Perfect Fluids and Stress–Energy

An ideal relativistic fluid is not defined by the absence of microscopic motion. It is defined by what a comoving observer can measure locally: one energy density, one isotropic pressure, and no preferred spatial direction, heat flux, or shear stress. Those statements determine the stress–energy tensor uniquely. The resulting decomposition is the starting point for stellar equilibrium, relativistic hydrodynamics, and every controlled extension that adds viscosity, magnetic fields, radiation, or multiple constituents.

Required background. Hilbert Stress–Energy from Metric Variation supplies the symmetric stress tensor and its conservation law; Events, Intervals, and Minkowski Geometry supplies timelike four-velocities and local rest frames; and Differential Equations supplies the conservation-law viewpoint used below.

Helpful background. The Einstein Equation explains how matter sources curvature; Relativistic Matter and Compact Objects locates the fluid model within the volume; and Computational Relativity develops numerical evolutions of the same conservation laws.

Local isotropy fixes the tensor

Let $u^a$ be the future-directed fluid four-velocity, normalized by

$$u^a u_a=-1.$$

The tensor

$$h^{ab}=g^{ab}+u^a u^b$$

projects into the three-dimensional rest space orthogonal to $u^a$: $h^{ab}u_b=0$ and $h^a{}_c h^c{}_b=h^a{}_b$. In a local orthonormal frame comoving with the matter, isotropy permits only

$$T^{\hat a\hat b}=\operatorname{diag}(\varepsilon,p,p,p),$$

where $\varepsilon$ is total rest-frame energy density, including rest mass, and $p$ is thermodynamic pressure. Reassembling this statement covariantly gives

$$T^{ab}=\varepsilon u^a u^b+p h^{ab} =(\varepsilon+p)u^a u^b+p g^{ab}.$$

The first form separates energy from spatial stress; the second displays the enthalpy density $\varepsilon+p$. The reason is simple: a spatial vector would select a preferred direction, and unequal eigenvalues of the spatial stress would select preferred axes. Isotropy forbids both. A constitutive equation of state such as $p=p(\varepsilon,s)$, with entropy per particle $s$, is additional physics, not part of the tensor decomposition. The observer projections, local-isotropy argument, and perfect-fluid result are developed in Andersson and Comer 2021, §§ 5.1–5.2, Eqs. (5.2)–(5.17).

What different observers measure

For any unit timelike observer $n^a$, define the observer’s spatial projector $\gamma^{ab}=g^{ab}+n^a n^b$. The measured energy density, momentum density, and spatial stress are

$$E=T_{ab}n^a n^b, \qquad S^a=-\gamma^a{}_c T^{cd}n_d, \qquad S^{ab}=\gamma^a{}_c\gamma^b{}_dT^{cd}.$$

Write $u^a=W(n^a+v^a)$, where $n_a v^a=0$ and $W=(1-v^2)^{-1/2}$. Direct substitution yields

$$E=(\varepsilon+p)W^2-p, \qquad S^a=(\varepsilon+p)W^2v^a,$$ $$S^{ab}=(\varepsilon+p)W^2v^a v^b+p\gamma^{ab}.$$

Thus a perfect fluid generally carries momentum and has anisotropic coordinate-frame stress. “Perfect” refers to the local rest frame, not to every observer’s decomposition. Setting $n^a=u^a$ gives $E=\varepsilon$, $S^a=0$, and $S^{ab}=p h^{ab}$, which independently checks every sign. Andersson and Comer give the relative-observer decomposition in § 5.4, Eqs. (5.41)–(5.50) and the Eulerian variables and primitive recovery in §§ 11.2–11.3, Eqs. (11.21)–(11.50).

Two invariant contractions recover the rest-frame scalars without choosing coordinates:

$$T_{ab}u^a u^b=\varepsilon, \qquad \frac13 h_{ab}T^{ab}=p.$$

The trace is $T^a{}_a=-\varepsilon+3p$. For an isotropic gas of massless particles, $p=\varepsilon/3$ and the trace vanishes; for pressureless dust, $p=0$ and $T^{ab}=\varepsilon u^a u^b$. An individual radiation beam is also trace-free but is not a perfect fluid: zero trace alone does not establish isotropy.

Conservation and the matter variables

The tensor alone does not close the dynamics. One normally combines

$$\nabla_aT^{ab}=0$$

with conservation of a particle or baryon current,

$$N^a=n u^a, \qquad \nabla_aN^a=0,$$

and an equation of state. Here $n$ is rest-frame number density. The rest-mass density $\rho=mn$ is useful when one conserved species dominates, while the specific internal energy $e$ is defined by $\varepsilon=\rho(1+e)$ in units with $c=1$. Confusing $\rho$ with $\varepsilon$ drops internal and binding contributions and therefore changes both inertia and gravity.

If the fluid is minimally coupled to general relativity, Einstein’s equation is

$$G_{ab}=8\pi G T_{ab}.$$

The Bianchi identity makes $\nabla_aT^{ab}=0$ necessary, but it does not supply the equation of state. Likewise, local conservation does not prove that a single perfect-fluid velocity exists. Mixtures can have relative currents; collisionless matter can have a broad momentum distribution; dissipative matter can support heat flux $q^a$ and anisotropic stress $\pi^{ab}$. The more general decomposition

$$T^{ab}=\varepsilon u^a u^b+p h^{ab}+2u^{(a}q^{b)}+\pi^{ab}$$

uses $q^au_a=0$, $\pi^{ab}u_b=0$, and $\pi^a{}_a=0$, with $p$ equal to one third of the spatial trace. It reduces to the perfect-fluid form only when $q^a=0$ and $\pi^{ab}=0$ in the chosen material frame. These are physical closure assumptions, not coordinate choices; setting one component to zero by changing coordinates does not remove the tensor itself.

Energy conditions and limits of the model

For a perfect fluid, the weak energy condition requires $\varepsilon\geq0$ and $\varepsilon+p\geq0$; the dominant energy condition adds $|p|\leq\varepsilon$. These inequalities constrain admissible stress tensors but do not by themselves guarantee thermodynamic stability or causal sound propagation. Those require derivatives of the equation of state, developed on Thermodynamics, Sound Speed, and Causality.

The model is local. It can remain accurate in curved spacetime if microscopic mean free paths and relaxation times are short compared with the relevant macroscopic scales. It is not a general closure for collisionless beams or for matter with important magnetic, elastic, radiative, or viscous stresses. A shock requires a more careful distinction: the microscopic shock layer is dissipative, but the fluid on its two sides can still be described by perfect-fluid stress and connected by conservation jump conditions and an entropy inequality. Failure to resolve that microscopic layer does not by itself invalidate the macroscopic weak solution.

Check your understanding

An observer sees a fluid moving with speed $v$ along one orthonormal spatial axis. Show that the measured longitudinal stress exceeds the transverse stress by $(\varepsilon+p)W^2v^2$, and explain why this does not violate rest-frame isotropy.

Solution

Choose $v^{\hat a}=(v,0,0)$ in the observer’s spatial frame. From the spatial projection,

$$S^{\hat x\hat x}=(\varepsilon+p)W^2v^2+p, \qquad S^{\hat y\hat y}=S^{\hat z\hat z}=p.$$

Therefore $S^{\hat x\hat x}-S^{\hat y\hat y}=(\varepsilon+p)W^2v^2$. This anisotropy is the momentum flux of moving matter. Boosting to the fluid frame sets $v=0$ and restores three equal spatial stresses, so the defining isotropy is intact.

References

• Andersson, Nils, and Gregory L. Comer. “Relativistic Fluid Dynamics: Physics for Many Different Scales.” Living Reviews in Relativity 24, 3 (2021). DOI.