= Solution
Affine geodesic motion gives $U^c\nabla_cU_a=0$. Commuting the two <covariant derivative>[covariant derivatives] and applying the product rule yields
$$
\begin{aligned}
U^c\nabla_cB_{ab}
&=U^c\nabla_c\nabla_bU_a\\
&=\nabla_b(U^c\nabla_cU_a)
-(\nabla_bU^c)(\nabla_cU_a)
+R_{cba}{}^dU_dU^c\\
&=-B^c{}_bB_{ac}+R_{cba}{}^dU_dU^c.
\end{aligned}
$$
This proves the required transport identity. The curvature term is symmetric after screen projection, so taking the antisymmetric screen part and inserting the optical decomposition gives the <null-twist propagation equation>
$$
\boxed{U^c\nabla_c\widehat\omega_{ab}
=-\theta\widehat\omega_{ab}
+2\widehat\sigma_{c[a}\widehat\omega_{b]}{}^c}.
$$
In particular, an initially twist-free congruence remains twist-free.
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