Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 9D (Cosmology)

Table of contents
- a 9D
  - Solution a
- b 9D
  - Solution b
- c 9D
  - Solution c

a

Solution

From

E = ρ V

,

P = ρ /3

, and

d E = T, d S - P, d V

,

T, d S = V, d ρ + \frac{4}{3} ρ, d V .

Using

ρ = 4 σ T^{4} / c

gives

d S = \frac{16 σ}{3 c} (T^{3} d V + 3 T^{2} V d T) = d (\frac{16 σ}{3 c} V T^{3}) .

Therefore

S = \frac{16 σ}{3 c} V T^{3}, s = \frac{S}{V} = \frac{16 σ}{3 c} T^{3},

up to an irrelevant additive entropy constant.

b

Solution

When

Γ ≫ H

, reactions maintain thermal and chemical equilibrium, so the expansion is quasistatic and adiabatic. With no change in the equilibrium degrees of freedom, comoving entropy and photon number are conserved. Since

s \propto T^{3}, s a^{3} = constant,

one gets

T^{3} a^{3} = constant

and hence

T \propto a^{- 1}

.

c

Solution

Before electron-positron annihilation, the electromagnetic plasma has

g_{* s} = 2 + \frac{7}{8} (2 + 2) = \frac{11}{2},

whereas afterward only the two photon polarizations remain, so

g_{* s} = 2

. Entropy conservation in the decoupled electromagnetic sector gives

\frac{11}{2} T_{before}^{3} a^{3} = 2 T_{γ}^{3} a^{3} .

Neutrinos receive none of this entropy and continue cooling as

T_{ν} \propto a^{- 1}

. Therefore

\frac{T _{ν}}{T _{γ}} = (\frac{4}{11})^{1/3} .

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