Comparing the stated curvature <power spectrum> with $P_{\mathcal R}=2\pi^2A_s/k^3$ at the pivot scale gives the <primordial scalar amplitude>
$$
A_s=\frac{H^2}{8\pi^2\epsilon_VM_{\rm Pl}^2}
\simeq
\boxed{\frac{\Lambda^{12}}
{12\pi^2M_{\rm Pl}^6m^4\phi^2}}.
$$
The <scalar spectral index> is
$$
n_s-1\simeq-6\epsilon_V+2\eta_V
\simeq
\boxed{-\frac{3M_{\rm Pl}^2m^4\phi^2}{\Lambda^8}
-\frac{2M_{\rm Pl}^2m^2}{\Lambda^4}}.
$$
Sufficiently close to the hilltop, $\epsilon_V$ is suppressed by $\phi^2$ and the $\eta_V$ term controls the <spectral tilt>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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