Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 310 / 2

For the hilltop inflation potential,The two potential slow-roll parameters are consequentlyand

Estimating the end of slow-roll inflation by ${ϵ}_V({\varphi }_e)\simeq 1$ and using the leading hilltop expression givesThis is a formal leading-order estimate. Its self-consistency requires $m^2{\varphi }_e^2\ll {\Lambda }^4$; if that condition fails, the full potential or the mechanism that ends inflation must be retained.

The homogeneous inflaton obeysUnder the slow-roll approximation, the acceleration and kinetic-energy corrections may be neglected, so the field equation and Friedmann equation becomeThus a positive field rolls away from the hilltop while $H$ is approximately constant.

The slow-roll e-fold count before the end of inflation isAt leading order near the hilltop, $V/V^{\prime }\simeq -{\Lambda }^4/(m^2\varphi )$, and henceThis has $N({\varphi }_e)=0$ and gives

Comparing the stated curvature power spectrum with $P_{\mathcal{R}}=2{\pi }^2{A}_s/k^3$ at the pivot scale gives the primordial scalar amplitudeThe scalar spectral index isSufficiently close to the hilltop, ${ϵ}_V$ is suppressed by ${\varphi }^2$ and the ${\eta }_V$ term controls the spectral tilt.

Neglecting the ${ϵ}_V$ contribution to the tilt, the observed value givesPart (d) then gives ${\varphi }_{60}=e^{-1}{\varphi }_e$. Using the formal end estimate from part (b), ${ϵ}_V({\varphi }_{60})\simeq e^{-2}$. The amplitude relation therefore yieldsThus the requested order-of-magnitude estimate isThe simultaneous approximations make ${ϵ}_V({\varphi }_{60})$ only moderately small, so this numerical result should be read as the order-of-magnitude estimate requested rather than a precision fit.