Taking the ratio of the supplied <primordial tensor power spectrum> and scalar spectrum cancels the common factor $(H/2\pi)^2$:
$$
r=\frac{8}{M_{\rm Pl}^2}
\frac{\dot\phi^2}{H^2}
=16\epsilon.
$$
To first slow-roll order $\epsilon=\epsilon_V$, so the <tensor-to-scalar ratio> is
$$
\boxed{r=16\epsilon_V},
\qquad
\boxed{C=16}.
$$

Solution

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

 Solution

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