Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 15C (Classical Dynamics)

The action variable for a regular periodic one-degree-of-freedom Hamiltonian orbit is $I(E)=(2\pi {)}^{-1}\oint pdq$, with orientation along the motion. Along its energy curve, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. Differentiating the loop integral branchwise givesAt simple turning points the two momentum branches join and their endpoint contributions cancel, so the integrable turning-point singularities do not spoil this differentiation. The formula concerns regular periodic levels, not a separatrix of infinite period.

The energy equation rearranges to $p^2+(E+1)q^2=E{\mu }^2$, an ellipse with semiaxes $p_0=\mu \sqrt{E}$ and $q_0=\mu \sqrt{E/(E+1)}$. At low energy it is nearly circular with both radii $\mu \sqrt{E}$; at high energy it is tall and thin, with $q_0\to \mu$ while $p_0$ grows. The area enclosed is $\pi p_0{q}_0$, soConsequently $T\sim \pi {\mu }^2/(2\sqrt{E})=\boxed{\pi {\mu }^3/(2p_0)}$ at high energy.