<Conformal time> obeys $dt=a\,d\eta$, so $\dot\phi=a^{-1}\phi'$. Therefore
$$
\boxed{S_{\rm int}=\frac{\lambda}{3!}\int d^3x\,d\eta\,a(\eta)(\phi')^3}.
$$
For expanding <de Sitter spacetime>, $a=-1/(H\eta)$ with $-\infty<\eta<0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 312 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-312/1/i/solution