Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 (Dynamics of Astrophysical Discs)

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

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In steady state, mass conservation makes the outward flux constant, $\mathcal{F}=\dot{M}$. Angular-momentum conservation then givesThe zero-torque condition at $r_{\mathrm{out}}$ fixes the constant to $\dot{M}{h}_{\mathrm{out}}$, soFor a Kepler orbit, $h=\sqrt{GMr}$ and $d\Omega /dr=-3\Omega /(2r)$. The viscous torque is consequently $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Equating the two expressions yields the steady decretion disk structure

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At the inner edge, the imposed torque isThereforeThe angular-momentum conservation law says that $\mathcal{F}h+\mathcal{G}$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\mathrm{out}}$ carries the injected angular momentum together with the angular momentum $\dot{M}{h}_{\mathrm{in}}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more specific angular momentum.

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The luminosity from both disk faces is the radial integral of the given viscous heating rate:Substituting the steady profile and ${\Omega }^2=GM/r^3$ givesUsing $\Theta =\dot{M}\sqrt{GM}(\sqrt{r_{\mathrm{out}}}-\sqrt{r_{\mathrm{in}}})$ turns this intoThe first term is mechanical power injected by the stellar magnetic torque. The second is $\dot{M}(E_{\mathrm{in}}-E_{\mathrm{out}})$, where $E=-GM/(2r)$ is the specific energy of a circular orbit; it is negative because outward motion makes the material less tightly bound and consumes some torque power. The remainder is radiated by viscous dissipation.

Matter reaching the outer edge still has binding energy $GM/(2r_{\mathrm{out}})$ per unit mass. A wind that reaches infinity with negligible terminal energy must therefore receivefrom the stellar radiation field. In a signed disk-energy balance, the material removed at the edge carries orbital energy flux $\dot{M}{E}_{\mathrm{out}}=-GM\dot{M}/(2r_{\mathrm{out}})$.

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For $r_{\mathrm{in}}\ll r\ll r_{\mathrm{out}}$, part (i) gives $\bar{\nu }\Sigma \propto r^{-1/2}$. Since ${\Omega }^2\propto r^{-3}$, the dissipative flux obeysIn the spectrum, introduce the dimensionless variableThen $rdr$ is proportional to ${\omega }^{-16/7}{x}^{9/7}dx$, soAt intermediate frequencies the inner limit is much smaller than one and the outer limit much larger than one, allowing them to be replaced by zero and infinity. Thus

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The first equation is vertical hydrostatic equilibrium. In a thin disk, stellar gravity has vertical component $-{\Omega }^{2}z$, giving $dP/dz=-\rho {\Omega }^{2}z$. The second is local energy conservation: Keplerian shear has $rd\Omega /dr=-3\Omega /2$, so viscous dissipation per unit volume isThe third equation is radiative diffusion. With radiation energy density $a_{\mathrm{R}}{T}^4=4\sigma T^4/c$, an optically thick medium carrieswhich rearranges to the stated gradient. Thomson scattering makes $\kappa$ approximately independent of density and temperature. The final equation adds perfect-gas pressure $k\rho T/({\mu }_m{m}_p)$ and radiation pressure $a_{\mathrm{R}}{T}^4/3=4\sigma T^4/(3c)$.

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The alpha disk prescription isEquivalently, the turbulent shear stress is of order $\alpha P$. Turbulent eddies cannot remain coherent on scales much larger than the vertical thickness $H$, and strongly supersonic eddies shock and dissipate, so their speed is at most of order $c_s$. The product of the largest plausible speed and length is therefore $c_{s}H$. Magnetorotational turbulence supplies a physical mechanism for correlated radial and azimuthal motions and magnetic stresses, while $\alpha$ parametrizes their uncertain efficiency.

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Let $\beta =P_{\mathrm{gas}}/P_{\mathrm{rad}}$. Constancy of $\beta$ givesThe total pressure isConsequently, if $T=T_{0}f$, then $\rho ={\rho }_0{f}^3$ and $P=P_0{f}^4$, whereSubstitution in hydrostatic balance shows that $f^{\prime }=-2z/H^2$ and

Finally,Absorbing the numerical factor $4/3$ and this dimensionless integral into $A$ gives the requested form

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The vertically averaged alpha prescription givesRadiative diffusion over thickness $H$ givesAt fixed $\Sigma$, part (c) gives $\beta H{T}_0^3=\text{constant}$.

In the gas-pressure limit $\beta \gg 1$, hydrostatic balance gives $H\propto T_0^{1/2}$ and hence $\beta \propto T_0^{-7/2}$. ThereforeA temperature increase raises cooling faster than heating, so this equilibrium is thermally stable.

In the radiation-pressure limit $\beta \ll 1$, the two structural relations give $H\propto T_0^4$ and $\beta \propto T_0^{-7}$. HenceHeating now rises faster, so this equilibrium is thermally unstable.

More generally, combining the structural relations givesBecause $\beta$ decreases monotonically with $T_0$,The ratio has one minimum, at $\beta =2/3$, and therefore a horizontal equilibrium level can be crossed at most twice. Thus there are at most two thermal equilibria: a stable gas-pressure branch and an unstable radiation-pressure branch.

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The Toomre stability criterion for a razor-thin isothermal gas disk usesSelf-gravity, represented by $G\Sigma$, amplifies overdensities. Pressure, represented by $c_s$, suppresses short wavelengths, while epicyclic motion with frequency $\kappa$ suppresses long-wavelength radial collapse. Axisymmetric disturbances are stable for $Q\ge 1$ and unstable for $Q<1$.

For a Keplerian disk, $\kappa =\Omega$, $H\simeq c_s/\Omega$, ${\Omega }^2\simeq G{M}_{\ast }/r^3$, and $\Sigma \sim M_D/(\pi r^2)$. ThereforeThe instability condition $Q\lesssim 1$ is equivalently

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After a radial Fourier transform, Poisson's equation away from the sheet becomesThe decaying, reflection-symmetric solution is $A{e}^{-|k||z|}$. Integrating Poisson's equation across $z=0$ gives the derivative jumpso $-2|k|A=4\pi G\tilde{\Sigma }$. Thus the midplane potential is

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Write the perturbations as $\delta \Sigma =\sigma$, radial and azimuthal velocities $u,v$, and pressure $\delta P=p$, all proportional to $e^{ikx+st}$. Linearization in the shearing sheet givesandThe continuity and pressure equations implyEliminating $\sigma ,u,v,p,\varphi$ then yields

When $\tau |s|\ll 1$, temperature relaxes during a disturbance and the pressure response is isothermal, $p/\sigma \simeq c_0^2$. When $\tau |s|\gg 1$, relaxation is negligible and the response is adiabatic, $p/\sigma \simeq \gamma c_0^2$, so the effective sound speed is $\sqrt{\gamma }{c}_0$. At finite relaxation time the phase lag between compression and pressure also damps stable waves.

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At marginal stability write $s=i\omega$ with real $\omega$. The imaginary part of the pressure factor isFor $\gamma >1$, $\tau >0$, and nonzero $k$, the dispersion relation can have zero imaginary part only when $\omega =0$. Thus every marginal mode hasThe marginal equation is consequentlyDefining the isothermal Toomre parameter $Q_0=c_0\Omega /(\pi G{\Sigma }_0)$, its two roots areThey exist when $Q_0\le 1$, and the interval between them is unstable when $Q_0<1$. Finite thermal relaxation therefore leaves the onset criterionequal to the isothermal criterion, regardless of $\gamma$ and the nonzero value of $\tau$.

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For $\tau \Omega =ϵ\ll 1$, expand the pressure factor:For a stable density wave defineThe dispersion relation becomesPerturbing the two isothermal roots givesThus the leading amplitude-damping rate is

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