Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 (Dynamics of Astrophysical Discs)

Inside the innermost stable circular orbit, nearly circular motion is unstable and gas enters the plunging region of a black-hole accretion disk. Its inflow time becomes shorter than the time on which internal stress can communicate angular momentum back to the disk, motivating the zero-torque inner boundary condition at $r_{\mathrm{ISCO}}$.

In a steady state, the given diffusion equation impliesThe constant mass supply fixes $C=\dot{M}/(6\pi )$, and a second integration givesZero torque means $\bar{\nu }\Sigma =0$ at $r=r_{\mathrm{ISCO}}$, so $C_0=r_{\mathrm{ISCO}}^{1/2}$. Thus the Keplerian accretion disk relation is

Vertical hydrostatic equilibrium is established by sound waves over the disk scale height, soThe local heating-cooling balance is established on the thermal timeThe surface density of a disk evolves only as angular momentum is redistributed, on the viscous timeFor a thin disk, $H/r\ll 1$, so $t_{\mathrm{vert}}\lesssim t_{\mathrm{th}}\ll t_{\nu }$. The vertical and thermal equations can therefore be treated as local equilibria while $\Sigma$ evolves.

Write $K=11a/12$, so $P=K{T}^4$. With $\rho \sim \Sigma /H$, vertical balance givesAt fixed $\Sigma$ and radius, the volumetric viscous heating isThe neutrino cooling isThermal stability of an accretion disk requires the cooling rate to have the larger logarithmic temperature slope:Since $\beta >0$, the stable range is

Local thermal equilibrium givessoSubstitute this into $dP/dz=-z{\Omega }^2\rho$ and use $P=K{T}^4$. The result integrates towithTherefore, for $|z|\le H$,The exponent requested for the temperature is $m=\beta$, and the midplane relations are

For $\beta =1/2$, the exact profiles areIntegration through both disk faces givesSince $\mu =\alpha P/\Omega$ and $\bar{\nu }\Sigma =\int \mu dz$,

For $\beta =1/2$, part (b) givesParts (a) and (c)(i) imply $P_{0}H/\Omega \propto \dot{M}f$. With constant $\dot{M}$,For Keplerian rotation, $\Omega \propto r^{-3/2}$, and consequentlyIt follows thatand

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 $R^{1/2}=R_0^{1/2}{\tau }^{1/5}$,

For the Keplerian shearing sheet velocity $\mathbf{U}=-(3/2)\Omega x{\mathbf{e}}_y$, its material acceleration vanishes because $\mathbf{U}\cdot \nabla \mathbf{U}=0$. The $x$ component of the Coriolis acceleration is $-3{\Omega }^{2}x$, which is cancelled by $-\nabla {\Phi }_t=+3{\Omega }^{2}x{\mathbf{e}}_x$ with the signs in the stated equation; constant $P_0$ supplies no force. The flow is also incompressible.

Let $\mathbf{u}=\mathbf{U}+{\mathbf{u}}^{\prime }$ and $h^{\prime }=P^{\prime }/\rho$. Axisymmetry removes $\mathbf{U}\cdot \nabla {\mathbf{u}}^{\prime }$, while ${\mathbf{u}}^{\prime }\cdot \nabla \mathbf{U}=-(3/2)\Omega u_x^{\prime }{\mathbf{e}}_y$. Keeping first-order terms givesDifferentiate the momentum equations in time and use incompressibility to eliminate $h^{\prime }$. The radial velocity obeysFor a plane wave proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, this gives the inertial wave dispersion relation

When $k_x=0$ and $k_z\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $h^{\prime }=0$. The remaining motion has $\omega =\pm \Omega$ and satisfies $u_y^{\prime }=\mp i{u}_x^{\prime }/2$: each horizontal layer executes an epicyclic motion, with the phase varying vertically but no pressure or vertical-velocity perturbation.

Write the vortex aspect ratio as $\chi$ to distinguish it from cylindrical radius. The Kida vortex core flow isAs $\chi \to \infty$, $u_x\to 0$ and $u_y\to -(3/2)\Omega x$, recovering Keplerian shear.

For a fluid particle,It therefore circulates around an ellipse with angular frequency $3\Omega /[2(\chi -1)]$ and period

For a perturbation depending only on $z,t$, horizontal pressure gradients vanish. Linearization givesTaking ${\mathbf{u}}^{\prime }\propto e^{-i\omega t}$ yieldsThe first bracket is positive for $\chi >3/2$, while the second is negative for $1<\chi <4$. Their product is therefore negative, so $\omega$ is imaginary and the mode grows precisely when