Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 321 / 1

Take $0<r_1<r_2$ and positive mean kinematic viscosity $\bar{\nu }$. The steady state of the supplied viscous evolution of an accretion disk equation satisfies ${\mathcal{F}}^{\prime }=-2\pi rS(r)$. The outer no-flux condition therefore givesPositive $S$ produces positive inward accretion rate in this sign convention. Set $Y=r^{1/2}\bar{\nu }\Sigma$. In a Keplerian disk, the viscous torque in an accretion disk is proportional to $Y$, so the zero-torque inner boundary condition gives $Y(r_1)=0$. Since $Y^{\prime }=\mathcal{F}/(6\pi r^{1/2})$, a second integration yieldsThis is the steady viscous disk supplied by a radial source. It does not require $\bar{\nu }$ to be independent of radius: the equation directly determines $\bar{\nu }\Sigma$. If viscosity depends on the local thermodynamic state, the displayed formula is an implicit relation completed by a separate closure. For nonnegative integrable $S$ it gives nonnegative surface density of a disk, and the inner accretion rate equals the total supplied mass per unit time, $2\pi {\int }_{r_1}^{r_2}rS(r)dr$. Differentiating both nested integrals recovers the flux and the differential equation.

Interpret the outer Dirac delta function as a boundary source carrying its full mass, or as the limit of a narrow source lying just inside $r_2$. Then ${\int }_u^{r_2}vS(v)dv=r_2{S}_0$ for $r_1<u<r_2$, andThus the interior is the usual steady Keplerian accretion disk with a constant inward accretion rate. The product $\bar{\nu }\Sigma$ vanishes at the stress-free inner edge, and the surface density of a disk does too if the viscosity has a finite nonzero boundary value; far from that edge $\bar{\nu }\Sigma$ approaches $\dot{M}/(3\pi )$. Its density need not itself be constant because the viscosity may vary.

The outer-edge source of a viscous disk requires care about the boundary trace. The flux just inside the injecting edge is $\mathcal{F}(r_2^{-})=\dot{M}$, while the no-flux value outside the source is $\mathcal{F}(r_2^{+})=0$; the source produces the intervening jump. A simultaneously continuous interior condition $\mathcal{F}(r_2^{-})=0$ would contradict nonzero steady injection. A symmetrically regularized delta integrated only up to its center can instead give half its full weight; that would mean a different supplied mass, not a different interior solution for the same $\dot{M}$. The physically normalized one-sided limiting source removes this convention ambiguity.

In the vertical equations, $\mu$ denotes dynamic viscosity, while ${\mu }_m$ is the mean molecular weight in units of the proton mass $m_p$. The symbol $k$ denotes the Boltzmann constant. For the intended hydrostatic equilibrium calculation, put $P=K{\rho }^{\gamma }$ with $K>0$, $\gamma >1$, and set $\mathcal{R}=k/({\mu }_m{m}_p)$. Dividing the hydrostatic equation by $\rho$ and integrating from the symmetric midplane givesThe specific enthalpy vanishes at the free surface. Hence the polytropic vertical structure in stellar gravity isOn $|z|<H$, the ideal gas relation gives $T=T_m(1-z^2/H^2)$ and $P=P_m(1-z^2/H^2{)}^{\gamma /(\gamma -1)}$; the vacuum extension sets $\rho =P=0$ outside. The formula needs $\gamma >1$ for a finite positive semi-thickness and assumes the harmonic vertical gravity remains valid through this height.

There is a genuine inconsistency if all four printed structure equations are imposed literally with $F=0$ and finite positive opacity: radiative diffusion then requires $dT/dz=0$, whereas the polytropic temperature above has a nonzero gradient away from the midplane. The intended result is valid for the hydrostatic polytropic equation of state with the radiative closure replaced or neglected, for example an adiabatic opaque limit. It is not an exact simultaneous solution of the stated finite-opacity radiative system. With that radiative system and a nonvacuum ideal gas, $\gamma =1$ instead gives an isothermal solutionwhich has no finite free surface. For $\gamma \ne 1$, constant $T$ and constant $K$ would force constant density, contradicting hydrostatic balance on a nontrivial vertical interval. This is the zero-radiative-flux obstruction for a polytropic ideal gas.

Use a gas-pressure-supported, optically thick alpha disk, with $\mathcal{R}=k/({\mu }_m{m}_p)$ and positive parameters. Here $\sigma$ is the Stefan–Boltzmann constant in the Stefan–Boltzmann law, and $F_{+}$ and $F_{-}$ below are heating and cooling flux estimates per disk face. Hydrostatic equilibrium and the ideal gas relation give $\mathcal{R}T\sim {\Omega }^2{H}^2$, while $\rho \sim \Sigma /H$. Integrating viscous heating over one semi-thickness and estimating radiative diffusion giveHere the stipulated negative hydrogen ion opacity exponent is $\kappa ={\kappa }_0{\rho }^{1/3}{T}^{10}$, which is a local approximation distinct from other empirical opacity fits. Substituting $T\sim {\Omega }^2{H}^2/\mathcal{R}$ and equating heating and cooling givesDropping numerical constants of order one is essential here. This is the negative-hydrogen-opacity alpha-disk thickness scaling. Its angular thickness estimate isFor a central mass $M$, Keplerian rotation gives $\Omega =(GM/r^3{)}^{1/2}$, so $H/r\propto r^{53/82}\Sigma (r{)}^{-7/41}$. The isothermal sound speed $c_s^2=\mathcal{R}T$ isAn adiabatic sound speed differs by the constant factor $\sqrt{\gamma }$ if the adiabatic index is fixed. For $\Sigma \propto r^{-p}$, constant $\alpha$, and fixed composition, the disk aspect ratio grows as $r^{(53+14p)/82}$, so there is disk flaring when $p>-53/14$; $c_s\propto r^{(6+7p)/41}$. The problem supplies no numerical parameters or specified $\Sigma (r)$, so no unique numerical $H/r$ or unconditional radial power can be inferred.

If the additional standard closure of a constant inward accretion rate well outside a zero-torque edge is intended, $\bar{\nu }\sim \alpha \Omega H^2$ and $\bar{\nu }\Sigma \simeq \dot{M}/(3\pi )$ imply $\Sigma \propto (\alpha \Omega H^2{)}^{-1}$. Eliminating it gives $H^9\propto {\Omega }^{-38/3}$ at fixed $\dot{M},\alpha$, hence $H/r\propto r^{10/9}$ and $c_s\propto r^{11/18}$. These stronger powers use that extra steady-flow assumption. All these local scalings cease to apply if the thin disk or the stated opacity regime fails.

Assume the vertical dynamical timescale of a disk is shorter than the thermal timescale of an accretion disk, which is itself shorter than the viscous timescale. This ordering is controlled in a thin disk with $\alpha \ll 1$: $t_{\mathrm{dyn}}\sim {\Omega }^{-1}$, $t_{\mathrm{th}}\sim (\alpha \Omega {)}^{-1}$ and $t_{\nu }\sim (r/H{)}^2{t}_{\mathrm{th}}$. Therefore compare heating and cooling at fixed surface density of a disk, radius, $\alpha$ and composition. Vertical hydrostatic equilibrium adjusts on the vertical dynamical timescale of a disk, giving $H\propto T^{1/2}$ and $\rho \propto T^{-1/2}$. Consequently the stipulated negative hydrogen ion opacity varies as $\kappa \propto T^{10-1/6}=T^{59/6}$. The rates per disk face then scale asThus heating increases while cooling decreases after a positive temperature perturbation. At equilibrium $Q^{+}=Q^{-}=Q_0$, a small change gives $\delta (Q^{+}-Q^{-})=(41/6)(Q_0/T)\delta T$, which drives the perturbation further from equilibrium for positive effective heat capacity. A negative temperature perturbation likewise increases the net cooling. This proves thermal instability from negative hydrogen ion opacity within the local closure; it is the temperature-slope test of thermal stability of an accretion disk. Holding volume density rather than surface density of a disk fixed would give a different cooling exponent, but still the same qualitative instability; the fixed-column comparison is the relevant thin-disk thermal test.