Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 1 / a

Solution

P = wρ

\overset{ρ}{˙} + 3 H (1 + w) ρ = 0.

Using

H = \overset{a}{˙} / a

\frac{d ρ}{ρ} = - 3 (1 + w) \frac{d a}{a} .

Taking

a_{0} = 1

ρ (a) = ρ_{0} a^{- 3 (1 + w)} = ρ_{0} (1 + z)^{3 (1 + w)} .

H (z) = H_{0} E (z), E (z) = [\sum_{i} Ω_{i, 0} (1 + z)^{3 (1 + w_{i})}]^{1/2} .

Along a radial light ray,

d χ = d t / a

d t = - d z / [(1 + z) H]

and

a = (1 + z)^{- 1}

d χ = - \frac{d z}{H ( z )} .

χ (z) = \frac{1}{H _{0}} \int_{0}^{z} \frac{d z ^{'}}{[ \sum _{i} Ω _{i, 0} ( 1 + z ^{'} ) ^{3 (1 + w_{i})} ] ^{1/2}} .

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