= Solution
Throughout this question dots denote <conformal time> derivatives, and $\mathcal H=\dot a/a$ is the <conformal Hubble parameter>. The monopole of the <photon Boltzmann hierarchy> gives $\dot\Theta_0+k\Theta_1/3=\dot\phi$. With the specified velocity convention, the <photon continuity equation> is therefore
$$
\boxed{\dot\delta_\gamma-\frac43kv_\gamma=4\dot\phi.}
$$
The dipole equation is $\dot\Theta_1+k(2\Theta_2/5-\Theta_0)=\dot\tau(\Theta_1+v_b)+k\psi$. Thus the <photon Euler equation> is
$$
\boxed{\dot v_\gamma+\frac{k}{4}\delta_\gamma-\frac{2k}{5}\Theta_2+k\psi=\dot\tau(v_\gamma-v_b).}
$$
In particular, $\dot\tau<0$ makes the scattering term damp the <photon-baryon velocity slip>; it does not amplify it.
The linear momentum density of the <photon-baryon fluid> is weighted by enthalpy, not merely energy density:
$$
q_{\mathrm{tot}}=\frac43\bar\rho_\gamma v_\gamma+\bar\rho_bv_b.
$$
Elastic <Thomson scattering> exchanges momentum internally. Its <photon> force density is $(4\bar\rho_\gamma/3)\dot\tau(v_\gamma-v_b)$, so the <baryon> force density is its negative. Dividing by $\bar\rho_b$ gives the <baryon Euler equation with Thomson drag>
$$
\boxed{\dot v_b+\mathcal H v_b+k\psi=\frac{\dot\tau}{R}(v_b-v_\gamma),\qquad R=\frac{3\bar\rho_b}{4\bar\rho_\gamma}.}
$$
This $R$ is the <baryon loading parameter>, the <baryon>-to-<photon> inertia ratio. The two collision terms cancel exactly when the equations are weighted by their enthalpies. Background conservation gives $\bar\rho_b\propto a^{-3}$ and $\bar\rho_\gamma\propto a^{-4}$, so $\dot R=\mathcal H R$.
Back to article page
