Action-angle coordinates

Remarks

Here the convention is $\omega = \sum_k dp_k \wedge dq_k$ on $\R^{2n}$, so that the actions play the role of momenta and the angles the role of positions. In these coordinates the Hamiltonian system reads $I_k' = 0$ and $\theta_k' = \partial K / \partial I_k (I)$, so $\theta(t) = \theta(0) + t \, \nabla K(I)$.

Example: the harmonic oscillator $H = \frac{1}{2}(p^2 + q^2)$ on $\R^2 \setminus \set{0}$. Put $q = \sqrt{2I} \sin \theta$ and $p = \sqrt{2I} \cos \theta$ with $I > 0$. A short computation gives $dp \wedge dq = dI \wedge d\theta$, and $H = I$, so both clauses hold with $K(I) = I$.

Non-example: polar coordinates $q = r \sin \theta$, $p = r \cos \theta$ with $r > 0$ give $H = r^2/2$, but $dp \wedge dq = r \, dr \wedge d\theta$, so clause 1 fails.

For an integrable system, the actions can be computed as $I_k = \frac{1}{2\pi} \oint_{\gamma_k} p \, dq$ over basic cycles $\gamma_k$ of an invariant torus, when $\omega = d(\sum_k p_k \, dq_k)$.