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For a flat single-fluid background with constant barotropic equation of state, $w\ne -1/3$, adiabatic perturbations and zero anisotropic stress, $a\propto {\eta }^{2/(1+3w)}$ and $a^2\bar{\rho }\propto {\eta }^{-2}$. The scalar potential obeys ${\varphi }^{\prime \prime }+[6(1+w)/(1+3w)]{\varphi }^{\prime }/\eta -w{\nabla }^2\varphi =0$. Substituting the Poisson constraint $\Delta \propto {\eta }^2{\nabla }^2\varphi$ gives the displayed equation. The homogeneous mode and the degenerate vacuum fluid $w=-1$ require separate interpretation.

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.

In a spherical self-gravitating equilibrium with positive mass density, $d\psi /dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho (r)$ can be expressed as functions of $\psi$. Hydrostatic equilibrium further givesSince $p$ decreases outward wherever $\rho >0$, eliminating $\psi$ gives a barotropic equation of state along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.

For $p=K{\rho }^{1+1/n}$, integrate the polytropic enthalpy balance $dp/\rho =-d\psi$:For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho (R)=0$ and therefore ${\psi }_T=\psi (R)$. With the gravitational potential zero at infinity, an isolated sphere has ${\psi }_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the gravitational potential.

Put $\theta =(\rho /{\rho }_c{)}^{1/n}$ and $\xi =\alpha r$, with ${\alpha }^2=4\pi G{\rho }_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi ={\psi }_T-(n+1)K{\rho }_c^{1/n}\theta$. Substitution into the spherical gravitational Poisson equation yields the Lane-Emden equationLet ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)>0$. Then the Lane-Emden mass formula and radius areFor a common fixed polytropic index with a finite first zero, their scalings areThe usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.

For an ideal gas of fixed composition, $T_c\propto K{\rho }_c^{1/n}$. Consequently homologous polytropes at fixed central temperature have $K\propto {\rho }_c^{-1/n}$ and both $M,R\propto {\rho }_c^{-1/2}$, recovering $M\propto R$. The common central temperature, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different polytropic mass-radius relation, and no finite index makes its generic fixed-$K$ exponent exactly one.

For adiabatic monatomic hydrogen, $\gamma =5/3$ and $n=3/2$. At a common fixed specific entropy, hence common $K$, the same relations giveHere $\bar{\rho }=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that mass sequence.

Efficient radiative cooling in molecular clouds prevents compression from forcing the temperature upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and temperature. It is therefore possible to assemble dense massive cold clouds while disposing of compressional heat. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.

For a common isothermal sound speed $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the homologous isothermal mass-radius relationFor example, the singular isothermal sphere has $M(R)=2c_T^{2}R/G$. A regular finite cloud requires a pressure-confined isothermal sphere or another outer boundary: an unconfined isothermal sphere does not end at a finite vacuum radius. Equal temperature alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.

Use the spatial-potential convention of this paper and assume $\bar{\rho }+\bar{P}\ne 0$. A passive time displacement changes the spatial scalar by $\tilde{\Psi }-\Psi =\bar{N}H{\xi }^0$ and changes the energy density byThe two contributions to the uniform-density curvature perturbation therefore cancel:A spatial scalar displacement changes $E$ but not this combination. Hence $\zeta$ is gauge invariant at linear order. Pure cosmological-constant matter, for which $\bar{\rho }+\bar{P}=0$, does not supply a energy density clock and this formula is not defined for that component alone.

Now take the definite constant-$w$ barotropic equation of state and use physical-time differentiation $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$. Put $A=\bar{\rho }+\bar{P}$. Background stress-energy conservation gives $\mathcal{D}A=-3H(1+w)A$. Differentiating $\zeta =-\Psi +\delta \rho /(3A)$, then substituting the two supplied perturbation equations, givesThus the lapse and expansion perturbations cancel without imposing a gauge. In terms of the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-w\delta \rho$,For the stipulated barotropic equation of state, $\delta P=w\delta \rho$, so the first term vanishes. For a Fourier mode, $\Delta =-k^2/a^2$. For regular long-wavelength solutions the remaining gradients are suppressed when $k\ll aH$. Consequently $\dot{\zeta }\simeq 0$ on superhorizon scales, with exact conservation in the leading zero-gradient limit. It is not an exact finite-$k$ identity. Entropy perturbations or a singular gradient expansion can invalidate that conclusion; the supplied adiabatic assumption is what removes the unsuppressed pressure source.

This superhorizon conservation of uniform-density curvature is useful because inflationary fluctuations leave the horizon during cosmic inflation and may re-enter only much later. In an adiabatic single-clock evolution, the conserved uniform-density curvature perturbation transports their initial amplitude through reheating and the later radiation in cosmology and matter eras without solving every short-distance process. It then determines the initial conditions for large-scale structure and Cosmic microwave background anisotropy. Extra fluctuating fields or fluids can introduce isocurvature perturbations and convert them into curvature, so conservation must be checked rather than assumed during such transitions.

A fixed comoving volume has physical volume proportional to $a^3$. The stretched domain walls within it have area proportional to $a^2$, and their energy per physical area is constant. The cosmological domain wall network therefore has $\rho \propto a^{-1}$, equivalently a barotropic equation of state with $w=-2/3$.

The flat Friedmann equation becomes ${\dot{a}}^2=Ca$ for a positive constant $C$. On the expanding branch, integration gives $a=(C/4)(t-t_B{)}^2$. Measuring age from the initial time $t_B=0$, $H=\dot{a}/a=2/t$. Thus

For constant barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation gives $\bar{\rho }\propto a^{-3(1+w)}$. In conformal time the flat Friedmann equation becomes ${\mathcal{H}}^2=(8\pi G/3)a^2\bar{\rho }$, where $\mathcal{H}=a^{\prime }/a$. Integration, for $w\ne -1/3$, givesChoose the conformal-time origin so that ${\eta }_B=0$ and work on the appropriate expansion branch. Write $\beta =2/(1+3w)$, so $\mathcal{H}=\beta /\eta$ and $2{\mathcal{H}}^{\prime }+(1+3w){\mathcal{H}}^2=0$.

Combining the supplied energy and momentum constraints gives the comoving-gauge density contrast Poisson relationFor this barotropic fluid $\delta P=w\bar{\rho }\delta$. Substitute the energy constraint into the pressure equation to obtainso the bracket vanishes and the coefficient of ${\varphi }^{\prime }/\eta$ is $A=6(1+w)/(1+3w)$.

Because $a^2\bar{\rho }=C{\eta }^{-2}$, the Poisson relation gives $\Delta =({\eta }^2/(4\pi GC)){\nabla }^2\varphi$. Apply the same potential equation to $F={\eta }^2{\nabla }^2\varphi$: differentiating ${\eta }^{-2}F$ givesSince $F$ is a constant multiple of $\Delta$, the constant-equation-of-state comoving density equation isThe primes here denote conformal-time derivatives, corresponding to the overdots in the question. A pure vacuum fluid $w=-1$ has zero enthalpy and no independent fluid velocity or nonzero-wave-number density perturbation; its formal equation should be interpreted with that degeneracy in mind.

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.