For the quadratic <hilltop inflation> potential while $V_0$ dominates,
$$
\eta_V\simeq-\frac{m^2M_{\rm Pl}^2}{V_0},
\qquad
\epsilon_V\simeq
\frac{M_{\rm Pl}^2m^4\phi^2}{2V_0^2}
=\frac12\eta_V^2\frac{\phi^2}{M_{\rm Pl}^2}.
$$
Since $\epsilon_V$ is of higher order near the hilltop, the <scalar spectral index> obeys $n_s-1\simeq2\eta_V$. The e-fold integral becomes
$$
\widetilde N(\phi)
=\frac1{|\eta_V|}\log\frac{\phi_e}{\phi}
=\frac2{1-n_s}\log\frac{\phi_e}{\phi},
$$
and hence
$$
\phi=\phi_e
e^{(n_s-1)\widetilde N/2}.
$$
Substitution into $r=C\epsilon_V$ yields the <quadratic hilltop slow-roll prediction>
$$
\boxed{r\simeq\frac18C(n_s-1)^2
\left(\frac{\phi_e}{M_{\rm Pl}}\right)^2
e^{(n_s-1)\widetilde N}},
$$
so $\boxed{f=1/8}$.

For $n_s-1=-0.04$ the expression decreases as $\widetilde N$ increases. Its largest allowed value therefore uses $\widetilde N=40$ and $\phi_e/M_{\rm Pl}<1$:
$$
r<\frac18(16)(0.04)^2e^{-0.04(40)}
=0.0032e^{-1.6}
\simeq\boxed{6.5\times10^{-4}}.
$$

Solution

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

 Solution

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