= Solution
In a stationary axisymmetric <gravitational potential>, <conservation of angular momentum> gives $L_z=R^2\dot\phi$. Eliminating $\dot\phi$ introduces the <effective potential>
$$
\Phi_{\rm eff}(R,z)=\Phi(R,z)+\frac{L_z^2}{2R^2},\qquad\ddot R=-\partial_R\Phi_{\rm eff},\quad\ddot z=-\partial_z\Phi.
$$
For an equatorial <circular orbit>, radial balance is $\partial_R\Phi_{\rm eff}(R_c,0)=0$, giving
$$
\boxed{\Phi_{,R}(R_c,0)=\frac{L_z^2}{R_c^3}.}
$$
Vertical balance also requires $\Phi_{,z}(R_c,0)=0$.
For <epicyclic motion>, assume a twice differentiable stationary <gravitational potential> symmetric under $z\mapsto-z$, small displacements $|x|,|z|\ll R_c$, and a stable <circular orbit>. Choose the <epicyclic guiding center> using the conserved $L_z$. Reflection symmetry makes $\Phi_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ gives
$$
\boxed{\ddot x=-\kappa^2x,\qquad\ddot z=-\nu^2z,\qquad\kappa^2=\Phi_{,RR}+\frac{3L_z^2}{R_c^4},\quad\nu^2=\Phi_{,zz}.}
$$
The derivatives are evaluated at the guiding centre, and stability requires $\kappa^2,\nu^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The <radial epicyclic frequency> and <vertical epicyclic frequency> are the frequencies of these independent linear oscillations.
Set $\Omega=L_z/R_c^2$. Along the family of equatorial <circular orbits>, $\Phi_{,R}=R\Omega^2$. Differentiating this relation and adding the centrifugal contribution gives
$$
\boxed{\kappa^2=R\frac{d\Omega^2}{dR}+4\Omega^2.}
$$
To interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has $\kappa^2=2(1+q)\Omega^2$. A <Keplerian disk> has $q=-1/2$ and $\kappa=\Omega$, a <flat galaxy rotation curve> has $q=0$ and $\kappa=\sqrt2\Omega$, and <solid-body rotation> has $q=1$ and $\kappa=2\Omega$. Typical galactic <rotation curves> lie between these slopes. Thus \b[$\Omega\lesssim\kappa\lesssim2\Omega$ is a useful galactic range], not a theorem for every possible axisymmetric potential. As a precise sufficient example, <epicyclic frequency bounds for monotone spherical density> follow from $\kappa^2=\Omega^2+4\pi G\rho$ and $0\leq\rho\leq3M/(4\pi R^3)$.
Use a Cartesian frame rotating with the <epicyclic guiding center>, with $x$ pointing radially outwards and $y=R_c(\phi-\Omega t)$ in the direction of rotation. To first order, <conservation of angular momentum> gives
$$
\dot\phi=\frac{L_z}{(R_c+x)^2}=\Omega\left(1-\frac{2x}{R_c}\right)+O(x^2/R_c^2),\qquad\dot y=-2\Omega x.
$$
The radial <harmonic oscillator> solution and its azimuthal integral are
$$
\boxed{x=X\cos(\kappa t+\chi),\qquad y=-\frac{2\Omega X}{\kappa}\sin(\kappa t+\chi)=-Y\sin(\kappa t+\chi),\qquad\frac XY=\frac\kappa{2\Omega}.}
$$
A constant in $y$ merely changes the azimuthal origin of the guiding centre. The <epicyclic ellipse> obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.
For the <Oort constants>, subtract the defining expressions to obtain $\Omega=A-B$ and insert $R\Omega'=-2A$ into the <radial epicyclic frequency> formula:
$$
\kappa^2=4\Omega(\Omega-A)=-4\Omega B,\qquad\frac XY=\sqrt{\frac{-B}{A-B}}.
$$
The solar-neighbourhood values give $\Omega=26.5\,\mathrm{km\,s^{-1}\,kpc^{-1}}$, $\kappa=\sqrt{1272}\,\mathrm{km\,s^{-1}\,kpc^{-1}}$, and
$$
\boxed{X/Y=\sqrt{24/53}\simeq0.673.}
$$
The solar <epicyclic ellipse> is therefore about $1.49$ times longer azimuthally than radially.
A complete radial oscillation takes $T_r=2\pi/\kappa$. The oscillatory part of $\dot\phi$ has zero average over this interval, so the <epicyclic azimuthal advance> is
$$
\boxed{\Delta\psi=\Omega T_r=\frac{2\pi\Omega}{\kappa}=2\pi\left(4+\frac{d\log\Omega^2}{d\log R}\right)^{-1/2}.}
$$
This describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,
$$
\boxed{\Delta\psi=\pi\sqrt{53/24}\simeq4.669\ \mathrm{rad}\simeq267.5^\circ.}
$$
The azimuthal advance estimate uses the same linear <epicyclic motion> approximation as the axis ratio.
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