The <radial action> is
$$
J_r=\frac1\pi\int_{r_-}^{r_+}
\sqrt{2[E-\Phi(r)]-\frac{L^2}{r^2}}\,dr.
$$
With the variable $x$ from part b this becomes
$$
J_r=\frac{b\sqrt{-2E}}{2\pi}
\int_{x_-}^{x_+}
\left(\frac1x+\frac1{x-2}\right)
\sqrt{(x-x_-)(x_+-x)}\,dx.
$$
Applying the supplied integral twice, with $c=0$ and $c=2$, and using the root sum and products from parts b and c gives
$$
\boxed{
J_r=\frac{GM}{\sqrt{-2E}}
-\frac12\left(L+\sqrt{L^2+4GMb}\right)}.
$$
This is the <radial action of the spherical isochrone model>. Rearranging,
$$
\frac{2GM}{\sqrt{-2E}}
=2J_r+L+\sqrt{L^2+4GMb},
$$
and squaring yields the <Hamiltonian> in <action-angle variables>:
$$
\boxed{
H(J_r,L)
=-\frac{2(GM)^2}
{\left(2J_r+L+\sqrt{L^2+4GMb}\right)^2}}.
$$
As a check, differentiating this Hamiltonian gives $\Omega_r=(-2E)^{3/2}/(GM)$ and the frequency ratio found in part c.

Solution

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

{title2=$J_r$}

For specific energy $E<0$ and specific angular momentum $L$,
$$
J_r=\frac{GM}{\sqrt{-2E}}
-\frac12\left(L+\sqrt{L^2+4GMb}\right).
$$
Consequently the Hamiltonian in actions is
$$
H(J_r,L)=-\frac{2(GM)^2}
{\left(2J_r+L+\sqrt{L^2+4GMb}\right)^2}.
$$

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 Incoming links: Radial action of the spherical isochrone model