Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 55 / 3

For fully ionized hydrogen, the outward radiation force on a coupled proton-electron pair at radius $r$ is ${\sigma }_{T}L/(4\pi r^{2}c)$, while its inward gravitational force is approximately $GM{m}_p/r^2$. Equating them yields the Eddington luminosity,This uses isotropic radiation, Thomson scattering and efficient momentum coupling between electrons and ions. Composition or other opacity sources alter the corresponding limit.

With radiative efficiency ${ϵ}_r$, a rest-mass supply rate gives $L={ϵ}_r{\dot{M}}_{\mathrm{in}}{c}^2$. In the approximation that supplied mass is added to the hole, continuous accretion at the Eddington luminosity givesHere $t$ is the elapsed time since seed formation. If the radiated rest mass is counted exactly, $\dot{M}=(1-{ϵ}_r){\dot{M}}_{\mathrm{in}}$ and the Salpeter time becomes $t_e={ϵ}_{r}c{\sigma }_T/[(1-{ϵ}_r)4\pi G{m}_p]$. The displayed approximation in the question neglects this correction; retaining it lengthens the growth time.

The required growth factor is $10^7$, corresponding to $\ln (10^7)=16.12$ e-folds. With the supplied efficiency,At $z=8$, the expansion is overwhelmingly matter-dominated. Using total ${\Omega }_{\mathrm{mat}}=0.25$ and $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ givesThe baryon density is part of the total matter density, so it is not added to ${\Omega }_{\mathrm{mat}}$ in this age calculation. Dark-energy and radiation corrections are small for this estimate. Even starting at the earliest possible time, the approximate growth time exceeds the age, and a real stellar seed forms later. A $30M_{\odot }$ seed cannot reach the stated mass through uninterrupted Eddington-limited growth with the stated efficiency. Exactly retaining the radiated mass gives about $806\mathrm{Myr}$ and strengthens this conclusion. A heavier seed, lower radiative efficiency or super-Eddington episodes can relieve the time constraint; interruptions make it harder.

Use the supplied cosmological cosmic baryon fraction $f_b={\Omega }_{\mathrm{bar}}/{\Omega }_{\mathrm{mat}}=0.2$, and assume the halo initially has that fraction. Its total baryon mass is then $M_{b,\mathrm{tot}}=6\times 10^{11}{M}_{\odot }$. The selected baryons are the lowest-angular-momentum fractionHere the given cumulative specific angular momentum distribution must be normalized separately for the baryon component: equal baryon and dark-matter distributions mean equal normalized fractions, not equal absolute masses.

The virial velocity and the largest specific angular momentum in the selected inner baryon population areEstimate the outer radius of the settled low-angular-momentum component using circular rotational support and an enclosed-mass approximation to its self-gravity. With negligible dark matter inside the component, $v_d^2\simeq G{M}_b/r_d$ and $j_d=r_d{v}_d$, soThe given rounded gravitational constant produces the same estimate. The very small radius follows from selecting a small low-$j$ fraction, rather than assigning all central baryons the halo-edge angular momentum.

These estimates assume that gas radiates energy, preserves each parcel's specific angular momentum, and settles with negligible pressure support and no strong redistribution or cancellation of its angular-momentum vectors. They also assume that the phrase “innermost baryons” selects the lowest-$j$ material, the original cosmic baryon supply is retained, and the central gas supplies the dominant gravity. A possible central black hole or an exact flattened disk potential changes the numerical coefficient; the stated baryonic mass is used for this estimate.

The low-angular-momentum baryonic disk estimate uses a cumulative distribution uniform in $j$ from zero to $j_d$, so its mean is $j_d/2$. The quoted $r_d$ is the outer radius based on the cutoff angular momentum, not a one-zone radius based on the mean. Within the same enclosed-mass approximation, $M(<j)=M_{b}j/j_d$ and circular balance give $r(j)=j{j}_d/(G{M}_b)$: the rotation curve is approximately flat and the half-mass radius is $r_d/2$. Treating every baryon as one shell with the mean $j$ would instead give $r_d/4$ and twice the velocity, a different radius convention rather than the outer edge of the supplied distribution.