Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 / 2

Write $\mu =GM$ and $s=\sqrt{b^2+r^2}$. The gravitational potential of the spherical isochrone model obeysThe spherical Poisson equation therefore givesIts small-radius asymptotic expansion isso the model has a finite-density core. At large radius,The $r^{-4}$ envelope has finite total mass $M$.

For a bound orbit let $E<0$ be the specific energy and $L$ the specific angular momentum. Introduce the dimensionless radius suggested in the question,The radial energy equation becomeswhere $x_{-}$ and $x_{+}$ are the two turning points. Comparing coefficients givesThe time from periapsis to apoapsis isThus the radial orbital period isIt depends on $E$ but not on $L$, which is the defining isochrone property.

During half a radial oscillation,Useand the two root productsThe stated elementary integrals then give the azimuthal advance over one radial period:Consequently the frequency ratio of the spherical isochrone model isor equivalently $T_{\varphi }/T_r=2/(1+L/\sqrt{L^2+4GMb})$.

Deep in the constant-density core, $L^2\ll 4GMb$ and ${\Omega }_{\varphi }/{\Omega }_r\to 1/2$, as for an isotropic harmonic oscillator: the radius completes two oscillations per revolution. Far outside the core, $L^2\gg 4GMb$ for the corresponding circular scale and the ratio tends to one, recovering the closed Kepler orbit. Intermediate orbits undergo apsidal precession because the ratio is generally not rational.

The radial action isWith the variable $x$ from part b this becomesApplying the supplied integral twice, with $c=0$ and $c=2$, and using the root sum and products from parts b and c givesThis is the radial action of the spherical isochrone model. Rearranging,and squaring yields the Hamiltonian in action-angle variables:As a check, differentiating this Hamiltonian gives ${\Omega }_r=(-2E{)}^{3/2}/(GM)$ and the frequency ratio found in part c.