= Solution
For the <shifted density perturbation in Newtonian gauge>, set $\Delta_R=\delta-4\Psi$. The radiation continuity equation gives $\Delta_R'=-4\theta/3$, and its Euler equation gives
$$
\Delta_R''=-\frac{k^2}{3}\Delta_R-\frac{8k^2}{3}\Psi.
$$
Writing $D_R=k^2+6\mathcal H^2$, the relativistic constraint becomes
$$
\Psi=-\frac{3\mathcal H^2}{2D_R}\left(\Delta_R-\frac{3\mathcal H}{k^2}\Delta_R'\right).
$$
Eliminating the potential therefore gives
$$
\boxed{\Delta_R''+\frac{12\mathcal H^3}{k^2+6\mathcal H^2}\Delta_R'
+\frac{k^2}{3}\left(1-\frac{12\mathcal H^2}{k^2+6\mathcal H^2}\right)\Delta_R=0}.
$$
This has the requested $\mathcal F=12\mathcal H^2/(k^2+6\mathcal H^2)$. Well inside the <sound horizon>, $k\gg\mathcal H$, the potential contribution is suppressed relative to the density term and $\delta\simeq\Delta_R$. Thus
$$
\boxed{\delta=C\cos(k\eta/\sqrt3)+D\sin(k\eta/\sqrt3)}.
$$
These are sound waves: radiation-pressure gradients restore a compression, while inertia carries the fluid through equilibrium. The perfect-fluid approximation neglects diffusion damping and freely streaming species' anisotropic stress. It gives the high-frequency limiting oscillations, not a claim that every radiation-era perturbation is an undriven sinusoid at horizon entry.
Back to article page
