Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 9D (Cosmology)

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

a

Solution

Use

E = m c^{2} + p^{2} / (2 m) + \dots

and replace the Fermi denominator by its Boltzmann form. The Gaussian integral gives

n_{X} = (\frac{2 π m _{X} k T}{h ^{2}})^{3/2} e^{- m_{X} c^{2} / k T} .

b

Solution

At

k T_{d} = m_{X} c^{2} / α

n_{γ} = 16 π ζ (3) (k T / h c)^{3}

gives

r (α) = 2 π α^{3/2} e^{- α} / [8 ζ (3)]

.

c

Solution

Today

ρ_{X} / ρ_{B} = (m_{X} / m p) r / η ≃ Ω_{D M} / Ω_{B} ≃ 5

. Hence

m_{X} ≃ \frac{5 η}{r ( α )} m_{p} = \frac{3 \times 1 0 ^{- 9}}{r ( α )} m_{p},

so

μ = 3 \times 1 0^{- 9}

. For suitable large

α

this is nonrelativistic and can match the observed abundance.

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