 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/14b/c/solution

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

Neglect the decaying mode from part a. A mode that enters during matter domination grows in amplitude by

\frac{a ( t _{0} )}{a ( t _{H} )} = (\frac{t _{0}}{t _{H}})^{2/3},

so its variance grows by

(t_{0} / t_{H})^{4/3}

. For

k < k_{eq}

, part b gives

P (k) = \frac{C}{k ^{3}} (\frac{k}{k _{0}})^{4} = \frac{C k}{k _{0}^{4}} .

For

k > k_{eq}

, the mode enters during radiation domination and is assumed not to grow significantly until equality. It then grows by

a (t_{0}) / a_{eq}

. Since the equality mode satisfies

\frac{t _{eq}}{t _{0}} = (\frac{k _{0}}{k _{eq}})^{3},

the post-equality variance growth is

(\frac{a ( t _{0} )}{a _{eq}})^{2} = (\frac{t _{0}}{t _{eq}})^{4/3} = (\frac{k _{eq}}{k _{0}})^{4} .

P (k) = V ⟨ ∣ δ_{k} (t_{0}) ∣^{2} ⟩ = ⎩ ⎨ ⎧ \frac{C k}{k _{0}^{4}}, \frac{C k _{eq}^{4}}{k ^{3} k _{0}^{4}}, k < k_{eq}, k > k_{eq} .

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