Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 347 / 3

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.