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?


1 Answer 1


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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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