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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 givesThere 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.

In the two null coordinates the inverse metric has $g^{UV}=g^{VU}=-f^{-1}$ and $g^{UU}=g^{VV}=0$. ThereforeBoth tangents are null and future-directed, since $f>0$ and both coordinate directions are future-directed. To prove the affine parameter property, use the null-gradient affine-geodesic identity. The one-form $k_a=-{\nabla }_{a}U$ is closed, so ${\nabla }_a{k}_b={\nabla }_b{k}_a$. ConsequentlyThe same argument applies to ${\ell }_a=-{\nabla }_{a}V$. Thus both vector fields satisfy the geodesic equation without a nonaffinity term, and generate affinely parametrized null geodesic congruences. The coordinate vectors alone need not be affine; the factors $f^{-1}$ are essential.

A null geodesic congruence is a smooth family of null geodesics filling a region without crossings there. Choose a smooth future-directed tangent $U^a$ and an affine parameter $\lambda$, soThe space orthogonal to $U$ is degenerate because it contains $U$ itself. To measure genuine transverse separation, introduce an auxiliary null vector $N^a$ with $U\cdot N=-1$ and the screen-space projectorIn four dimensions $q$ has a positive-definite two-dimensional image and annihilates $U$ and $N$. The optical tensor is the transverse part of the velocity derivative,Its irreducible decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a small transverse area, $\theta =d\log \delta A/d\lambda$. The symmetric trace-free null shear distorts the shape of the transverse beam without changing its area at first order. The antisymmetric null twist, also called rotation of the congruence, describes the rotational part of its transverse deformation. These are properties of the screen quotient $U^{\perp }/⟨U⟩$; the auxiliary vector chooses a convenient representative of that quotient. For an affinely parametrized null geodesic congruence, the contraction constraints also give $\theta ={\nabla }_a{U}^a$.

Parallel transport the auxiliary null vector $N$ and a screen basis along each generator of the null geodesic congruence. This keeps the screen-space projector fixed under the corresponding transported derivative. Use the curvature convention $[{\nabla }_c,{\nabla }_b]V_a=-R^d{}_{acb}{V}_d$.

With $B_{ab}={\nabla }_b{U}_a$, differentiation of the affine geodesic equation givesHere the first term comes from commuting the derivative past $U^c$, and the second from the curvature commutator. The constraints $B_{ab}{U}^a=B_{ab}{U}^b=0$ follow respectively from the null normalization and affine geodesic equation. Projecting onto the screen therefore yields the optical matrix equationThe trace of the tidal matrix is $R_{ab}{U}^a{U}^b$; extra terms from replacing the screen trace by the spacetime trace vanish by curvature antisymmetry. Taking the trace givesDecompose the optical tensor as $\hat{B}=\frac{1}{2}\theta I+\hat{\sigma }+\hat{\omega }$. The null shear is symmetric and trace-free and the null twist is antisymmetric. Their cross traces vanish, andThe negative sign in the last term is the identity $\mathrm{tr}({\hat{\omega }}^2)=-{\sum }_{I,J}{\hat{\omega }}_{IJ}^2$ on the positive-definite screen. Substitution proves the four-dimensional Null Raychaudhuri equation:The coefficient is $1/(D-2)$ in $D$ spacetime dimensions; the factor $1/2$ here belongs to the four-dimensional congruence in this question, not the five-dimensional geometry of the next question.

Let $k$ generate the Killing horizon and let $\ell =fk$ be an affinely parametrized generator wherever $k\ne 0$. Choose an auxiliary null vector $n$ with $n\cdot \ell =-1$ and the screen-space projector $q_{ab}=g_{ab}+{\ell }_a{n}_b+n_a{\ell }_b$. The optical tensor is the screen projection $B_{ab}=q_a{}^c{q}_b{}^d{\nabla }_d{\ell }_c$.

Since the projector annihilates $k$, the symmetric part isby the Killing equation. Its trace is the null expansion and its trace-free symmetric part is the null shear, so both vanish. Also $\ell$ is normal to the null hypersurface. Frobenius theorem gives ${\ell }_{[a}{\nabla }_b{\ell }_{c]}=0$, which implies $q_a{}^c{q}_b{}^d{\nabla }_{[d}{\ell }_{c]}=0$: the null twist vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth null geodesic congruence containing its generators, independently of its extension off the horizon. At a regular bifurcation surface, use the smooth affine generator and continuity. ThusThis is the vanishing of the optical scalars on a Killing horizon.

A null geodesic congruence is a smooth family of nonintersecting null geodesics filling an open spacetime region. For its affinely parametrized tangent $U^a$,Writing $B_{ab}={\nabla }_b{U}_a$, differentiation of the null norm and the geodesic equation give respectively

Because $U$ is a null tangent, metric compatibility givesBecause $U$ is tangent to an affinely parametrized null geodesic congruence,by the geodesic equation. Thus $B$ has no component along $U$ in either index.