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A future trapped surface is a smooth compact spacelike two-surface without boundary whose two future-directed orthogonal null expansions are strictly negative. In four-dimensional general relativity, the Penrose singularity theorem states: a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a trapped surface, and the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector is future null-geodesically incomplete. With the Einstein field equations, the null energy condition implies this null convergence condition; a cosmological constant drops out of the null contraction.

The Kruskal spacetime is an example. In its black hole interior, use future null normals in Ingoing Eddington-Finkelstein coordinates:For a round sphere of areal radius $r$, its area is $4\pi r^2$ and its null expansions areBoth are negative for $0<r<2M$. The vacuum Einstein field equations give $R_{ab}=0$, and a two-ended Kruskal Cauchy hypersurface is noncompact. The future radial null geodesics reaching $r=0$ in finite affine parameter provide precisely the incompleteness predicted by the Penrose singularity theorem. A trapped surface at $r<2M$ is strictly trapped; the horizon sphere has one zero null expansion instead.

The second law of black-hole mechanics says that the total area of future event horizon cross-sections cannot decrease toward the future, assuming the classical Einstein field equations, the null energy condition, suitable global predictability and regularity, and here the stipulated future completeness of its generators.

A future event horizon is an achronal boundary generated by null geodesics. If a smooth generator had $\theta <0$, the null focusing theorem would produce a conjugate point to a spacelike surface within finite affine parameter. Future completeness lets the generator reach it, but beyond such a point it could no longer remain on the achronal boundary, a contradiction. Hence $\theta \ge 0$ wherever the null expansion is defined. A transported area element obeysEvery generator present on an early cut continues to a later cut without decreasing its area contribution. New generators can join at past endpoints, adding area; future endpoints in a regular spacetime are excluded. ThereforeThis argument also covers the sum of initially disconnected event horizons of the black holes in a merger. Nonsmooth crease sets require the standard regularity treatment; they do not provide a classical area-decrease loophole under these hypotheses. Quantum negative-energy flux can violate the classical null energy condition, so this is a classical black-hole area theorem, not an unconditional law for evaporating black holes.

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 $\theta ={\nabla }_a{U}^a$.

The Einstein field equations and the null energy condition imply the null convergence condition:The scalar-curvature and any cosmological constant terms vanish because $U^a{U}_a=0$. For generators of the null hypersurface, the supplied twist-free property gives $\hat{\omega }=0$. The null shear squared is nonnegative on the screen. Thus the Null Raychaudhuri equation impliesStarting from $\theta ({\lambda }_0)={\theta }_0<0$, the null expansion remains negative for as long as the regular congruence exists. HenceBefore the right-hand side reaches zero, inversion of the negative quantities givesSince the inverse of a finite negative null expansion cannot be nonnegative, the regular congruence cannot continue through the proposed upper limit. The null focusing theorem therefore givesProvided the geodesic itself extends this far, the transverse area collapses and $\theta \to -\infty$ at or before this bound. An earlier end of the affine geodesic would instead be geodesic incompleteness. A divergence of the null expansion marks a caustic or a conjugate point to a spacelike surface; it does not by itself establish a curvature singularity.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

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 Physical-process first law for a rotating black hole applies to a small perturbation of a stationary nonextremal vacuum black hole by infalling neutral matter, followed by relaxation to stationarity. Its first-order form isHere the background horizon generator is $\xi =t+{\Omega }_H\phi$, normalized by the unit time translation at infinity, and its surface gravity is constant, $\kappa >0$, as in the Zeroth law of black-hole mechanics. The formula neglects terms quadratic in the perturbation. Choose the process sufficiently weak that the horizon generators used below have no large caustics, and assume appropriate early and late stationary limits and convergent fluxes.

On the background Killing horizon, choose an affine generator $\ell =d/d\lambda$ such that $\xi =\kappa \lambda \ell$; the bifurcation or early stationary limit corresponds to $\lambda =0$. The background null expansion and null shear vanish. The null twist is zero because the perturbed event horizon is a null hypersurface. Linearizing the Null Raychaudhuri equation and using the Einstein field equations givesThe expansion squared and shear squared are second order, and the metric-trace part of the Einstein equations vanishes in this null contraction. Future stationarity sets $\delta \theta \to 0$. To first order $\delta \theta$ is the derivative of the fractional area change, so, using the background area element $d{A}_0$,The integration by parts uses $[\lambda \delta \theta {]}_0^{\infty }=0$, guaranteed by the stated endpoint conditions.

The conserved stress-energy current from a Killing vector gives the absorbed mass and angular momentum, with signsConsequentlywhich proves the law. Any other escaping flux must be accounted for separately when identifying these absorbed charges with the changes in the black hole. Electric-charge variations would require the corresponding potential-charge term; they are excluded in this version. The convention $16\pi G=1$ in Question 1 is not needed here.

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.

When $r_{-}=0$, $\Omega =0$ and $f=(r^2-r_{+}^2)/L^2$. A circle $S$ at fixed $v,r$ has circumference $2\pi r$. Choose its future null normals asFor a one-dimensional transverse surface, each null expansion is the logarithmic derivative of its length:Inside the horizon, $0<r<r_{+}$ implies $f<0$, so both expansions are negative. Every such circle is therefore a trapped surface, specifically a Trapped circle inside a nonrotating BTZ black hole.

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.

On $r=0$, the screen metric is $h_{ij}$ and $U={\partial }_{\lambda }$. Therefore the null expansion isThe derivative formula for the determinant givesand henceThus $\theta$ is the fractional rate of change of an infinitesimal transverse area carried along the generators: positive expansion enlarges the beam and negative expansion focuses it.

The second law of black-hole mechanics states that the area of spatial cross-sections of a future event horizon cannot decrease toward the future, provided the null energy condition and the standard global assumptions hold.

Let $k^a$ be an affinely parametrized horizon generator. It is hypersurface-orthogonal, so its null twist vanishes. The Null Raychaudhuri equation in $D$ dimensions and the Einstein equation giveIf $\theta ({\lambda }_0)<0$, integration implies that $\theta$ diverges to $-\infty$ within affine distance at most $(D-2)/|\theta ({\lambda }_0)|$. The resulting focal point would make the generator leave the achronal boundary that defines the event horizon, contradicting its assumed future completeness. Hence $\theta \ge 0$ everywhere. Since null expansion obeysevery horizon area element, and therefore every complete cross-section area, is nondecreasing.

Decompose the optical tensor on the two-dimensional screen space asThe null expansion, null shear, and null twist are respectivelyThe shear is symmetric and trace-free, while the twist is antisymmetric.