Paper 2, Section I, D

Cosmology
Part II, 2009

(a) The equilibrium distribution for the energy density of a massless neutrino takes the form

ϵ=4πch30p3dpexp(pc/kT)+1.\epsilon=\frac{4 \pi c}{h^{3}} \int_{0}^{\infty} \frac{p^{3} d p}{\exp (p c / k T)+1} .

Show that this can be expressed in the form ϵ=αT4\epsilon=\alpha T^{4}, where the constant α\alpha need not be evaluated explicitly.

(b) In the early universe, the entropy density ss at a temperature TT is s=s= (8σ/3c)NST3(8 \sigma / 3 c) \mathcal{N}_{S} T^{3} where NS\mathcal{N}_{S} is the total effective spin degrees of freedom. Briefly explain why NS=N+NSD\mathcal{N}_{S}=\mathcal{N}_{*}+\mathcal{N}_{S D}, each term of which consists of two separate components as follows: the contribution from each massless species in equilibrium (Ti=T)\left(T_{i}=T\right) is

N=bosons gi+78fermions gi\mathcal{N}_{*}=\sum_{\text {bosons }} g_{i}+\frac{7}{8} \sum_{\text {fermions }} g_{i}

and a similar sum for massless species which have decoupled,

NSD=bosons gi(TiT)3+78fermions gi(TiT)3\mathcal{N}_{S D}=\sum_{\text {bosons }} g_{i}\left(\frac{T_{i}}{T}\right)^{3}+\frac{7}{8} \sum_{\text {fermions }} g_{i}\left(\frac{T_{i}}{T}\right)^{3}

where in each case gig_{i} is the degeneracy and TiT_{i} is the temperature of the species ii.

The three species of neutrinos and antineutrinos decouple from equilibrium at a temperature T1MeVT \approx 1 \mathrm{MeV}, after which positrons and electrons annihilate at T0.5MeVT \approx 0.5 \mathrm{MeV}, leaving photons in equilibrium with a small excess population of electrons. Using entropy considerations, explain why the ratio of the neutrino and photon temperatures today is given by

TνTγ=(411)1/3\frac{T_{\nu}}{T_{\gamma}}=\left(\frac{4}{11}\right)^{1 / 3}