= Solution
For <geodesic motion of uncharged dust>, use the pressureless stress-energy form, $T^{\mu\nu}=\rho u^\mu u^\nu$ and $u_\nu u^\nu=c^2$. <stress-energy conservation> expands to
$$
0=\nabla_\mu T^{\mu\nu}
=u^\nu\nabla_\mu(\rho u^\mu)+\rho u^\mu\nabla_\mu u^\nu.
$$
Contracting with $u_\nu$ kills the second term, since $u_\nu\nabla_\mu u^\nu=\tfrac12\nabla_\mu(c^2)=0$. Hence $\nabla_\mu(\rho u^\mu)=0$, and wherever $\rho>0$,
$$
\boxed{u^\mu\nabla_\mu u^\nu=0}.
$$
On a flow line $u^\mu=dx^\mu/d\tau$, so this becomes $d^2x^\nu/d\tau^2+\Gamma^\nu{}_{\alpha\beta}(dx^\alpha/d\tau)(dx^\beta/d\tau)=0$. Each <relativistic dust> element therefore follows a <timelike geodesic> with <proper time> as an <affine parameter>. <pressure> gradients or external nongravitational <forces> would add an <acceleration> term.
Back to article page
