= Solution
For <relativistic dust>, $T^{\alpha\beta}=\rho u^\alpha u^\beta$, with $u^\alpha u_\alpha=-c^2$. Expanding <stress-energy conservation> gives
$$
0=\nabla_\alpha(\rho u^\alpha u^\beta)
=u^\beta\nabla_\alpha(\rho u^\alpha)
+\rho u^\alpha\nabla_\alpha u^\beta.
$$
Contract with $u_\beta$. <Metric compatibility> and the fixed normalization imply
$$
u_\beta u^\alpha\nabla_\alpha u^\beta
=\frac12u^\alpha\nabla_\alpha(u_\beta u^\beta)=0.
$$
Consequently
$$
-c^2\nabla_\alpha(\rho u^\alpha)=0.
$$
This is the covariant <mass conservation> equation. Substituting it back, wherever $\rho>0$, gives
$$
\boxed{u^\alpha\nabla_\alpha u^\beta=0.}
$$
Thus the dust's fluid elements follow <timelike geodesics>. With $u^\alpha=dx^\alpha/d\tau$, the <proper time> $\tau$ is an <affine parameter>, and explicitly
$$
\frac{d^2x^\beta}{d\tau^2}
+\Gamma^\beta{}_{\alpha\gamma}
\frac{dx^\alpha}{d\tau}\frac{dx^\gamma}{d\tau}=0.
$$
The absence of <pressure> removes the force produced by pressure gradients; gravity is encoded by the <Levi-Civita connection>. At a point with $\rho=0$ this calculation makes no claim about a fluid velocity.
Back to article page
