Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1

For the Paczyński-Wiita potential, circular balance givesThereforeThe squared radial epicyclic frequency isCircular orbits change from stable to unstable where ${\kappa }_r^2$ vanishes, so

The steady disk should end at the innermost stable circular orbit, $r_{\mathrm{in}}=3r_S$. Inside it, matter plunges inward on an orbital timescale and cannot maintain the nearly circular shear flow assumed by the viscous-disk equations. If the plunging region communicates negligible stress back across the ISCO, the natural boundary condition is the zero-torque inner boundary condition

Take $\dot{M}>0$ for inward accretion, so the outward radial mass flux is $F=-\dot{M}$. Since $F=-(dh/dr{)}^{-1}d\mathcal{G}/dr$ and $\mathcal{G}(r_{\mathrm{in}})=0$,Using $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^{3}d\Omega /dr$ and $x=r/r_S$ givesThus $\boxed{A=3\sqrt{3}/2}$. Far from the hole this approaches the Keplerian accretion disk result $\bar{\nu }\Sigma =\dot{M}[1-(r_{\mathrm{in}}/r{)}^{1/2}]/(3\pi )$, although the pseudo-Newtonian boundary factor retains a different finite-radius shape.

Let $c_s^2\sim P_g/\rho$. The assumed magnetic pressure givesso the plasma beta obeys $\boxed{\beta =P_g/P_m\sim {\mathrm{Ma}}^{-1}}$. In the magnetically dominated regime, vertical hydrostatic equilibrium gives $P_m/\rho \sim {\Omega }^2{H}^2$. Combining this with $P_m/\rho \sim r\Omega c_s$ yields

Order-of-magnitude vertical balance givesMagnetic support also gives ${\Omega }^2{H}^2\sim r\Omega T^{1/2}$, so $T\sim {\Omega }^2{H}^4/r^2$. Eliminating $P,F,\rho ,T$ givesFor Keplerian $\Omega \propto r^{-3/2}$,The alpha disk stress implies $\bar{\nu }\sim \alpha P/(\rho \Omega )\sim \alpha \Omega H^2$, and therefore