Perfect fluid in general relativity ($T_{ab}=(\rho +p)u_a{u}_b+p{g}_{ab}$)

A relativistic perfect fluid has energy density $\rho$, isotropic pressure $p$ and unit timelike four-velocity $u$. Projecting ${\nabla }_a{T}^{ab}=0$ along and orthogonally to $u$ gives its energy and Euler equations.

For a perfect fluid, the relativistic stress tensor is $T_{ab}=(\rho +p)u_a{u}_b+p{g}_{ab}$ in signature $(-,+,+,+)$ with $u^a{u}_a=-1$.

Relativistic dust is a continuum of noninteracting massive particles with one local four-velocity $u^{\mu }$ and negligible pressure. With signature $(+---)$, $u^{\mu }{u}_{\mu }=c^2$ and rest mass density $\rho$, its stress-energy tensor has the displayed form. Contracting stress-energy conservation with $u_{\nu }$ and using metric compatibility gives $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting back yields $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$ wherever $\rho >0$: each particle follows a geodesic. The cosmological pressureless matter model is a homogeneous application.

For identical particles of rest mass $m$ and electric charge $q$, the four-current is $j^{\mu }=(q/m)\rho u^{\mu }$. The matter stress-energy tensor obeys ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, with the opposite force density in the electromagnetic stress-energy tensor. Antisymmetry of the electromagnetic field tensor makes its contraction with two four-velocities vanish, so mass continuity still holds. Removing that term leaves the displayed covariant Lorentz force law. The matter tensor alone is not divergence-free when an electromagnetic force acts.

For a canonical scalar with timelike gradient, its stress-energy tensor has perfect-fluid form. In a homogeneous background, $\rho ={\dot{\varphi }}^2/2+V$ and $p={\dot{\varphi }}^2/2-V$. Potential domination gives negative pressure, while the scalar's physical rest-frame sound speed is one. A constant negative background $w$ therefore does not force its perturbations to obey the adiabatic barotropic closure $\delta p=w\delta \rho$.

// Target: cosmology.bigb

For $d{s}^2=a^2(d{\tau }^2-d{\mathbf{x}}^2)$ and a homogeneous canonical scalar, $\rho ={\varphi }^{\prime 2}/(2a^2)+V$ is the physical density, whereas $T_{00}=a^2\rho$. At $w=-1/3$, the kinetic and potential energies are $K=\rho /3$ and $V=2\rho /3$, so ${\varphi }^{\prime 2}=2a^2\rho /3=2T_{00}/3$. A single power of ${\varphi }^{\prime }$ cannot replace the square in this energy identity.

// Target: cosmology.bigb

The canonical vorticity two-form is ${\omega }_{ab}={\nabla }_a(ϵu_b)-{\nabla }_b(ϵu_a)$, with specific enthalpy per particle $ϵ$. Potential flow $ϵu_a={\nabla }_a\chi$ has $\omega =0$ for a torsion-free connection. Contracting this identity with $u^a$ yields $ϵu^a{\nabla }_a{u}_b={\nabla }_bϵ-u_b{u}^a{\nabla }_aϵ$, the isentropic relativistic Euler equation in signature $(+---)$. This canonical condition does not mean the unweighted four-velocity one-form is closed.

In signature $(+---)$, the matter action $S=\int \sqrt{-g}[p(ϵ)+n((\nabla \chi {)}^2-{ϵ}^2)/(2ϵ)]$ uses independent $n,ϵ,\chi$. Variations of $n,ϵ,\chi$ give $(\nabla \chi {)}^2={ϵ}^2$, $p^{\prime }(ϵ)=n$, and particle number conservation in a relativistic fluid for $u_a={\nabla }_a\chi /ϵ$. Its Hilbert stress-energy tensor is $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The action describes isentropic potential flow, with vanishing canonical vorticity of a relativistic fluid, rather than every vortical perfect-fluid solution.

The number current is $J^a=n{u}^a$, where $n$ is the rest-frame number density and $u^a$ the unit four-velocity. Its covariant continuity equation is ${\nabla }_a{J}^a=0$, equivalently $u^a{\nabla }_{a}n+n{\nabla }_a{u}^a=0$. Integration over a spacetime volume and the divergence theorem equate the particle numbers crossing its boundary. This conservation law is separate from stress-energy conservation.

For number density $n$, rest-frame energy density $\rho$ and pressure $p$, the enthalpy per particle is $ϵ=(\rho +p)/n$. In an isentropic barotropic fluid, $d\rho =ϵdn$ and $dp=ndϵ$. It differs from enthalpy per unit mass by the particle mass when that mass is fixed. Using $ϵ$ as thermodynamic variable gives $n=p^{\prime }(ϵ)$ and $\rho =ϵp^{\prime }(ϵ)-p(ϵ)$.

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 $c_s^2\le 1$.

For metric signature $(+---)$, a perfect fluid has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The Einstein field equations and static Killing field is a Ricci eigenvector imply $(\rho +p)(k\cdot u)u_{[a}{k}_{b]}=0$. Two nonzero timelike vectors cannot be orthogonal, so $k\cdot u\ne 0$. If $\rho +p\ne 0$, the four-velocity is parallel to the static Killing vector. Choosing both future-directed gives the displayed normalization. For $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not determine a preferred fluid velocity; the strict enthalpy hypothesis is therefore material.

A perfect fluid with equation of state $P=\rho$ is called a stiff fluid. For positive energy density, its constant equation-of-state parameter is $w=1$. The cosmological continuity equation gives $\rho \propto a^{-6}$. In a flat universe dominated by this fluid, the Friedmann equation then gives $a(t)\propto (t-t_B{)}^{1/3}$. This statement assumes a nonzero fluid density and negligible other components.

For a spatially homogeneous minimally coupled scalar field with zero scalar potential, the stress-energy tensor has $\rho =P={\dot{\varphi }}^2/2$. Its field equation is $\ddot{\varphi }+3H\dot{\varphi }=0$, so $a^3\dot{\varphi }$ is constant and its energy density decreases as $a^{-6}$. Thus its homogeneous stress is precisely that of a stiff fluid. Spatial gradients would introduce directional stresses, so this equivalence requires homogeneity.

In units $8\pi G=1$, a free homogeneous scalar has $\rho =P={\dot{\varphi }}_0^2/2$ and therefore a stiff equation of state. Its equation gives $a^2{\varphi }_0^{\prime }=\text{constant}$, while the Friedmann equation gives $3{\mathcal{H}}^2={\varphi }_0^{\prime 2}/2$. The displayed contracting branch satisfies both equations for negative conformal time. Constant normalizations and an additive scalar shift can be restored.

In conformal Newtonian gauge, a free scalar has no linear anisotropic stress. Eliminating its perturbation with the momentum constraint gives ${\Phi }_k^{\prime \prime }+3{\Phi }_k^{\prime }/\tau +k^2{\Phi }_k=0$. Substitution of ${\Phi }_k=(-\tau {)}^{-1}y(-k\tau )$ gives the order-one Bessel equation. The displayed Hankel function has positive incoming frequency $e^{-ik\tau }$ at early times.

The momentum constraint relates the incoming scalar mode $a^{-1}\sqrt{\hslash /(2k)}{e}^{-ik\tau }$ to the Newtonian-potential Hankel modes in stiff contraction. Large-argument Hankel asymptotics give $|N_k{|}^2=3\pi \hslash /(32k^2)$. The small-argument behavior $H_1^{(1)}(z)\sim -2i/(\pi z)$ then gives the displayed spectrum. Its dimensional power is $k^{-4}$ and its power per logarithmic interval is proportional to $k^{-1}$. These are Newtonian-potential spectra; a curvature-perturbation spectrum need not have the same exponent.

The timelike projection of stress-energy conservation is $u^a{\nabla }_a\rho +(\rho +p){\nabla }_a{u}^a=0$. It is the local adiabatic first law for a perfect fluid.

The spatial projection of stress-energy conservation is $(\rho +p)u^b{\nabla }_b{u}^a=-(g^{ab}+u^a{u}^b){\nabla }_{b}p$.

For $d{s}^2=(1+2\varphi )d{t}^2-d{\mathbf{x}}^2$ with static weak $\varphi$, normalization gives $u^0=1-\varphi +O({\varphi }^2,v^2)$ and ${\Gamma }^i{}_{00}={\partial }_i\varphi$. Around a homogeneous state, the first-order equations are ${\partial }_t\delta n+n_0\nabla \cdot \mathbf{v}=0$, ${\partial }_t\delta \rho +({\rho }_0+p_0)\nabla \cdot \mathbf{v}=0$, and $({\rho }_0+p_0){\partial }_t\mathbf{v}=-\nabla \delta p-({\rho }_0+p_0)\nabla \varphi$. The inertial density is enthalpy density even though the flow is slow. Only a further nonrelativistic equation of state, $p\ll \rho \simeq mn$, replaces it by mass density.

For metric signature $(+---)$, a stationary perfect fluid follows a timelike Killing vector field $K$ with four-velocity $U=K/N$. Its Killing-flow four-acceleration is $U^b{\nabla }_b{U}_a=-{\nabla }_a\log N$. The relativistic Euler equation and stationarity of the pressure give the displayed hydrostatic equilibrium equation. This uses stationarity along $K$ and does not require hypersurface orthogonality.

The equation of state of isotropic blackbody radiation is $p=\rho /3$. The barotropic hydrostatic first integral gives $\rho N^4=C$. The Stefan–Boltzmann law $\rho \propto T^4$ then yields the Tolman–Ehrenfest relation $TN=\mathrm{constant}$. Here $N=\sqrt{|K^a{K}_a|}$ for the stationary timelike Killing vector field; its normalization only changes the constants.

A connected nonvacuum perfect fluid with equation of state $p=\rho /3$ obeys the stationary radiation density law $\rho =C/N^4$ with $C>0$. On a compact region of a regular spacetime, a smooth Killing vector field has bounded norm. Consequently $\rho$ cannot tend to zero at a finite regular free surface. A vanishing timelike norm instead makes the density diverge. A radiation-filled cavity can have a material wall with nonzero boundary pressure; this is a different boundary condition.

For a barotropic fluid in hydrostatic equilibrium along a Killing vector, assume $\rho +p>0$. Integrating $dp/(\rho +p)=-d\log N$ gives the displayed first integral, constant on each connected fluid region. For the equation of state $p=w\rho$ with constant $w>0$, it gives $\rho N^{(1+w)/w}=C_1$.

The Tolman–Oppenheimer–Volkoff equation governs a static, spherically symmetric perfect fluid in general relativity. In geometrized units, if $m(r)$ is the mass inside areal radius $r$, then

The ratio of enclosed gravitational mass to areal radius in units $G=c=1$. In a spherical star the local value is $m(r)/r$, and the surface value is $M/R$. Bounds on this ratio constrain relativistic stellar models; it is unrelated to topological compactness.

For a smooth centre with density ${\rho }_c$ and pressure $p_c$, the Tolman–Oppenheimer–Volkoff equation gives $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and $p(r)=p_c-2\pi ({\rho }_c+p_c)({\rho }_c/3+p_c)r^2+O(r^4)$. A fixed barotropic stellar equation of state leaves one physical central parameter; the additive lapse constant fixes time normalization.

The interior Schwarzschild metric is the exact static solution for a spherical fluid of constant density. If the star has mass $M$ and radius $R$, its pressure is

The Interior Schwarzschild metric has central pressureIt diverges when $M/R=4/9$, the limiting compactness in Buchdahl's theorem.

The dominant energy condition for the Interior Schwarzschild metric reduces to its largest pressure satisfying $p_c\le \rho$. Substitution of the exact central pressure gives $\sqrt{1-2M/R}\ge 1/2$, hence $M/R\le 3/8$. Equality has finite central pressure and is allowed by the non-strict condition.

Buchdahl's theorem gives $2M/R\le 8/9$ for a static spherical perfect-fluid star whose density is nonincreasing outwards, subject to its regularity assumptions. Equality would require divergent central pressure, so every regular such star satisfies $M/R<4/9$.

Let ${\rho }_0>0$ be a density threshold and let $a,\mu$ be the radius and enclosed mass where a monotonically decreasing stellar density reaches that threshold. Then $\mu \ge 4\pi {\rho }_0{a}^3/3$. A local compactness bound $\mu /a<4/9$ implies $a<(3\pi {\rho }_0{)}^{-1/2}$ and $\mu <4/(9\sqrt{3\pi {\rho }_0})$, independently of the equation of state above the threshold. A total-mass bound also requires suitable control of the known low-density envelope.