 Solution

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.