= Solution
For the <spherical null expansions in double-null coordinates>, the cross-sectional area of a symmetry sphere is $A=4\pi r^2$. For the two affine <null geodesic congruences>, the definition of <null expansion> therefore gives
$$
\boxed{\theta_k=\frac{2r_{,V}}{fr},\qquad\theta_\ell=\frac{2r_{,U}}{fr}.}
$$
There is no extra derivative of $f$ in these formulas: the <null expansion> differentiates the transverse area, not the normalization of the tangent. Equivalently, the projected metric is $q_{AB}=r^2\gamma_{AB}$ and its <optical tensor> is $(k(r)/r)q_{AB}$, or $(\ell(r)/r)q_{AB}$ for the other congruence. It is purely trace, so both <null shears> vanish.
The <null twist> vanishes because the covector tangents are gradients of $-U$ and $-V$, and hence their <exterior derivatives> vanish. Each congruence is orthogonal to a family of null hypersurfaces. This also identifies the <Frobenius theorem> reason for zero <null twist>.
Back to article page
