Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 321 / 1

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

Here $\beta =p_{\mathrm{rad}}/p_{\mathrm{gas}}$ is the radiation-to-gas pressure ratio, not plasma beta. Write the opacity as ${\kappa }_{\mathrm{op}}$ to distinguish it from the radial epicyclic frequency. The radiation pressure is $p_{\mathrm{rad}}=4\sigma T^4/(3c)$, so radiative diffusion can be written asSince $\beta$ is independent of height, $p_{\mathrm{rad}}=\beta p/(1+\beta )$. Substitution of hydrostatic equilibrium, $p^{\prime }=-\rho {\Omega }^{2}z$, givesDifferentiating and using the viscous heating equation, together with $r{\Omega }^{\prime }=-q\Omega$, yieldsThus the effective dynamic viscosity $\rho \nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated kinematic viscosity satisfiesUse the previous result and the specified Eddington accretion rate convention,to obtain the full thicknessThe constant-pressure-ratio vertical disk model therefore has thickness proportional to the inward accretion rate. This is a full surface-to-surface thickness, rather than a density disk scale height. Its use as a thin disk requires $2H\ll r$.