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Use metric signature $(-,+,+,+)$ and a bulk four-velocity normalized by $u^{\alpha }{u}_{\alpha }=-c^2$. A perfect fluid in general relativity is isotropic in its local rest frame, with no rest-frame energy flux or anisotropic stress. In an orthonormal frame in spacetime comoving with that fluid,Here $\rho$ is rest-frame energy density divided by $c^2$, including random kinetic energy, and $p$ is isotropic pressure. For a particle gas, $p=\frac{1}{3}\int \mathbf{p}\cdot \mathbf{v}dn$ and $\rho c^2=\int Edn$, with $dn$ the rest-frame particle distribution. Isotropy makes the off-diagonal spatial stresses and net energy flux vanish.

Introduce the spatial projector of a timelike congruenceThe only isotropic decomposition with the preceding components isSince both sides are tensors, equality in the comoving orthonormal frame in spacetime establishesCoordinate-independent definitions of the two scalar fields areThus $\rho$ is not, in general, only the sum of the particle rest masses per volume.

The microscopic averaging notation needs a normalization qualification. The normalization of an averaged four-velocity matters because a literal average of normalized microscopic four-velocities is not normally a normalized bulk four-velocity. For example, six equally weighted streams with the same ordinary speed $v$, directed along the positive and negative coordinate axes, have an isotropic velocity distribution andThe physical rest-frame bulk four-velocity is $(c,0,0,0)$, not this unnormalized average. Even taking the source's schematic kinetic expression literally with prefactor ${\rho }_s$ givesSubstituting $u=⟨v⟩$ and the same ${\rho }_s$ into the claimed fluid expression would instead increase $T^{00}$ by $p({\gamma }^2-1)$. The standard result therefore requires a normalized bulk velocity and a carefully defined rest-frame total energy density; it does not follow with both microscopic identifications interpreted literally. More generally the bulk velocity can be defined as the future-directed timelike energy eigenvector of the stress-energy tensor.

The equation ${\nabla }_{\alpha }{T}^{\alpha \beta }=0$ is local stress-energy conservation. In a weak gravitational field and the nonrelativistic limit its leading energy component gives mass conservation,while its spatial components give momentum conservation, or the inviscid Euler equationsThe next energy order gives conservation of internal plus kinetic energy, including pressure work and gravitational work. These are local conservation laws, not an assertion that arbitrary curved spacetimes possess a global conserved total energy.

Write $D=u^a{\nabla }_a$, $\theta ={\nabla }_a{u}^a$ for the expansion scalar, $a_b=u^a{\nabla }_a{u}_b$ for the four-acceleration, and $P_b{}^c={\delta }_b{}^c+u_b{u}^c$ for the spatial projector of a timelike congruence. The normalization of the four-velocity gives $u^b{a}_b=0$ and $u^b{\nabla }_a{u}_b=0$. The product rule and metric compatibility give the covariant derivative of the perfect fluid in general relativity stress-energy tensor:Its tensor contraction with the four-velocity isand its projection with the spatial projector of a timelike congruence isThus stress-energy conservation implies the relativistic perfect-fluid energy equation and the relativistic Euler equation:Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^c{F}_c-u_b{u}^c{F}_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho +p=0$, without dividing by that quantity.

First differentiate the unit normalization $X^{\alpha }{X}_{\alpha }=-1$. Metric compatibility givesThe timelike geodesic vector field equation gives the other contraction,For the spatial projector of a timelike congruence,and symmetry also gives $h_{\alpha \beta }{X}^{\beta }=0$. Thus

Splitting the deformation tensor of a timelike geodesic congruence into symmetric and antisymmetric parts and substituting the definitions givesMoreover, the orthogonality of $B_{\alpha \beta }$ implieswhile in four dimensionsIt follows thatHencewhich is the kinematic decomposition of a timelike geodesic congruence.

Use Minkowski metric $g_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ and units with the speed of light equal to one. Thus $u^{\mu }{u}_{\mu }=-1$, and $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }{u}_{\nu }$ is the spatial projector of a timelike congruence. Keep the medium velocity fixed when varying the electromagnetic four-potential. Since the electromagnetic field tensor is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, its variation givesBoth tensors are antisymmetric. Integration by parts, with variations vanishing at the boundary, therefore givesThe Euler-Lagrange field equation is ${\partial }_{\mu }(F^{\mu \nu }-M^{\mu \nu })=0$. Renaming the free index and using antisymmetry givesThese are the inhomogeneous Maxwell equations with the bound charge and bound current encoded by the polarization-magnetization tensor.