Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 347 / 3

Vertically integrating the continuity equation and imposing a steady axisymmetric flow givesThus the inward-positive mass accretion rate isThe zero-torque condition at $R_{\mathrm{ISCO}}$ gives the standard Shakura--Sunyaev thin disk relationCombining this relation with mass conservation, or substituting it into the stated drift equation, yieldsFar outside the inner edge this reduces to $u_R\simeq -3\nu /(2R)$.

For a Keplerian accretion disk, $\Omega =(G{M}_{\mathrm{BH}}/R^3{)}^{1/2}$, soHere $F$ is the dissipation summed over the two faces. The total luminosity isThis is the binding energy per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The zero-torque inner boundary condition ensures that no additional mechanical work enters from smaller radii.

The luminosity emitted outside radius $R$ isWriting $y=(R_{\mathrm{ISCO}}/R{)}^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and henceHalf of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.

Far from the inner edge, use $u_R\simeq -3\nu /(2R)$ and the alpha-viscosity prescriptionMass conservation then givesAt the onset of disk self-gravity, $\rho =M_{\mathrm{BH}}/R^3$. Since ${\Omega }^2=G{M}_{\mathrm{BH}}/R^3$, this condition reduces to

The blackbody approximation and local radiative balance on one face givewhile the stated ideal-gas estimate is $c_s^2=\gamma k_B{T}_{\mathrm{eff}}/\mu$. Inserting these expressions into the self-gravity condition givesSolving for $R$ producesBeyond this radius, the disk's own gravity rivals the central vertical gravity; fragmentation and star formation can then invalidate the smooth steady thin-disk model.