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The squared adiabatic sound speed is the ratio of background pressure and energy-density time derivatives. For a barotropic equation of state it is $dP/d\rho$. The non-adiabatic pressure perturbation is the difference $\delta P-c_a^2\delta \rho$; in a general field or multicomponent model this background quantity need not equal the rest-frame propagation speed.

For a flat background dominated by one constant-$w$ barotropic equation of state, assume $\delta P=w\delta \rho$ and no scalar anisotropic stress. The Einstein field equations then reduce the common Newtonian-gauge potential to the stated Fourier transform equation, with $\mathcal{H}=2/[(1+3w)\tau ]$. For radiation the regular solution is $3{\Phi }_0(\sin x-x\cos x)/x^3$, $x=k\tau /\sqrt{3}$, and oscillates with a decaying envelope after sound-horizon entry. For pressureless matter the two solutions are a constant and ${\tau }^{-5}$. A non-adiabatic pressure perturbation would supply an additional source, so constant background $w$ alone does not justify the homogeneous equation.

For $W=\bar{\rho }+\bar{P}\ne 0$, the conserved uniform-density curvature in the spatial-metric convention $1-2\Phi$ isThe source's plus sign in front of both terms is not gauge invariant and cannot yield the claimed conservation. Choosing the opposite overall sign, $\Phi -\delta \rho /(3W)$, is equally valid, but changing only one term is necessary. A passive time displacement $t\mapsto t+T$ gives $\tilde{\Phi }=\Phi +\bar{N}HT$ and $\widetilde{\delta \rho }=\delta \rho +3\bar{N}HWT$, so their combination above is unchanged. This is the uniform-density curvature sign test.

The lapse perturbation in the perturbed energy-conservation source is $\Psi$, not the spatial-curvature perturbation $\Phi$. Indeed the perturbation of the proper-time density derivative is $\mathcal{D}\delta \rho -\Psi \mathcal{D}\bar{\rho }$, while the normal expansion changes by $-\kappa$. The consistent equation isFor a constant barotropic equation of state, $\delta P=w\delta \rho$ and $\mathcal{D}W=-3H(1+w)W$. Differentiate ${\zeta }_{\mathrm{ud}}$ and cancel the density terms:More generally, $\delta P_{\mathrm{nad}}=\delta P-(\mathcal{D}\bar{P}/\mathcal{D}\bar{\rho })\delta \rho$ givesRegular scalar modes have gradient terms suppressed by $(k/aH{)}^2$. Thus adiabatic uniform-density curvature is conserved to leading order on superhorizon scales, ${\dot{\zeta }}_{\mathrm{ud}}\simeq 0$. This is a long-wavelength limit, not exact finite-wavenumber conservation; it assumes a regular gradient expansion and excludes $W=0$, where this variable is undefined. In contrast the literal printed variable obeys $\mathcal{D}{\zeta }_{\mathrm{printed}}=2\mathcal{D}\Phi +O(k^2)$ even after repairing the energy equation, and generally evolves across changes of the background equation of state.

There are three practical benefits. First, uniform-density curvature is gauge invariant, so matching it does not depend on coordinate slicing. Second, superhorizon conservation of uniform-density curvature transports primordial amplitudes through changes in the expansion history without solving each short-scale dynamical equation. Third, when a mode re-enters the horizon it supplies initial conditions for density growth and Cosmic microwave background anisotropy; non-adiabatic pressure is also an explicit diagnostic of when that simple transport fails.

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

The cosmological perfect-fluid continuity equation and the constant barotropic equation of state $P=w\rho$ giveOn every interval where $a$ is monotone, separation of variables yieldsHence the constant-equation-of-state density scaling iswhere $\mu$ is a constant. The same expression extends through a regular turning point by continuity.

A perfect fluid in general relativity has stress-energy tensorwhere $u^{\mu }{u}_{\mu }=-1$. Besides the metric tensor, one needs the energy density $\rho$, the pressure $P$, and three independent components of the normalized four-velocity $u^{\mu }$: five scalar functions of spacetime in total.

The four equations of stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, split parallel and perpendicular to $u^{\mu }$ intoAn equation of state, such as a barotropic equation of state $P=P(\rho )$, supplies the fifth relation needed to close the evolution.

Take a fixed comoving region, whose physical volume is $V=V_0{a}^3$. Its energy is $E=\rho V$, so the first-law relation $dE=-PdV$ givesDifferentiating with respect to cosmic time and using the Hubble parameter $H=\dot{a}/a$ yields the cosmological perfect-fluid continuity equation

For the barotropic equation of state $P=w\rho$, separation gives the constant-equation-of-state density scalingSubstituting this into the flat Friedmann equation and integrating the expanding branch gives, for $w\ne -1$,For $w=-1$, the density and Hubble parameter are constant and instead $a(t)\propto e^{Ht}$.

Conformal time is defined byAt radiation–string equality, let each component have density ${\rho }_{\mathrm{eq}}$ at $a=a_{\mathrm{eq}}$. Since radiation in cosmology has $w=1/3$ and a cosmic string network has $w=-1/3$,ConsequentlyThusOn the expanding branch, integration givesChoosing the Big Bang to occur at $\tau =0$ gives the radiation--cosmic-string Friedmann solution

For a component with constant barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation givesHence ${\rho }_i={\rho }_{i,0}{a}^{-3(1+w_i)}$. Choose the present scale factor to be $a_0=1$, so the cosmological redshift obeys $1+z=a^{-1}$, and define the present cosmological density parameter bySubstitution in the spatially flat Friedmann equation then yields

For a component with the barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation isUsing $H=\dot{a}/a$ makes this a separable differential equation:Taking $a_0=1$ gives the constant-equation-of-state density scalingThe spatially flat Friedmann equation and the cosmological density parameters therefore giveAlong a radial light ray, $d\chi =dt/a$. Since the redshift-time relation is $dt=-dz/[(1+z)H]$ and $a=(1+z{)}^{-1}$,Hence the comoving radial distance is

For the barotropic equation of state $P=w\rho$, the continuity equation separates asSince $a(t_0)=1$,which is the constant-equation-of-state density scaling.

When $Q=0$, the expanding branch satisfieswhere $H_0=\dot{a}(t_0)>0$. Because $w>-1$, integration givesand hence the flat constant-equation-of-state scale factorIt vanishes atas $|w|<1$ ensures $1+w>0$.

For a barotropic equation of state, $\delta P=c_s^2\delta \rho =c_s^2\bar{\rho }\delta$. A spatial Fourier mode of comoving wavenumber $k$ therefore satisfiesThe Jeans wavenumber is defined by equality of the pressure and self-gravity terms:For $k\gg k_J$, pressure dominates and produces acoustic oscillations whose amplitude is affected by Hubble friction. For $k\ll k_J$, self-gravity dominates and a growing mode develops: this is Jeans instability.