Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 347 (Astrophysical Black Holes)

At distance $r$, isotropic luminosity $L$ produces radiative acceleration $a_{\mathrm{rad}}=\kappa L/(4\pi r^{2}c)$ in material of opacity $\kappa$. In fully ionized hydrogen, one Electron supplies a Thomson scattering cross-section ${\sigma }_T$ for approximately one Proton mass of material, so $\kappa ={\sigma }_T/m_p$. Balancing radiation against the Newtonian gravitational field of the black hole gives the Eddington luminosityBelow this luminosity gravity can confine optically thin ionized hydrogen; above it, electron scattering alone accelerates the gas outward.

An electron--positron pair has mass $2m_e$ and total scattering cross-section $2{\sigma }_T$, so its opacity is ${\sigma }_T/m_e$. ThereforeThe limit is lower because the radiative force per unit inertial mass is larger in a pair plasma.

For a cloud with arbitrary constant opacity per unit mass,Suppose the luminosity jumps to a constant $L>L_{\mathrm{Edd}}$ while the cloud is at rest at $r_0$. With $\Gamma =L/L_{\mathrm{Edd}}$, its radial acceleration isMultiplication by $\dot{r}$ and integration from $r_0$ to infinity gives conservation of energyThis result assumes that $L$ and $\kappa$ remain constant and that the cloud stays optically thin.

Use cylindrical coordinates $(R,\varphi ,z)$, put $s=(R^2+z^2{)}^{1/2}$, and write the purely azimuthal velocity as $\mathbf{u}=R\Omega {\mathbf{e}}_{\varphi }$. The steady radial, azimuthal, and vertical momentum equations areThus $\nabla p=\rho {\mathbf{g}}_{\mathrm{eff}}$, where the effective gravity in a rotating frame isWhen radiation pressure supplies the pressure support, force balance gives

On the upper conical surface $z=R\tan \alpha$, the tangent and outward normal areBecause a pressure surface is an equipotential surface, ${\mathbf{g}}_{\mathrm{eff}}\cdot \mathbf{t}=0$. Since $s=R/\cos \alpha$, this fixesThe lower face gives the reflected result. One face has area element $dA=2\pi R,dR/\cos \alpha$. Adding both faces and using $F=c|{\mathbf{g}}_{\mathrm{eff}}|/\kappa$ yieldsConsequentlyFor a broad radial range, each logarithmic interval can radiate an Eddington-scale contribution because rotation reduces the surface-normal gravity. The factor $\sin \alpha$ makes a thin wedge faint and lets a geometrically thick funnel expose a larger effective gravity and emitting area.

A singular isothermal sphere has total density and circular speedIf gas traces the gravitating matter, ${\rho }_g=f_{\mathrm{gas}}{\rho }_{\mathrm{tot}}$. A moving black hole gravitationally focuses that gas, so the relevant estimate is the Bondi--Hoyle--Lyttleton accretion rateHydrostatic gas in the isothermal potential has $c_s$ of order $\sigma$, and thereforeup to an order-one coefficient. Taking $c_s=\sigma$ and $v_c=\sqrt{2}\sigma$ gives $2/(3\sqrt{3})$ multiplying the displayed expression. The same $G^2{M}^2\rho /v^3$ scaling that produces Chandrasekhar dynamical friction therefore sets the likely capture rate as the hole spirals inward. Feedback, angular momentum, a gas core, or a multiphase medium can reduce the actual rate.

Well inside the Bondi accretion radius, pressure is negligible and the inflow approaches the free-fall speedSteady spherical mass conservation givesEstimating the outward optical depth across a radial scale $r$ with electron-scattering opacity $\kappa ={\sigma }_T/m_p$,The integral optical depth to infinity differs only by a factor of two in this free-fall approximation.

At the Schwarzschild radius, $\tau =1$ whenIf the conventional Eddington accretion rate is ${\dot{M}}_{\mathrm{Edd}}=L_{\mathrm{Edd}}/({\eta }_0{c}^2)$, then ${\dot{M}}_{\mathrm{crit}}=2{\eta }_0{\dot{M}}_{\mathrm{Edd}}$. This comparison depends on the reference efficiency ${\eta }_0$, while ${\dot{M}}_{\mathrm{crit}}$ itself follows directly from optical depth.

Writing $\dot{m}=\dot{M}/{\dot{M}}_{\mathrm{crit}}$ givesFor $\dot{m}>1$, the inner flow is optically thick and its photosphere lies at

The diffusion speed is $v_{\mathrm{diff}}\sim c/\tau$. Henceand equality defines the photon-trapping radiusInside $r_t$, inward advection outruns outward diffusion, so photon trapping in an accretion flow carries most released radiation into the black hole.

Only the binding energy released outside $r_t$ escapes efficiently. Its order of magnitude isThe radiative efficiency of black-hole accretion therefore decreases inversely with supply rate:Order-one coefficients depend on the optical-depth and inner-boundary conventions, but the luminosity saturation and $eta\propto {\dot{M}}^{-1}$ are robust.

Vertically integrating the continuity equation and imposing a steady axisymmetric flow givesThus the inward-positive mass accretion rate isThe zero-torque condition at $R_{\mathrm{ISCO}}$ gives the standard Shakura--Sunyaev thin disk relationCombining this relation with mass conservation, or substituting it into the stated drift equation, yieldsFar outside the inner edge this reduces to $u_R\simeq -3\nu /(2R)$.

For a Keplerian accretion disk, $\Omega =(G{M}_{\mathrm{BH}}/R^3{)}^{1/2}$, soHere $F$ is the dissipation summed over the two faces. The total luminosity isThis is the binding energy per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The zero-torque inner boundary condition ensures that no additional mechanical work enters from smaller radii.

The luminosity emitted outside radius $R$ isWriting $y=(R_{\mathrm{ISCO}}/R{)}^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and henceHalf of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.

Far from the inner edge, use $u_R\simeq -3\nu /(2R)$ and the alpha-viscosity prescriptionMass conservation then givesAt the onset of disk self-gravity, $\rho =M_{\mathrm{BH}}/R^3$. Since ${\Omega }^2=G{M}_{\mathrm{BH}}/R^3$, this condition reduces to

The blackbody approximation and local radiative balance on one face givewhile the stated ideal-gas estimate is $c_s^2=\gamma k_B{T}_{\mathrm{eff}}/\mu$. Inserting these expressions into the self-gravity condition givesSolving for $R$ producesBeyond this radius, the disk's own gravity rivals the central vertical gravity; fragmentation and star formation can then invalidate the smooth steady thin-disk model.