Steady <mass conservation> gives $\dot M=-2\pi R\Sigma u_R$. Substituting this into the angular-momentum equation and integrating from the <innermost stable circular orbit> with zero torque gives
$$
\nu\Sigma R^3\frac{d\Omega}{dR}
=-\frac{\dot M}{2\pi}(l-l_{\rm ISCO})
=R\Sigma u_R(l-l_{\rm ISCO}).
$$
Therefore
$$
u_R=\frac{\nu R^2\,d\Omega/dR}{l-l_{\rm ISCO}}.
$$
Using $\nu=\alpha c_sH$, $c_s\simeq H\Omega$, $l=Ru_\phi$, and $R\,d\Omega/dR$ of order $-\Omega$ gives
$$
\boxed{|u_R|\simeq
\alpha\frac{H^2}{R^2}
\frac{u_\phi l}{l-l_{\rm ISCO}},}
$$
up to the order-unity Keplerian factor $3/2$.

At the sonic transition, $|u_R|\simeq c_s\simeq(H/R)u_\phi$. Hence
$$
\frac{l-l_{\rm ISCO}}l\simeq\alpha\frac HR\ll1
$$
for a geometrically thin disk with $\alpha\ll1$. The <specific angular momentum> therefore differs only fractionally from $l_{\rm ISCO}$ before the gas enters the <plunging region of a black-hole accretion disk>. Its much shorter inflow time then prevents appreciable viscous transport, justifying angular-momentum conservation across the ISCO and the zero-torque boundary condition.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 347 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-347/2/c/solution