Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 1

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