Perfect Fluids in General Relativity

A perfect fluid is defined by isotropy in its local rest frame. Its stress–energy has an energy density $\rho$, an isotropic pressure $p$, and no rest-frame energy flux or anisotropic stress. Covariant conservation splits into an energy equation along the flow and a relativistic Euler equation orthogonal to it. Pressure therefore enters twice but in different ways: $\rho+p$ is the inertial coefficient accelerated by a pressure gradient, while $\rho+3p$ controls the comoving Ricci-curvature contraction in four-dimensional Einstein gravity without a cosmological term.

Required background. Classical Fields and Their Actions provides local field equations and variational reasoning; Flat-Background Hilbert Stress-Energy from Metric Variation provides observer projections of stress–energy; Affine Connections and Covariant Derivatives provides covariant divergence and acceleration.

Helpful background. Events, Intervals, and Minkowski Geometry reviews local rest frames; Tensor Fields and Natural Operations reviews tensor decompositions; Continuum Kinematics and Balance Laws supplies continuum balance and constitutive closure; Presymplectic Potentials and Currents provides useful action-based background.

Rest-frame isotropy fixes the tensor

Let $u^\mu$ be the fluid four-velocity and $h_{\mu\nu}=g_{\mu\nu}+u_\mu u_\nu$ the projector onto its local rest space. Isotropy permits only a scalar multiple of $h_{\mu\nu}$ as spatial stress. Thus

$$\boxed{ T_{\mu\nu} =\rho u_\mu u_\nu+p h_{\mu\nu} =(\rho+p)u_\mu u_\nu+pg_{\mu\nu}.}$$

The invariant projections recover the fluid variables:

$$T_{\mu\nu}u^\mu u^\nu=\rho, \qquad h_\alpha{}^\mu T_{\mu\nu}u^\nu=0, \qquad \frac13h^{\mu\nu}T_{\mu\nu}=p.$$

In an orthonormal comoving frame, $T_{\hat\mu\hat\nu}=\operatorname{diag}(\rho,p,p,p)$. These formulas are not an equation of state: they specify tensor form, while a relation such as $p=p(\rho,s)$ supplies material physics. Reall presents the same rest-frame interpretation and tensor in Reall 2013, §5.2, pp. 53–54, PDF. Kontou and Sanders use it as the model-independent perfect-fluid form in Kontou–Sanders 2020, §§2–2.1, pp. 6–10, equations (4) and (21), PDF.

Parallel projection gives energy balance

Define the derivative along the flow, expansion, and four-acceleration by

$$\dot\rho=u^\mu\nabla_\mu\rho, \qquad \theta=\nabla_\mu u^\mu, \qquad a^\nu=u^\mu\nabla_\mu u^\nu.$$

Expanding the divergence gives a useful decomposition:

$$\nabla_\mu T^{\mu\nu} =u^\nu\left[\dot\rho+(\rho+p)\theta\right] +(\rho+p)a^\nu +h^{\nu\mu}\nabla_\mu p.$$

Contract with $u_\nu$. Orthogonality $u_\nu a^\nu=0$ and $u_\nu h^{\nu\mu}=0$ leave

$$\boxed{ \dot\rho+(\rho+p)\theta=0.}$$

This is local energy balance for an adiabatic perfect fluid. Expansion lowers the energy density not only by diluting rest energy but also through pressure work. A separate conserved particle current $n u^\mu$ and thermodynamic relation are needed if $\rho$ is to be decomposed into rest mass and internal energy.

Orthogonal projection gives relativistic Euler flow

Project instead with $h^\alpha{}_{\nu}$:

$$\boxed{ (\rho+p)a^\alpha =-h^{\alpha\mu}\nabla_\mu p.}$$

Only the pressure gradient within the instantaneous rest space accelerates the fluid. The factor $\rho+p$ is relativistic inertial density. If $p=0$ and $\rho>0$, the equation reduces to $a^\alpha=0$, reproducing dust. If $\rho+p=0$, division by this coefficient is invalid: the tensor becomes $T_{\mu\nu}=-\rho g_{\mu\nu}$, the velocity drops out, and conservation requires $\nabla_\mu\rho=0$ for an isolated component.

In a local comoving inertial frame at one event, the spatial equation reduces to $(\rho+p)d\mathbf v/dt=-\boldsymbol\nabla p$. This is a local check, not a global Newtonian approximation; curvature and connection effects return away from the chosen event.

A static pressure-balance check

Consider a static region with metric $ds^2=-N^2(\mathbf x)dt^2+h_{ij}dx^i dx^j$, where $N>0$, and a fluid at rest in these coordinates, $u=N^{-1}\partial_t$. Its worldlines are generally accelerated even though their spatial coordinates are constant. Directly computing $u^\mu\nabla_\mu u_i$ gives $a_i=\partial_i\ln N$. The Euler equation therefore becomes

$$\partial_i p=-(\rho+p)\partial_i\ln N.$$

In a weak field, $N\simeq1+\Phi$; for nonrelativistic matter with $p\ll\rho$, this reduces to $\boldsymbol\nabla p=-\rho\boldsymbol\nabla\Phi$. Thus an upward pressure force can hold a fluid at rest while freely falling dust would accelerate. This is a local consistency check of the conservation law on a specified background; solving for a self-gravitating star also requires Einstein’s equation and boundary data.

Pressure as a source of Ricci curvature

In four dimensions the trace is

$$T=-\rho+3p.$$

With $\Lambda=0$, trace reversal of Einstein’s equation gives

$$R_{\mu\nu} =8\pi G \left[ (\rho+p)u_\mu u_\nu +\frac12(\rho-p)g_{\mu\nu} \right].$$

Contracting with the comoving velocity yields

$$R_{\mu\nu}u^\mu u^\nu =4\pi G(\rho+3p).$$

This equation is the precise local sense in which isotropic pressure contributes to timelike Ricci focusing. It does not say that pressure is an independent scalar “gravitational charge”: the full tensor and the observer direction matter. Kontou and Sanders give the corresponding perfect-fluid energy-condition combinations, including $\rho+p$ and $\rho+3p$ in four dimensions, in Kontou and Sanders 2020, § 2.1, pp. 8–10, PDF.

Closure and validity

The variables are $\rho$, $p$, and three independent components of $u^\mu$ after normalization; four conservation equations do not close the system. An equation of state, and often particle-number or entropy evolution, supplies the missing information. The perfect-fluid approximation also assumes a coarse-graining scale on which heat flux, viscosity, anisotropic pressure, and microscopic mean-free-path effects are negligible.

For a simple nondissipative material, introduce particle density $n$, entropy per particle $s$, and the first-law form

$$d\rho=\frac{\rho+p}{n}\,dn+nT\,ds.$$

Combining $\nabla_\mu(n\,u^\mu)=0$ with the energy equation gives $nT\dot s=0$. Where $nT\ne0$, entropy per particle is constant along each flow line. This does not force $s$ to be spatially uniform; it expresses adiabatic transport by a perfect fluid. Shocks or dissipative stresses instead produce entropy and require a weak or extended thermodynamic formulation. An equation of state such as $\rho=\rho(n,s)$ then determines $p=n(\partial\rho/\partial n)_s-\rho$ and closes the ideal system locally.

At shocks, the fields need not be differentiable and the equations must be imposed weakly with jump conditions. At multistreaming, no single rest velocity exists. Detailed equations of state, dissipative relativistic hydrodynamics, kinetic closure, and compact-star microphysics are outside this chapter.

Exercise: expansion and pressure work

Let $\mathcal V(\tau)$ be an infinitesimal comoving volume with $\dot{\mathcal V}/\mathcal V=\theta$. For a constant equation-of-state parameter $p=w\rho$, integrate the energy equation. Compare dust and radiation.

Solution

The parallel projection becomes

$$\frac{\dot\rho}{\rho} =-(1+w)\frac{\dot{\mathcal V}}{\mathcal V}.$$

Integration gives

$$\rho\,\mathcal V^{1+w}=\text{constant}.$$

For dust, $w=0$ and $\rho\propto\mathcal V^{-1}$, so the energy in a comoving volume is constant. For radiation, $w=1/3$ and $\rho\propto\mathcal V^{-4/3}$; the additional factor beyond dilution is pressure work, equivalently the redshifting of each quantum’s energy in an expanding flow.

References

• Kontou, Eleni-Alexandra, and Ko Sanders. “Energy Conditions in General Relativity and Quantum Field Theory.” Classical and Quantum Gravity 37 (2020): 193001. DOI. Open PDF.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes, 2013. PDF.