Toomre's stability criterion ($Q$)

For a razor-thin isothermal gas disk,Axisymmetric disturbances are stable when $Q\ge 1$: pressure stabilizes short wavelengths, epicyclic motion stabilizes long wavelengths, and self-gravity drives intermediate wavelengths.

At the marginal radial wavenumber, the forced axisymmetric density mode with vortensity has zero squared frequency. Its mass density amplitude is $C_0+C_{1}t-\Omega {\Sigma }_0^2{q}^{\prime }{t}^2$. Thus there is no exponential instability but nonzero conserved vortensity can force quadratic growth, and a homogeneous generalized mode can grow linearly. The spectral boundary $Q=1$ of the Toomre stability criterion must not be mistaken for boundedness of every initial-value solution.

Linear axisymmetric compressive modes of a barotropic closure of a razor-thin disk combine radial epicyclic restoration, softened self-gravity and pressure. The off-plane razor-thin Poisson kernel supplies the exponential gravity reduction. Here $k$ is the nonnegative radial wavenumber magnitude. The full linear system also has an axisymmetric geostrophic mode.

The most unstable softened disk wavenumber minimizes the squared mode frequency. Instability is $Q<Q_c$, while $Q\ge Q_c$ is spectrally stable for this compressive pair. Since $s_{\ast }<(1+\delta {)}^{-1}$, $Q_c^2<(1+2\delta )/(1+\delta {)}^2\le 1$. At zero gravitational softening both bounds become equality. The physical path with fixed $\kappa ϵ/c_s$ obeys $\delta =(\kappa ϵ/c_s)/Q$.

With $s=Q{c}_{s}k/\kappa$ and $\delta =\kappa ϵ/(Q{c}_s)$, maximize $F(s)=2s{e}^{-\delta s}-s^2$. Its unique positive critical point satisfies the displayed equation. For positive gravitational softening, $e^{-\delta s_{\ast }}<1$ and $1-\delta s_{\ast }>0$, giving the strict bound. Unsoftened gravity has $s_{\ast }=1$.

For a razor-thin fluid disk with sound speed or effective random speed $c$, surface density $\Sigma$ and radial epicyclic frequency $\kappa$, $Q=c\kappa /(\pi G\Sigma )$ compares pressure and rotational support with self-gravity. The local axisymmetric fluid Toomre stability criterion is $Q\ge 1$. For a Keplerian shearing sheet, $\kappa =\Omega$. Collisionless stellar disks use a different numerical normalization; gas drag can permit secular growth even when this fluid parameter exceeds unity.

For $\Omega ={\Omega }_0{r}_0/r$, surface density $\Sigma ={\Sigma }_0{e}^{1-r/r_0}$ and constant sound speed $a_0$, the Toomre parameter is $Q(x)=Q_0{e}^{x-1}/x$, where $x=r/r_0$ and $Q_0=\sqrt{2}{\Omega }_0{a}_0/(\pi G{\Sigma }_0)$. Its logarithmic derivative is $1-1/x$, so the minimum is $Q_0$ at $r_0$. The formal inner and outer limits both diverge; if $Q_0<1$, local axisymmetric instability occupies a bounded intermediate annulus. A central departure from the assumed flat rotation law can change the inner conclusion.

For the razor-thin fluid disk dispersion relation, set $Q=\kappa a_0/(\pi G\Sigma )$ and $k_T={\kappa }^2/(2\pi G\Sigma )$. When $Q<1$, the unstable band isThe quadratic dispersion has its minimum at $|k{|}_{\ast }=\pi G\Sigma /a_0^2$, with fastest growth ${\gamma }_{\ast }=\kappa \sqrt{Q^{-2}-1}$. At $Q=1$ the two band edges coincide and the mode is marginal. These are local axisymmetric fluid results, not a global stability theorem.

In the razor-thin fluid disk dispersion relation, the cold limit is ${\omega }^2={\kappa }^2-2\pi G\Sigma |k|$. Rotation therefore suppresses wavelengths above ${\lambda }_T=2\pi /k_T=4{\pi }^{2}G\Sigma /{\kappa }^2$. With this convention, $k_T$ is the cold-disk boundary of instability, not the fastest-growing wave number when finite pressure is present. At marginal fluid stability $Q=1$, the minimum occurs at $|k|=2k_T$.

For constant circular speed $v_0$, angular velocity is $v_0/r$ and radial epicyclic frequency is $\sqrt{2}{v}_0/r$. Constant effective random speed and surface density then give $Q=r_Q/r$. The local axisymmetric fluid model is stable inside $r_Q$ and unstable outside it. This statement is conditional on the thin-disk local approximation, rather than a proof of global or collisionless stability.

At fluid Toomre parameter $Q=1$, the local spiral-wave dispersion relation becomes $n^2({\Omega }_P-\Omega {)}^2=(a_0|k|-\kappa {)}^2$. The minus-sign, longer-wavelength root is nonnegative only between the two Lindblad resonance conditions. The plus-sign, shorter-wavelength root remains positive outside that interval. Thus this dispersion relation alone does not prove that every density wave is confined between Lindblad resonances; an additional physical restriction is needed.

In a star-dominated Keplerian disk with $H\simeq c_s/\Omega$ and characteristic disk mass $M_D\sim \pi r^2\Sigma$, the Toomre stability criterion becomes $Q\sim (H/r)(M_{\ast }/M_D)$. Thus instability can occur at a disk-to-star mass ratio comparable to the small disk aspect ratio. The numerical coefficient depends on whether $M_D$ means a local annular estimate or integrated enclosed mass.

In a cooling self-gravitating disk, $Q<1$ excites gravitational turbulence whose dissipation raises the sound speed, while $Q>1$ suppresses that heating and lets cooling lower the sound speed. A sustained state can therefore regulate itself near $Q\simeq 1$.

A mixing-length estimate for gravitational turbulence takes the largest unstable wavelength $\ell$ and orbital growth rate $\Omega$, giving turbulent velocity $u_{\mathrm{turb}}\sim \Omega \ell$ and kinematic viscosity $\nu \sim u_{\mathrm{turb}}\ell \sim \Omega {\ell }^2$.

For a locally plane-parallel thin disk, assume a polytropic equation of state $P=K{\rho }^{1+1/n}$ with $K>0$ and polytropic index $n>0$. Let ${\rho }_0$ be the midplane mass density and $c_0^2=(1+1/n)K{\rho }_0^{1/n}$ the midplane squared sound speed. Take the external vertical gravitational acceleration to be $-{\Omega }^{2}z$, as for a local Keplerian disk.

The specific enthalpy is $\mathcal{H}=(n+1)K{\rho }^{1/n}$. Define $w=\mathcal{H}/c_0^2=n(\rho /{\rho }_0{)}^{1/n}$ and $\zeta =\Omega z/c_0$. Hydrostatic equilibrium gives $d\mathcal{H}/dz=-{\Omega }^{2}z-d{\Phi }_d/dz$, while the plane-parallel Poisson equation gives $d^2{\Phi }_d/d{z}^2=4\pi G\rho$. Combining them yieldsThis $Q_0$ is a midplane-density parameter; it is not the surface-density Toomre parameter. Midplane symmetry imposes $w(0)=n$ and $w^{\prime }(0)=0$. The first positive zero of $w$ defines the model's surface, where the mass density vanishes.

In the negligible-self-gravity limit $Q_0\to \infty$, $w=n-{\zeta }^2/2$ and the half-thickness is $\sqrt{2n}{c}_0/\Omega$. This finite polytropic surface differs from an isothermal Gaussian atmosphere with no sharp edge.^[ref]

The self-gravitating radius of an accretion disk is the radius at which its Toomre stability criterion reaches $Q=1$. Beyond it, a smooth thin disk is vulnerable to spiral structure, gravitational transport, or fragmentation.