 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 toContracting with $u_{\nu }$ kills the second term, since $u_{\nu }{\nabla }_{\mu }{u}^{\nu }=\frac{1}{2}{\nabla }_{\mu }(c^2)=0$. Hence ${\nabla }_{\mu }(\rho u^{\mu })=0$, and wherever $\rho >0$,On a flow line $u^{\mu }=d{x}^{\mu }/d\tau$, so this becomes $d^2{x}^{\nu }/d{\tau }^2+{\Gamma }^{\nu }{}_{\alpha \beta }(d{x}^{\alpha }/d\tau )(d{x}^{\beta }/d\tau )=0$. 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.