Take $\dot M>0$ for inward accretion, so the outward radial mass flux is $F=-\dot M$. Since $F=-(dh/dr)^{-1}d\mathcal G/dr$ and $\mathcal G(r_{\rm in})=0$,
$$
\mathcal G=\dot M(h-h_{\rm in}),\qquad
h_{\rm in}=\frac{3\sqrt3}{2}\sqrt{GMr_S}.
$$
Using $\mathcal G=-2\pi\bar\nu\Sigma r^3d\Omega/dr$ and $x=r/r_S$ gives
$$
\boxed{\bar\nu\Sigma=
\frac{\dot M}{\pi}\frac{x-1}{3x-1}
\left[1-\frac{3\sqrt3}{2}\left(x^{-1/2}-x^{-3/2}\right)\right]}.
$$
Thus $\boxed{A=3\sqrt3/2}$. Far from the hole this approaches the <Keplerian accretion disk> result $\bar\nu\Sigma=\dot M[1-(r_{\rm in}/r)^{1/2}]/(3\pi)$, although the pseudo-Newtonian boundary factor retains a different finite-radius shape.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-321/1/b/ii/solution