Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 321 / 1

In the Toomre stability criterionthe disk's gravitational potential supplies the destabilizing attraction, gas pressure represented by $c_s$ suppresses short wavelengths, and the Keplerian shear represented by the angular frequency $\Omega$ prevents coherent collapse on long scales through epicyclic motion. Axisymmetric perturbations are stable for $Q>1$ and gravitationally unstable for $Q<1$.

The state $Q\sim 1$ is a self-regulating fixed point. If cooling lowers $c_s$ until $Q<1$, gravitational instability produces shocks and turbulence that heat the disk and raise its effective velocity dispersion. If $Q$ becomes appreciably larger than one, the instability and its heating switch off, while turbulent dissipation and radiative cooling reduce $c_s$. A sustained gravito-turbulent state therefore remains close to marginal stability.

The first equation is vertical hydrostatic equilibrium: the pressure gradient balances the vertical gravity of the central mass, ${\Omega }^{2}z$, and the disk's own potential ${\Phi }_d$. The second is the plane-parallel Poisson equation for disk self-gravity. The third balances the vertical increase of radiative flux against local viscous heating in a Keplerian alpha disk, and the fourth is the optically thick radiative diffusion law.

With $P/\rho \sim c_s^2$, $\Sigma \sim \rho H$, and all terms in hydrostatic balance comparable,The first comparison gives $H\sim c_s/\Omega$; inserting it into the second gives $\Omega c_s\sim G\Sigma$. Henceup to the order-one constants deliberately omitted by the scaling argument.

For a perfect gas, $c_s^2\sim kT/({\mu }_m{m}_p)$. At the margin of the Toomre stability criterion,Integrating viscous heating through the disk and applying radiative diffusion givesEquating these fluxes yieldsThe kinematic viscosity is $\bar{\nu }\sim \alpha c_s^2/\Omega$, soBecause a Keplerian accretion disk has $\Omega \propto r^{-3/2}$,

For $P=K{\rho }^{1+1/n}$,The pseudo-enthalpy definition therefore integrates towhere the constant makes $w=0$ at the surface. Thus

Hydrostatic balance is $c_0^{2}dw/dz=-{\Omega }^{2}z-d{\Phi }_d/dz$. Differentiate it, use the Poisson equation, and set $\zeta =\Omega z/c_0$:Since the definitions in the question give $Q_0={\Omega }^2/(\pi G{\rho }_0)$,

For $n=1$, put $\lambda =2/\sqrt{Q_0}$. The solution isThe surface $z=H$ is the first zero, so, using $c_0^2=2K{\rho }_0$,