 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-310/4/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For

V = λ ϕ^{4}

ϵ_{V} = \frac{8 M _{Pl}^{2}}{ϕ ^{2}}, η_{V} = \frac{12 M _{Pl}^{2}}{ϕ ^{2}} = \frac{3}{2} ϵ_{V} .

Thus

n_{s} - 1 = - 6 ϵ_{V} + 2 η_{V} = - 3 ϵ_{V}

. Combining this with the tensor-to-scalar ratio

r = 16 ϵ_{V}

n_{s} - 1 = - \frac{3}{16} r .

The measured tilt

n_{s} - 1 = - 0.04

would require

r ≃ 0.213

, whereas

r = 0.001

predicts

n_{s} - 1 ≃ - 1.88 \times 1 0^{- 4}

. Therefore quartic slow-roll inflation is inconsistent with the stated measurements.

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