Inside the <innermost stable circular orbit>, nearly circular motion is unstable and gas enters the <plunging region of a black-hole accretion disk>. Its inflow time becomes shorter than the time on which internal stress can communicate <angular momentum> back to the disk, motivating the <zero-torque inner boundary condition> at $r_{\rm ISCO}$.

In a <steady state>, the given diffusion equation implies
$$
r^{1/2}\frac d{dr}\left(r^{1/2}\bar\nu\Sigma\right)=C.
$$
The constant mass supply fixes $C=\dot M/(6\pi)$, and a second integration gives
$$
\bar\nu\Sigma=\frac{\dot M}{3\pi}
\left[1-C_0r^{-1/2}\right].
$$
Zero torque means $\bar\nu\Sigma=0$ at $r=r_{\rm ISCO}$, so $C_0=r_{\rm ISCO}^{1/2}$. Thus the <Keplerian accretion disk> relation is
$$
\boxed{\bar\nu\Sigma=\dot M f(r)},
\qquad
\boxed{f(r)=\frac1{3\pi}
\left[1-\left(\frac{r_{\rm ISCO}}r\right)^{1/2}\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-321/1/a/solution