= Solution
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,
$$
u^{\hat\alpha}=(c,0,0,0),\qquad
T^{\hat\alpha\hat\beta}
=\operatorname{diag}(\rho c^2,p,p,p).
$$
Here $\rho$ is rest-frame <energy density> divided by $c^2$, including random kinetic energy, and $p$ is isotropic <pressure>. For a particle gas, $p=\tfrac13\int\mathbf p\cdot\mathbf v\,dn$ and $\rho c^2=\int E\,dn$, 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 congruence>
$$
P^{\alpha\beta}=g^{\alpha\beta}+\frac{u^\alpha u^\beta}{c^2}.
$$
The only isotropic decomposition with the preceding components is
$$
T^{\alpha\beta}=\rho\,u^\alpha u^\beta+pP^{\alpha\beta}.
$$
Since both sides are <tensors>, equality in the comoving <orthonormal frame in spacetime> establishes
$$
\boxed{
T^{\alpha\beta}
=\left(\rho+\frac p{c^2}\right)u^\alpha u^\beta
+p\,g^{\alpha\beta}.
}
$$
Coordinate-independent definitions of the two <scalar fields> are
$$
\boxed{
\rho=\frac{T^{\alpha\beta}u_\alpha u_\beta}{c^4},
\qquad
p=\frac13P_{\alpha\beta}T^{\alpha\beta}.
}
$$
Thus $\rho$ is not, in general, only the sum of the particle rest masses per volume.
\b[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 and
$$
\langle v^\alpha\rangle=(\gamma c,0,0,0),\qquad
\langle v^\alpha\rangle\langle v_\alpha\rangle=-\gamma^2c^2,
\qquad
\gamma=(1-v^2/c^2)^{-1/2}>1.
$$
The 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 $\rho_s$ gives
$$
T^{00}=\rho_s\gamma^2c^2,\qquad
p=\frac{\rho_s\gamma^2v^2}{3}.
$$
Substituting $u=\langle v\rangle$ 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>,
$$
\partial_t\rho+\boldsymbol\nabla\cdot(\rho\mathbf v)=0,
$$
while its spatial components give <momentum conservation>, or the inviscid <Euler equations>
$$
\rho(\partial_t+\mathbf v\cdot\boldsymbol\nabla)\mathbf v
=-\boldsymbol\nabla p-\rho\boldsymbol\nabla\Phi.
$$
The 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.
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