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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.

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.

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.