 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/9d/b/solution

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

From

3 H \dot{ϕ} = - c^{2} V^{'}

and

H^{2} = 8 π G V / (3 c^{2})

\frac{ϕ ˙ ^{2}}{2 c ^{2}} = \frac{c ^{2} V ^{'2}}{18 H ^{2}} = \frac{c ^{4} n ^{2} V ( ϕ )}{48 π G ϕ ^{2}} .

The slow-roll approximation fails when this kinetic contribution is comparable with

V

ϕ_{end} ≃ \frac{n c ^{2}}{48 π G} .

The number of e-folds is

N_{e} = lo g \frac{a _{end}}{a _{i}} = \frac{4 π G}{c ^{4} n} (ϕ_{i}^{2} - ϕ_{end}^{2}) .

Using the estimate above,

N_{e} ≃ \frac{n}{12} [(\frac{ϕ _{i}}{ϕ _{end}})^{2} - 1] .

For

n = 6

and the roughly

60

e-folds needed for the flatness problem,

ϕ_{i} ≃ 11 ϕ_{end} .

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