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For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.