Congruence of geodesics

Remarks

For such a congruence set $B_{ab} = \nabla_b u_a$ and $h_{ab} = g_{ab} + u_a u_b$, the metric on vectors orthogonal to $u$. Then $B_{ab} u^a = B_{ab} u^b = 0$, and $B_{ab}$ splits into the expansion $\theta = \nabla_a u^a$, the shear $\sigma_{ab} = B_{(ab)} - \frac13 \theta h_{ab}$ and the rotation $\omega_{ab} = B_{[ab]}$, with $B_{ab} = \frac13 \theta h_{ab} + \sigma_{ab} + \omega_{ab}$. The rotation vanishes exactly when $u$ is orthogonal to a family of hypersurfaces. Congruences of null geodesics are defined in the same way with $g(u, u) = 0$, using a two-dimensional transverse space.

Example: in Minkowski space, inside the future light cone of the origin, let $\tau = \sqrt{-x_\mu x^\mu}$ and $u^\mu = x^\mu / \tau$. The curves are straight lines through the origin, so they are geodesics, and $g(u, u) = x_\mu x^\mu / \tau^2 = -1$. Here $\theta = \partial_\mu u^\mu = 4/\tau + x^\mu x_\mu / \tau^3 = 3/\tau$, so this congruence of free-fall world lines expands.

Non-example: the world lines of uniformly accelerated observers in Rindler coordinates form a congruence of timelike curves with $g(u, u) = -1$. But $\nabla_u u \ne 0$, so clause 2 fails and it is not a congruence of geodesics.