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Transport two connecting vectors between neighbouring rays so that they commute with $U$, and project them onto the screen. Their infinitesimal transverse lengths and angles evolve according to the symmetric part of the optical tensor. The trace part gives uniform fractional dilation, while the trace-free null shear stretches one transverse direction and contracts the other, changing a circular beam to an ellipse without changing its area to first order. With zero null twist, there is no antisymmetric rotation of the beam.

If $q_{AB}$ is the metric on an infinitesimal transverse element, then $d{q}_{AB}/d\lambda =2{\hat{B}}_{(AB)}$. Its area $A$ therefore obeysThus positive null expansion means a beam's cross-sectional area increases, and negative expansion means it decreases; null shear records the directional distortion rather than this overall area change.

The second law of black-hole mechanics states that the total area of a classical future event horizon cannot decrease toward the future, under the appropriate causal and energy assumptions. Use the null convergence condition $R_{ab}{U}^a{U}^b\ge 0$; in general relativity the Einstein field equations imply it from the null energy condition. Assume, as allowed here, that the horizon's null generators remain on the horizon and are complete to the future. The horizon is an achronal boundary, and its smooth generators are hypersurface-orthogonal, so their null twist is zero. The Null Raychaudhuri equation therefore givesIf some smooth horizon point had ${\theta }_0<0$, then while the optical description remains regular, $\theta$ stays negative andThusforcing focusing no later than ${\lambda }_0+2/|{\theta }_0|$.

The assumed geometrical result is that this focusing produces a conjugate point to a spacelike surface, namely a focal point relative to a small spacelike horizon cross-section through the original point. A null geodesic beyond such a focal point ceases to generate an achronal boundary: a timelike variation connects the cross-section to points just beyond it. Since the cross-section and the generator lie on the achronal horizon, this is impossible. Future completeness ensures that a singular endpoint cannot terminate the generator before the finite focusing parameter. This future-complete horizon focusing proof of the area theorem rules out ${\theta }_0<0$.

Therefore $\theta \ge 0$ on smooth horizon pieces, and the transverse-area identity givesIntegrating over generators proves area nondecrease between later cross-sections. At crease sets, new generators may enter the horizon from the past, adding area; horizon generators do not leave it toward the future under the stated assumptions. Hence the statement also applies to total horizon area through black-hole mergers. Hawking's area theorem is classical: quantum stress-energy can violate the null energy condition, so it does not forbid a shrinking evaporating black hole.

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 $-{\theta }^2/(D-2)$.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there. For affine tangent $k$, the null expansion is the trace of the optical tensor: $\theta =d(\log A)/d\lambda$ for an infinitesimal pencil's area. Positive expansion means spreading; negative expansion means focusing.

For hypersurface-orthogonal generators in four dimensions, the Null Raychaudhuri equation readsThe null twist vanishes. The null energy condition implies $R_{ab}{k}^a{k}^b\ge 0$ through the Einstein field equations. An initial ${\theta }_0<0$ would force a focal point within affine distance at most $2/|{\theta }_0|$. A future-complete generator cannot develop such a point while remaining on an achronal event horizon. Thus $\boxed{\theta \ge 0}$ under the energy and global predictability/future-completeness hypotheses of Hawking's area theorem. Quantum energy-condition violations or failure of those global assumptions remove the conclusion.

The event horizon is the global boundary $\partial J^{-}({\mathcal{I}}^{+})$ and depends on the whole future. An apparent horizon on a chosen spacelike slice is the outermost marginally outer trapped surface, ordinarily ${\theta }_{\mathrm{out}}=0$, ${\theta }_{\mathrm{in}}<0$. It is quasi-local and slice-dependent; its tube need not be null.

For a shell arriving at $v=0$, the final horizon is $r=2M$. Before arrival, outgoing flat-space rays obey $dr/dv=1/2$. Tracing this horizon backward gives $r=2M+v/2$ until its beginning at $r=0$, $v=-4M$. No apparent horizon exists in that earlier flat region, explicitly illustrating the event horizon's dependence on future collapse.

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.

One standard form of the Penrose singularity theorem assumes a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector $k$, and a closed future trapped surface. It concludes future null geodesic incompleteness: some future-inextendible null geodesic has a finite upper endpoint of its affine parameter.

The focusing mechanism is the Null Raychaudhuri equation. The normal generators have zero null twist, soAn initial ${\theta }_0<0$ therefore gives a conjugate point to a spacelike surface within affine distance at most $2/|{\theta }_0|$, assuming the generator can be continued that far. Such a generator ceases to lie on the achronal boundary after its first focal point. Compactness of the trapped surface supplies a uniform bound for all normalized initial null normals. Future completeness would consequently make its future boundary compact. Projection along timelike curves to a connected Cauchy hypersurface is injective on the achronal boundary and has open image. Compactness makes the image closed as well; the nonempty image must therefore be the entire hypersurface, contradicting its noncompactness. This explains why the global assumptions supplement local focusing.

For Reissner-Nordstrom spacetime, the Maxwell stress-energy tensor satisfies the null energy condition; the Einstein field equations imply the required null convergence condition. The spheres in the band just found are closed and trapped. Apply the theorem to a maximal Cauchy development with a noncompact Cauchy hypersurface and containing one such sphere. That globally hyperbolic spacetime must be future null-geodesically incomplete.

The Penrose theorem at a Cauchy horizon needs care: the full maximal analytic Reissner-Nordstrom spacetime has inner Cauchy horizons and is not globally hyperbolic. It does not satisfy every hypothesis of the displayed theorem. In the exact solution some incomplete geodesics of the globally hyperbolic development reach a smoothly extendible Cauchy horizon in finite affine parameter. The theorem asserts incompleteness of the development, not that every such endpoint is a curvature singularity. The separate curvature singularity at $r=0$ does not justify silently dropping the theorem's global hypothesis.

For an affinely parametrized null geodesic congruence, put $B^a{}_b={\nabla }_b{U}^a$. Differentiating along a ray and commuting covariant derivatives givesusing $[{\nabla }_c,{\nabla }_b]V^a=R^a{}_{dcb}{V}^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_b{U}^b=0$ and $U_a{B}^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the optical tensor squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yieldsNow decompose the optical tensor into null expansion, null shear and null twist. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while ${\hat{\omega }}_{ab}{\hat{\omega }}^{ba}=-{\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$. HenceSubstitution proves the Null Raychaudhuri equation:The squares use the positive-definite screen metric. Thus null shear contributes to focusing, whereas nonzero null twist has the opposite sign.

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.

Use the Null Raychaudhuri equation for $\ell =f^{-1}{\partial }_U$. Its null shear and null twist vanish, and $R_{ab}{\ell }^a{\ell }^b=f^{-2}{R}_{UU}$. Substitution of ${\theta }_{\ell }=2r_{,U}/(fr)$ givesThe quadratic terms cancel. Thus the double-null focusing identity isHere the last equality uses the Einstein field equations in units $G=1$; terms proportional to the metric vanish in this null component. The inequality follows from the null energy condition.

For any point of the future domain of dependence $D^{+}(\Sigma )$, the past continuation of its $\ell$ generator must meet the Cauchy hypersurface $\Sigma$: every past-inextendible causal curve from that point meets $\Sigma$. Along its future continuation, $U$ increases, and $f^{-1}{r}_{,U}$ is nonincreasing by the focusing identity. Its initial value is negative, so it stays negative. Since $f>0$, $r_{,U}<0$ throughout the regular part of $D^{+}(\Sigma )$. The argument is used only where the smooth double-null chart has $r>0$; it does not extend a congruence beyond a singular endpoint.

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.

The Physical-process first law for a rotating black hole, in four-dimensional general relativity with $G=1$, isIt applies to a small neutral matter influx into an initially stationary, nonextremal black hole that settles to another stationary state. The background surface gravity and angular velocity are $\kappa$ and ${\Omega }_H$, and $\delta E,\delta J$ are the Killing energy and angular momentum delivered through the event horizon. Other conserved charges are held fixed. The perturbation must be small enough that its horizon generators do not develop caustics, and all equalities below are to first order.

Let $t^a$ be the stationary Killing vector field normalized at infinity, and ${\psi }^a$ the axial Killing vector field with $2\pi$-periodic orbits. The horizon generator is ${\xi }^a=t^a+{\Omega }_H{\psi }^a$. Choose an affine tangent $k^a$ and parameter $\lambda$ with zero at the past stationary limiting section, so affine horizon-generator scaling gives, on the background horizonThe background null expansion and null shear vanish, and null twist vanishes for the hypersurface-orthogonal horizon generators. Linearizing the Null Raychaudhuri equation and using the Einstein field equations givesThe quadratic null expansion and null shear terms are second order. With background cross-sectional measure $d{A}_0$, the first-order area change is $\delta A={\int }_{\mathcal{H}}\delta \theta d\lambda d{A}_0$. Multiplying the evolution equation by $\lambda$ and integrating by parts gives the first-order horizon-area response to matter fluxThe boundary term $[\lambda \delta \theta ]$ vanishes: $\lambda =0$ at the past limiting section, and the final stationary boundary condition gives zero expansion in the settled future. Equivalently, assume a sufficiently localized influx with the required late-time decay.

The energy and angular-momentum fluxes have signsThese correspond to the particle conventions $E=-p\cdot t$ and $J=p\cdot \psi$. Their combination isThis proves the requested Physical-process first law of black-hole mechanics. For charged infall the corresponding law contains an additional horizon-potential term ${\Phi }_H\delta Q$; the energy-minus-angular-momentum version assumes that work term is absent.

Use an affine parameter $\lambda$ along a generator of a null hypersurface. Its null twist vanishes by hypersurface orthogonality. The screen metric is positive definite, so ${\hat{\sigma }}^{ab}{\hat{\sigma }}_{ab}\ge 0$. The Einstein field equations and null energy condition imply $R_{ab}{U}^a{U}^b\ge 0$, giving from the Null Raychaudhuri equationStarting with ${\theta }_0<0$, the null expansion stays negative as long as the congruence remains regular. Set $z=-1/\theta >0$. Then $z^{\prime }={\theta }^{\prime }/{\theta }^2\le -1/2$, soA finite negative null expansion cannot persist to $\lambda -{\lambda }_0=2/|{\theta }_0|$. Thus the null focusing theorem givesThis is a focal breakdown of the transverse congruence, provided the generator and regular geometry extend that far; it is not by itself a curvature singularity or a statement that the individual null geodesic terminates there.

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.