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

Table of contents
- Solution i

Solution

For fully ionized hydrogen, the outward radiation force on a coupled proton-electron pair at radius

r

is

σ_{T} L / (4 π r^{2} c)

, while its inward gravitational force is approximately

GM m_{p} / r^{2}

. Equating them yields the Eddington luminosity,

L_{Edd} = \frac{4 π GM m _{p} c}{σ _{T}} .

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}_{in} c^{2}

. In the approximation that supplied mass is added to the hole, continuous accretion at the Eddington luminosity gives

\dot{M} ≃ \frac{L _{Edd}}{ϵ _{r} c ^{2}} = \frac{M}{t _{e}}, M (t) = M_{0} e^{t / t_{e}}, t_{e} ≃ ϵ_{r} \frac{c σ _{T}}{4 π G m _{p}} ≃ 4.5 \times 1 0^{8} ϵ_{r} yr .

Here

t

is the elapsed time since seed formation. If the radiated rest mass is counted exactly,

\dot{M} = (1 - ϵ_{r}) \dot{M}_{in}

and the Salpeter time becomes

t_{e} = ϵ_{r} c σ_{T} / [(1 - ϵ_{r}) 4 π G m_{p}]

. The displayed approximation in the question neglects this correction; retaining it lengthens the growth time.

The required growth factor is

1 0^{7}

, corresponding to

ln (1 0^{7}) = 16.12

e-folds. With the supplied efficiency,

t_{grow} ≃ 45 Myr \times 16.12 ≃ 725 Myr .

At

z = 8

, the expansion is overwhelmingly matter-dominated. Using total

Ω_{mat} = 0.25

and

H_{0} = 70 km s^{- 1} Mp c^{- 1}

gives

t (z) ≃ \int_{z}^{\infty} \frac{d z ^{'}}{H _{0} Ω _{mat} ( 1 + z ^{'} ) ^{5/2}} = \frac{2}{3 H _{0} Ω _{mat}} (1 + z)^{- 3/2}, t (z = 8) ≃ 690 Myr .

The baryon density is part of the total matter density, so it is not added to

Ω_{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 $30 M_{⊙}$ seed cannot reach the stated mass through uninterrupted Eddington-limited growth with the stated efficiency. Exactly retaining the radiated mass gives about

806 Myr

and strengthens this conclusion. A heavier seed, lower radiative efficiency or super-Eddington episodes can relieve the time constraint; interruptions make it harder.

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