Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 3 / i / Solution

For a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.