Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 2 / c

Table of contents
- Solution c

Solution

When

μ_{ν} / T_{ν} ≫ 1

, the Fermi-Dirac distribution approaches the step function

f (p) = 1

for

p < μ_{ν}

and zero for

p > μ_{ν}

. For one internal state and negligible mass, the degenerate relic neutrino energy density is therefore

ρ_{ν} ≃ \frac{1}{2 π ^{2}} \int_{0}^{μ_{ν}} p^{3} d p = \frac{μ _{ν}^{4}}{8 π ^{2}},

so

A = 1/ (8 π^{2})

.

At thermal equilibrium, the antineutrino chemical potential is

- μ_{ν}

. Its occupation is approximately

e^{- (p + μ_{ν}) / T_{ν}}

ρ_{\overset{ν}{ˉ}} ≃ \frac{3 T _{ν}^{4}}{π ^{2}} e^{- μ_{ν} / T_{ν}} .

It is exponentially smaller than the neutrino density; this particle-antiparticle imbalance is a cosmological lepton asymmetry.

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