Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 347 (Astrophysical Black Holes)

Let the mean particle mass be $\mu m_p$. For a monatomic ideal gas with $\gamma =5/3$,A uniform sphere of mass $M=(4\pi /3){\rho }_0{r}^3$ has Newtonian gravitational energyAt the virial threshold, the pressure term $3\int PdV=3M{k}_{B}T/(\mu m_p)$ balances $|W|$. ThereforeEliminating $r$ gives the Jeans massThe numerical coefficient depends on the convention used to identify a finite cloud with a Jeans mode. For example, assigning the mass inside a sphere of radius half the standard Jeans instability wavelength givesBoth conventions have the physically invariant scaling

For adiabatic collapse, $T{\rho }^{1-\gamma }$ is constant. With $\gamma =5/3$,The rising Jeans mass produces adiabatic suppression of fragmentation: smaller subregions become more pressure-supported as density increases.

For isothermal fragmentation, $T$ stays approximately constant, andThe instability scale then falls during collapse, allowing hierarchical fragmentation until cooling fails, opacity rises, or another source of support intervenes.

Primordial metal-free gas cools inefficiently, principally through molecular hydrogen, and remains relatively hot. It therefore has a larger Jeans mass and tends toward a top-heavy initial mass function of massive Population III stars. Metal lines and dust let enriched gas remain cool to higher density, so Population II stars extend to much lower birth masses.

These alternatives map directly onto black-hole seed channels. Massive Population III remnants produce light Population III remnant black-hole seeds. If cooling and fragmentation are strongly suppressed while a primordial halo supplies rapid inflow, near-monolithic collapse can produce a heavy direct-collapse black-hole seed. Intermediate cooling and fragmentation in a dense cluster can instead permit a runaway stellar-collision black-hole seed. The Jeans argument selects plausible mass scales; angular momentum, feedback, chemistry, and accretion determine which channel actually operates.

The axisymmetric razor-thin astrophysical disk equations areFor the stationary background, mass and azimuthal momentum conservation are automatic, while radial force balance and the Poisson equation giveHere $d{p}_0/dR=0$ under the stated constant-pressure assumption.

Retaining first-order perturbations and using $c_s^2=dp/d\Sigma$ giveswhereis the squared radial epicyclic frequency. In the local $|k|R\gg 1$ limit, a Fourier mode $e^{i(kR-\omega t)}$ obeysEliminating ${\Sigma }^{\prime }$, $u_{\varphi }^{\prime }$, and ${\Phi }^{\prime }$ yields the local dispersion relation

For a Keplerian orbit, $\Omega \propto R^{-3/2}$, soInstability requires ${\omega }^2<0$, equivalently a mode with positive imaginary frequency. Treating the right-hand side as a quadratic in $x=|k|$, its two roots areReal distinct roots, and hence an unstable interval $x_{-}<|k|<x_{+}$, exist exactly when the Toomre stability criterion has $Q<1$.

With $h=c_s/\Omega$,The constant positive term is epicyclic restoration by rotation and stabilizes long wavelengths. The negative term is the razor-thin disk's self-gravity and drives collapse. The positive $k^2$ term is gas-pressure restoration and stabilizes short wavelengths. Gravitational instability can therefore survive only on an intermediate band of scales.

Steady mass conservation with inward-positive accretion rate givesWrite the specific angular momentum as $l=R^2\Omega$. Multiplying the azimuthal equation by $2\pi R$ and using the mass equation shows that the sum of advected and viscous angular-momentum flux is constant:The right-hand side implements the zero-torque inner boundary condition. For a Keplerian accretion disk, $l\propto R^{1/2}$ and $R^{3}d\Omega /dR=-(3/2)l$. ThereforeandFar outside the inner edge, $\Sigma \simeq \dot{m}/(3\pi \nu )$ and $u_R\simeq -3\nu /(2R)$.

The Shakura--Sunyaev thin disk model replaces poorly resolved turbulent angular-momentum transport by a stress proportional to pressure,Equivalently, an eddy viscosity has $\nu \sim v_{\mathrm{eddy}}{\ell }_{\mathrm{eddy}}$. Turbulent motions much faster than the sound speed $c_s$ would shock, while eddies much larger than the disk scale height $H$ would not fit within the disk. Writing $v_{\mathrm{eddy}}{\ell }_{\mathrm{eddy}}=\alpha c_{s}H$ therefore gives the alpha disk prescriptionwhere vertical hydrostatic balance gives $H\simeq c_s/{\Omega }_K$. The parameter $\alpha$ summarizes the correlation and efficiency of turbulent or magnetic stresses; it is not a molecular viscosity.

Let $\epsilon =H/R\ll 1$. Vertical hydrostatic balance gives $c_s\sim H{\Omega }_K$, whereas leading radial force balance gives $u_{\varphi }\sim R{\Omega }_K$. HenceUsing part (a) and $\nu =\alpha c_{s}H$ far from the inner edge,Thus subsonic radial drift in a thin disk coexists with highly supersonic orbital motion. Gas follows nearly circular, pressure-coherent orbits and loses angular momentum only slowly, completing of order $(\alpha {\epsilon }^2{)}^{-1}$ revolutions during its viscous inflow.

At fixed radiative efficiency, the Eddington ratio implies $\dot{m}\propto f_{\mathrm{Edd}}M$. Far outside the innermost stable circular orbit, part (a) gives $\Sigma \propto \dot{m}/\nu$. The supplied viscosity law therefore gives, in physical radius,The disk mass follows by radial integration:Since the Schwarzschild radius satisfies $R_S\propto M$, replacing $R$ by $(R/R_S)R_S$ contributes another factor $M^{5/4}$. Thusso

The alpha disk relation and Keplerian angular frequency giveCombining this with $\Omega \propto M^{1/2}{R}^{-3/2}$ and the surface-density scaling from part (d), the Toomre stability criterion becomesIt decreases strictly with radius, so it crosses unity at a unique self-gravitating radius of an accretion disk. Expressing radius in units of $R_S\propto M$ givesSolving $Q(R_{\mathrm{sg}})=1$ therefore yieldsand

Let $x_{\mathrm{ISCO}}=R_{\mathrm{ISCO}}/R_S$. Requiring a non-self-gravitating annulus outside the innermost stable circular orbit gives $R_{\mathrm{sg}}>R_{\mathrm{ISCO}}$, and equality definesFor a nonspinning hole, $x_{\mathrm{ISCO}}=3$, and with $C_2\simeq 10^5$, $\alpha \simeq 0.1$, and $f_{\mathrm{Edd}}\simeq 1$, this is approximately $8\times 10^9{M}_{\odot }$, conventionally quoted as order $10^{10}{M}_{\odot }$. Above this mass the disk would become self-gravitating essentially as soon as stable circular orbits begin, so the assumed smooth Shakura--Sunyaev thin disk cannot provide a broad luminous accretion region.

The corresponding Eddington luminosity is of order $10^{48}\mathrm{erg}{\mathrm{s}}^{-1}$, comparable to the upper envelope of quasar luminosities. This supports self-gravity as one contributor to the observed luminous-mass ceiling. It is not an absolute upper bound on black-hole mass: mergers, radiatively inefficient growth, nonstandard gas supply, spin-dependent inner radii, and fragmented or episodic accretion can all build a more massive hole without maintaining this particular steady thin disk.

A radiatively inefficient accretion flow radiates only a small fraction of the energy released before the gas crosses the inner boundary. At low Eddington ratio, an optically thin flow has such low density that radiative cooling, commonly proportional to density squared, is slower than inflow; the gas remains hot and forms an advection-dominated accretion flow. At high Eddington ratio, an optically thick slim accretion disk can instead undergo photon trapping in an accretion flow: diffusion is slower than inward motion, so radiation is advected into the hole.

The local accretion-flow advection balancehas three sign classes. If $q_{\mathrm{adv}}=0$, local heating equals local radiative cooling and the flow is a radiatively efficient thin disk. If $q_{\mathrm{adv}}>0$, heating exceeds cooling and inward advection removes the excess; low-rate ADAFs and high-rate slim disks are the two principal realizations. If $q_{\mathrm{adv}}<0$, radiation exceeds local dissipation and compressive advection supplies heat, producing a luminous hot accretion flow branch.

Denote the four terms byso the radial or poloidal momentum equation is $I=P+G+C$.

i) In a thin Keplerian accretion disk, radial inertia and pressure are higher-order in $H/R$, leaving $G+C\simeq 0$.

ii) In a nearly static stellar atmosphere, $I=C=0$ and hydrostatic pressure balance gives $P+G=0$.

iii) In pressureless gravitational collapse, rotation and pressure are negligible, so $I=G$; this is free fall.

iv) A slim accretion disk retains radial inertia and radial pressure together with gravity and centrifugal support, so all four terms generally survive: $I=P+G+C$.

v) A stationary geometrically thick disk or torus has negligible poloidal inertia but order-one pressure support, giving $P+G+C=0$.

vi) Nonrotating Bondi accretion has $C=0$ and $I=P+G$. In a highly supersonic Bondi--Hoyle limit the pressure term is also negligible, reducing this to ballistic $I=G$.

vii) A sub-Keplerian advection-dominated accretion flow has significant pressure support and radial inflow as well as rotation, so again $I=P+G+C$, with $C$ smaller than the Keplerian value and the remaining inward gravity balanced by $P$ and $I$.

The stated standard thin-disk dissipation flux is summed over the two faces, so the total luminosity isThis equals the Newtonian standard thin-disk luminosity and the orbital binding energy delivered per unit time at the inner edge. It is half the magnitude $GM\dot{m}/R_{\mathrm{ISCO}}$ of the potential-energy decrease because the other half appears as orbital kinetic energy. In the zero-torque model that remaining mechanical energy passes through the inner edge rather than being dissipated at larger radii. The associated Newtonian radiative efficiency of black-hole accretion is $GM/(2R_{\mathrm{ISCO}}{c}^2)$.

Steady mass conservation gives $\dot{m}=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed viscous torque in an accretion diskThenIntegration from $R_1$ to $R_2$ gives

The viscous power generated in an annulus is $2\pi R{F}_{\mathrm{diss}}dR=\mathcal{G}d\Omega$. Taking the inner torque to vanish and writing $l_{\mathrm{in}}=l(R_{\mathrm{in}})$ gives $\mathcal{G}=-\dot{m}(l-l_{\mathrm{in}})$, henceIntegration by parts yieldsFor steady circular force balance, $de=\Omega dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\mathrm{out}}-e_{\mathrm{in}}$. Therefore

When $R_{\mathrm{out}}\gg R_{\mathrm{in}}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\mathrm{in}}=-R_{\mathrm{in}}^2{\Omega }_{\mathrm{in}}^2/2$, soComparison with part (c) givesA Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because photon trapping in an accretion flow carries part of that generated energy through the inner edge.