The collisionless equation states that $f$ is constant along a photon geodesic:
$$
\frac{df}{d\eta}
=\partial_\eta f+\frac{dx^i}{d\eta}\partial_if
+\epsilon'\partial_\epsilon f
+\widehat p^{i\prime}\frac{\partial f}{\partial\widehat p^i}=0.
$$
At linear order $dx^i/d\eta=\widehat p^i$. The last term is second order because the background distribution is isotropic. Inserting the stated temperature parametrization into this <Free-streaming photon Boltzmann equation> and using the energy equation gives
$$
\boxed{\Theta'+\widehat p^i\partial_i\Theta
=-\widehat p^i\partial_i(\delta N+\psi')}.
$$
For a Fourier mode, with $\mu=\widehat{\mathbf k}\mathbin\cdot\widehat{\mathbf p}$, this is
$$
\boxed{\Theta_{\mathbf k}'+ik\mu\Theta_{\mathbf k}
=-ik\mu(\delta N_{\mathbf k}+\psi_{\mathbf k}')}.
$$

Solution

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

{c}

In the absence of collisions, a photon phase-space density is constant along its geodesic. Linearizing this statement gives a first-order transport equation for the direction-dependent temperature perturbation.


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 Incoming links: Free-streaming photon Boltzmann equation