= Solution
For the <radial epicyclic frequency>, introduce the <effective potential> at fixed <specific angular momentum> $l$: $\Phi_{\rm eff}=\Phi+l^2/(2r^2)$. A <circular orbit> at $r_c$ satisfies $\Phi'(r_c)=l^2/r_c^3=r_c\Omega^2(r_c)$. A small radial displacement $\xi$ obeys $\ddot\xi=-\Phi_{\rm eff}''(r_c)\xi$. Differentiate at fixed $l$ before inserting the circular-orbit relation:
$$
\begin{aligned}
\kappa^2&=\frac{3l^2}{r_c^4}+\Phi''(r_c)
=3\Omega^2+\frac{d}{dr}(r\Omega^2)\\
&=\boxed{4\Omega^2+2r\Omega\Omega'
=4\Omega^2\left(1+\frac r{2\Omega}\Omega'\right)}.
\end{aligned}
$$
For a <flat rotation curve>, $\Omega\propto r^{-1}$, so \b[$\kappa=\sqrt2\,\Omega$].
In a perturbation proportional to $\exp[i(\int k\,dr+n\phi-\omega t)]$, $n$ is the integer azimuthal mode number, giving $|n|$ arms for a nonaxisymmetric pattern. Its angular <pattern speed> is $\Omega_p=\omega/n$ for a real <frequency> and $n\ne0$; $a_0$ is the unperturbed <gas> <sound speed>, or the specified effective fluid pressure-response speed. For growing patterns the angular phase speed uses $\operatorname{Re}\omega/n$. Axisymmetric modes have $n=0$: their <pattern speed> is undefined, and one must use $(\omega-n\Omega)^2\to\omega^2$, not set the pattern-speed expression to zero.
Define the <Toomre parameter> and <Toomre wavenumber> by
$$
\boxed{k_T=\frac{\kappa^2}{2\pi G\sigma},\qquad
Q=\frac{\kappa a_0}{\pi G\sigma}.}
$$
Then $q=|k|/k_T$ gives
$$
\frac{\omega^2}{\kappa^2}
=1-q+\frac{Q^2q^2}{4}
=1-\frac1{Q^2}+\frac{Q^2}{4}
\left(q-\frac2{Q^2}\right)^2.
$$
For $a_0>0$, the minimum is negative precisely when \b[$Q<1$]. Thus this condition is necessary and sufficient for an exponentially growing axisymmetric mode within the given local razor-thin fluid model; $Q=1$ is marginal. It is not a guarantee against global or nonaxisymmetric instabilities, and the collisionless stellar-disk criterion has a different pressure-response normalization.
The <Toomre unstable band and fastest growth> follow by minimizing $\omega^2$:
$$
\boxed{|k|_*=\frac{\pi G\sigma}{a_0^2}
=\frac{2k_T}{Q^2},\qquad
\lambda_*=\frac{2a_0^2}{G\sigma},\qquad
\gamma_*=\kappa\sqrt{Q^{-2}-1}.}
$$
Here the growing branch has $\omega=i\gamma_*$ under the stated Fourier convention. Both signs of the radial <wave number> have the same <growth rate>. The unstable band is $q_-<q<q_+$, where $q_\pm=2(1\pm\sqrt{1-Q^2})/Q^2$.
For the <minimum Toomre parameter in an exponential flat-rotation disk> with constant $a_0$, put $x=r/r_0$:
$$
Q(r)=Q_0\frac{e^{x-1}}x,\qquad
Q_0=\frac{\sqrt2\,\Omega_0a_0}{\pi G\sigma_0}.
$$
The <logarithmic derivative> is $d\log Q/dx=1-1/x$, so the unique minimum occurs at $r=r_0$. Both $Q\to\infty$ as $r\downarrow0$ and $Q\to\infty$ as $r\to\infty$. Hence inner and outer regions are stable within this idealized <rotation> law, even when an intermediate annulus is unstable. The whole disk is strictly stable in this local axisymmetric test if
$$
\boxed{\Omega_0>\frac{\pi G\sigma_0}{\sqrt2\,a_0}.}
$$
Equality makes the ring at $r_0$ marginal. The central conclusion uses the assumed <flat rotation curve> all the way inward; real central <rotation> laws need not retain that limit.
\Image[/past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2012/paper-1-disk-stability.png]
{title=Local disk stability}
Back to article page
