Understanding the Change of Variables in the Hannay Angle Proof

[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

 

[andresB]

what author are you following? Also, I'm not sure exactly which the step you are struggling with.

 

[wrobel]

Look like
$$\int_{\{I=const\}}\omega=\int_{\{I\le const\}}d\omega,\quad \omega=I\frac{\partial \theta}{\partial\lambda^i}d\lambda^i=Id\theta$$

 

[etotheipi]

what author are you following? Also, I'm not sure exactly which the step you are struggling with.

This is Tong, page 120 and equation (4.174) of this document. I'm struggling mainly to understand the second equality, which changes the integral from an integral over $\theta$ to an integral over phase space, with an added delta function. I figured it might have something to do with Stokes, but haven't gotten their RHS

I'm not sure how to show that$$ \int_{\{ I \leq \text{constant} \}} \frac{\partial A}{\partial I} dq' dp' = \int_{\{ I \leq \text{constant} \}} A(q', p') \delta(I - I') \frac{dq' dp'}{2\pi}$$or otherwise if I made a mistake with Stokes

 

[wrobel]

I'm struggling mainly to understand the second equality, which changes the integral from an integral over # to an integral over phase space, with an added delta function.

I suspect the author hardly think about accurate sense of this formula with delta function and actually this formula is not used in the sequel and does not influence anything :)
If it is not so, you have to replace argument based on this formula with mathematically correct one.

 

[etotheipi]

Ah okay, maybe Tong might have done a little bit of fudging . 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.

https://iopscience.iop.org/article/10.1088/0305-4470/18/2/011/pdf

 

[wrobel]

Everything depends on whether formula (6) is the final result or it is used to prove some other theorem. In first case this is just another strange math. slang from the physics textbook. In the second case give me the final result and I will try to obtain it by normal mathematical language

 

[wrobel]

but instead he says that is defined like that.

what does prohibit to define <f> by the first contour integral and ignore the second integral with delta function?

 

[etotheipi]

I believe, (6) is only used to prove results later on in the paper, i.e. he uses it as his definition of the $\langle \cdot \rangle$ notation.

I found another discussion in this paper by Berry:
https://pdfs.semanticscholar.org/425d/786cd1e3b58dcab757951df11a1572d6bf87.pdf
In section 2, author derives a similar equation (15), but maybe in a more rigorous way?

The theorem Tong tries to prove is:$$\int_S \left( \frac{\partial}{\partial \lambda_a} \left\langle \frac{\partial \theta}{\partial \lambda_b} \right\rangle - \frac{\partial}{\partial \lambda_b} \left\langle \frac{\partial \theta}{\partial \lambda_a} \right\rangle\right) dS_{ab} = \frac{d}{dI} \int_S \left\langle \frac{\partial \theta}{\partial \lambda_a} \frac{\partial I}{\partial \lambda_b} - \frac{\partial \theta}{\partial \lambda_b} \frac{\partial I}{\partial \lambda_a} \right\rangle dS_{ab}$$

 

[wrobel]

Equation (15) is completely clear; the last formula in your post is also clear. Both follow from #3

 

[andresB]

note that the equality
$$\oint A(I,\theta)\mathrm{d}\theta=\int\frac{\partial A}{\partial I}\mathrm{d}I\wedge\mathrm{d}\theta$$
cannot be true because the LHS depends on $I$ while on the RHS $I$ has been integrated away. (this is maybe easier to see in the more straightforward definition of the average $\left\langle f\right\rangle =\frac{1}{2\pi}\int_{0}^{2\pi}f(I,\theta)d\theta$)

Also, the Dirac delta cannot be so easily ignored as it is indeed used (see for example the appendix in the oringial paper of Hannay).

Now, As remarked by Hannay in the appendix of his paper, we have to distinguish the action as a function of q, p, and the external parameters $I'=I'(q,p,\lambda)$ and the constant value of $I$ for a given torus.
I think what happens is more or less as following:

$$\begin{align*}
\oint A(I,\theta)\mathrm{d}\theta =\oint\left[\int A(I',\theta)\delta(I'-I)dI'\right]\mathrm{d}\theta
=\int dI'\oint A(I',\theta)\mathrm{\delta(I'-I)}d\theta
=\int dI'\int\frac{d}{dI'}\left[A(I',\theta)\mathrm{\delta(I'-I)}dI'\wedge d\theta\right]
=\int A(I',\theta)\mathrm{\delta(I'-I)}dI'\wedge d\theta
=\int A(q,p)\mathrm{\delta(I'(q,p,\lambda)-I)}dp\wedge dq
\end{align*}$$

 

[wrobel]

cannot be true because the LHS depends on while on the RHS has been integrated away

nevertheless it is true because one should understand what was written:
$$\int_{I=const}A(I,\theta)d\theta=\int_{I\le const}\frac{\partial A}{\partial I}(I,\theta)dI\wedge d\theta$$