 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/9e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

A mode that crosses during matter domination subsequently grows by

\frac{a ( τ _{0} )}{a ( τ _{H} )} = (\frac{τ _{0}}{τ _{H}})^{2} .

Its present amplitude is consequently

∣ δ (k, τ_{0}) ∣ = (\frac{τ _{0}}{τ _{H}})^{2} τ_{H}^{2} A k^{1/2} = A τ_{0}^{2} k^{1/2} .

Now

τ_{0}^{2} = D_{eq} τ_{eq}^{2}, τ_{eq} = \frac{2 π}{k _{eq} c} .

Squaring therefore gives

∣ δ (k, τ_{0}) ∣^{2} = \frac{C}{k _{eq}^{4}} k, C = A^{2} D_{eq}^{2} (\frac{2 π}{c})^{4} .

Since

τ_{H} \propto k^{- 1}

, modes crossing chronologically between equality and today satisfy

k_{0} \leq k \leq k_{eq}

. The reverse ordering printed with the target formula is inconsistent with the stated definitions of

k_{0}

and

k_{eq}

; it does not affect the spectrum calculation.

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