument? The original paper by Hannay uses a similar expression [equation (6)], but instead he says that ##\langle f \rangle## is defined like that.
