Vertically integrating the <continuity equation> and imposing a steady <axisymmetric flow> gives
$$
\frac1R\frac d{dR}(R\Sigma u_R)=0.
$$
Thus the inward-positive <mass accretion rate> is
$$
\boxed{\dot m=-2\pi R\Sigma u_R=\text{constant}}.
$$
The zero-torque condition at $R_{\rm ISCO}$ gives the standard <Shakura--Sunyaev thin disk> relation
$$
\nu\Sigma=\frac{\dot m}{3\pi}
\left[1-\left(\frac{R_{\rm ISCO}}R\right)^{1/2}\right].
$$
Combining this relation with mass conservation, or substituting it into the stated drift equation, yields
$$
\boxed{u_R=-\frac{3\nu}{2R}
\left[1-\left(\frac{R_{\rm ISCO}}R\right)^{1/2}\right]^{-1}}.
$$
Far outside the inner edge this reduces to $u_R\simeq-3\nu/(2R)$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-347/3/a/solution