 Solution

Write $T=\overline{T}(1+\Theta )$, with $\Theta =\Delta T/\overline{T}$ at linear order. Since $\overline{T}\propto a^{-1}$, the background Planck function depends only on comoving momentum. Expanding its temperature-shifted argument givesUsing $T$ rather than $\overline{T}$ in the perturbation denominator changes only higher-order terms.

The photon brightness perturbation is the energy-weighted angular perturbationIntegration by parts gives $\int q^4{f}_0^{\prime }dq=-4\int q^3{f}_{0}dq$; the boundary term vanishes at both endpoints for a Planck spectrum. Therefore $\Delta =4\Theta$. Multiplying the transport equation by $q^3$ and integrating yields the synchronous photon brightness equationThe PDF's definition is $4\Delta T/T$; the TeX's denominator $\hslash$ is a transcription error.

Expand $\Delta$ in Legendre moments in $\mu$. The monopole is the density/temperature perturbation, and the dipole is the bulk velocity. Before decoupling, rapid Thomson scattering isotropizes photons in the electron rest frame. In the tight-coupling approximation, the scattering rate is much larger than $k$ and $\mathcal{H}$, so the quadrupole and higher moments are suppressed by the mean-free-path parameter, with the quadrupole typically of order $k/\dot{\kappa }$ times the dipole. To leading order only monopole and dipole are needed. This suppression uses the collision term restored before decoupling; it does not follow from the collisionless equation by itself. Finite quadrupole corrections generate polarization and diffusion damping.