Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 (Dynamics of Astrophysical Discs)

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

For $|z|\ll r$, the gravitational potential has the Taylor expansionHence the vertical gravity is $-{\partial }_z\Phi \simeq -{\Omega }_K^{2}z$, where ${\Omega }_K^2=G{M}_{\star }/r^3$. Vertical hydrostatic equilibrium balances a pressure gradient of order $P/H$ against $\rho {\Omega }_K^{2}H$. Since $c_s^2\sim P/\rho$,The thin disk condition $H/r\ll 1$ is therefore equivalent to a highly supersonic orbital speed.

Put $K(r)=e^{s(r)}$. The perfect gas relation in the question becomes the polytropic equation of stateAt fixed $r$, vertical hydrostatic balance isAfter one integration,Defining the midplane adiabatic sound speed by $c_{s0}^2=\gamma P_0/{\rho }_0=\gamma K{\rho }_0^{\gamma -1}$ givesThis is a vertically truncated polytropic atmosphere; the physical perfect-gas case has $\gamma >1$.

At the midplane, radial hydrostatic equilibrium givesIf $d{P}_0/dr<0$, then $\Omega <{\Omega }_K$: the pressure gradient provides part of the inward force and the gas is Sub-Keplerian. For radial pressure scale comparable to $r$,Using $H\sim c_s/{\Omega }_K$ and ${\Omega }_K-\Omega \simeq ({\Omega }_K^2-{\Omega }^2)/(2{\Omega }_K)$,

The linear term $-(3/2){\Omega }_{K}x{\mathbf{e}}_y$ is the local Keplerian shear in the shearing sheet. The remaining constantis the gas's azimuthal velocity relative to the local Keplerian frame. The outwardly decreasing midplane pressure found in part (a)(iii) makes the gas Sub-Keplerian, so

Write $\Delta \mathbf{u}=(U,W,0)$. The background shear givesThe constant radial and azimuthal components of the dust equation are thereforeWith the Stokes number $\mathrm{St}=\tau {\Omega }_K$, their solution is

The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. Drag force removes its angular momentum, and it drifts radially inward because $U<0$. Its radial speed isDifferentiation with respect to $\mathrm{St}>0$ shows that the unique maximum occurs atwhere $|U|=|V|$.

For the axisymmetric perturbation, defineLinearization about the uniform dust density and drifting equilibrium givesThe factors $2{\Omega }_K$ and ${\Omega }_K/2$ are the Coriolis acceleration and Keplerian-shear couplings. Applying ${\mathcal{D}}_{\tau }({\mathcal{D}}_{\tau }^2+{\Omega }_K^2)$ to the continuity equation and using the three momentum equations eliminates ${\mathbf{u}}^{\prime }$. All remaining forcing terms are proportional to $e^{-i\omega t}$, sowhere $F$ is independent of time.

On the forcing $e^{-i\omega t}$,The other three factors have roots with real decay rate $-1/\tau$, so a real-frequency gas wave can resonate only with the undamped factor $\mathcal{D}$. The resonance condition is thereforeAt resonance, $\mathcal{D}$ annihilates the forcing, so a particular solution acquires one power of time:This is a streaming instability resonance: the gas wave's phase velocity along $\mathbf{k}$ matches the dust drift projected along the same wavevector.

Letbe the vertical absolute vorticity. Taking the vertical curl of the momentum equation removes both the tidal potential and the isothermal pressure force, because each is a gradient field. The two-dimensional vorticity equation isThe surface-density equation isThe quotient rule now givesSince $\zeta =W/\Sigma$ is the vortensity,

For $u_x=0$ and $u_y=u_y(x)$, the material acceleration vanishes. The radial component of force balance isThereforeThe first term is the background Keplerian shear; the second is the geostrophic balance correction produced by the surface-density gradient.

Differentiating the equilibrium velocity givesOn the other hand, the definition of vortensity and $\Sigma ={\Sigma }_0{e}^{\sigma }$ giveMultiplying by $2/\Omega$, and introducing the dimensionless variablesyields

Linearizing $e^{\sigma }=1+\sigma +O({\sigma }^2)$ givesThe prescribed $q$ and the decay conditions are even in $\xi$, so the localized solution is even. Continuity of $\sigma$ and ${\sigma }^{\prime }$ at $\xi =\pm d$ determines all constants:Thus $\Sigma ={\Sigma }_0{e}^{\sigma }\simeq {\Sigma }_0(1+\sigma )$ is a symmetric density bump: it has a concave parabolic core, reaches its maximum at $\xi =0$, and has exponential tails approaching ${\Sigma }_0$. The linearization is self-consistent when $d+d^2/2\ll 1$.

Let $\delta ={\Sigma }^{\prime }/{\Sigma }_0$ and define the derivative following the background Keplerian shear byThe linearized continuity and momentum equations areThe coefficient $\Omega /2$ combines the background shear with the Coriolis acceleration.

Applying ${\mathcal{D}}_0$ to the shearing-wave ansatz produces an extra phase termThe perturbation equations have spatially uniform amplitude coefficients only when this term vanishes. HenceFor $k_y\ne 0$, the radial wavenumber changes linearly. A leading disturbance with $k_x/k_y<0$ first opens until $k_x=0$, then becomes an increasingly tightly wound trailing disturbance with $k_x/k_y>0$. This is the geometric swing of a shearing wave.

The linearized vortensity amplitude isSince the exact vortensity obeys $D_t\zeta =0$ and the time-dependent $k_x$ cancels the background shear in the phase, linearization gives

On the zero-vortensity branch,Substitution in the two momentum equations yields the linear system of differential equationswithwhere $k_x=k_x(t)$ is the result from part (ii).

When $k_y=0$, $k_x$ is constant and the system becomesThusFor time dependence $e^{-i\omega t}$, the dispersion relation isExplicitly,and $\tilde{\delta }=2i{k}_x{\tilde{u}}_y/\Omega$. This is an axisymmetric inertial-acoustic wave: pressure supplies the $c_s^2{k}_x^2$ restoring term and epicyclic motion supplies the ${\Omega }^2$ term.