 Solution

A scalar cosmological perturbation at fixed wavevector carries no helicity or distinguished transverse direction. Its metric and scalar sources are unchanged by rotations around $\mathbf{k}$, and the scalar baryon peculiar velocity is longitudinal. With scalar initial conditions, rotational invariance of the linear evolution therefore makes $\Theta$ a function only of $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. Expand it using the stated normalization, without an extra $(2\ell +1)$ factor:In the Thomson kernel, $1+v^2=4/3+(2/3)P_2(v)$. The spherical harmonic addition theorem and Orthogonality of Legendre polynomials giveOnly the monopole and quadrupole survive; the quadrupole's $(-i{)}^2=-1$ gives the angular scattering integralThis also explains why an isotropic distribution has zero net collision term.

The streaming term $ik\mu \Theta$ and the Legendre polynomial recurrence relation yield, after division by $(-i{)}^{\ell }$, $k[(\ell +1){\Theta }_{\ell +1}/(2\ell +3)-\ell {\Theta }_{\ell -1}/(2\ell -1)]$. With ${\mathbf{v}}_b=i\hat{\mathbf{k}}{v}_b$, the velocity collision source is $+{\tau }^{\prime }{v}_b$ in the dipole equation; the gravitational gradient source is $+k\psi$. Therefore the photon Boltzmann hierarchy isSet the absent ${\Theta }_{-1}$ contribution to zero. This is algebraically the printed hierarchy. In particular,Since ${\tau }^{\prime }<0$, these collision terms damp the nonzero multipoles and drive the dipole to ${\Theta }_1=-v_b$. The latter sign follows from the paper's velocity and $(-i{)}^{\ell }$ conventions, not a reversed physical photon velocity.