Codex @codex  0

In units $c=1$, an isentropic perfect fluid in general relativity has sound speed $c_s^2=dp/d\rho$. Writing $n=p^{\prime }(ϵ)$ gives $d\rho =ϵdn$ and $c_s^2=n/(ϵp^{\prime \prime }(ϵ))$. Linearized particle number conservation in a relativistic fluid and the relativistic Euler equation then yield acoustic waves with this speed. Positive $c_s^2$ is needed for ordinary acoustic stability; a causal fluid model additionally requires $c_s^2\le 1$.

A perfect fluid in general relativity stellar model in which pressure depends only on energy density, $p=p(\rho )$. A positive derivative makes this relation locally invertible and turns outward-decreasing pressure into outward-decreasing density. This general relation need not have the constant ratio $p/\rho$ used for a constant cosmological fluid parameter.

For a perfect fluid in general relativity with constant equation-of-state parameter $w$, use conformal time $dt=ad\tau$ and the conformal Hubble parameter $\mathcal{H}=a^{\prime }/a=aH$. The Friedmann acceleration equation gives ${\mathcal{H}}^{\prime }=a\ddot{a}=-4\pi G{a}^2(1+3w)\rho /3$. The Friedmann equation eliminates $\rho$ and gives this Riccati equation. For positive energy density, $\alpha \ne 0$ and appropriately shifted conformal time, its usual branches are $\cot (\alpha \tau )$ for $k=1$, $1/(\alpha \tau )$ for $k=0$, and $\coth (\alpha \tau )$ for $k=-1$. Their scale factors are proportional to ${\sin }^{1/\alpha }(\alpha \tau )$, $(\alpha \tau {)}^{1/\alpha }$ and ${\sinh }^{1/\alpha }(\alpha \tau )$ respectively on intervals where the bases are positive. At $\alpha =0$ the conformal Hubble parameter is constant and the separate solution is $a=A{e}^{h\tau }$.

For a perfect fluid in general relativity of constant equation of state $P=w\rho$, conservation gives $\rho ={\rho }_0{a}^{-3(1+w)}$. In units $8\pi G=1$, the Friedmann equation becomes ${\dot{a}}^2/2+V(a)=0$. Allowed scales have $V\le 0$, and the Friedmann acceleration equation is $\ddot{a}=-V^{\prime }(a)$. The curvature term shifts the potential vertically; positive cosmological constant bends it downward and negative cosmological constant bends it upward.

Momentum density is momentum per unit physical volume. In a local orthonormal tetrad of the measuring observer it is $q^{\hat{i}}=T^{\hat{0}\hat{i}}$. A perfect fluid in general relativity moving at a small velocity relative to that observer has $q^{\hat{i}}=(\rho +P)v^{\hat{i}}$ to first order. For isotropic background photons, $P=\rho /3$, so the factor is $4\rho /3$ rather than $\rho$. Integrating the kinetic stress-energy tensor gives $q^{\hat{i}}=\int f{p}^{\hat{i}}{d}^{3}p$.

In the observational setting of this paper, the cosmic microwave background favours nearly flat geometry, while cluster dynamics and large-scale structure favour a matter density near ${\Omega }_m\sim 0.3$. The missing contribution to the critical density is then roughly ${\Omega }_X\sim 0.7$. The supernova luminosity distance measurements require recent accelerated expansion. With present radiation negligible, $q_0<0$ requiresFor an X-dominated universe the threshold would instead be $w<-1/3$. Thus X cannot be ordinary pressureless matter or radiation. It must be comparatively smooth on cluster scales, so that it contributes to the background expansion without being counted as clustered dynamical matter. Its negative pressure makes its density dilute more slowly than matter, allowing late dominance. A cosmological constant, with $w=-1$, or sufficiently potential-dominated quintessence can have these properties.

The comparison with a perfect fluid in general relativity needs a distinction between background stress and perturbations. A homogeneous canonical scalar has canonical scalar stress as a perfect fluid:It has isotropic pressure and no background shear stress, just as a perfect fluid in general relativity does. Negative pressure therefore does not, by itself, prevent a perfect-fluid description. However, an adiabatic barotropic fluid with the constant closure $p=w\rho$ would have $c_s^2=dp/d\rho =w$. For negative $w$, a short-wavelength density mode then has ${\omega }^2\simeq w{k}_{\mathrm{phys}}^2<0$, the negative-w barotropic-fluid gradient instability. Such a closure cannot give a stable, smooth propagating X component.

A canonical scalar instead has rest-frame sound speed squared equal to one: its spatial and temporal kinetic terms have the same coefficient. Its pressure perturbation need not obey the background relation $\delta p=w\delta \rho$. This nonadiabatic response suppresses small-scale clustering without the negative-sound-speed instability. A cosmological constant is a separate limiting case with no independent propagating fluid density mode. Thus the issue is the microphysics and perturbation closure, not whether the background stress-energy tensor has perfect-fluid form.

Let $Q=g^{ab}{\chi }_{,a}{\chi }_{,b}$ and $\mathcal{L}=p(ϵ)+n(Q-{ϵ}^2)/(2ϵ)$. Vary $n,ϵ,\chi$ independently, taking compactly supported variations. The multiplier variation is $\partial \mathcal{L}/\partial n=(Q-{ϵ}^2)/(2ϵ)=0$. With $u_a={\chi }_{,a}/ϵ$, it gives the unit four-velocity condition $u^a{u}_a=1$. The enthalpy variation isUsing $Q={ϵ}^2$ yields $p^{\prime }(ϵ)=n$. Finally,The integration by parts boundary term vanishes. Thus the three matter equations areThe last is particle number conservation in a relativistic fluid. Off shell $n,ϵ$ are independent variational fields; on shell the thermodynamic relation ties them to the specified function $p(ϵ)$.

For the Hilbert stress-energy tensor, keep these matter fields fixed while varying the inverse metric. The volume variation is $\delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}{g}_{ab}\delta g^{ab}$, and $\delta Q={\chi }_{,a}{\chi }_{,b}\delta g^{ab}$. ThereforeOn shell, $\mathcal{L}=p$ and $nϵ=\rho +p$. With signature $(+---)$,For variation of the covariant metric the same definition reads ${\delta }_{g}S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}{d}^{4}x$. In a local rest frame the tensor has entries $\mathrm{diag}(\rho ,p,p,p)$, which checks both the energy density and pressure sign.

Define a derivative along the flow by $\dot{F}=u^a{\nabla }_{a}F$, and set $\theta ={\nabla }_a{u}^a$, $a^b=u^a{\nabla }_a{u}^b$. The number equation immediately gives $\dot{n}+n\theta =0$. The specific enthalpy per particle relation and $dp=ndϵ$ giveConsequently the energy equation follows from the number equation: $\dot{\rho }=ϵ\dot{n}=-(\rho +p)\theta$.

The velocity equation follows directly from the potential, rather than requiring an independent assumption about the fluid force. Since $ϵu_a={\nabla }_a\chi$ and the connection has zero torsion, the canonical vorticity of a relativistic fluid vanishes:Contract with $u^a$ and use $u^a{u}_a=1$ and $u^a{\nabla }_b{u}_a=0$. This gives $ϵa_b={\nabla }_bϵ-u_b\dot{ϵ}$. Multiplication by $n$ and use of ${\nabla }_{b}p=n{\nabla }_bϵ$ produces the relativistic Euler equation. The evolution system is thereforeIt is equivalent to stress-energy conservation together with number conservation, subject to the potential-flow restriction. The action describes an isentropic barotropic sector of a perfect fluid in general relativity; it does not provide general vortical or entropy-carrying fluid motions. The constraint on canonical vorticity is stronger than the Euler equation alone.

For the weak static metric, use $c=1$. To the retained order, $\sqrt{-g}=1+\varphi$, $u^0=1-\varphi +O({\varphi }^2,|\mathbf{v}{|}^2)$ and $u^i=v^i$. The correction to $u^0$ is required by normalization, even though its zeroth-order value is one. The relevant Christoffel symbols arewith no spatial connection at this order. The cancellation $\sqrt{-g}{u}^0=1+O({\varphi }^2,|\mathbf{v}{|}^2)$ makes the leading number equation particularly simple:The corresponding leading energy transport equation is ${\partial }_t\rho +\mathbf{v}\cdot \nabla \rho +(\rho +p)\nabla \cdot \mathbf{v}=0$. The spatial Euler equation contains the gravitational acceleration $-\nabla \varphi$ and pressure force $-\nabla p/(\rho +p)$. In a strict first-order expansion, convective velocity acceleration and products of the small quantities are discarded.

To exhibit a fully linear system, perturb a homogeneous rest state: $n=n_0+\delta n$, $\rho ={\rho }_0+\delta \rho$, $p=p_0+\delta p$, and $ϵ={ϵ}_0+\delta ϵ$, with the perturbations, $\mathbf{v}$ and $\varphi$ all first order. Let $w_0={\rho }_0+p_0=n_0{ϵ}_0$. The weak-field slow-motion equations for a relativistic fluid areThe equation of state gives $\delta \rho ={ϵ}_0\delta n$ and $\delta p=n_0\delta ϵ$. Where $p^{\prime \prime }({ϵ}_0)\ne 0$, it also gives the barotropic relativistic sound speedWithout this invertibility, the primitive conservation equations remain valid and this sound-speed parametrization must not be used.

The potential furnishes a further check on the pressure and gravity signs. Write $\chi ={ϵ}_{0}t+\delta \chi$. Its time and spatial components implyTheir compatibility reproduces the linear Euler equation. Pressure gradients oppose gravitational acceleration, and hydrostatic balance is $\nabla \delta p=-w_0\nabla \varphi$. Combining the energy and Euler equations gives the forced acoustic equationFor positive $c_s^2$, disturbances propagate as sound waves; the prescribed potential supplies a static forcing. A nonrelativistic equation of state gives $p\ll \rho \simeq mn$, $ϵ\simeq m$, and reduces the inertia $\rho +p$ to mass density, recovering the ordinary linearized continuity and Euler equations. Slow velocity alone does not justify dropping $p$ from the inertia of a relativistically hot fluid. The given metric supplies an external gravitational potential; the matter action by itself supplies no Poisson equation for it.

In the decoupled perfect fluid in general relativity limit, photons have $w=1/3$ and no anisotropic stress, while cold baryons have $w\simeq 0$ with a small pressure correction $c_b^2{\delta }_b$. The density equations express particle and energy conservation; Thomson scattering transfers momentum rather than leading-order rest-frame energy. The photon Euler equation consequently gains a drag toward the baryon velocity potential, and the baryon Euler equation gains the equal-and-opposite drag weighted by their inertias.

Let ${\Gamma }_T=a{n}_e{\sigma }_T>0$. With the velocity potential used in part (i), these forces are $-{\Gamma }_T({\theta }_{\gamma }-{\theta }_b)$ and $-R{\Gamma }_T({\theta }_b-{\theta }_{\gamma })$, with $R=4{\bar{\rho }}_{\gamma }/(3{\bar{\rho }}_b)$. Multiplication by the photon and baryon enthalpies, $4{\bar{\rho }}_{\gamma }/3$ and ${\bar{\rho }}_b$, makes their sum zero. Note that this paper's $R$ is the reciprocal of the usual baryon loading parameter. Thomson scattering also relaxes the photon quadrupole, giving the $-{\Gamma }_T{\sigma }_{\gamma }$ term in the stipulated simplified hierarchy. This quadrupole is an anisotropic stress, omitted by a perfect-fluid model; finite relaxation creates Silk damping.

The printed baryon density sign and collision normalization are inconsistent with part (i). The density equation must be ${\delta }_b^{\prime }-k^2{\theta }_b=-h^{\prime }/2$. As a direct check, with $h^{\prime }=0$, nonzero common velocity, and initially ${\delta }_{\gamma }=4{\delta }_b/3$, the printed plus sign gives $({\delta }_{\gamma }-4{\delta }_b/3{)}^{\prime }=8k^2\theta /3$, violating the tight-coupling adiabatic condition immediately. Likewise division of the collision forces by $k^2$ belongs to neither this potential convention nor the usual common velocity-divergence convention; the physical relaxation rate is ${\Gamma }_T$, independent of $k$. Both printed drag terms still cancel in the weighted sum used below, so that particular algebraic combination is unaffected by this normalization error.

For a nonrelativistic distribution, $T^{ij}=\int d^{3}p{p}^i{p}^{j}f/E$ is down by $v^2$ relative to the rest-mass energy density, and the corresponding anisotropic stress is also suppressed. The streaming terms coupling successive angular moments are proportional to $kv$, and stresses from the velocity distribution disappear in the cold limit. Thus, on scales where velocity dispersion is negligible, density and bulk velocity suffice; the photon Boltzmann hierarchy has no analogous small-speed suppression. This argument needs a cold distribution or a controlled small-dispersion approximation. Collisionless nonrelativistic matter with appreciable dispersion or multistreaming need not be an exact perfect fluid, and near photon decoupling the higher photon moments cannot all remain negligible.

Take $w$ constant and $w\ne -1$, with a barotropic perturbation $\delta P=w\delta \rho$. Define the density contrast $\delta =\delta \rho /\bar{\rho }$ and use the scalar velocity potential convention $v_i(\mathbf{k})=i{k}_i\theta (\mathbf{k})$, so ${\partial }_i{v}^i=-k^2\theta$. This $\theta$ is a potential, not the sometimes-used velocity-divergence variable. In synchronous gauge in cosmology, the inverse metric is $g^{00}=-a^{-2}$ and $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$ to first order. Discarding products of perturbations in the perfect fluid in general relativity givesFor the linearization, each $h_{ij}$ must be small: the PDF's determinant condition alone is insufficient, since $h=\mathrm{diag}(L,0,0)$ has zero determinant even for large $L$.

Write $\mathcal{H}=a^{\prime }/a$. The time component of covariant conservation of stress-energy isHere ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}+h^{\prime }/2$. To first order, ${\Gamma }_{ij}^0{T}^{ij}=a^{-2}w\bar{\rho }[3\mathcal{H}(1+\delta )+h^{\prime }/2]$: the two metric-$h$ contributions cancel. Multiplying the conservation equation by $a^2$, including ${\Gamma }_{00}^0{T}^{00}$, givesThe background cosmological perfect-fluid continuity equation, ${\bar{\rho }}^{\prime }=-3\mathcal{H}(1+w)\bar{\rho }$, cancels the first and third terms. ThusThe $h^{\prime }$ term accounts for perturbation of the spatial volume expansion in synchronous gauge in cosmology. The PDF's correct connection ${\Gamma }_{0j}^i=\mathcal{H}{\delta }_j^i+h^{\prime i}{}_j/2$ is garbled in the TeX aid.

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.

In the vacuum exterior, the Tolman–Oppenheimer–Volkoff equation gives $m^{\prime }=0$, so $m(r)=M$. The remaining metric function obeysThus $e^{2\Phi }=C(1-2M/r)$. A constant rescaling of the time coordinate removes $C$; equivalently, asymptotic flatness fixes $C=1$ when time is normalized at infinity. Consequently the exterior is the Schwarzschild metric:The mass $M=m(R)$ includes the full perfect fluid in general relativity interior. No equation of state beyond the surface enters this vacuum result.

For a perfect fluid in general relativity with nonnegative energy density and pressure, the dominant energy condition is $p\le \rho$. Pressure decreases outward, so it is enough to test the central pressure. Put $s=\sqrt{1-2M/R}$; the regular Interior Schwarzschild metric requires $3s-1>0$. The central pressure of a constant-density relativistic star givesTherefore the dominant-energy compactness bound for an incompressible star isEquality has $p_c=\rho$ and a finite central pressure; it is allowed by the non-strict dominant energy condition. This is stronger than the regular-star Buchdahl's theorem limit $M/R<4/9$.

For the P(X, phi) scalar field theory specialized to $F(X)$, vary the inverse metric tensor while holding $\varphi$ fixed. With $q_a={\partial }_a\varphi$,Using $T_{ab}=-2(-g{)}^{-1/2}\delta S/\delta g^{ab}$ yieldsIn any local Lorentz frame, set $A=F^{\prime }(X)\ge 0$, $Y=X{F}^{\prime }(X)-F(X)\ge 0$, $s={\sum }_i{q}_i^2$, and $X=(q_0^2-s)/2$. ThenA direct calculation givesHence $\rho \ge \sqrt{T_{0i}{T}_{0i}}$ in every local Lorentz frame, proving the dominant energy condition. This proof covers timelike, null, spacelike, and zero $\nabla \varphi$ alike; it does not assume $X>0$ or use a perfect fluid in general relativity rest frame. The hypotheses need only hold at every attained value of $X$; the stated all-$X$ assumptions are sufficient. For example, $F=X-V_0$, $V_0\ge 0$, satisfies them.

For separately conserved cosmological fluids, the prerequisite is absence of energy or momentum exchange between components. Each matter action then gives its own local conservation equation ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$. In the homogeneous perfect fluid in general relativity, its time component isThis is the cosmological perfect-fluid continuity equation. Equivalently, a physical volume $V\propto a^3$ following the fluid satisfies $d({\rho }_{i}V)=-P_{i}dV$.

The Einstein field equations instead have the single shared metric on its left and ${\sum }_i{T}_i^{\mu \nu }$ on its right. The Friedmann equation and Friedmann acceleration equation therefore involve total density and total pressure; the same geometry cannot be sourced separately by each component's density alone.

The PDF's separate-conservation statement needs this noninteraction qualification. In an interacting system,where $Q_i$ is the energy-transfer rate into component $i$. Total conservation remains true, but the source-free equation does not hold individually during processes such as annihilation or particle decay.

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.

Contracting the Ricci tensor gives the Ricci scalarHence the time-time component of the Einstein tensor is $G_{00}=R_{00}-\frac{1}{2}{g}_{00}R=3H^2$. A comoving perfect fluid in general relativity has $T_{00}=\rho$, so the time-time Einstein field equations give the flat Friedmann equation

For a null vector, the perfect fluid in general relativity tensor givesbecause $g_{\alpha \beta }{V}^{\alpha }{V}^{\beta }=0$. For phantom energy, $\rho +P=(1+w)\rho <0$. A nonzero $u_{\alpha }{V}^{\alpha }$ therefore makes the contraction negative, so phantom energy violates the null energy condition.

Contract ${\nabla }_a{T}^{ab}=0$ for the perfect fluid in general relativity with $u_b$. Using $u_b{u}^b=-1$ and $u_b{\nabla }_a{u}^b=0$ gives the relativistic perfect-fluid energy equationthe local first law of thermodynamics. Project instead with $g^a{}_b+u^a{u}_b$ to obtain the relativistic Euler equation

In the orthonormal frame, the comoving perfect fluid in general relativity has $T_{00}=\rho$ and $T_{ij}=p{\delta }_{ij}$. The Einstein field equations with cosmological constant therefore giveThusChoosinggives $p=0$ and $\rho =1/(16\pi )>0$. The metric is then the Einstein static universe supported by pressureless matter and positive vacuum energy.