 Solution

The alpha disk relation and Keplerian angular frequency giveCombining this with $\Omega \propto M^{1/2}{R}^{-3/2}$ and the surface-density scaling from part (d), the Toomre stability criterion becomesIt decreases strictly with radius, so it crosses unity at a unique self-gravitating radius of an accretion disk. Expressing radius in units of $R_S\propto M$ givesSolving $Q(R_{\mathrm{sg}})=1$ therefore yieldsand

Let $x_{\mathrm{ISCO}}=R_{\mathrm{ISCO}}/R_S$. Requiring a non-self-gravitating annulus outside the innermost stable circular orbit gives $R_{\mathrm{sg}}>R_{\mathrm{ISCO}}$, and equality definesFor a nonspinning hole, $x_{\mathrm{ISCO}}=3$, and with $C_2\simeq 10^5$, $\alpha \simeq 0.1$, and $f_{\mathrm{Edd}}\simeq 1$, this is approximately $8\times 10^9{M}_{\odot }$, conventionally quoted as order $10^{10}{M}_{\odot }$. Above this mass the disk would become self-gravitating essentially as soon as stable circular orbits begin, so the assumed smooth Shakura--Sunyaev thin disk cannot provide a broad luminous accretion region.

The corresponding Eddington luminosity is of order $10^{48}\mathrm{erg}{\mathrm{s}}^{-1}$, comparable to the upper envelope of quasar luminosities. This supports self-gravity as one contributor to the observed luminous-mass ceiling. It is not an absolute upper bound on black-hole mass: mergers, radiatively inefficient growth, nonstandard gas supply, spin-dependent inner radii, and fragmented or episodic accretion can all build a more massive hole without maintaining this particular steady thin disk.