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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.

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.

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.

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.

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.

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.

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.

For a four-dimensional spherical metric $d{s}^2=-fd{v}^2+2dvdr+r^{2}d{\Omega }^2$, a radial null vector normal $K$ to a round sphere has null expansion ${\theta }_K=\frac{1}{2}{q}^{AB}{\mathcal{L}}_K{q}_{AB}=2K(r)/r$, where $q_{AB}=r^2{\gamma }_{AB}$. The angular metric changes only by a common scale, so radial congruences have zero null shear; hypersurface orthogonality also gives zero null twist. The convention here is the trace, not the screen-averaged expansion.

For a four-dimensional spherically symmetric metric $d{s}^2=-2fdUdV+r^{2}d{\Omega }^2$, the affine null gradients are $f^{-1}{\partial }_V$ and $f^{-1}{\partial }_U$. Their null expansions are $2r_{,V}/(fr)$ and $2r_{,U}/(fr)$. Their null shear and null twist vanish because the transverse metric changes only by a common scale and the covectors are exact gradients.