= Solution
{c}
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 $\alpha=u^\flat$ be the <differential one-form> obtained by the <musical isomorphism>. Its <exterior derivative> satisfies
$$
(\iota_u d\alpha)_b=u^a(\nabla_a u_b-\nabla_bu_a)
=a_b-\frac12\nabla_b(u^au_a)=0.
$$
Here $\iota_u$ is the <interior product>. By <hypersurface orthogonality> and the <Frobenius theorem>, $\alpha\wedge d\alpha=0$. Taking the <interior product> and using $\alpha(u)=-1$ gives
$$
0=\iota_u(\alpha\wedge d\alpha)
=-d\alpha-\alpha\wedge\iota_u d\alpha=-d\alpha.
$$
Thus $\boxed{du^\flat=0}$. 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 $dt(\partial_i)=0$,
$$
g_{tt}=g(u,u)=-1,\qquad g_{ti}=\alpha(\partial_i)=-dt(\partial_i)=0.
$$
The metric on the orthogonal complement of a <timelike vector> is <positive-definite>, so the <synchronous coordinates for irrotational dust> have
$$
\boxed{u=\partial_t,\qquad g=-dt^2+g_{ij}(t,\mathbf x)\,dx^i\otimes dx^j.}
$$
This construction is local; crossing <geodesics> can prevent these <synchronous coordinates> from covering a larger region.
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