The <slow-roll e-fold count> before the end of inflation is
$$
N(\phi)=\int_t^{t_e}H\,dt
=\frac1{M_{\rm Pl}^2}\int_{\phi_e}^{\phi}\frac{V}{V'}\,d\phi.
$$
At leading order near the hilltop, $V/V'\simeq-\Lambda^4/(m^2\phi)$, and hence
$$
N(\phi)\simeq
\frac{\Lambda^4}{M_{\rm Pl}^2m^2}
\log\frac{\phi_e}{\phi}.
$$
This has $N(\phi_e)=0$ and gives
$$
\boxed{\phi_{60}=\phi_e
\exp\!\left(-\frac{60M_{\rm Pl}^2m^2}{\Lambda^4}\right)}.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-310/2/d/solution