8
$\begingroup$

Given a 1 d.o.f Hamiltonian $H(q,p)$ what is the general procedure for finding action angle variables $(I, \theta)$?

I have read the Wikipedia page on action angle variables and canonical transforms but have difficulty applying the general methods to specific problems. Can someone explain the method to me using a simple general example?

$\endgroup$
1

1 Answer 1

7
$\begingroup$

In local coordinates the canonical transformation to action angle coordinates $(q,p)\rightarrow (Q,P)$ can be related by, \begin{equation} \boxed{P_i=\frac{1}{2\pi}\oint p_idq^i \ \ \ \ \ \text{and}\ \ \ \ \ Q^i=\frac{\partial }{\partial P_i}\int p_idq^i} \end{equation} For Example:

Consider the one dimensional harmonic oscillator with the following Hamiltonian $H=\frac 1{2m}\big[p^2+m^2\omega ^2q^2\big]$. Rearrange this for $p$ and take the hypersurface $H=E$. \begin{equation} p=\pm \sqrt{2mE-m^2\omega ^2q^2} \end{equation} Then use the above equation to compute $P$. \begin{equation} P=\frac{1}{2\pi }\oint \sqrt{2mE-m^2\omega ^2q^2}dq \end{equation} The integral is now over $0$ to $2\pi$ which is easier to handle. This works out as, \begin{equation} \frac {1}{2\pi}\oint ^{2\pi}_{0}\cos^2Q\ dQ\cdot \frac {2E}{\omega} =\frac{E}{\omega} \end{equation} Therefore we have used the quoted formula to compute the action variable for the harmonic oscillator.

$\endgroup$
3
  • 1
    $\begingroup$ I have a small question. Why is the limit taken from 0 to $2\pi$? $\endgroup$
    – Icchyamoy
    Oct 24, 2017 at 11:56
  • $\begingroup$ This could become a truly great answer if you expanded on the subtleties of the integrations, noting that one is a countour integral whereas the other is an indefinite integral. $\endgroup$ Oct 28, 2018 at 23:37
  • 1
    $\begingroup$ The limits are for one complete period. It has to be taken as the turning points if the motion is a closed loop on the phase space. If on the other hand the system has energy greater than the binding energy, then the period can be scaled to $2\pi$. $\endgroup$ Nov 29, 2019 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.