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A spherical trapped surface has both future null expansions negative. Thus the initial sphere has $r_{,U}<0$ and $r_{,V}<0$. The result of the previous part already gives $r_{,U}<0$ at every regular future sphere in $D^{+}(\Sigma )$.

Apply the same Null Raychaudhuri equation to $k=f^{-1}{\partial }_V$. The exchanged-coordinate double-null focusing identity isAlong $U=U_0$, increasing $V$ is future-directed. ThereforeBoth null expansions remain negative, proving persistence of spherical trapping along this future null direction. As usual, the conclusion concerns every $V_1>V_0$ for which the sphere exists in the regular chart in $D^{+}(\Sigma )$; the inequality does not assert existence of such spheres beyond a focusing singularity.

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.

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.

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 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.

Commuting covariant derivatives and using $U^b{\nabla }_b{U}_a=0$ gives the optical evolution equationTaking its screen trace yieldsThe optical decomposition and the symmetry or antisymmetry of its pieces implyThus the four-dimensional Null Raychaudhuri equation is

The hypersurface generators have zero null twist. Contracting the Einstein field equations twice with the null tangent removes the trace term, and the null energy condition gives $R_{ab}{U}^a{U}^b=8\pi T_{ab}{U}^a{U}^b\ge 0$. Since the squared null shear is nonnegative, the Null Raychaudhuri equation impliesWhile $\theta <0$ this is equivalent toStarting from $\theta (0)={\theta }_0<0$, the right-hand side reaches zero no later than $\lambda =2/|{\theta }_0|$. The reciprocal expansion must therefore vanish andThis is the null focusing theorem.

The Physical-process first law for a rotating black hole states that a small flux of matter through an initially and finally stationary horizon obeysin units $G=1$. Let $k^a$ be an affinely parametrized horizon tangent with affine parameter $\lambda =0$ on a bifurcation surface. Constancy of the surface gravity from the Zeroth law of black-hole mechanics and Gaussian null coordinates give the horizon generatorTo first order about a stationary horizon, the squared expansion and shear in the Null Raychaudhuri equation are second order. The Einstein field equations reduce it toThe final stationary condition $\theta (\infty )=0$ givesSince $d(\delta A)/d\lambda =\int \theta dA$, reversing the order of integration yieldsThe stress-energy current from a Killing vector gives the horizon Killing-energy fluxwhich proves the stated law.

The second law of black-hole mechanics, or Hawking's area theorem, states that the total area of future-event-horizon cross-sections cannot decrease toward the future under the null energy condition and the usual predictability assumptions. If any horizon generator had $\theta <0$, the twist-free Null Raychaudhuri equation and the null focusing theorem would force a conjugate point at finite affine parameter. A generator complete to the future cannot pass through such a point while remaining on the achronal boundary of $J^{-}({\mathcal{I}}^{+})$. Hence $\theta \ge 0$ everywhere on the regular future horizon, and $dA/d\lambda =\theta A\ge 0$ proves the area law.

Affine geodesic evolution and the Ricci identity give the screen-projected optical equationTaking the screen trace yieldsThe optical decomposition and the antisymmetry of $\hat{\omega }$ implyTherefore the $D$-dimensional Null Raychaudhuri equation isand the requested constant is $\boxed{A=D-2}$.

For a small amount of matter falling into an initially and finally stationary rotating black hole, the Physical-process first law for a rotating black hole isLet $q^a$ be an affinely parametrized horizon generator, with affine parameter $\lambda =0$ at the background bifurcation surface. The horizon Killing vector field iswhere constancy of $\kappa$ is the Zeroth law of black-hole mechanics. The flux of the conserved Killing current through the horizon is

On the stationary background, expansion and shear vanish. To first order in the perturbation, the quadratic optical terms in the Null Raychaudhuri equation may be dropped, and the Einstein equation givesImpose the teleological final condition $\theta (\infty )=0$. Integration followed by reversal of integration order givesComparison with the Killing-energy flux proves the stated law.

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.

Differentiate $B_{ab}={\nabla }_b{U}_a$ along an affinely parametrized generator. Commuting covariant derivatives and using ${\nabla }_{U}U=0$ gives the optical evolution equationTaking its screen trace and usingproduces the Null Raychaudhuri equationThe generators lie in a null hypersurface, so they are hypersurface orthogonal. The Frobenius theorem therefore gives ${\hat{\omega }}_{ab}=0$, leaving

Contracting the Einstein field equations with the null tangent eliminates the trace term and givesby the null energy condition. The screen metric is positive definite, so ${\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}\ge 0$. The Null Raychaudhuri equation consequently impliesWhile $\theta <0$ this is equivalent toIf $heta({\lambda }_0)={\theta }_0<0$, integration givesThe right-hand side reaches zero after affine distance $2/|{\theta }_0|$. A finite negative expansion cannot pass through this value, so $heta\to -\infty$ no later than that point. This is the null focusing theorem.

Let $k^a=(d/d\lambda {)}^a$ be an affine horizon generator, normalized so that the horizon Killing field is ${\xi }^a=\kappa \lambda k^a$. To first order in the small total injected energy, expansion and shear-squared terms are second order, and the Null Raychaudhuri equation on the horizon becomesThe flux of Killing energy through all the shells isSubstitute the linearized focusing equation and integrate by parts. Stationarity before and after the process makes the endpoint term $\lambda \theta$ vanish, while $dA/d\lambda =\theta A$ to first order. ThereforeUsing the Hawking temperature $T_H=\kappa /(2\pi )$ and Bekenstein-Hawking entropy $S_{BH}=A/4$ gives the Physical-process first law of black-hole mechanicsThe derivation is linear in the stress tensor, so separated shells simply contribute additively.

The physical-process first law derives the area change caused by a small flux of matter through an initially and finally stationary horizon. Linearized Null Raychaudhuri equation evolution converts the Killing-energy flux into $\Delta M=\kappa \Delta A/(8\pi )$.

Under the null energy condition and the usual global assumptions excluding naked pathologies, the area of every cross-section of a future event horizon is nondecreasing toward the future. The Null Raychaudhuri equation and future completeness rule out negative expansion of its generators.