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Gaussian null coordinates adapt one coordinate to the generators of a null hypersurface, one affine coordinate to transverse null geodesics, and the remaining coordinates to a spacelike cross-section. If the generator parameter is affine, the metric can locally be written

A Killing horizon is a null hypersurface whose null generators are orbits of a Killing vector field $K$. On the horizon, $K^b{\nabla }_b{K}^a=\kappa K^a$ defines its surface gravity.

A null coordinate is a coordinate function $u$ whose gradient is null, $g^{ab}{\partial }_{a}u{\partial }_{b}u=0$. Its level sets are null hypersurfaces. Pairs of such coordinates straighten radial null directions in two-dimensional orbit metrics; a monotone reparametrization of either function preserves its null level sets. Examples include the retarded and advanced null coordinates and the radial Kruskal–Szekeres coordinates.

Hawking's area theorem states that the area of later cross-sections of a classical future event horizon cannot be smaller than that of earlier cross-sections, including the sum over initially disconnected components. Its hypotheses include the null energy condition, the Einstein equations, and the global regularity/predictability conditions that exclude future naked pathologies of the horizon generators. A convenient sufficient version assumes those generators are future complete. Without the energy and global hypotheses the assertion is not unconditional.

Let $k^a$ be an affinely parametrized null generator, with expansion $\theta$. Because the horizon is a null hypersurface, the generator congruence has zero null twist. The Null Raychaudhuri equation in four dimensions givesThe null shear square is nonnegative and the null energy condition makes $R_{ab}{k}^a{k}^b=8\pi G{T}_{ab}{k}^a{k}^b\ge 0$; a cosmological term has zero null contraction. Suppose $\theta ({\lambda }_0)<0$. Integration of this inequality forces $\theta \to -\infty$ within affine distance at most $2/|\theta ({\lambda }_0)|$. This produces a focal point to the initial horizon cross-section. Beyond it the null geodesic cannot remain on an achronal boundary, whereas an event horizon is precisely such a boundary. Future completeness, or the corresponding standard predictability condition ensuring continuation over this interval, gives the contradiction. HenceThis proves the local area increase along every smooth generator bundle. Generators can enter the horizon at past crease points, but cannot leave through future endpoints on a regular horizon under these hypotheses. New generators add area, rather than subtracting it, so the result extends across the nonsmooth merger set and to whole horizon cross-sections. This is the future-complete horizon focusing proof of the area theorem.

Choose an early cross-section with the two separate components of areas $A_1,A_2$, and a later cross-section of the settled merged hole. Following the old generators forward maps their area elements injectively into the later horizon; a future crossing/focal point would violate the same achronality argument. Each element has nondecreased area. Any additional generators contribute nonnegative area, givingThis is a comparison of event-horizon cross-sections, not an assertion that set containment alone bounds the areas of arbitrary apparent horizons.

For a Schwarzschild black hole, $r_h=2Gm/c^2$ and $A=4\pi r_h^2=16\pi G^2{m}^2/c^4$. Applying the merger inequality to the two equal initial masses and the final Schwarzschild mass givesFor initially well-separated holes at rest, with no extra incoming energy or external work, energy conservation gives $E_{\mathrm{rad}}=(2m-M)c^2$. The area bound on equal-mass Schwarzschild merger radiation is consequentlyThis is the maximum permitted by the area/energy inequalities, not a demonstration that a dynamical merger attains it. A reversible limiting area change would be needed to saturate the bound. If initial separation, binding energy or external agents change the initial total energy, replace $2m{c}^2$ by that energy when computing escaped radiation. Quantum evaporation does not contradict the theorem: the classical energy-condition hypotheses are not retained unchanged in that setting.

Let $f(r)=1-2M/r+Q^2/r^2$. A positive-radius horizon requires a root of $r^2-2Mr+Q^2=0$, givingFor the usual asymptotically flat Reissner-Nordstrom spacetime, the black-hole conditions areThe strict inequality gives a nonextremal outer event horizon at $r_{+}$; equality gives an extremal black hole. If $M<|Q|$ with $M>0$, there is no horizon and the central curvature singularity is naked. Nonpositive $M$ cannot supply a positive-radius black-hole horizon; $M=Q=0$ is Minkowski spacetime. For $Q=0$, the zero root $r_{-}=0$ is the singularity, not an additional regular horizon.

In the static coordinates the vector $k={\partial }_t$ has constant components and every metric component is independent of $t$. Its Lie derivative therefore satisfiesFor the torsion-free Levi-Civita connection, ${\mathcal{L}}_{k}g=2{\nabla }_{(a}{k}_{b)}$, proving the Killing equation directly. This also normalizes the Killing vector field to unit time translation at infinity.

At $r=r_{+}$, $k^2=-f=0$. Use the regular advanced chart $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, in whichOn the horizon, $k_a=(dr{)}_a$. Thus the Killing vector field is both null and normal to the horizon, and its orbits generate that Killing horizon. The normal lines of a null hypersurface are null geodesics, possibly with nonaffine parameter. To see this without assuming a separate geodesic property, the Killing equation givesRestriction to the future horizon therefore givesThe acceleration is proportional to the tangent, which is exactly the geodesic equation with a nonaffine parameter. An affinely parametrized tangent can be obtained by rescaling $k$ along each generator; that rescaled vector need not itself extend to a spacetime Killing vector field. At a nonextremal bifurcation surface, $k$ vanishes, so a nonzero generator tangent must be described in a regular affine parametrization there.

On the past branch, use the retarded chart with radial-time cross term $-2dudr$. Now $k_a=-(dr{)}_a$ at the horizon, so the same calculation yields the opposite signed nonaffinity. Since $r_{+}$ is constant and the spacetime is spherically symmetric, $f^{\prime }(r_{+})/2$ is constant over the future horizon. This gives the two signs in the horizon relation without assuming their value beforehand.

Factor $f=(r-r_{+})(r-r_{-})/r^2$. Its derivative at the outer horizon gives the Reissner-Nordstrom horizon surface gravityIt is positive for $M>|Q|$, and at $Q=0$ reduces to the Schwarzschild surface gravity $1/(4M)$. For $M=|Q|>0$, the root is double and $\kappa =0$. Thus the horizon remains present when its surface gravity vanishes: it is a degenerate Killing horizon, and $k$ itself has affine parametrization along its nonzero horizon generators. The strictly positive-constant description applies only to the nonextremal case; the zero value is the legitimate extremal limit.

In advanced coordinates, $K={\partial }_v$ is a Killing vector field and $K^2=-f$. The normal covector to $r=2m$ is $dr$; raising it gives $g^{ar}{\partial }_a={\partial }_v+f{\partial }_r$, which equals $K$ on the horizon. Thus $r=2m$ is a null hypersurface normal to $K$, and hence a Killing horizon.

The surface gravity is defined by ${\nabla }_{K}K=\kappa K$ on this horizon, equivalently ${\nabla }_a(K^2)=-2\kappa K_a$. Here $K_a=(-f,1,0,0)$, so the radial component gives $\kappa =f^{\prime }(2m)/2$. Because $F(2m)=0$,It is positive for all finite real $\alpha$ and $m>0$; the stationary Killing vector field retains its asymptotic normalization.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

Assume $M>0$ and use geometric units. The stationary Killing vector field $K={\partial }_t$ gives the conserved Killing energy $E=-g(K,U)=f\dot{t}$. For radial motion, normalization of the four-velocity gives $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$. Combining these identities yieldsAt rest at infinity, $f\to 1$ and $\dot{r}\to 0$, so $E=1$. Choose the inward branch $\dot{r}=-\sqrt{2M/r}$. Its proper time from the Schwarzschild event horizon to the curvature singularity isIn ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the spacetime. This is the Schwarzschild horizon-to-singularity proper time for fall from infinity, rather than a universal time for all infall energies.

For this geodesic congruence, set $w(r)=\sqrt{2M/r}$. Its four-velocity and metric dual areThe right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-{\partial }_a\tau$, with free-fall proper time in Schwarzschild spacetime determined byIndeed $U^a{\partial }_a\tau =-U^a{U}_a=1$, so this scalar advances by the actual proper time on every member of the infalling geodesic congruence, with a synchronized choice of additive offsets.

Substitution of $dt=d\tau -w{f}^{-1}dr$ in the Schwarzschild metric givesUsing $f=1-w^2$, the Painlevé–Gullstrand coordinates therefore haveAll coefficients are finite at $r=2M$; the determinant of the $\tau ,r$ block is $-f-w^2=-1$. Hence the horizon is a coordinate singularity of Schwarzschild time, not a degeneracy of the metric. The new spatial slices have $d{\ell }^2=d{r}^2+r^{2}d{\Sigma }^2$, the flat Euclidean metric, and $g^{ab}{\partial }_a\tau {\partial }_b\tau =-1$ makes their normals the freely falling four-velocities. By contrast, the curvature scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.

For Ingoing Eddington-Finkelstein coordinates, use $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, for example $r_{\ast }=r+2M\log |r/(2M)-1|$. ThenBoth constructions cross the future horizon. Their time functions have different meanings: $v=\text{constant}$ is an ingoing null hypersurface, whereas $\tau =\text{constant}$ is a flat spacelike hypersurface orthogonal to the infallers. Thus the former is adapted to ingoing null geodesics, and the latter to a timelike free-fall clock congruence.

For the Kruskal diagram, choose Kruskal–Szekeres coordinates with $U_K=-e^{-(t-r_{\ast })/(4M)}$, $V_K=e^{(t+r_{\ast })/(4M)}$ in the right exterior and extend them across its horizon. They obeyThe four regions of Kruskal spacetime are I, the right exterior $(U_K<0,V_K>0)$; II, the future black hole $(U_K>0,V_K>0)$; III, the other exterior $(U_K>0,V_K<0)$; and IV, the past white hole $(U_K<0,V_K<0)$. The null axes are horizons, their intersection is the bifurcation two-sphere, and the spacelike boundaries $U_K{V}_K=1$ are the future and past curvature singularities. A point in this radial diagram represents a symmetry two-sphere.

The domain of the ingoing Painleve-Gullstrand chart is region I together with region II and their common future horizon. More precisely, it is $V_K>0$, $U_K{V}_K<1$, or $r>0$ with finite $\tau$. To verify the boundary rather than infer it from the locally regular metric, put $\rho =\sqrt{r/(2M)}$ and writeThis follows by subtracting $r_{\ast }$ from the preceding radial primitive, and $S$ is finite at $r=2M$. Thus $V_K=\exp [(\tau -S(r))/(4M)]>0$ throughout the chart. Conversely every point with $V_K>0$ and $r>0$ gives a finite $\tau$. The past horizon $V_K=0$, the bifurcation sphere, region III and region IV are absent. Reversing the sign of the radial term produces the outgoing chart through the white hole instead.

Put $r_{+}=\ell \sqrt{M}$ and $f=(r^2-r_{+}^2)/{\ell }^2$. The apparent failure of the nonrotating BTZ black hole coordinates at $r=r_{+}$ is a coordinate singularity. Define the tortoise coordinate and ingoing time byThe Ingoing BTZ coordinates giveAll coefficients are analytic at $r_{+}>0$, and the determinant is $-r^2$, so the metric tensor extends nondegenerately through that radius. The Killing vector field $K={\partial }_v$ agrees with ${\partial }_t$ in the exterior. On $r=r_{+}$, $K^2=-f=0$ and $K_a=(dr{)}_a$. It is therefore both normal and tangent to this null hypersurface, proving that it is a Killing horizon.

One may include both horizon branches and their bifurcation surface explicitly. Set $\kappa =r_{+}/{\ell }^2$, $u=t-r_{\ast }$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. ThenThis analytic extension of a spacetime is regular near $UV=0$, and $K=\kappa (V{\partial }_V-U{\partial }_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the extended Killing horizon at $r=r_{+}$; $K$ vanishes only at their regular bifurcation circle.

An event horizon is a null hypersurface. For its spherical defining function $H(v,r)=r-r_H(v)$, the inverse radial metric tensor has $g^{vv}=0$, $g^{vr}=1$ and $g^{rr}=f=1-2m(v)/r$. The normal's null condition isIt follows that the event horizon follows the outgoing radial null geodesics of the Vaidya metric:This ordinary differential equation also describes other outgoing null hypersurfaces; the global event horizon is selected by its future boundary condition, not by this local equation alone.

Locally write the null hypersurface as $S=0$ with $dS\ne 0$, and extend its normal as $n_a={\nabla }_{a}S$. On the hypersurface, $n^a{n}_a=0$, so $n^a{\nabla }_{a}S=n^2=0$: the raised normal is itself tangent. The Levi-Civita connection and symmetry of the scalar Hessian giveAlthough $n^2$ vanishes on the hypersurface, it need not vanish off it. Its tangential derivatives vanish, so ${\nabla }_a(n^2)$ must be proportional to the normal $n_a$. Hence ${\nabla }_{n}n=\kappa n$ on the hypersurface for some smooth local coefficient $\kappa$. The normal's integral curves of a vector field remain in the hypersurface and satisfy the nonaffine geodesic equation. Rescale to $k=hn$, where $n(\log h)=-\kappa$ along each generator. Direct substitution gives ${\nabla }_{k}k=0$. Thus the normal generates null geodesics within the hypersurface, with an affine parameter after this rescaling. This proves the null hypersurface normal generates null geodesics property without incorrectly setting a transverse derivative of $n^2$ to zero.

A Killing horizon is a null hypersurface with a Killing vector field $\xi$ as its normal and generator. Write $\xi =d/d\alpha$ along its orbits. These are null geodesics, but their Killing parameter need not be affine:The coefficient is the surface gravity for the chosen normalization; nondegeneracy means $\kappa \ne 0$. If $\lambda$ is affine and $q=d\lambda /d\alpha$, then $\xi =qd/d\lambda$ and $d(\log q)/d\alpha =\kappa$. For constant $\kappa$, $\lambda ={\lambda }_0+C{e}^{\kappa \alpha }$.

The Killing equation yields ${\partial }_{\mu }({\xi }^2)=2{\xi }^{\nu }{\nabla }_{\mu }{\xi }_{\nu }=-2{\xi }^{\nu }{\nabla }_{\nu }{\xi }_{\mu }$. On the horizon, $\boxed{{\partial }_{\mu }{\xi }^2=-2\kappa {\xi }_{\mu }}$. Multiplying $\xi$ by a positive constant multiplies $\kappa$ by that constant; physical temperature therefore needs a stated time normalization.

Locally write the null hypersurface as $u=0$, $du\ne 0$, with normal ${\ell }_{\mu }=f{\partial }_{\mu }u$, $f\ne 0$. A vector $V$ is tangent exactly when $V(u)=0$. Nullness gives $\ell (u)={\ell }^2/f=0$ on the hypersurface. Hence $\boxed{\ell \in T\mathcal{N}}$: its normal direction lies inside its tangent space.

Put $n=du$ and $q=n_{\mu }{n}^{\mu }$. Torsion freedom gives $n^{\mu }{\nabla }_{\mu }{n}_{\nu }=\frac{1}{2}{\partial }_{\nu }q$. Since $q$ vanishes on $\mathcal{N}$, all tangential derivatives vanish there and $dq=Cdu$ there for a scalar $C$. ConsequentlyThis is the unparametrized null geodesic equation. The integral curves are therefore generators of $\mathcal{N}$. An affine rescaling of a null normal removes the proportionality coefficient locally: if ${\nabla }_{\ell }\ell =\kappa \ell$, set $K=h\ell$ with $\ell (\log h)=-\kappa$.

Locally write the null hypersurface as $F=0$. Its generators are tangent to its raised normal, so on it $U_a=h{\nabla }_{a}F$ for a nonzero scalar $h$. The antisymmetric derivative issince the Hessian of $F$ is symmetric for the torsion-free Levi-Civita connection. Each term contains $\nabla F$, which is proportional to $U$ and is killed by the screen-space projector. HenceThis is the null version of hypersurface orthogonality in the Frobenius theorem. A null hypersurface has no independent normal direction outside its tangent space: its null vector normal also generates it. That is why the same argument applies to the generators' null twist.

The inverse Kerr metric in ingoing Kerr coordinates gives the raised normal to a surface of constant $r$:On $r=r_{+}$, $\Delta =0$, so the normal is null and tangent to the horizon. It reduces toA constant linear combination of the two Killing vector fields is again a Killing vector field. Therefore $k+{\Omega }_{H}m$ is normal to this null hypersurface, establishing that it is a Killing horizon, with Kerr horizon angular velocityThe second equality uses $\Delta (r_{+})=0$. This normal calculation proves hypersurface orthogonality on the horizon, rather than merely proving that the proposed vector happens to have zero norm there.

Locally write a smooth hypersurface as $\Phi =0$ with $n_a={\nabla }_a\Phi \ne 0$. It is a null hypersurface when $n_a{n}^a=0$ on it. Then $n^a$ is both normal and tangent, and the induced metric is degenerate along $n^a$.

Because covariant derivatives commute on a scalar,The scalar $n^2$ vanishes on the hypersurface, so its derivative annihilates every tangent direction there. Its derivative is consequently proportional to $n_a$: locally $n^2=\Phi q$ for a smooth function $q$, givingThus the normal generates null pregeodesics. Rescale $k^a=\alpha n^a$, choosing $n^b{\nabla }_b\log \alpha =-\kappa$, to get $k^b{\nabla }_b{k}^a=0$. These are the affinely parametrized generators of the null hypersurface. The proportionality need not vanish: setting $n^2=0$ only on the hypersurface does not set its full transverse derivative to zero.

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.