Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 (Dynamics of Astrophysical Discs)

For the Paczyński-Wiita potential, circular balance givesThereforeThe squared radial epicyclic frequency isCircular orbits change from stable to unstable where ${\kappa }_r^2$ vanishes, so

The steady disk should end at the innermost stable circular orbit, $r_{\mathrm{in}}=3r_S$. Inside it, matter plunges inward on an orbital timescale and cannot maintain the nearly circular shear flow assumed by the viscous-disk equations. If the plunging region communicates negligible stress back across the ISCO, the natural boundary condition is the zero-torque inner boundary condition

Take $\dot{M}>0$ for inward accretion, so the outward radial mass flux is $F=-\dot{M}$. Since $F=-(dh/dr{)}^{-1}d\mathcal{G}/dr$ and $\mathcal{G}(r_{\mathrm{in}})=0$,Using $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^{3}d\Omega /dr$ and $x=r/r_S$ givesThus $\boxed{A=3\sqrt{3}/2}$. Far from the hole this approaches the Keplerian accretion disk result $\bar{\nu }\Sigma =\dot{M}[1-(r_{\mathrm{in}}/r{)}^{1/2}]/(3\pi )$, although the pseudo-Newtonian boundary factor retains a different finite-radius shape.

Let $c_s^2\sim P_g/\rho$. The assumed magnetic pressure givesso the plasma beta obeys $\boxed{\beta =P_g/P_m\sim {\mathrm{Ma}}^{-1}}$. In the magnetically dominated regime, vertical hydrostatic equilibrium gives $P_m/\rho \sim {\Omega }^2{H}^2$. Combining this with $P_m/\rho \sim r\Omega c_s$ yields

Order-of-magnitude vertical balance givesMagnetic support also gives ${\Omega }^2{H}^2\sim r\Omega T^{1/2}$, so $T\sim {\Omega }^2{H}^4/r^2$. Eliminating $P,F,\rho ,T$ givesFor Keplerian $\Omega \propto r^{-3/2}$,The alpha disk stress implies $\bar{\nu }\sim \alpha P/(\rho \Omega )\sim \alpha \Omega H^2$, and therefore

The viscous time isWith zero stellar torque, total angular momentum is conserved and $J\sim M(G{M}_{\star }R{)}^{1/2}$, so $M\propto R^{-1/2}$. HenceandAccreting material loses specific angular momentum; conservation requires a diminishing fraction of the disk to spread outward and carry that angular momentum.

If no mass is accreted, $M$ is constant and $\Sigma \sim M/R^2$. The viscous-time estimate becomes $t\propto R^{2-m+2n}$, while $J\sim M(G{M}_{\star }R{)}^{1/2}$. ThusThe central torque supplies the angular momentum that allows the fixed disk mass to spread.

Set $r=r_0{x}^2$, $\Sigma ={\Sigma }_0{x}^{-3}S$, and $\bar{\nu }={\bar{\nu }}_0{x}^{2m-3n}{S}^n$. Then ${\partial }_r=(2r_{0}x{)}^{-1}{\partial }_x$. Substitution into the viscous evolution of an accretion disk equation, with $\tau =3{\bar{\nu }}_{0}t/(4r_0^2)$, cancels the common dimensional factors and gives

For $m=4,n=2$, the equation is the one-dimensional porous medium equation $S_{\tau }=(S^3{)}_{xx}$. Use a similarity solution $S={\tau }^{-1/4}f(\xi )$ and $\xi =x{\tau }^{-1/4}$. ThenThe no-mass-flux condition at $x=0$ sets the integration constant to zero:With $f(0)=1$,ConsequentlyIts edge is $R=12r_0{\tau }^{1/2}\propto t^{1/2}$. Moreover,is time-independent. This agrees with part (b), since $2-m+2n=2$.

Use a local Cartesian frame rotating at $\Omega =\sqrt{GM/r_0^3}$ about a circular orbit at $r_0$, with $x=r-r_0\ll r_0$, and neglect curvature beyond leading order. The radial gravitational-plus-centrifugal potential isIts first derivative vanishes at $r_0$ and its second derivative there is $-3{\Omega }^2$. Dropping a constant gives the shearing-sheet tidal potential

For ${\mathbf{u}}_0=-\frac{3}{2}\Omega x{\mathbf{e}}_y$, advection and viscosity vanish, while its Coriolis acceleration cancels $-\nabla {\Phi }_t$; constant pressure and $\theta =0$ complete the equilibrium. For axisymmetric perturbations,

For a mode $e^{i\mathbf{k}\cdot \mathbf{x}+st}$, put $D=s+\nu k^2$. The linear equations becomewith $s\theta =u_x$ and $k_x{u}_x+k_z{u}_z=0$. Eliminating $u_y,u_z,P,\theta$ gives the dispersion relationExpanding,

When $N=0$, the factorized relation is $s(D^2+{\varpi }^2)=0$, soThe oscillatory roots are inertial waves restored by rotation; viscosity leaves their leading oscillation frequency unchanged and damps their amplitude at rate $\nu k^2$. The zero root is the decoupled buoyancy-variable mode.

For $\nu =0$,An exponentially growing root exists exactly when $1+n^2<0$, orThus an adverse radial entropy gradient $N^2<0$ must overcome the epicyclic restoring effect of rotation. Rotation stabilizes gradients with $-{\Omega }^2<N^2<0$.

Let $ϵ=\Omega /(\nu k^2)\ll 1$ and $s=ϵ\Omega s_1+\cdots$. Since $D=\Omega /ϵ+O(ϵ\Omega )$, the leading terms ofgiveThe mode is unstable for every $\boxed{N^2<0}$, a weaker criterion than the inviscid condition. Strong viscosity damps the velocity response and removes the rapid epicyclic restoration, permitting a slow viscous-convective instability; its growth rate nevertheless tends to zero on very small scales.