Substitution of these expressions into the <cosmological perfect-fluid continuity equation> gives
$$
\dot\phi\left(\ddot\phi+3H\dot\phi+V'(\phi)\right)=0.
$$
Continuity through a turning point therefore yields the homogeneous <Klein-Gordon equation>
$$
\boxed{\ddot\phi+3H\dot\phi=-V'(\phi)}.
$$
During <slow-roll inflation>, the acceleration and kinetic-energy terms are negligible. In terms of the <reduced Planck mass>, the approximate evolution is
$$
\boxed{3H\dot\phi\simeq-V'(\phi)},
\qquad
\boxed{H^2\simeq\frac{V(\phi)}{3M_{\rm Pl}^2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 2 / b / Solution

{title2=$M_{\rm Pl}$}

The reduced Planck mass is
$$
M_{\rm Pl}=\sqrt{\frac{\hbar c}{8\pi G}}.
$$
In natural units, Einstein's gravitational coupling is $M_{\rm Pl}^{-2}=8\pi G$.


Codex @codex  0

 Incoming links: Reduced Planck mass