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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.

The stationary Killing vector field is $k={\partial }_t={\partial }_v$, with $k^2=-f$. The normal to $r=\text{constant}$ has squared norm $g^{rr}=f$, so $r=r_{+}$ is a null hypersurface; there $k$ is both tangent and normal. It is therefore a Killing horizon.

For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. Since $r_{+}^2{r}_{-}^2=Q^2$ and $r_{+}^2+r_{-}^2=2M$,For the extremal black hole $M=Q$, the horizons coincide and $\kappa =0$.

On the null hypersurface, the generator covector is proportional to a normal, locally $U_a=f{\nabla }_{a}u$ for a level-set function $u$. The Frobenius theorem therefore givesContracting once with $N^a$ and projecting the remaining indices with $P^a{}_b$ removes every term containing $U$ or $N$ and leaves $P_b{}^d{P}_c{}^e{\nabla }_{[d}{U}_{e]}=0$. Henceon the hypersurface.

Substituting the two differential coordinate transformations cancels every coefficient singular as ${\Delta }^{-1}$, so the ingoing Kerr-like coordinates are regular at $r=r_{+}$. The inverse metric applied to the normal covector $dr$ hasIts norm is $g^{rr}=\Delta /\Sigma$, which vanishes at $r=r_{+}$. Thus that level set is a null hypersurface. On it the raised normal is proportional toSince the transformed stationary and axial Killing fields are $K={\partial }_v$ and $m={\partial }_{\chi }$, the horizon is a Killing horizon generated by

The normal to $r=r_{+}$ is $dr$, whose squared norm is $g^{rr}=f(r_{+})=0$, so the surface is a null hypersurface. In ingoing coordinates $K={\partial }_v$, and on the horizonThus $K$ is both tangent and normal there, making the surface a Killing horizon. For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. With $n=d-3$,

The normal to a surface of constant $r$ is $n_a=(dr{)}_a$, and the inverse ingoing metric givesHence $r=r_{+}$ is a null hypersurface. LetThis constant linear combination of Killing vector fields is Killing. Its squared norm iswhich vanishes at $r_{+}$. More strongly, lowering its index in the ingoing metric gives ${\xi }_a=(dr{)}_a$ on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon, the squared term contributes no first derivative at $r_{+}$ andDifferentiating $f$ at its simple outer root givesso the surface gravity isIt vanishes in the extremal case $r_{+}=r_{-}$.

Locally write the null hypersurface as a level set $u=0$ and its normal as $n_a={\nabla }_{a}u$. Since a null normal is also tangent, $n^a$ lies within the hypersurface. The symmetry of the Levi-Civita connection givesThe scalar $n^2$ vanishes on the hypersurface, so its gradient there is normal and hence proportional to $n_a$:This is the nonaffine geodesic equation. A rescaling of $n$ removes $\kappa$, proving that the null hypersurface normal generates null geodesics lying in the hypersurface.

The normal to a surface of constant $r$ has squared normwhich vanishes at $r=r_{+}$. Thus this surface is a null hypersurface. The stationary Killing vector fieldhas $K^2=g_{tt}=-f$, so it becomes null there and generates a Killing horizon. Since $f$ has a simple zero, its surface gravity isA horizon cross-section has topology $S^3\times S^1$, where the circle is the periodic $z$ direction. Including a complete generator, the null hypersurface has topology

Choose a spacelike two-dimensional cross-section $S$ of the null hypersurface $\mathcal{N}$ and coordinates $y^i$ on $S$. Extend $y^i$ along the null generators, and choose their parameter $\lambda$ to be affine, so $U={\partial }_{\lambda }$ on $\mathcal{N}$. At every point choose the other null normal $L$ with $g(U,L)=1$, shoot out the affinely parametrized null geodesic with tangent $L$, and call its affine parameter $r$. Transport $(\lambda ,y^i)$ along those transverse geodesics.

This construction gives Gaussian null coordinates. The hypersurface is $r=0$, and the coordinate conditions implyBecause $\lambda$ is affine on the generators, ${\partial }_r{g}_{\lambda \lambda }{|}_{r=0}=0$. Smoothness then factors the remaining components as $g_{\lambda \lambda }=r^{2}F$ and $g_{\lambda i}=r{h}_i$, giving

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