Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 61 (Dynamics of Astrophysical Discs)

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; the pressure gradient balances it. The second is an equation of state combining ideal gas pressure with isotropic radiation pressure. Write $\mathcal{R}=k/({\mu }_m{m}_p)$, so $p_g=\mathcal{R}\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the Stefan-Boltzmann constant and ${\mu }_m$ the mean molecular weight in proton-mass units.

The third is local energy balance: the increase of upward radiative flux with height equals viscous dissipation. Here $\mu$ is dynamic viscosity, not mean molecular weight. A Keplerian disc has shear $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. The fourth is optically thick radiative diffusion, with opacity per unit mass $\kappa$ and diffusion directed down the temperature gradient. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

Take the positive constant ratio $\lambda =p_r/p_g$. The equation of state impliesThis is a polytropic equation of state of index three. Differentiating the pressure and using vertical hydrostatic equilibrium gives $4(1+\lambda )\mathcal{R}\rho T_z=-\rho {\Omega }^{2}z$. HenceDefine the semi-thickness $H$ by $T(H)=0$ and put $s=1-z^2/H^2$. Then the constant gas-radiation ratio vertical disc structure iswithThe surface density is the integral across both faces. The supplied polynomial integral gives ${\int }_0^1(1-x^2{)}^{3}dx=16/35$, so $\Sigma =(32/35){\rho }_{0}H$. Substitution yieldsAll coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.

For constant opacity, the ratio $T^3/\rho =3c\lambda \mathcal{R}/(4\sigma )$ cancels the height dependence in the radiative diffusion coefficient. Since $T_z=-{\Omega }^{2}z/[4(1+\lambda )\mathcal{R}]$, the radiative flux isConsequently $F_z=c\lambda {\Omega }^2/[\kappa (1+\lambda )]$. Equating this to the viscous heating rate givesThus the dynamic viscosity is uniform in height even though pressure and mass density decrease. Its units are mass per length per time, since $c/\kappa$ has those units. A pointwise pressure-proportional alpha prescription would be incompatible with this exact profile, which is why the subsequent alpha relation is vertically averaged.

The required pressure integral uses ${\int }_0^1(1-x^2{)}^{4}dx=128/315$. With $\Sigma =(32/35){\rho }_{0}H$ and $p_0/{\rho }_0={\Omega }^2{H}^2/8$,The vertically averaged alpha closure for a mixed-pressure layer therefore givesThe height-integrated dynamic viscosity is $2H\mu =8cH\lambda /[9\kappa (1+\lambda )]$. Equate it to $\bar{\nu }\Sigma$ to obtainThese are integrated equalities, not an imposed pointwise relation $\mu (z)\propto p(z)$.

Now keep $\alpha$, the opacity and molecular constants independent of radius, and use Keplerian rotation $\Omega \propto r^{-3/2}$. In the gas-dominated limit, $\lambda \ll 1$ but positive, the last equality gives $\lambda \propto H\Omega \Sigma$. The column-density relation gives $\lambda \Sigma \propto H^7{\Omega }^6$. Eliminating $\lambda$ yields $H^6\propto {\Sigma }^2{\Omega }^{-5}$, whenceThis is the gas-pressure branch of a mixed-pressure alpha disk. In the radiation-dominated limit, $\lambda /(1+\lambda )\to 1$, so $H\propto (\Omega \Sigma {)}^{-1}$ andFor viscous stability of an accretion disk, linearize the Keplerian viscous diffusion equation at fixed radius: a local mass density perturbation has diffusion coefficient $3{\partial }_{\Sigma }(\bar{\nu }\Sigma )$. The gas branch has $\bar{\nu }\Sigma \propto {\Sigma }^{5/3}$ and a positive derivative, so it smooths perturbations. The radiation branch has $\bar{\nu }\Sigma \propto {\Sigma }^{-1}$ and a negative derivative, giving radiation-pressure viscous instability. The gas-pressure branch is viscously stable; the radiation-pressure branch is viscously unstable in this closure. This concerns radial mass-transport stability on wavelengths where the vertically averaged thin-disc model applies.

The boundary of a steady Kida vortex must be material. Put $E=x^2/b^2+y^2/a^2$. Requiring $\mathbf{u}\cdot \nabla E=0$ on the ellipse givesThe relative vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=-\beta -\alpha =-S+{\zeta }_0$, so the supplied vorticity relation impliesCombining these independent conditions yieldsFor the usual positive orbital shear both are positive, the flow circulates clockwise and its streamlines have the same aspect ratio as the patch. The velocity coefficients in this question are unrelated to the alpha viscosity parameter in Question 1.

The steady incompressible shearing sheet equation isThe final term follows from the local tidal potential $-\Omega S{x}^2$. For the given linear velocity, the advective acceleration is $-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$, while the Coriolis acceleration is $2\Omega \beta x{\mathbf{e}}_x+2\Omega \alpha y{\mathbf{e}}_y$. Equating components givesThe cross derivatives vanish, so the pressure Hessian of an elliptical shearing-sheet vortex integrates toThe additive constant is fixed by exterior matching or the pressure reference. No zero-pressure free-boundary condition is imposed on an incompressible vortex patch: it is embedded in a surrounding fluid.

A pressureless test particle feels tidal gravity, Coriolis acceleration and linear drag toward the prescribed gas velocity. Its planar equations areThere is no gas-pressure force on this test particle. For $(x,y)=(X,Y)e^{\lambda t}$, the coefficient matrix isNontrivial solutions require a zero determinant. Expansion givesEach root has a nonzero modal vector in the matrix kernel; real motions are obtained from conjugate pairs. Repeated roots may also supply polynomially weighted exponential solutions. Negative real parts make the centre linearly asymptotically attracting for trajectories governed by this interior linear flow.

Take every finite positive drag coefficient $\gamma >0$ and defineThe coefficients of the characteristic polynomial are $c_3=2\gamma$, $c_2=K+{\gamma }^2$, $c_1=2L\gamma$ and $c_0=P{\gamma }^2$. The nontrivial Routh-Hurwitz criterion reduces exactly toThe printed strict chain $K>L>P>0$ makes every coefficient positive and both bracket terms positive, so it proves attraction for every positive drag coefficient.

There is, however, a boundary error in the asserted necessity. To have the bracket positive for every $\gamma >0$, its constant and slope as a function of ${\gamma }^2$ must be nonnegative, and cannot both vanish. Together with coefficient positivity this gives the exact drag-polynomial stability for all positive stopping rates:Necessity follows by taking arbitrarily small and arbitrarily large positive $\gamma$; if one bracket coefficient is negative, the inequality fails in the corresponding limit. Sufficiency follows because at least one coefficient is strictly positive. Thus $K=L>P$ and $K>L=P$ still give strict decay at each finite positive $\gamma$. They need not give a uniform stability margin in the zero- or infinite-drag limit.

In a Keplerian shearing sheet, $S=3\Omega /2$ andThe differences areHence $r>3$ establishes the requested strict chain and trapping conclusion. At $r=3$, however, $K=L={\Omega }^2$ and $P=9{\Omega }^2/16$, soAll coefficients remain positive. This is an explicit counterexample within the stated vortex model: $r=3$ also attracts particles for every finite positive drag coefficient, so the literal strict “if and only if” is false. The exact Keplerian all-positive-drag condition is $r\ge 3$. If $\gamma =0$ is included, the polynomial has zero and purely imaginary roots and there is no asymptotic attraction; “all drag coefficients” must mean positive drag. The local attraction claim presumes the particle remains in the patch, as is assured for sufficiently small perturbations of its centre.

Let ${\Sigma }_0$ be the uniform background surface density, ${\mathbf{u}}_0=-Sx{\mathbf{e}}_y$, and write $\Sigma ={\Sigma }_0+{\Sigma }^{\prime }$, $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$, ${\Phi }_d={\Phi }_{d0}+{\Phi }_d^{\prime }$. The background sheet has no horizontal gravitational gradient; its vertical field is irrelevant to the stated planar equations. Put $D_0={\partial }_t-Sx{\partial }_y$. Keeping first-order terms givesThe $-S{v}_x$ contribution comes from $\mathbf{v}\cdot \nabla {\mathbf{u}}_0$, while the Coriolis acceleration supplies the remaining coupling. The prescribed tidal potential has no perturbation, and its background force already balances the background Coriolis term. The isothermal pressure linearizes to $c_s^2\nabla {\Sigma }^{\prime }/{\Sigma }_0$.

Apply the shearing-wave ansatz to each disturbance. Acting on its phase givesThus a spatially uniform amplitude system exists whenFor a nonzero horizontal wavenumber $k=|\mathbf{k}|$, the razor-thin disk Poisson kernel follows from $({\partial }_z^2-k^2){\tilde{\Phi }}_d^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }\delta (z)$. The potential decays above and below the sheet as ${\tilde{\Phi }}_{d,m}^{\prime }{e}^{-k|z|}$. Integrating across the midplane gives the jump $-2k{\tilde{\Phi }}_{d,m}^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }$, henceSubstitute this potential and the phase evolution into the linearized equations. The barotropic shearing-wave gravitational amplitude system isThe self-gravity sign is attractive and opposes the pressure response. The uniform $k=0$ mass density shift requires its own gravitational reference and is not obtained by dividing by $k$.

Write $B=2\Omega -S$, $A(t)=c_s^2-2\pi G{\Sigma }_0/k(t)$, and $w=i{k}_x{\tilde{v}}_y-i{k}_y{\tilde{v}}_x$. Differentiation, including ${\dot{k}}_x=S{k}_y$, givesThe two pressure/self-gravity cross terms cancel. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{\Sigma }_0(k_x{\tilde{v}}_x+k_y{\tilde{v}}_y)$, so $\dot{w}=B{\dot{\tilde{\Sigma }}}^{\prime }/{\Sigma }_0$. ConsequentlyThe conserved amplitude is linearized vortensity: the perturbation of the absolute vertical vorticity divided by surface density,A two-dimensional inviscid barotropic fluid materially conserves this potential vorticity. Both pressure enthalpy and gravitational forces are gradients, so they create no baroclinic vorticity source. Since $q_0$ is uniform, its linear perturbation obeys $D_0{q}^{\prime }=0$, explaining why a shearing wave has constant vortensity amplitude.

For $k_y=0$ and nonzero $k_x$, the wavenumber is constant. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{k}_x{\Sigma }_0{\tilde{v}}_x$, and the conserved vortensity relation givesDifferentiate continuity and substitute radial momentum and this identity. The forced axisymmetric density mode with vortensity obeysThe constant forcing represents the stationary vortical or balanced component, which is lost if every mode is assumed to have nonzero frequency. Let $k_{\ast }=\pi G{\Sigma }_0/c_s^2=\kappa /(Q{c}_s)$. Completing the square givesFor $Q>1$, every nonzero radial wavenumber has positive ${\omega }_k^2$. The solution is a constant balanced offset $-2\Omega {\Sigma }_0^2{\tilde{q}}^{\prime }/{\omega }_k^2$ plus bounded sine and cosine oscillations. Pressure stabilizes short wavelengths, rotation stabilizes long wavelengths, and self-gravity is strongest at intermediate wavelengths.

For $Q<1$, the Toomre stability criterion fails in the bandThere ${\omega }_k^2<0$ and the homogeneous solutions grow or decay exponentially. The fastest rate is $s_{\max }=\kappa \sqrt{Q^{-2}-1}$ at $|k_x|=k_{\ast }$. Outside the band the modes oscillate; at its edges the frequency vanishes.

At $Q=1$, ${\omega }_k^2\ge 0$ and it vanishes only at $|k_x|=\kappa /c_s$. At that critical wavenumber,This Toomre marginal algebraic growth is spectrally marginal, not bounded for all data: a nonzero conserved vortensity gives quadratic secular growth, and even zero vortensity can allow the linear term. Other wavenumbers still oscillate about a balanced offset. $Q>1$ gives bounded axisymmetric modes, $Q<1$ allows exponential instability, and $Q=1$ has a zero-frequency marginal mode with possible algebraic growth. These statements are axisymmetric; nonaxisymmetric shearing disturbances can have transient swing behavior. For $k_x=k_y=0$, continuity instead fixes the uniform mass density shift, while the velocities execute epicyclic motion; no $1/k$ gravitational formula is used.