Near the minimum, the <Taylor expansion> of the potential is
$$
V(\phi)\simeq V_0b^2\phi^2=\frac12m^2\phi^2,
\qquad
m^2=\frac{4V_0}{3M_{\rm Pl}^2}.
$$
Neglecting <Hubble friction> during one short oscillation reduces the <Klein-Gordon equation> to the <harmonic oscillator equation>
$$
\boxed{\ddot\phi+m^2\phi\simeq0}.
$$
The <virial theorem> gives $\langle\dot\phi^2/2\rangle=\langle V\rangle$, so $\langle P_\phi\rangle=0$. Restoring the slow cosmological damping in the averaged <cosmological perfect-fluid continuity equation> gives
$$
\dot{\langle\rho_\phi\rangle}+3H\langle\rho_\phi\rangle=0,
\qquad
\boxed{\langle\rho_\phi\rangle\propto a^{-3}}.
$$
The coherently oscillating inflaton therefore behaves as <pressureless matter>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-310/2/d/solution