At horizon exit, the <primordial tensor power spectrum> is proportional to $H^2$. Therefore
$$
n_T=\frac{d\log\Delta_t^2}{d\log k}
=\frac{-2\epsilon}{1-\epsilon}
=-2\epsilon+O(\epsilon^2).
$$
Thus
$$
\boxed{C=2}.
$$
The ratio of the stated tensor and scalar spectra is
$$
r=\frac{8M_{\rm Pl}^{-2}(H/2\pi)^2}
{(2\epsilon M_{\rm Pl}^2)^{-1}(H/2\pi)^2}
=16\epsilon.
$$
Eliminating $\epsilon$ gives the <single-field inflation consistency relation>
$$
\boxed{r=-8n_T}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 4 / c / Solution

{title2=$\Delta_t^2$}

For the two graviton polarizations in canonical single-field inflation,
$$
\Delta_t^2(k)=\frac8{M_{\rm Pl}^2}
\left(\frac{H}{2\pi}\right)^2
$$
at horizon exit.


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