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The Bondi--Sachs formalism uses outgoing null hypersurfaces to describe asymptotically flat radiating spacetimes.

For $0<|Q|<M$, the charged-particle radial equation in Reissner-Nordstrom spacetime implies that any radial turning point obeysSince $1+l^2/r^2>0$, it follows that $F\ge 0$. Thus no massive test particle, regardless of its electric charge, can have a radial turning point in $r_{-}<r<r_{+}$, where $F<0$. At a regular horizon crossing, a timelike vector cannot lie in the tangent space of a null hypersurface, so a massive particle has $\dot{r}\ne 0$ there as well. The algebraic inequalities $r\ge r_{+}$ or $r\le r_{-}$ are necessary conditions, not an existence theorem for radial turning points.

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.

A regular hypersurface $h=0$, with $dh\ne 0$, is a null hypersurface if its normal is null:The raised normal is then also tangent to the hypersurface, and its induced metric is degenerate. This is a local geometric property.

In an asymptotically flat spacetime, choose its future null infinity ${\mathcal{I}}^{+}$. The black hole region and future event horizon areHere $J^{-}({\mathcal{I}}^{+})$ is the causal past of future null infinity: the region from which a future-directed causal curve can escape to arbitrarily distant observers. An event horizon therefore depends on the global future causal structure, and an arbitrary null hypersurface need not be an event horizon.

In the extremal case, the squared norm of the Killing vector field isFor $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended spacetime. ThusThe static chart cannot itself test the geometry at its horizon. In the regular Eddington-Finkelstein coordinates, the normal to $h=r-M=0$ is $n=dr$, andAt $r=M$ this gives $n^{♯}={\partial }_v=K$, a nonzero null vector tangent to the surface. Therefore $r=M$ is a null hypersurface, even though $F$ has the same sign on its two sides.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

Take the expanding flat branch $a(t)=A{e}^{H_{\Lambda }t}$, with $A>0$. A radial null geodesic in FLRW spacetime satisfies $d\rho /dt=\pm 1/a(t)$. For the incoming past light cone of the spacetime event $(t_0,\rho =0)$, integration giveswith arbitrary angular direction. Equivalently, conformal time $\eta =-1/(H_{\Lambda }a)$ gives $\rho ={\eta }_0-\eta$.

Let the reception time tend to infinity. The largest comoving radial distance from which a signal emitted at time $t$ can reach the central comoving observer isThus the cosmological event horizon has constant proper distance in cosmology:An emitted spacetime event is seen at a finite time if and only if $a(t)\rho <H_{\Lambda }^{-1}$. Equality would require $t_0=\infty$, while an event with $a(t)\rho >H_{\Lambda }^{-1}$ never reaches the observer. This is a bound on events at their emission time; a fixed comoving observer can have earlier visible events and later invisible ones.

The boundary really is a null hypersurface. For $f(t,\rho )=\rho -[H_{\Lambda }a(t){]}^{-1}$, its normal has components ${\partial }_{t}f=1/a$ and ${\partial }_{\rho }f=1$, soThe finite future integral, and hence this observer event horizon, uses the expanding branch. For the contracting flat solution $a=A{e}^{-H_{\Lambda }t}$, the same integral diverges and there is no future observer event horizon of this kind; the static zero-density solution likewise has none.

On the extension in part (i), the hypersurface $\mathcal{N}$ has nonzero normal covector $n_a={\partial }_a\lambda$. Its squared norm isThis vanishes on $\mathcal{N}$, which is therefore a null hypersurface. Raising the index givesSince ${\partial }_t={\partial }_v$ at fixed $\lambda$, the original Killing vector field becomes the smooth nonzero field $K={\partial }_v$. The transformed metric tensor is independent of $v$, so $K$ remains a Killing vector field, andThus $K$ is both tangent and normal to $\mathcal{N}$. This is exactly a Killing horizon:The Killing vector field itself does not vanish there; a vanishing squared norm means nullness, not a zero vector.

A regular hypersurface $f=0$, with $df\ne 0$, is a null hypersurface if its normal covector has zero squared norm:Its normal vector is then also tangent to the hypersurface; the induced metric tensor is degenerate.

For the Kerr metric the normal to a constant-$r$ surface has squared normFor $0<a<M$ the two roots are positive and distinct. To verify that they are genuine regular null hypersurfaces, rather than trusting singular Boyer-Lindquist coordinates, introduce ingoing Kerr coordinatesSubstitution givesThere are no ${\Delta }^{-1}$ terms. At either root $\Sigma >0$, and the determinant is $-{\Sigma }^2{\sin }^2\theta$ away from the angular coordinate poles. Thus the metric tensor extends smoothly across each root. The raised normal on either surface isThis is a nonzero null Killing vector field normal to the surface. Both surfaces are Killing horizons, with horizon angular velocities ${\Omega }_{\pm }$.

The event-horizon terminology in the paper needs qualification. In the usual asymptotically flat Kerr black hole extension, relative to a chosen exterior's future null infinity, the future branch of $r=r_{+}$ is the event horizon, the boundary $\partial J^{-}({\mathcal{I}}^{+})$. In the intervening black-hole region $\Delta <0$, the coordinate $r$ has timelike gradient and future-directed curves move toward smaller $r$; they cannot return across the outer boundary to that exterior. The inner surface $r=r_{-}$ is a Cauchy horizon, not a second event horizon for the same asymptotic end. Nullness is a local property; identifying an event horizon additionally requires this global causal structure.^[ref]

The allowed endpoint $a=0$ provides an especially concrete counterexample to the unrestricted description of both roots. The Kerr metric then becomes Schwarzschild spacetime, $r_{+}=2M$ is its regular event horizon, but $r_{-}=0$ is a curvature singularity:It is not a regular inner null hypersurface. Thus the two-horizon calculation above assumes $0<a<M$.

The inverse of the $(v,\lambda )$ metric tensor block isThe normal covector $d\lambda$ has squared norm ${\lambda }^2/m^2$. ThereforeMore explicitly, the induced metric tensor is $-{\lambda }_0^{2}d{v}^2+m^{2}d{\Omega }_2^2$; at zero its sole zero direction is ${\partial }_v$. The coefficients are independent of $v$, so $K={\partial }_v$ is a Killing vector field. On $\lambda =0$,Hence $K$ is both tangent and normal to this null hypersurface, proving it is a Killing horizon. The definition of surface gravity is ${\nabla }_{K}K=\kappa K$ on the Killing horizon, or equivalently $d(K^2)=-2\kappa K^{♭}$. Since $d(K^2)=-2\lambda d\lambda$ vanishes there while $K^{♭}\ne 0$,Thus this is a degenerate Killing horizon, and $v$ is an affine parameter on each of its null geodesics.

For a concrete horizon frequency comparison, fix the same angular position for a narrow high-frequency light beam and two freely falling radial observers. Work in geometric optics, with the beam's affinely transported wavevector tangent to the Killing horizon, $k=qK$. Here $q>0$ is constant because ${\nabla }_{K}K=0$. Choose the observers as translates of one timelike geodesic by the isometry generated by $K$, crossing at $v_1$ and $v_2$. Give them the same positive Killing energy per unit rest mass, $E=-K\cdot W$. At their crossings their normalized four-velocities havewhich satisfy $g(W,W)=2m{W}^v{W}^{\lambda }=-1$. They can be realized by local geodesics with precisely these initial data. The measured angular frequency isTo connect this measurement to surface gravity, if ${\nabla }_{K}K=\kappa K$ with constant $\kappa$, affine transport of $k=q(v)K$ would give $dq/dv=-\kappa q$. For observers related by the same Killing vector field translation, $-K\cdot W$ agrees at the crossings, giving ${\omega }_2/{\omega }_1=e^{-\kappa (v_2-v_1)}$. Here there is no accumulated redshift along the Killing horizon for these equally prepared observers. Arbitrary observers with different four-velocities can still measure different frequencies through their different Killing energies.

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.