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

The line-of-sight solution integrates the gravitational source along the unperturbed photon trajectory. In Fourier space the integrating factor $e^{ik\mu\eta}$ accounts for free streaming between emission and observation.


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