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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 14B / b / Solution

Normalize

a (t_{0}) = 1

and put

k_{0} = 2 π / (c t_{0})

. The cosmological horizon crossing condition is

\frac{t _{H}}{t _{0}} = \frac{k _{0}}{k} a (t_{H}) .

For

t_{H} > t_{eq}

, matter domination gives

a (t_{H}) = (t_{H} / t_{0})^{2/3}

, and therefore

\frac{t _{H}}{t _{0}} ≃ (\frac{k _{0}}{k})^{3} .

For

t_{H} < t_{eq}

, radiation domination and continuity at equality give

a (t) = a_{eq} (\frac{t}{t _{eq}})^{1/2}, a_{eq} = (\frac{t _{eq}}{t _{0}})^{2/3} = \frac{1}{1 + z _{eq}} .

Substitution in the crossing condition yields

\frac{t _{H}}{t _{0}} ≃ \frac{1}{1 + z _{eq}} (\frac{k _{0}}{k})^{2} .

These are the two branches of the horizon-crossing time across matter-radiation equality.

Solved by gpt-5.6-sol high.

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