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If a null vector tangent satisfies ${\nabla }_{K}K=\kappa K$, then $U=hK$ is affine when $K(\log h)=-\kappa$. Choose $h=1$ on the initial transverse surface. The screen-space projector kills derivative terms proportional to $K$, so the initial optical tensor and its null expansion agree with those of $K$. In ingoing spherical coordinates, $K={\partial }_v+(f/2){\partial }_r$ has $\kappa =f^{\prime }/2$.

The null expansion, null shear and null twist of a congruence tangent to the generators of a Killing horizon all vanish there. If $k$ is its Killing generator and $\ell =fk$ is an affine generator, the screen-space projector kills every term proportional to $k$. The symmetric part of the projected derivative of $\ell$ therefore vanishes by the Killing equation. The antisymmetric part vanishes by Frobenius theorem, since $\ell$ is hypersurface-orthogonal. These conclusions extend through a regular bifurcation surface by continuity.

For an affine null geodesic congruence with future tangent $U$, choose a future unit timelike vector $T$ on a transverse section, put $E=-U\cdot T$, and set $N=T/E-U/(2E^2)$ there. It has $N^2=0$ and $U\cdot N=-1$. Parallel transport along each generator preserves these contractions by metric compatibility and ${\nabla }_{U}U=0$. The construction is local up to caustics and supplies the auxiliary normal for a screen-space projector.

Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.

A null geodesic congruence is a smooth family of null geodesics filling a region, with one curve through each point before caustics form. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^2=0$ and ${\nabla }_{U}U=0$. Choose a null partner $N^a$ with $U\cdot N=-1$, transported parallel to $U$ along each generator. The screen-space projectorprojects onto the positive-definite two-dimensional plane transverse to $U,N$. Define the optical tensor and its decomposition bywhereThese are respectively the null expansion, null shear and null twist, the latter also called rotation. In the Null Raychaudhuri equation, ${\hat{\sigma }}^2={\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}$ and ${\hat{\omega }}^2={\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$ use the screen metric. The trace convention for $\theta$ is important: it is not half the trace. For an affine null tangent it also equals ${\nabla }_a{U}^a$.

For the radial null pair above, the screen-space projector selects the angular tangent plane, with metric $q_{AB}=r^2{\gamma }_{AB}$. The ingoing covector is $U_a=-E(dv{)}_a$, and ${\Gamma }^v{}_{AB}=-r{\gamma }_{AB}$. Hence the optical tensor isIt is purely symmetric and proportional to the screen metric, givingFor $E=1$, the null expansion is $-2/r$. The area check $A\propto r^2$ and $dr/d\lambda =-E$ gives $A^{\prime }/A=-2E/r$, in agreement with the spherical null expansion in ingoing coordinates. In vacuum, $d\theta /d\lambda =-2E^2/r^2=-{\theta }^2/2$ also satisfies the Null Raychaudhuri equation; the ingoing beam focuses at the singularity rather than remaining future complete.

Choose the affine tangent $k^a$ to a null geodesic congruence, and an auxiliary null vector ${\ell }^a$ with $k\cdot \ell =-1$. The screen-space projectorannihilates $k$ and $\ell$ and restricts to a positive-definite two-dimensional metric in four spacetime dimensions. Neighboring transverse rays have relative displacement governed by the optical tensorIts irreducible decomposition defines the three requested optical quantities:The null expansion $\theta$ is the fractional rate of change of transverse area, $d\log A/d\lambda$. The null shear $\sigma$ changes a beam's shape at fixed area to first order; it is symmetric and trace-free. The null twist $\omega$ is antisymmetric and measures local rotation of the transverse rays; it vanishes for a hypersurface-orthogonal null geodesic congruence. In $D$ dimensions replace $1/2$ by $1/(D-2)$. These definitions use the trace convention for null expansion, rather than the alternative screen-averaged convention.

Take an affinely parametrized null geodesic congruence, ${\nabla }_{k}k=0$. Choose the auxiliary null vector and a transverse orthonormal screen basis to undergo parallel transport along each ray. For the unprojected tensor $C_{ab}={\nabla }_b{k}_a$, the product rule and the Ricci identity yieldHere the Riemann curvature tensor convention is the one in Question 1; its last-index antisymmetry converts the covector commutator to the displayed curvature term. Because $C_{ab}{k}^a=C_{ab}{k}^b=0$, inserting the screen-space projector in the product term adds only terms killed by these contractions. Parallel propagation makes differentiation commute with the screen projection. Therefore, in screen indices $A,B$,This is the optical matrix evolution equation. Its trace has curvature contraction $q^{ab}{R}_{acbd}{k}^c{k}^d=R_{cd}{k}^c{k}^d$: the additional terms in $q^{ab}-g^{ab}$ vanish by curvature antisymmetry. Write $B=(\theta /2)q+\sigma +\omega$ using the null expansion, null shear, and null twist. Symmetric--antisymmetric cross terms have zero trace, $\sigma$ is trace-free, andThe minus sign in the last trace comes from ${\omega }_{ab}=-{\omega }_{ba}$ and the positive screen metric. Taking the trace therefore derives the Null Raychaudhuri equation:For a nonaffine tangent obeying ${\nabla }_{k}k=\kappa k$, the same product-rule derivation adds $+\kappa \theta$ on the right; the projected derivative of $\kappa$ multiplies $k_a$ and vanishes. In $D$ dimensions the quadratic null expansion term is $-{\theta }^2/(D-2)$.

Away from a bifurcation set, the normal to a Killing horizon is its nonzero Killing vector field $K$. Any affine tangent to the same generators has the form $k=hK$. In the optical tensor, derivatives of $h$ contribute a factor of $K_a$ and disappear under the screen-space projector. Consequentlyby the Killing equation. The trace and trace-free symmetric part are therefore zero: $\theta =0$ and ${\sigma }_{ab}=0$.

For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its exterior derivative and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the Frobenius theorem. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the optical scalars on a Killing horizon satisfyNo affine normalization of the Killing vector field itself was assumed. At a regular bifurcation surface, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.

The screen-space projector annihilates both null vectors: $P^a{}_b{U}^b=P^a{}_b{N}^b=0$, and satisfies $P^2=P$. Its image is the two-dimensional spacelike screen orthogonal to $U,N$. The screen metric isIt is positive definite on that screen. Projecting both indices of $B_{ab}={\nabla }_b{U}_a$ gives the optical tensor ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$.

Define its three parts byThus ${\hat{B}}_{ab}=\frac{1}{2}\theta q_{ab}+{\sigma }_{ab}+{\omega }_{ab}$. The null expansion $\theta$ is its trace and equals $d\log \mathcal{A}/d\lambda$ for the infinitesimal beam area $\mathcal{A}$. The null shear $\sigma$ is symmetric and trace-free, measuring shape change at fixed first-order area. The null twist $\omega$, also called rotation, is antisymmetric and measures failure of screen directions to remain hypersurface-orthogonal. In $D$ dimensions replace $1/2$ by $1/(D-2)$. Here and below expansion means the trace, rather than its average over the screen dimensions.

Locally write the null hypersurface as $F=0$. Its generators are tangent to its raised normal, so on it $U_a=h{\nabla }_{a}F$ for a nonzero scalar $h$. The antisymmetric derivative issince the Hessian of $F$ is symmetric for the torsion-free Levi-Civita connection. Each term contains $\nabla F$, which is proportional to $U$ and is killed by the screen-space projector. HenceThis is the null version of hypersurface orthogonality in the Frobenius theorem. A null hypersurface has no independent normal direction outside its tangent space: its null vector normal also generates it. That is why the same argument applies to the generators' null twist.

For each of the two normal directions, launch a null geodesic congruence orthogonally from the sphere with initial tangent $U=X$ or $U=Y$. For $U=X$ take initial auxiliary $N=Y$; for $U=Y$ take initial $N=X$, and then parallel transport $N$ as in part (a). On the sphere the resulting screen-space projector projects onto its angular tangent space, whose metric is $q_{AB}=r^2{\gamma }_{AB}$.

For any radial normal $K$, the spherical null expansion in ingoing coordinates follows directly from area variation:Equivalently $\mathcal{A}\propto r^2$ gives $K(\log \mathcal{A})=2K(r)/r$. ConsequentlySpherical symmetry also makes the null shear and null twist zero for these radial congruences.

There is a small parametrization distinction. The smooth field $X=-{\partial }_r$ is affine, since ${\Gamma }^a{}_{rr}=0$. The natural smooth field $Y={\partial }_v+(f/2){\partial }_r$ obeys ${\nabla }_{Y}Y=(f^{\prime }/2)Y$. To meet part (a)'s affine convention, use an affine rescaling of a null normal along the outgoing generators, with scaling equal to one on $S$. Its angular derivatives on $S$ then leave the projected derivative unchanged. The displayed null expansions are therefore precisely those for the affinely launched congruences, even though that convenient global expression for $Y$ is nonaffine away from its initial sphere.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there, with one tangent direction assigned smoothly at every point. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$. Choose a second null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isIt annihilates $U,N$ and restricts to a positive-definite metric on the two-dimensional transverse screen. Project the covariant derivative of the tangent to obtain the optical tensorIts decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a transverse area, $\theta =A^{-1}dA/d\lambda$. The null shear changes a beam's shape without changing its area to first order. The null twist is its antisymmetric transverse deformation. The factor $1/2$ uses four spacetime dimensions.

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.

The screen-space projectordefines the optical tensor ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$. In four spacetime dimensions its irreducible decomposition isThese are the null expansion, null twist, and null shear.

Let $U^a$ be the affine null tangent and choose another null vector $N^a$ with $U\cdot N=-1$. The screen-space projectorprojects onto the $(D-2)$-dimensional transverse space. The optical tensor isIts irreducible decomposition definesthe null expansion,the null shear, andthe null twist, also called rotation.

Choose an auxiliary null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isand the optical tensor is ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$ for $B_{ab}={\nabla }_b{U}_a$. Because the screen is two-dimensional, its irreducible decomposition iswhereThese are respectively the null expansion, null shear, and null twist, also called rotation.

With $U^2=N^2=0$ and $U\cdot N=-1$, define the screen-space projectorDirect contraction gives $P^a{}_b{U}^b=P^a{}_b{N}^b=0$, while a vector orthogonal to both $U$ and $N$ is unchanged. Moreover $P^a{}_c{P}^c{}_b=P^a{}_b$ and $P^a{}_a=2$, so its image is precisely the two-dimensional transverse space.