= Solution
To test <radial stability of an axisymmetric circular orbit>, let $\Phi(r)$ be the time-independent <gravitational potential> per unit stellar mass in the midplane. Axial symmetry conserves the <specific angular momentum> $j=r^2\dot\varphi$. The radial equation is
$$
\ddot r=\frac{j^2}{r^3}-\Phi'(r)
=-\frac{\mathrm d\Phi_{\rm eff}}{\mathrm dr},
\qquad
\Phi_{\rm eff}(r)=\Phi(r)+\frac{j^2}{2r^2}.
$$
A <circular orbit> at $r_0$ obeys $j_0^2=r_0^3\Phi'(r_0)$. Choose the guiding <circular orbit> with the same conserved <angular momentum> as the perturbed <star>. Setting $r=r_0+x$ and expanding at fixed $j=j_0$ gives
$$
\ddot x=
\left[-\frac{3j_0^2}{r_0^4}-\Phi''(r_0)\right]x+O(x^2)
=-\kappa^2x+O(x^2),
$$
where the <radial epicyclic frequency> is
$$
\boxed{\kappa^2=\Phi''(r_0)+\frac3{r_0}\Phi'(r_0)
=4\Omega_0^2+
r_0\left.\frac{\mathrm d\Omega^2}{\mathrm dr}\right|_{r_0}
=\left.\frac1{r^3}\frac{\mathrm d j_c^2}{\mathrm dr}\right|_{r_0}.}
$$
Here $\Omega^2(r)=\Phi'(r)/r$ and $j_c=r^2\Omega$ refer to the family of <circular orbits>. They are differentiated as functions of radius only after the perturbation equation has been expanded at conserved stellar $j$.
For a radially stable <circular orbit>, $\kappa^2>0$, so to linear order
$$
\boxed{x(t)=C\cos(\kappa t)+D\sin(\kappa t).}
$$
Thus the orbit performs small simple harmonic radial oscillations about its guiding radius.
\Image[/past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2016/paper-1-epicycle.png]
{title=Original linear epicycle calculation for stable circular motion with a flat circular-speed curve}
Symmetry alone is insufficient to guarantee <stability>: $\kappa^2<0$ gives exponentially growing displacement, and $\kappa^2=0$ is marginal. The intended statement assumes a stable unperturbed orbit.
For a Kepler potential $\kappa=\Omega$; for a flat circular-speed curve $\kappa=\sqrt2\,\Omega$; for <solid-body rotation> $\kappa=2\Omega$. Conservation of $j$ also gives $\dot\varphi\simeq\Omega_0(1-2x/r_0)$, explaining the accompanying azimuthal epicycle rather than a purely radial motion in the inertial plane.
Back to article page
