The subspace spanned by $e_4,e_5,e_6$ is the center of a Lie algebra. Each of their tensor squares is therefore invariant. On tensor products the infinitesimal Adjoint representation of a Lie algebra acts by a derivation, $D_x(u\otimes v)=[x,u]\otimes v+u\otimes [x,v]$. Write $u\odot v=u\otimes v+v\otimes u$ and let $S=e_1\odot e_5+e_3\odot e_6-e_2\odot e_4$. The only potentially nonzero checks areThe remaining generators are central, so they act trivially. This proves every claimed tensor invariant.

These tensors are contravariant. To obtain an ordinary metric tensor, take an invertible linear combination and invert it. For arbitrary real $a_4,a_5,a_6$ and $t\ne 0$, setIn the ordered basis $(e_1,e_2,e_3;e_4,e_5,e_6)$ its matrix isSince $K^{T}K=I$, $\det Q=-t^6\ne 0$, andLet ${\lambda }^a$ be the dual left-invariant coframe. The resulting neutral invariant metrics on the free two-step Lie algebra on three generators areHere $({\lambda }^a{)}^2={\lambda }^a\otimes {\lambda }^a$. Infinitesimal invariance of $Q$ implies that its inverse is an adjoint-invariant bilinear form. Its left-translated metric is right-invariant on a connected $G$, giving the desired four-parameter family of bi-invariant metrics. A change of variables removes the central $A$ block from the quadratic form, leaving three off-diagonal two-dimensional blocks. Each has one positive and one negative eigenvalue. Thus the metric signature is $(3,3)$; these are bi-invariant pseudo-Riemannian metrics, not positive-definite ones.

Connectedness, conventionally understood here, matters for the passage from infinitesimal to full group invariance. If $G$ is disconnected, its other components must preserve $Q$ separately. For example the Lie algebra automorphism reversing $e_1,e_4,e_6$ and fixing $e_2,e_3,e_5$ preserves all brackets but sends $S$ to $-S$. In the semidirect product of the simply connected group with this order-two automorphism, none of these nondegenerate $t\ne 0$ tensors is invariant under that component. Thus the Lie-algebra data alone give the stated family on the identity component, rather than an unconditional extension to every disconnected realization.

Take $M=|Q|>0$ and use $G=c=1$. The case $M=Q=0$ is flat spacetime and is not a black hole. In the nontrivial extremal Reissner-Nordstrom metric,The horizon is a double zero at $r=M$, with surface gravity $\kappa =f^{\prime }(M)/2=0$. The function $f$ is positive on both sides of the horizon. Thus the intermediate nonstatic region of a nonextremal charged hole disappears; one must not draw that region as a strip of nonzero width in the extremal Penrose diagram.

To construct the Penrose diagram, first integrate the radial tortoise coordinate:The additive constant has been chosen so that $r_{\ast }(0)=0$ in the inner block. In the exterior, $r_{\ast }\to -\infty$ at $M^{+}$ and $r_{\ast }\to +\infty$ at infinity. In the inner static region $0<r<M$, it increases from $0$ to $+\infty$. Set $u=t-r_{\ast }$ and $v=t+r_{\ast }$. The radial metric is $-fdudv$, so its radial null directions are $u=\mathrm{constant}$ and $v=\mathrm{constant}$.

Compactify each static block usingThis conformal compactification puts radial null geodesics at $45$ degrees. The exterior is a diamond: its right-hand edges are past and future null infinity, while its left-hand edges are the past and future degenerate horizons. At the left vertex, finite static $t$ and $r\to M^{+}$ give $u\to +\infty$, $v\to -\infty$. This is an end of the infinite spatial throat, not a regular bifurcation surface where the two horizon branches meet. A degenerate Killing horizon has no such bifurcation sphere. The inner block occupies the half-diamond $X>0$, because $r_{\ast }>0$ there. Its vertical edge $X=0$ is the timelike curvature singularity $r=0$; its two diagonal edges are the two horizon branches at $r=M$. Its outer vertex likewise represents an excluded infinite-throat end.

The static-coordinate compactification identifies the blocks but does not by itself establish the smooth horizon extension. For that, use ingoing Eddington-Finkelstein coordinates on the future horizon:Its radial determinant is $-1$, so it is nonsingular at $r=M$. Glue the exterior future edge, $u\to +\infty$ with $v$ finite, to the inner past edge, $u\to -\infty$ with the same finite $v$. The outgoing coordinate $u$ similarly gives $d{s}^2=-fd{u}^2-2dudr+r^{2}d{\Omega }^2$ and extends the other horizon branch. Continue adjoining these regular blocks across every extendible null edge. The construction repeats indefinitely to the future and past, producing the maximal analytic diagram with infinitely many asymptotic exterior regions and inner static regions bounded by timelike singularities. Horizons are null edges, never the excluded throat vertices. Identical edges in different copies must be glued with their time orientations preserved; they are not all the same exterior infinity.

The figure shows the two building blocks, rather than compressing the whole infinite extension into a misleading finite diagram. The red edges glue across a future horizon at finite advanced time. Outgoing coordinates extend the next edge to another exterior copy. The thin diagonal lines illustrate the radial light-cone directions.

For spatially conformally flat coordinates, compare the radial and angular parts of $f^{-1}d{r}^2+r^{2}d{\Omega }^2$ with $V(\rho {)}^2(d{\rho }^2+{\rho }^{2}d{\Omega }^2)$. The angular coefficient requires $V\rho =r$, and the radial coefficient then gives, in the exterior,Choose the integration constant to make $\rho =r-M$ and $V\to 1$ at infinity. In Cartesian coordinates $x=\rho \sin \theta \cos \phi$, $y=\rho \sin \theta \sin \phi$, $z=\rho \cos \theta$, put $\rho =\sqrt{x^2+y^2+z^2}$. The Extremal Reissner-Nordstrom isotropic coordinates giveThese coordinates have $\rho >0$ and describe the exterior only. The locus $\rho =0$ is the limit $r=M$, not an ordinary point included in this chart. An independent inner chart is obtained with $\rho =M-r$, $0<\rho <M$, giving $V=M/\rho -1$ and $W=(M/\rho -1{)}^{-1}$; it must be attached using the regular null coordinates, not by continuing the exterior radial coordinate through a Cartesian origin.

The infinite proper distance is real. Along a constant-$t$ exterior radius,It is a property of a static spacelike slice. It does not bound the proper time of a freely falling observer. Indeed, a neutral radial timelike geodesic has conserved energy per unit rest mass $E=f\dot{t}>0$ and normalization $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$, givingAt the horizon $\dot{r}\to -E$, so the remaining proper time is finite. Moreover,The observer crosses the regular future horizon with finite $v$ and finite proper time. This is the Extremal Reissner-Nordstrom causal crossing, which the singular static time coordinate conceals.

There is also a direct causal test for the event horizon. Inside the future horizon, outgoing radial null curves in the ingoing chart satisfy $dr/dv=f/2$. To reach $r=M$ from below they would requireNear the horizon, $M-r\sim 2M^2/v$. A future timelike or nonradial null curve cannot increase $r$ faster than the outgoing radial null curve, since $d{s}^2\le 0$ and $\dot{v}>0$ imply $dr/dv\le f/2$ after dropping the nonnegative angular term. Hence a signal that crossed inward cannot return to the original exterior's future null infinity. Curves can encounter other blocks in the maximal analytic extension, but their infinities are not the original observer's infinity. Relative to that asymptotic end, $r=M$ is the boundary of the causal past of future null infinity: it is an event horizon.

Finally, the Reissner-Nordstrom trapped spheres test explains why the proposed observation is unsurprising. Normalize future radial null normals by $\ell ={\partial }_v+(f/2){\partial }_r$, $n=-{\partial }_r$, with $g(\ell ,n)=-1$. The round-sphere null expansions areThere is no region of strictly future-trapped round spheres, while the horizon spheres are marginal with ${\theta }_{\ell }=0$. Absence of those particular trapped surfaces does not imply absence of an event horizon. The geometry has a nonzero horizon area $4\pi M^2$ and a genuine timelike singularity at $r=0$ in its extension. The infinite static throat does not remove the black hole: infall crosses its horizon in finite proper time, and no signal returns to the same exterior infinity.

Use signature $(-,+,+,+)$ and $c=1$. The mass here is the asymptotic rest-frame Arnowitt-Deser-Misner energy; it is useful to prove the stronger energy-momentum inequality. Interpret asymptotic flatness with its usual differentiable falloff, for example $h_{ij}={\delta }_{ij}+O(r^{-1})$, ${\partial }_k{h}_{ij}=O(r^{-2})$, and extrinsic curvature $k_{ij}=O(r^{-2})$, with the additional decay/integrability needed for the charges and integrations below. The complete nonsingular slice has no inner boundary. An oriented three-dimensional slice admits a spin structure. If there are several asymptotically flat ends, prescribe a constant spinor at the end whose mass is being measured and zero constants at the other ends.

Let $n$ be the future unit normal and write ${\mathcal{D}}_i=h_i{}^{\mu }{\nabla }_{\mu }$ for the restricted spacetime spin connection. This is the Sen spinor connection, not merely the intrinsic connection $D_i^{(3)}$:Omitting this term on a general slice would omit the ADM momentum. Choose ${\gamma }^{\hat{0}}$ anti-Hermitian, the spatial gamma matrices Hermitian, and $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$. The commuting-spinor Dirac current $V^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$ is future causal and $V^{\hat{0}}=|ϵ{|}^2$.

The Nester two-form supplies the needed integration identity. DefineHere multiple gamma indices denote antisymmetrized products. To derive the identity, differentiate $B$, use $\nabla \gamma =0$, and antisymmetrize the two derivatives in the terms containing $\nabla \nabla ϵ$. Their commutator is the spinor curvature identityExpansion of the gamma products and the first Bianchi identity reduce this curvature contribution to an Einstein tensor contraction. It vanishes for the Vacuum Einstein equations. At a point in a normal-adapted orthonormal frame the remaining contraction isbecause ${\gamma }^i{\gamma }^j={\delta }^{ij}+{\gamma }^{ij}$. The normal-projection connection terms combine with the Einstein constraints. Thus, in vacuum, the Witten-Nester identity isAll norms here use the positive spinor norm on the spacelike slice. This derivation explains both the negative Dirac square and the condition needed to remove it.

Solve the Witten spinor equation $\mathcal{D}ϵ=0$ with $ϵ\to {ϵ}_{\infty }$. The existence step is part of the argument. Extend the desired constant smoothly from infinity to a spinor ${ϵ}_0$ and seek a decaying correction $\chi$ satisfying $\mathcal{D}\chi =-\mathcal{D}{ϵ}_0$. The principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}^{2}I$, so this is an elliptic differential operator. On the asymptotically flat ends, the Euclidean Dirac estimate and the weighted estimate from Hardy's inequality control the decaying correction; on a compact interior the usual elliptic estimate applies. Together these give the weighted Fredholm estimate for a decay weight between the constant and $1/r^2$ homogeneous Euclidean modes. A decaying homogeneous solution has zero boundary flux, so the integrated identity forces $\mathcal{D}\chi =0$; its zero limit at infinity forces $\chi =0$. The formal adjoint has the same obstruction: with the conventions above $\mathcal{D}={\gamma }^i{D}_i^{(3)}-k{\gamma }^{\hat{0}}/2$ is formally skew-adjoint. Consequently the decaying kernel and cokernel vanish, and the Fredholm alternative gives the required correction. Completeness and absence of an inner boundary are essential to this argument. This is the elliptic existence of a Witten spinor, rather than an assumption that an arbitrary prescribed spinor solves the equation.

For completeness, identify the asymptotic flux. Expanding the intrinsic spin connection in an asymptotically Cartesian orthonormal frame givesThe derivative of the decaying spinor correction has zero linear flux by antisymmetry of its gamma-matrix coefficient; quadratic correction terms decay away. Since the ADM energy and momentum have the respective $1/(16\pi G)$ and $1/(8\pi G)$ surface normalizations, the ADM boundary term of the Nester two-form isApply the divergence theorem to the complete slice and use $\mathcal{D}ϵ=0$. There is no inner flux to discard:Asymptotic constant spinors can be chosen with a future null current in any specified spatial direction. Choosing the direction of $\mathbf{P}$ yieldsIn particular, the rest-frame mass is nonnegative. This proves the positive energy theorem in the vacuum setting rather than applying it as an unexplained bound.

If this mass is zero, then $E=0$ in the asymptotic rest frame and the inequality gives $\mathbf{P}=0$. The same identity with any nonzero ${ϵ}_{\infty }$ has zero right-hand side. Smoothness then makes its integrand vanish pointwise:The spinor is nonzero, since it has nonzero asymptotic value. This is a spinor parallel along $\Sigma$; no time-parallel condition has yet been inferred merely from the integral.

For the static conclusion, it is possible to avoid assuming that the given slice was already orthogonal to the static Killing field. Because $E=\mathbf{P}=0$, repeat the zero-energy construction for four linearly independent asymptotic Dirac spinors. Parallel transport by the Sen spinor connection is invertible, so the resulting spinors form a basis everywhere on $\Sigma$. Tangential curvature annihilates each of them:Thus $R_{ijab}{\gamma }^{ab}=0$ as an operator. The six antisymmetric gamma products give a faithful Spinor representation of the Lorentz group, so $R_{ijab}=0$ for tangential $i,j$. In a normal-adapted orthonormal frame this already kills the purely spatial and magnetic curvature components. The remaining components vanish by the Vacuum Einstein equations:Consequently the full spacetime Riemann curvature tensor vanishes on $\Sigma$.

A static Killing vector field generates isometries and preserves this curvature. Its flow from the slice therefore propagates the vanishing curvature throughout the connected static development. Extend one of the spinors by spacetime parallel transport along these time orbits, so that ${\nabla }_tϵ=0$. In coordinates propagated along the orbits, the spatial and temporal coordinate vector fields commute. Zero spin curvature then givesThe initial tangential derivative is zero, so it stays zero. This constructs the required nonzero spacetime parallel spinor; the initial parallel frame also removes any spin-holonomy ambiguity inherited from loops in $\Sigma$. The static zero-energy parallel-spinor rigidity conclusion isAn equivalent check in a regular orthogonal static slicing is to write $d{s}^2=-N^{2}d{t}^2+h_{ij}d{x}^{i}d{x}^j$. The slice has zero extrinsic curvature. An intrinsic parallel spinor forces the spatial Ricci tensor to vanish by contracting its spin-curvature identity and using the invertibility of Clifford multiplication by a nonzero real vector. By three-dimensional curvature from the Ricci tensor, $h$ is flat. The static vacuum equation $D_i{D}_{j}N=N{R}_{ij}(h)$ then makes the lapse gradient parallel, and asymptotic normalization forces $N=1$. The temporal spin connection vanishes as well.

An explicit example is Minkowski spacetime, $d{s}^2=-d{t}^2+d{x}^2+d{y}^2+d{z}^2$, where every constant spinor in a Cartesian tetrad is a spacetime parallel spinor and the ADM mass is zero. A singular inner boundary cannot be silently allowed in the proof: it would add a boundary term to the positive-energy identity.

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.

In a globally static region, choose the time coordinate along the future timelike Killing vector field $K={\partial }_t$. A positive-frequency solution of the Klein-Gordon equation has $i{\mathcal{L}}_K{p}_i={\omega }_i{p}_i$, with ${\omega }_i>0$. Assume appropriate boundary conditions and a Cauchy surface so that the conserved Klein-Gordon inner product isFor a real scalar, take the negative-frequency modes to be $n_i=p_i^{\ast }$. Choose a complete normalized basis withThe negative norm of the conjugate modes is crucial; this is not a positive-definite inner product on all classical solutions.

Promote the real Klein-Gordon field to the operatorThese canonical commutation relations implement the canonical field and conjugate-momentum commutator. The static vacuum state in a stationary spacetime satisfies $a_i|0⟩=0$, and the Hamiltonian is a sum of oscillators, ${\sum }_i{\omega }_i(a_i^{†}{a}_i+1/2)$ before zero-point regularization. Continuous labels replace sums by correctly normalized integrals. For a complex charged scalar there are separate particle and antiparticle oscillators, with the same mode-mixing argument in each sector.

Choose independently normalized past and future bases, and use the matrix indices of the question. Complex conjugation of the future positive-frequency expansion immediately givesConservation of the Klein-Gordon inner product gives $A^{†}A-B^{†}B=I$ and $A^{†}{B}^{\ast }-B^{†}{A}^{\ast }=0$ in this column-index convention. These ensure preservation of the canonical commutation relations.

Let $a_j$ annihilate past modes and $b_i$ annihilate future modes. Extracting the future coefficient from the field gives the Bogoliubov transformationThe minus sign follows from $(n_j^{-},n_k^{-})=-{\delta }_{jk}$. In the in-vacuum, only $⟨0_{\mathrm{in}}|a_j{a}_k^{†}|0_{\mathrm{in}}⟩={\delta }_{jk}$ contributes to the expectation of the future number operator. ThereforeThis is particle number from Bogoliubov coefficients: the intervening time-dependent geometry makes a past positive-frequency mode contain a future negative-frequency component. In infinite volume, localized wave packets avoid delta-function normalization artifacts; a common Fock-space implementation needs the usual square-summability condition on the creation-mixing coefficients.

For gravitational collapse, the initial vacuum can be defined before the hole forms, while late outgoing modes are defined using the asymptotic stationary time at future null infinity. The complete future basis also includes modes falling through the event horizon; exterior outgoing modes alone are not a complete basis of all propagated data. Tracing an outgoing mode backwards near a nonextremal horizon gives the exponential ray relation $v=v_H-C{e}^{-\kappa u}$, where $u$ is late retarded time and $\kappa$ is the surface gravity. An outgoing factor $e^{-i\omega u}$ becomes proportional to $(v_H-v{)}^{i\omega /\kappa }$ in the early advanced coordinate, and contains both frequency signs when Fourier decomposed there. Their squared Bogoliubov-coefficient ratio is $e^{-2\pi \omega /\kappa }$; the canonical normalization then gives a Bose factor $(e^{2\pi \omega /\kappa }-1{)}^{-1}$, multiplied by the greybody factor, the transmission probability through the exterior potential. Thus Hawking radiation has temperature $T_H=\kappa /(2\pi )$ in units $\hslash =k_B=c=1$, with greybody factor modifications to the spectrum. This is a free-field calculation on a fixed or slowly evolving background; eventual backreaction is an additional problem, and the nonextremal exponential map must not be blindly used when $\kappa =0$.

A horizon separates regions according to a specified causal, dynamical or geometrical criterion. The criterion must be stated: an event horizon, an apparent horizon, a Killing horizon and a Cauchy horizon answer different questions. Some horizons describe a black hole, others an observer's acceleration or the universe's expansion. A vanishing coordinate metric component alone does not establish any of these definitions.

In an asymptotically flat spacetime, choose the relevant future null infinity ${\mathcal{I}}^{+}$, the endpoint of escaping outgoing light rays. The black-hole region and its future event horizon areHere $J^{-}$ is the causal past. Events in $\mathcal{B}$ cannot send a causal signal to the chosen infinity. This is a global definition: identifying the horizon requires the full future development, so a local measurement need not reveal whether an event lies on it. Smooth portions of an event horizon are null hypersurfaces generated by null geodesics. Generators may join at past caustics or crease sets, so the horizon need not be smooth everywhere. A past event horizon is defined by the time-reversed escape criterion; in the maximally extended Schwarzschild solution it bounds a white-hole region.

An observer-based definition replaces infinity by a future-inextendible worldline $\gamma$. Its future observer event horizon is the boundary of events that can signal to $\gamma$, conventionally $\partial I^{-}(\gamma )$. A past observer horizon similarly bounds events reachable from its past worldline. Such horizons depend on the observer and need not imply a black hole.

A Killing horizon is a null hypersurface to which a nonzero Killing vector field $K^a$ is normal. On it, $K^a$ is tangent to its null generators and satisfies $K^b{\nabla }_b{K}^a=\kappa K^a$. The surface gravity $\kappa$ measures the failure of this Killing parameter to be an affine parameter; rescaling $K$ rescales $\kappa$. A degenerate Killing horizon has $\kappa =0$. Stationary black-hole horizons often are Killing horizons, whereas a general evolving event horizon need not have any Killing symmetry. The condition $K^a{K}_a=0$ alone is insufficient: the null vector must also be normal to a null hypersurface.

The Schwarzschild spacetime illustrates several of these notions. In units $G=c=1$, let $f=1-2M/r$. Its metric is $d{s}^2=fd{t}^2-f^{-1}d{r}^2-r^{2}d{\Omega }^2$, and $r=2M$ is the future event horizon of the black-hole region in the appropriate extension. The apparent coordinate singularity disappears in ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, $d{r}_{\ast }/dr=f^{-1}$:At $r=2M$ this metric is regular and the normal $dr$ is null. Future outgoing radial rays satisfy $dr/dv=f/2$; inside the horizon their areal radius decreases even on the outgoing family. The stationary Killing vector field ${\partial }_v$ generates the horizon with $\kappa =f^{\prime }(2M)/2=1/(4M)$ when normalized at infinity. The Kretschmann scalar is $48M^2/r^6$, finite at the horizon and divergent at $r=0$. Thus the event horizon is not the curvature singularity. The photon sphere at $r=3M$, which supports circular null orbits, is also not the horizon.

Quasi-local horizon notions use null expansions. For a closed spacelike two-surface, choose future outgoing and ingoing null normals $\ell ,n$, with $\ell \cdot n=1$ in signature $(+---)$. Each expansion is the logarithmic rate of its positive area element along the corresponding null normal. A future trapped surface has ${\theta }_{\ell }<0$ and ${\theta }_n<0$: both future light fronts initially shrink. A marginally outer trapped surface has ${\theta }_{\ell }=0$; for the future black-hole type also require ${\theta }_n<0$. On a chosen spacelike slice, the outer boundary of the future trapped region is an apparent horizon, ordinarily an outermost marginally outer trapped surface when regularity permits this characterization. It depends on the slice, and a slicing need not display an apparent horizon even when the global spacetime contains an event horizon.

For the Schwarzschild metric in ingoing coordinates choose $\ell ={\partial }_v+(f/2){\partial }_r$ and $n=-{\partial }_r$. The spherical null expansion in ingoing coordinates givesThe round spheres inside $2M$ are future trapped, while $r=2M$ is marginal. On standard horizon-penetrating slices its sections give the apparent horizon, coinciding with the stationary event and Killing horizon. This coincidence in a symmetric stationary example does not identify the notions in a general spacetime.

A trapping horizon is a hypersurface foliated by marginal surfaces. Its future outer type obeys ${\theta }_{\ell }=0$, ${\theta }_n<0$ and ${\mathcal{L}}_n{\theta }_{\ell }<0$. Unlike an event horizon, this can be diagnosed from local expansion data and their variation. The null energy condition explains its causal character. At a marginal surface, the Null Raychaudhuri equation, with the present curvature sign and vanishing null twist, givesin units $c=1$. A possible nonaffinity term vanishes here because it is proportional to the zero expansion. Write a horizon evolution vector as $V=\ell -Cn$. Preserving ${\theta }_{\ell }=0$ gives $C=({\mathcal{L}}_{\ell }{\theta }_{\ell })/({\mathcal{L}}_n{\theta }_{\ell })\ge 0$. Hence $V^2=-2C\le 0$: the future outer horizon is spacelike or null. Its area expansion is ${\theta }_V=-C{\theta }_n\ge 0$. These conclusions use the outer condition and the energy condition, not merely marginality.

For the global event horizon, the future-complete horizon focusing proof of the area theorem uses the same Null Raychaudhuri equation. Under the null energy condition, a negative generator expansion forces a future caustic in finite affine parameter. Future completeness and the usual regularity or predictability assumptions exclude such a caustic on the future event-horizon boundary. Its generator expansion is therefore nonnegative and its classical area cannot decrease. These assumptions matter: this is not an unconditional result for every null surface or every matter model.

Rotating and charged solutions provide further distinctions. A nonextremal Kerr black hole has $r_{\pm }=M\pm \sqrt{M^2-a^2}$, with $a=J/M$. Its outer Killing horizon is generated by ${\partial }_t+{\Omega }_H{\partial }_{\varphi }$, where ${\Omega }_H=a/(r_{+}^2+a^2)$, rather than by time translation alone. The ergosphere begins at the stationary-limit surface $g_{tt}=0$, whose radius is $M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the axis this lies outside the event horizon: it forbids static observers but does not forbid escape to infinity. In Reissner-Nordstrom spacetime, the nonextremal radii are $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer surface is an event horizon, whereas the inner surface is a Cauchy horizon in the exact analytic extension.

A Cauchy horizon concerns predictability from initial data. The future domain of dependence $D^{+}(\Sigma )$ of a partial Cauchy surface consists of events for which every past-inextendible causal curve intersects $\Sigma$. Its future Cauchy horizon isBeyond it, causal influences can arrive without meeting the specified initial surface; data on that surface no longer determine the solution there. This is distinct from the inability to send signals to infinity. The inner horizon of the exact charged black hole supplies an example, and the definition necessarily specifies which initial surface is being used.

Flat spacetime itself has acceleration horizons. In the right Rindler wedge, set $t=\xi \sinh (a\tau )$ and $x=\xi \cosh (a\tau )$, with $\xi >0$. The Minkowski spacetime metric becomes $d{s}^2=a^2{\xi }^{2}d{\tau }^2-d{\xi }^2-d{y}^2-d{z}^2$. The null boundaries $t=x$ and $t=-x$ are Rindler horizons of the boost Killing flow. A uniformly accelerated observer cannot receive signals from one side of its future horizon, even though the curvature vanishes and inertial observers encounter no such boundary. These are observer and Killing horizons, not a black-hole event horizon defined using the full Minkowski null infinity.

In a spatially flat Friedmann-Lemaître-Robertson-Walker metric with noncompact spatial sections, radial light rays have $d\chi /dt=\pm 1/a(t)$. The particle horizon measures how far signals could have travelled since the initial cosmic time; the cosmological event horizon measures how far signals emitted now could ever travel toward a comoving observer. Their comoving and proper radii areThe respective horizon is present only if its integral is finite. The event-horizon integral uses the actual future endpoint, often $t_f=\infty$, so it is global and observer-dependent. For $a(t)=t^q$ on $0<t<\infty$, a finite particle horizon exists for $0<q<1$, while a finite event horizon exists for $q>1$. Neither integral is finite for $q=1$. In the expanding flat patch of de Sitter spacetime, $a=e^{Ht}$ and the proper event-horizon radius is $H^{-1}$. The static form $d{s}^2=(1-H^2{r}^2)d{t}^2-(1-H^2{r}^2{)}^{-1}d{r}^2-r^{2}d{\Omega }^2$ identifies it as a cosmological Killing horizon; its surface-gravity magnitude is $H$ with the central static time normalization.

Finally, the Hubble radius must not automatically be called an event horizon. For a flat expanding cosmology, the two future radial null expansions of a sphere of proper radius $R$ are ${\theta }_{\pm }=2(H\pm 1/R)$. At $R=H^{-1}$ the ingoing expansion vanishes, while outside it both expansions are positive: this is the cosmological marginal, past-trapped type rather than a future black-hole trapped surface. A dust universe with $a\propto t^{2/3}$ has such a Hubble sphere but no future event horizon. Global causal boundaries, slice-dependent trapped-surface boundaries, symmetry horizons and initial-data boundaries must therefore be distinguished, even when several coincide in a special example.

Let $Q=g^{ab}{\chi }_{,a}{\chi }_{,b}$ and $\mathcal{L}=p(ϵ)+n(Q-{ϵ}^2)/(2ϵ)$. Vary $n,ϵ,\chi$ independently, taking compactly supported variations. The multiplier variation is $\partial \mathcal{L}/\partial n=(Q-{ϵ}^2)/(2ϵ)=0$. With $u_a={\chi }_{,a}/ϵ$, it gives the unit four-velocity condition $u^a{u}_a=1$. The enthalpy variation isUsing $Q={ϵ}^2$ yields $p^{\prime }(ϵ)=n$. Finally,The integration by parts boundary term vanishes. Thus the three matter equations areThe last is particle number conservation in a relativistic fluid. Off shell $n,ϵ$ are independent variational fields; on shell the thermodynamic relation ties them to the specified function $p(ϵ)$.

For the Hilbert stress-energy tensor, keep these matter fields fixed while varying the inverse metric. The volume variation is $\delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}{g}_{ab}\delta g^{ab}$, and $\delta Q={\chi }_{,a}{\chi }_{,b}\delta g^{ab}$. ThereforeOn shell, $\mathcal{L}=p$ and $nϵ=\rho +p$. With signature $(+---)$,For variation of the covariant metric the same definition reads ${\delta }_{g}S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}{d}^{4}x$. In a local rest frame the tensor has entries $\mathrm{diag}(\rho ,p,p,p)$, which checks both the energy density and pressure sign.

Define a derivative along the flow by $\dot{F}=u^a{\nabla }_{a}F$, and set $\theta ={\nabla }_a{u}^a$, $a^b=u^a{\nabla }_a{u}^b$. The number equation immediately gives $\dot{n}+n\theta =0$. The specific enthalpy per particle relation and $dp=ndϵ$ giveConsequently the energy equation follows from the number equation: $\dot{\rho }=ϵ\dot{n}=-(\rho +p)\theta$.

The velocity equation follows directly from the potential, rather than requiring an independent assumption about the fluid force. Since $ϵu_a={\nabla }_a\chi$ and the connection has zero torsion, the canonical vorticity of a relativistic fluid vanishes:Contract with $u^a$ and use $u^a{u}_a=1$ and $u^a{\nabla }_b{u}_a=0$. This gives $ϵa_b={\nabla }_bϵ-u_b\dot{ϵ}$. Multiplication by $n$ and use of ${\nabla }_{b}p=n{\nabla }_bϵ$ produces the relativistic Euler equation. The evolution system is thereforeIt is equivalent to stress-energy conservation together with number conservation, subject to the potential-flow restriction. The action describes an isentropic barotropic sector of a perfect fluid in general relativity; it does not provide general vortical or entropy-carrying fluid motions. The constraint on canonical vorticity is stronger than the Euler equation alone.

For the weak static metric, use $c=1$. To the retained order, $\sqrt{-g}=1+\varphi$, $u^0=1-\varphi +O({\varphi }^2,|\mathbf{v}{|}^2)$ and $u^i=v^i$. The correction to $u^0$ is required by normalization, even though its zeroth-order value is one. The relevant Christoffel symbols arewith no spatial connection at this order. The cancellation $\sqrt{-g}{u}^0=1+O({\varphi }^2,|\mathbf{v}{|}^2)$ makes the leading number equation particularly simple:The corresponding leading energy transport equation is ${\partial }_t\rho +\mathbf{v}\cdot \nabla \rho +(\rho +p)\nabla \cdot \mathbf{v}=0$. The spatial Euler equation contains the gravitational acceleration $-\nabla \varphi$ and pressure force $-\nabla p/(\rho +p)$. In a strict first-order expansion, convective velocity acceleration and products of the small quantities are discarded.

To exhibit a fully linear system, perturb a homogeneous rest state: $n=n_0+\delta n$, $\rho ={\rho }_0+\delta \rho$, $p=p_0+\delta p$, and $ϵ={ϵ}_0+\delta ϵ$, with the perturbations, $\mathbf{v}$ and $\varphi$ all first order. Let $w_0={\rho }_0+p_0=n_0{ϵ}_0$. The weak-field slow-motion equations for a relativistic fluid areThe equation of state gives $\delta \rho ={ϵ}_0\delta n$ and $\delta p=n_0\delta ϵ$. Where $p^{\prime \prime }({ϵ}_0)\ne 0$, it also gives the barotropic relativistic sound speedWithout this invertibility, the primitive conservation equations remain valid and this sound-speed parametrization must not be used.

The potential furnishes a further check on the pressure and gravity signs. Write $\chi ={ϵ}_{0}t+\delta \chi$. Its time and spatial components implyTheir compatibility reproduces the linear Euler equation. Pressure gradients oppose gravitational acceleration, and hydrostatic balance is $\nabla \delta p=-w_0\nabla \varphi$. Combining the energy and Euler equations gives the forced acoustic equationFor positive $c_s^2$, disturbances propagate as sound waves; the prescribed potential supplies a static forcing. A nonrelativistic equation of state gives $p\ll \rho \simeq mn$, $ϵ\simeq m$, and reduces the inertia $\rho +p$ to mass density, recovering the ordinary linearized continuity and Euler equations. Slow velocity alone does not justify dropping $p$ from the inertia of a relativistically hot fluid. The given metric supplies an external gravitational potential; the matter action by itself supplies no Poisson equation for it.