The massless field equation is
$$
\boxed{\phi''+2\mathcal H\phi'-\nabla^2\phi=0},
\qquad \mathcal H=\frac{a'}a.
$$
For a homogeneous solution this becomes
$$
(a^2\bar\phi')'=0.
$$
Therefore
$$
\boxed{\bar\phi'\propto a^{-2}},
\qquad \boxed{p=-2}.
$$
Since $\dot{\bar\phi}=\bar\phi'/a$, the cosmic-time velocity decays as $a^{-3}$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-312/2/i/solution