The integrating factor for the Fourier-space transport equation is $e^{ik\mu\eta}$. Therefore the <line-of-sight solution for free-streaming photons> between an initial time $\eta_i$ and observation at $\eta_0$ is
$$
\boxed{
\Theta_{\mathbf k}(\eta_0,\mu)
=e^{-ik\mu(\eta_0-\eta_i)}\Theta_{\mathbf k}(\eta_i,\mu)
-ik\mu\int_{\eta_i}^{\eta_0}d\eta\,
e^{-ik\mu(\eta_0-\eta)}
[\delta N_{\mathbf k}(\eta)+\psi_{\mathbf k}'(\eta)]}.
$$
The first term freely streams the initial angular distribution; the integral accumulates the lapse and shift source along the unperturbed photon path.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 312 / 4 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-312/4/iv/solution