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

Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

With $\mathcal{G}=0$ and $\mathcal{T}=2\pi v_m{h}^{\prime }\Sigma$, the evolution equation reduces to the radial continuity equationor, for the mass per unit radius $M_r=2\pi r\Sigma$,The outward mass flux is $F_M=2\pi r{v}_m\Sigma$. If $\dot{M}>0$ denotes the inward accretion rate, steady mass conservation requires $F_M=-\dot{M}$. HenceThis is positive because the magnetic drift velocity satisfies $v_m<0$.

For constant $v_m$, $M_r$ obeys the linear transport equationThe method of characteristics therefore givesThe localized profile translates inward without changing shape: material initially concentrated near $r_0$ reaches the central star after

Each annulus loses specific angular momentum as its characteristic moves to smaller $r$. The negative magnetic torque transmits that angular momentum along the large-scale field to the anchored interstellar medium. Thus the disk's angular momentum decreases even before mass crosses $r=0$; angular momentum is conserved only after the field and its anchoring medium are included.

Now the mass per unit radius satisfies the nonlinear scalar conservation lawIts characteristic curves have speedBecause $v_m<0$ and $|q^{\prime }|$ increases with $M_r$, the high central part of a localized pulse travels inward faster than its low-density edges. On the inner, rising side, faster high-density characteristics catch slower low-density ones, causing characteristic crossing and an inward-facing shock wave. On the outer, falling side, the characteristics separate and form a rarefaction wave. The initial bump therefore develops a sharp inner front and a broad outer tail while moving toward the star.

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 $O({\lambda }^{-1})$.