Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 347 (Astrophysical Black Holes)

The radiative efficiency of black-hole accretion isFor a steady thin disk with negligible stress at the innermost stable circular orbit, matter radiates the binding energy lost before plunging, so $\eta =1-E_{\mathrm{ISCO}}$. Black-hole spin changes both the ISCO radius and its specific orbital energy. A prograde disk around a rapidly rotating Kerr black hole reaches deeper into the potential and is more efficient than a retrograde disk; representative ideal values run from $\eta \simeq 0.057$ for a Schwarzschild hole toward $0.42$ for an extremal prograde Kerr hole, reduced to about $0.3$ when photon capture limits astrophysical spin-up.

In adiabatic Bondi accretion, spherical compression raises the gas's internal energy reversibly, but much of that energy is advected through the horizon. There is no sustained shear stress that converts orbital binding energy into heat at a sequence of radii. A Shakura--Sunyaev thin disk, by contrast, must transport angular momentum outward. Its differential rotation stores free energy, local stress dissipates that energy as heat, and the short cooling time lets an optically thick disk radiate it before accretion. High efficiency therefore requires irreversible heating beyond adiabatic compression.

The likely source is MRI-driven magnetohydrodynamic turbulence. A weak magnetic field couples neighboring annuli; when angular velocity decreases outward, magnetic tension transfers angular momentum outward and amplifies the displacement. The alpha disk prescription replaces the unresolved turbulent stress by $T_{r\varphi }=\alpha P$, or equivalently $\nu \sim \alpha c_{s}H$. It captures the correct dimensional scale because subsonic turbulent motions are bounded by $c_s$ and their largest local eddies by $H$.

Its limitations include the following.

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Uniform ${\rho }_0,p_0,$ and ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$ obey the unperturbed ideal magnetohydrodynamics equations whenIndeed $({\mathbf{v}}_0\cdot \nabla ){\mathbf{v}}_0=0$, while the Coriolis and tidal terms cancel:

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations areEliminating $B_x,B_y,v_x,v_y$ and writing the Alfvén speed as $v_A^2=B_0^2/(4\pi {\rho }_0)$ givesSince $d{\Omega }_0^2/d\log r=-2q{\Omega }_0^2$, a root has ${\omega }^2<0$ precisely when the constant term is negative:This is the magnetorotational instability criterion.

For a circular Kepler orbit, $q=3/2$. With $K=k^2{v}_A^2$, the unstable branch isMinimizing it givesand henceThe e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2{v}_A^2<3{\Omega }_0^2$, making the critical wavelength of the magnetorotational instabilityFor a thin isothermal disk, $H=c_{s,0}/{\Omega }_0$. Setting ${\lambda }_{\mathrm{crit}}=2H$ gives $v_A^2=3c_{s,0}^2/{\pi }^2$, and thereforeAbove this field strength the shortest unstable vertical MRI wavelength exceeds the full disk thickness, so no such vertical mode fits inside the disk. Magnetic tension then stabilizes this local mode; the result corresponds to a magnetic-to-gas pressure ratio $B^2/(8\pi p)=3/(2{\pi }^2)$.

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During a weak encounter, approximate one star's trajectory by a straight line with speed $v$ and impact parameter $b$. At longitudinal coordinate $z=vt$, the transverse acceleration isIntegrating from $t=-\infty$ to $\infty$ givesHere $Gm/b^2$ sets the gravitational acceleration, $b/v$ is the encounter duration, and the geometric projection produces the displayed transverse impulse.

In one crossing, the number of encounters with impact parameters in $(b,b+db)$ is, up to the system-geometry convention,Uncorrelated impulses add in mean square, soThe virial theorem gives $v^2\sim GNm/R$. Taking $b_{\max }\sim R$ and the strong-deflection scale $b_{\min }\sim Gm/v^2\sim R/N$ gives the Coulomb logarithm in stellar dynamics $\log \Lambda \sim \log N$ andVelocity memory is lost when the cumulative change reaches $v^2$, afterConsequently the two-body relaxation time iswith an order-unity prefactor depending on density profile and convention. The system is a collisional stellar system when $t_{\mathrm{rel}}$ is shorter than its age or the evolutionary timescale being studied.

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Equating stellar surface gravity to the differential black-hole acceleration givesso the tidal disruption radius isA nonrotating hole swallows the star without a visible disruption when this lies inside its capture scale, here approximated by the Schwarzschild radius $r_h=2G{M}_{\mathrm{BH}}/c^2$. Equating the two radii yields the Hills massFor a solar-type star this is of order $10^8{M}_{\odot }$.

The threshold does depend on spin. A Kerr black hole has spin- and inclination-dependent horizon, marginally bound, and capture radii. Prograde orbits around a rapidly rotating hole can approach more closely, allowing disruption by masses above the Schwarzschild Hills mass, whereas retrograde capture occurs farther out.

An intermediate-mass black hole lies well below this threshold for ordinary stars, so stars entering its loss cone are disrupted outside the horizon. The returning debris can grow the hole and produces a tidal disruption event that may reveal an otherwise quiescent cluster black hole through a flare. Dense clusters can supply repeated disruptions, although the rate depends on two-body relaxation, stellar collisions, binary interactions, and whether gravitational recoil or cluster dynamics ejects the hole.

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Insert the self-similar ansatz into the height-integrated equations. Mass conservation is already satisfied because $H\sim c_s/{\Omega }_K\propto R$ and $\rho RH{u}_R$ is constant. The angular-momentum and energy equations reduce towhile radial momentum givesDefineSolving the quadratic gives the exact advection-dominated accretion flow coefficientsFor ${\alpha }^2\ll 1$,and therefore

Efficient cooling means $f\to 0$ and hence ${ϵ}^{\prime }=ϵ/f\to \infty$ for fixed $\gamma <5/3$. ThenThe flow is therefore cold, nearly Keplerian, slowly accreting, and geometrically thin: the standard thin-disk limit.

For significant advection, $f=O(1)$ and all three deviations are explicit:The gas is hot and thick, pressure supplies part of the radial support, rotation is sub-Keplerian, and dissipated entropy is carried inward. As $\gamma \to 5/3$, ${ϵ}^{\prime }\to 0$, so $\Omega \to 0$, $c_s^2\to (2/5)v_K^2$, and $u_R\to -(3\alpha /5)v_K$: the self-similar rotating solution approaches a hot Bondi-like inflow. Sagittarius A* is the standard supermassive example: its luminosity is tiny compared with its Eddington luminosity despite an available gas supply, and its hot optically thin spectrum and low radiative efficiency are described by an ADAF or the broader radiatively inefficient accretion-flow family.

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The three principal black-hole seed channels occupy different mass ranges.

For radiative efficiency $\eta =0.1$, growth at Eddington ratio $f_{\mathrm{Edd}}$ obeyswhere $t_E=0.45\mathrm{Gyr}$. Between $z=25$ and GN-z11 at $z=10.6$,so the exponent is $6.04f_{\mathrm{Edd}}$. Equivalently, the Salpeter time is about $0.05\mathrm{Gyr}/f_{\mathrm{Edd}}$.

Reaching $10^6{M}_{\odot }$ requiresRepresentative values areContinuous Eddington-limited growth supplies only $e^{6.04}\simeq 420$, so a $100M_{\odot }$ seed reaches about $4\times 10^4{M}_{\odot }$, whereas a seed above roughly $2.4\times 10^3{M}_{\odot }$ can reach the target at a unit duty cycle. Light seeds require sustained mildly super-Eddington accretion, an earlier start, mergers, or lower effective efficiency. Heavy direct-collapse seeds need only a moderate time-averaged Eddington ratio and are therefore easier to reconcile with the short available time. No channel is ruled out by the mass alone because seed masses, obscuration, duty cycles, super-Eddington episodes, mergers, and the observational mass estimate are uncertain.

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For a swept shell, momentum conservation may be writtenThe first term is direct ultraviolet absorption plus trapped-infrared radiation pressure, the second is gravity from the black hole and host, and the last is external pressure. The derivative also accounts for the inertia of newly swept-up gas.

For a singular isothermal sphere,Outside the black hole's sphere of influence and with external pressure neglected, gravity is the constant force $4f_{\mathrm{gas}}{\sigma }^4/G$. The shell optical depths areThus the dust transparency radius and dimensionless optical depths areWith $R^{\prime }=R/R_{\tau }$, $t^{\prime }=t/t_{\tau }$, $t_{\tau }=R_{\tau }/(\sqrt{2}\sigma )$, andthe shell equation becomes

In the optically thick single-scattering regime, $1-e^{-{\tau }_{\mathrm{UV}}}\simeq 1$ and ${\tau }_{\mathrm{IR}}$ is neglected. Integrating the constant right-hand side givesThe zero-radius member has $R^{\prime }=\sqrt{\xi -1}{t}^{\prime }$ and hence the physical constant speed

As the shell becomes ultraviolet-thin,The net force becomes negative beyond the force-balance radiusbut inertia carries the shell farther before it stalls. Match the thick solution at $R^{\prime }=1$, where ${\dot{R}}^{\prime 2}=\xi -1$. In the thin approximation,Using $d/d{t}^{\prime }={\dot{R}}^{\prime }d/d{R}^{\prime }$ and integrating givesThe outer zero is the stalling radiusIncluding the optically thick travel time from a negligible launch radius, the dimensionless stalling time isSolving $d{t}_{\mathrm{stall}}^{\prime }/d\xi =0$ gives $\xi \simeq 1.248$, soFor $1<\xi <1+ϵ$, a stronger mildly supercritical source reaches transparency so much sooner that its total stalling time decreases even though it travels farther. Above this range, the increasing coasting distance dominates. Stalling means that loss of ultraviolet optical depth reduces radiation coupling below gravity; without renewed driving, the swept gas falls back or remains bound rather than escaping the halo.

In the infrared multi-scattering regime, the dominant force is $\xi {\tau }_{\mathrm{IR}}=a/R^{\prime }$ with $a=\xi {\kappa }_{\mathrm{IR}}/{\kappa }_{\mathrm{UV}}$. Seeking $R^{\prime }=A{t}^{\prime 2/3}$ ingives $A^3=9a/2$ and thereforeInfrared trapping supplies more than the single-scattering momentum while the shell is compact and optically thick. The speed nevertheless decreases as $t^{\prime -1/3}$ because the shell sweeps up mass and its infrared optical depth falls. Such driving can launch a powerful dusty radiation-pressure-driven shell, but propagation to halo scales requires enough integrated momentum before the shell becomes transparent; otherwise gravity eventually stalls it.

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