During half a radial oscillation,
$$
\Delta\phi_{1/2}
=\frac{L}{b\sqrt{-2E}}
\int_{x_-}^{x_+}
\frac{x-1}{x(x-2)\sqrt{(x-x_-)(x_+-x)}}\,dx.
$$
Use
$$
\frac{x-1}{x(x-2)}=\frac12\left(\frac1x+\frac1{x-2}\right)
$$
and the two root products
$$
x_-x_+=\frac{L^2+4\mu b}{-2Eb^2},
\qquad
(x_--2)(x_+-2)=\frac{L^2}{-2Eb^2}.
$$
The stated elementary integrals then give the azimuthal advance over one radial period:
$$
\boxed{\Delta\phi
=\pi\left(1+\frac{L}{\sqrt{L^2+4GMb}}\right)}.
$$
Consequently the <frequency ratio of the spherical isochrone model> is
$$
\boxed{
\frac{T_r}{T_\phi}
=\frac{\Omega_\phi}{\Omega_r}
=\frac12\left(1+\frac{L}{\sqrt{L^2+4GMb}}\right)},
$$
or equivalently $T_\phi/T_r=2/(1+L/\sqrt{L^2+4GMb})$.

Deep in the constant-density core, $L^2\ll4GMb$ and $\Omega_\phi/\Omega_r\to1/2$, as for an isotropic <harmonic oscillator>: the radius completes two oscillations per revolution. Far outside the core, $L^2\gg4GMb$ 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.

Solution

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

The radial and azimuthal frequencies obey
$$
\frac{\Omega_\phi}{\Omega_r}
=\frac12\left(1+\frac{L}{\sqrt{L^2+4GMb}}\right).
$$
The ratio tends to one half in the harmonic core and to one in the Keplerian exterior.


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