The right-hand side of the <slow-roll curvature power spectrum> is evaluated at <cosmological horizon exit>, $k=aH$. Using
$$
\dot\phi^2=2\epsilon M_{\rm Pl}^2H^2
$$
rewrites it as
$$
\Delta_{\mathcal R}^2
=\frac{H^2}{8\pi^2M_{\rm Pl}^2\epsilon}.
$$
At horizon exit, $d\log k=d\log(aH)=(1-\epsilon)dN$, 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>,
$$
n_s-1=\frac{d\log\Delta_{\mathcal R}^2}{d\log k}
=-2\epsilon-\eta.
$$
The supplied relations $\epsilon=\epsilon_V$ and $\eta_V=2\epsilon-\eta/2$ imply $\eta=4\epsilon_V-2\eta_V$. Therefore
$$
\boxed{n_s-1=-6\epsilon_V+2\eta_V}.
$$

Solution

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

 Solution

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