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

The right-hand side of the slow-roll curvature power spectrum is evaluated at cosmological horizon exit, $k=aH$. Usingrewrites it asAt horizon exit, $d\log k=d\log (aH)=(1-ϵ)dN$, so $d/d\log k=d/dN$ to first slow-roll order. With the first Hubble slow-roll parameter and second Hubble slow-roll parameter,The supplied relations $ϵ={ϵ}_V$ and ${\eta }_V=2ϵ-\eta /2$ imply $\eta =4{ϵ}_V-2{\eta }_V$. Therefore

For a monotonic inflaton trajectory, the potential slow-roll parameter satisfiesThe remaining slow-roll e-fold count is the integral from the present field value to its end value along the direction of motion:

Taking the ratio of the supplied primordial tensor power spectrum and scalar spectrum cancels the common factor $(H/2\pi {)}^2$:To first slow-roll order $ϵ={ϵ}_V$, so the tensor-to-scalar ratio is

For the quadratic hilltop inflation potential while $V_0$ dominates,Since ${ϵ}_V$ is of higher order near the hilltop, the scalar spectral index obeys $n_s-1\simeq 2{\eta }_V$. The e-fold integral becomesand henceSubstitution into $r=C{ϵ}_V$ yields the quadratic hilltop slow-roll predictionso $\boxed{f=1/8}$.

For $n_s-1=-0.04$ the expression decreases as $\tilde{N}$ increases. Its largest allowed value therefore uses $\tilde{N}=40$ and ${\varphi }_e/M_{\mathrm{Pl}}<1$: