Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 4 / d

Solution

For the quadratic hilltop inflation potential while

V_{0}

dominates,

η_{V} ≃ - \frac{m ^{2} M _{Pl}^{2}}{V _{0}}, ϵ_{V} ≃ \frac{M _{Pl}^{2} m ^{4} ϕ ^{2}}{2 V _{0}^{2}} = \frac{1}{2} η_{V}^{2} \frac{ϕ ^{2}}{M _{Pl}^{2}} .

Since

ϵ_{V}

is of higher order near the hilltop, the scalar spectral index obeys

n_{s} - 1 ≃ 2 η_{V}

. The e-fold integral becomes

N (ϕ) = \frac{1}{∣ η _{V} ∣} lo g \frac{ϕ _{e}}{ϕ} = \frac{2}{1 - n _{s}} lo g \frac{ϕ _{e}}{ϕ},

and hence

ϕ = ϕ_{e} e^{(n_{s} - 1) N /2} .

Substitution into

r = C ϵ_{V}

r ≃ \frac{1}{8} C (n_{s} - 1)^{2} (\frac{ϕ _{e}}{M _{Pl}})^{2} e^{(n_{s} - 1) N},

f = 1/8

For

n_{s} - 1 = - 0.04

the expression decreases as

N

increases. Its largest allowed value therefore uses

N = 40

and

ϕ_{e} / M_{Pl} < 1

r < \frac{1}{8} (16) (0.04)^{2} e^{- 0.04 (40)} = 0.0032 e^{- 1.6} ≃ 6.5 \times 1 0^{- 4} .

There are no discussions about this article yet.

There are currently no matching articles.