Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / c

Write $y=\nu \Sigma$. A steady inward accretion rate has $F_M=-\dot{M}$, so the integrated angular-momentum equation isFor $\beta =0$, $\mathcal{T}=0$ and thereforeafter imposing the zero-torque inner boundary condition at $r_{\mathrm{in}}$. In a Keplerian accretion disk, $h\propto r^{1/2}$ and $\mathcal{G}=3\pi hy$, so

For a Keplerian orbit, write $h=K{r}^{1/2}$, so $h^{\prime }=h/(2r)$. Substituting the stated torque laws intogivesWiththe integrating factor is $x^{1/2}{e}^{\lambda \sqrt{x}}$. The boundary condition $y(1)=0$ then givesThusand

As $\lambda \to 0$, the exponential function satisfiesConsequentlywhich is the purely viscous result from part (c)(i).

For $\lambda (\sqrt{x}-1)\gg 1$, the exponential is negligible andSince the magnetic prescription is $v_m=\beta r^{-1/2}$, this is precisely $\nu$ times the magnetically driven profile in part (b)(i).

The dimensionless cumulative competition in the exact profile is $\lambda (\sqrt{x}-1)$. The surface magnetic torque dominates wherewhereas the viscous torque in an accretion disk controls the inner transition where this quantity is at most order one. Thus the crossover is approximatelyFor $\lambda \ll 1$ this is $r/r_{\mathrm{in}}\sim {\lambda }^{-2}$; for $\lambda \gg 1$, magnetic transport dominates outside a narrow inner layer of fractional width $O({\lambda }^{-1})$.