= Solution
Work on $I>0$, with a local choice of the periodic <angle> $\theta$. The inverse formulas are
$$
I=\frac12\left(\alpha q^2+\frac{p^2}{\alpha}\right),\qquad
\theta=\operatorname{atan2}(\alpha q,p)\pmod{2\pi}.
$$
In particular, $\theta_q=p/(2I)$, $\theta_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their <Poisson bracket> is
$$
\boxed{\{\theta,I\}=\frac{p^2}{2I\alpha}+\frac{\alpha q^2}{2I}=1.}
$$
The other fundamental <Poisson brackets> vanish by antisymmetry. Thus this is a <canonical transformation> for every $\alpha>0$, on local charts away from the origin. At $I=0$ the <angle> is undefined.
In the new coordinates the <harmonic oscillator> <Hamiltonian> is
$$
H_0(\theta,I)=I\left(\frac{\alpha}{m}\cos^2\theta
+\frac{m\omega^2}{\alpha}\sin^2\theta\right).
$$
For $I$ to be a <conserved quantity> throughout <phase space>, <Hamilton's equations> require $\dot I=-\partial_\theta H_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega>0$, positivity of $\alpha$ selects
$$
\boxed{\alpha=m\omega,\qquad H_0=\omega I,\qquad
I(t)=I(0),\qquad\theta(t)=\theta(0)+\omega t\pmod{2\pi}.}
$$
These are <harmonic oscillator action-angle variables>; the <action variable> is $I=H_0/\omega$.
Back to article page
