Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 347 / 2

The standard Shakura--Sunyaev thin disk assumes: a steady state; axial symmetry; a geometrically thin disk $H/R\ll 1$; Newtonian, nearly Keplerian circular motion outside the inner edge; subsonic radial drift $|u_R|\ll u_{\varphi }$; vertical hydrostatic equilibrium; negligible disk self-gravity; an optically thick, locally thermal spectrum; local radiative balance between viscous heating and cooling; and a local alpha-viscosity stress with a zero-torque inner boundary near the innermost stable circular orbit. These assumptions also exclude dynamically dominant winds and large-scale external torques from the standard solution.

Start from the continuity equationIntegrate vertically, use axial symmetry and steadiness, and define the surface density of a disk $\Sigma =\int \rho dz$. Thenso the inward-positive accretion rate is

In a Keplerian accretion disk, $u_{\varphi }=R\Omega =(GM/R{)}^{1/2}$. Substitution into the steady azimuthal Navier--Stokes equation and cancellation of the common Keplerian factors givesSince $R^2\Omega \propto R^{1/2}$ and $d\Omega /dR=-3\Omega /(2R)$, this reduces to the standard viscous drift formulaThe minus sign describes inward drift when $\nu \Sigma R^{1/2}$ increases outward.

Let $l(R)=R^2\Omega =\sqrt{GMR}$ be the Keplerian specific angular momentum. In a source-free steady interval, the sum of advected and viscously transported angular momentum is constant:Inside the injection radius, $\dot{m}={\dot{m}}_0$. The zero-torque inner boundary condition at $R_{\mathrm{ISCO}}$ sets $C={\dot{m}}_0{l}_{\mathrm{ISCO}}$, and therefore

Outside $R_0$ there is no net mass flow in the stated steady distribution, but it must carry outward the angular momentum deposited by matter moving from $R_0$ to the ISCO. Its constant viscous torque in an accretion disk is thereforeThusThe two expressions agree at $R_0$; the jump in mass flux there is exactly the injected rate.

For Keplerian angular velocity, $R^2(d\Omega /dR{)}^2=9GM/(4R^3)$. The stated $F_{\mathrm{diss}}$ is the dissipation summed over both disk faces, so $F_{\mathrm{diss}}=2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. Inside $R_0$ this givesDifferentiating the factor $R^{-3}[1-(R_{\mathrm{ISCO}}/R{)}^{1/2}]$ shows that the maximum occurs atwhen this radius lies below $R_0$. In the ordinary inflowing region far from its inner edge, $T_{\mathrm{eff}}\propto R^{-3/4}$.

For the static angular-momentum sink outside $R_0$, part c instead givesso the genuinely large-radius behavior of the complete injected disk is $T_{\mathrm{eff}}\propto R^{-7/8}$.

Integrating $2\pi R{F}_{\mathrm{diss}}dR$ over both regions givesandHencewhich is exactly the loss of Keplerian orbital energy as matter moves from its injection orbit to the inner edge.

For a standard disk extending through a radius $R\gg R_{\mathrm{ISCO}}$,This is three times the binding-energy release $GM\dot{m}/(2R)$ available outside $R$. The excess is energy carried outward by the viscous torque in an accretion disk and dissipated at larger radii. In the injected model the nonaccreting outer disk is an especially direct example: it radiates despite having zero mean radial mass flux because it absorbs the angular momentum and mechanical work exported by the inner disk.