For this metric, $\sqrt{-g}=a^2$, while $g^{\eta\eta}=-a^{-2}$, $g^{zz}=a^{-2}$ and $g^{ij}=\delta^{ij}$. The massless <Klein-Gordon equation> is consequently
$$
\Box\phi
=\frac1{a^2}(-\partial_\eta^2+\partial_z^2)\phi
+\delta^{ij}\partial_i\partial_j\phi=0.
$$
Insert the spatial <Fourier transform> given in the question. Each <wavenumber> mode then satisfies
$$
\boxed{\Phi_{k_z\mathbf k}''
+\omega_{k_z\mathbf k}^2(\eta)\Phi_{k_z\mathbf k}=0},
\qquad
\boxed{\omega_{k_z\mathbf k}^2(\eta)=k_z^2+a(\eta)^2|\mathbf k|^2}.
$$
Thus every field mode is a <simple harmonic oscillator> with a time-dependent <angular frequency>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 4 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-311/4/b/i/solution