For a monotonic inflaton trajectory, the <potential slow-roll parameter> satisfies
$$
\left|\frac{d\phi}{dN}\right|
=M_{\rm Pl}\sqrt{2\epsilon_V(\phi)}.
$$
The remaining <slow-roll e-fold count> is the integral from the present field value to its end value along the direction of motion:
$$
\boxed{\widetilde N(\phi)
=\int_\phi^{\phi_e}
\frac{d\widetilde\phi}
{M_{\rm Pl}\sqrt{2\epsilon_V(\widetilde\phi)}}}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-310/4/b/solution