Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 2

The three terms represent competing physical effects:

Without self-gravity and for $k>0$,The phase velocity and group velocity areso

With self-gravity restored,The product changes sign atFor $k>k_{\mathrm{crit}}$, crests and a localized wave packet travel in the same radial direction. For $k<k_{\mathrm{crit}}$, the group velocity and phase velocity have opposite signs: the envelope and its wave energy propagate opposite to the individual crests. At $k_{\mathrm{crit}}$ the group velocity vanishes and neighboring wave components cause the packet to spread.

Put $x=kH$, with $H=c_s/\Omega$ and $Q=c_s\Omega /(\pi G\Sigma )$. The dispersion relation becomesThe two neutral roots areReal roots enclosing a range with ${\omega }^2<0$ exist precisely whenwhich is the Toomre stability criterion for axisymmetric instability. For $Q\ll 1$, the Taylor expansion of the square root gives

The longest unstable disturbance has mixing length $\ell \sim k_1^{-1}$ and grows on the orbital timescale ${\Omega }^{-1}$. Its characteristic turbulent velocity is therefore $u_{\mathrm{turb}}\sim \Omega \ell$, so the gravitoturbulent viscosity estimate isFor $Q\ll 1$,up to an unimportant numerical factor. Hencein a Keplerian disk.

For a disk of characteristic radius $R$ and mass $M_D$, $\Sigma \sim M_D/R^2$. The viscous timescale isUsing ${\Omega }^2=G{M}_{\ast }/R^3$ gives

The specific angular momentum of a circular Keplerian orbit is $\sqrt{G{M}_{\ast }r}$. Therefore the disk angular momentum iswhereThe absence of an external or inner-boundary torque makes $J$, and hence $B$, constant.

From $\bar{\nu }=A{r}^{9/2}{\Sigma }^2$,The combination $A{B}^{2}t$ has dimension $L^{5/2}$. Dimensional analysis therefore gives

For the supplied similarity solution, put $\tau =t/t_0$ and $u=\sqrt{r/R(t)}$. Then $r=R{u}^2$ and $dr=2Rudu$. The total mass isSince $R^{1/2}=R_0^{1/2}{\tau }^{1/5}$,