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For the velocity convention $v_{\gamma }=-{\Theta }_1$ and $\dot{\tau }<0$, momentum-conserving Thomson scattering gives ${\dot{v}}_b+\mathcal{H}{v}_b+k\psi =(\dot{\tau }/R)(v_b-v_{\gamma })$. Here $R$ is the baryon loading parameter. Its collision force is opposite to the photon force when both are weighted by their enthalpies.

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 ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$. With the specified velocity convention, the photon continuity equation is thereforeThe 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 isIn 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: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 dragThis $R$ 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 $\dot{R}=\mathcal{H}R$.

With $v_b=v_{\gamma }$ at leading order and $c_s^2=1/[3(1+R)]$, the photon continuity equation and total Euler equation give ${\ddot{\delta }}_{\gamma }+\dot{R}{\dot{\delta }}_{\gamma }/(1+R)+k^2{\delta }_{\gamma }/[3(1+R)]=4\ddot{\varphi }+4\dot{R}\dot{\varphi }/(1+R)-4k^2\psi /3$. The baryon loading parameter changes inertia and shifts the gravitational equilibrium. The leading equations omit diffusion damping.