For the <monomial inflation potential> $V=\lambda M_{\rm Pl}^4(\phi/M_{\rm Pl})^\alpha$,
$$
\epsilon\simeq\frac{M_{\rm Pl}^2}{2}\left(\frac{V_{,\phi}}V\right)^2
=\frac{\alpha^2M_{\rm Pl}^2}{2\phi^2},
$$
and
$$
\eta=4\epsilon-2M_{\rm Pl}^2\frac{V_{,\phi\phi}}V
=\frac{2\alpha M_{\rm Pl}^2}{\phi^2}.
$$
Slow-roll inflation is possible when both are much smaller than one:
$$
\frac{\phi^2}{M_{\rm Pl}^2}\gg
\max\!\left(\frac{\alpha^2}{2},2\alpha\right).
$$
Inflation ends once one of these conditions fails, commonly near $|\phi|\simeq\alpha M_{\rm Pl}/\sqrt2$ when $\epsilon\simeq1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 310 / 4 / c / Solution

{title2=$V(\phi)=\lambda M_{\rm Pl}^4(\phi/M_{\rm Pl})^\alpha$}

A monomial inflation potential is a positive power of the inflaton. Its potential slow-roll parameter is $\epsilon_V=\alpha^2M_{\rm Pl}^2/(2\phi^2)$.


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 Incoming links: Monomial inflation potential