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A nonzero causal vector is timelike or null: $g(v,v)\le 0$ with metric signature $(-,+,+,+)$. The zero vector is commonly included when stating the dominant energy condition.

On a scalar, the covariant wave operator is $\square \varphi =|g{|}^{-1/2}{\partial }_a(|g{|}^{1/2}{g}^{ab}{\partial }_b\varphi )$. For the metric signature $(-+++)$ it has the flat-space form $-{\partial }_t^2+{\nabla }^2$. This divergence expression is convenient for separation of variables and for integrating the Klein-Gordon equation by parts.

A free Klein-Gordon field is a scalar obeying $(\square -m^2-\xi R)\varphi =0$ in the mostly-plus metric signature. Global hyperbolicity provides a well-defined initial-value problem. Quantization additionally needs a state or positive-norm mode splitting; the equation alone does not select particles.

The metric signature records the signs of the diagonalized nondegenerate metric tensor, equivalently its numbers of negative and positive directions. For a four-dimensional Lorentzian manifold, both $(-,+,+,+)$ and its overall negative convention are common. The numerical signs of contractions and component Hodge star operator identities must be consistent with that choice and the orientation.

A real scalar field is minimally coupled to a metric tensor when its kinetic term uses the metric contraction of first derivatives and its scalar potential depends only on the field, without an explicit scalar curvature coupling. For metric signature $(-,+,+,+)$ the displayed Lagrangian and volume density $\sqrt{-g}$ give $\square \varphi -V^{\prime }(\varphi )=0$. Its stress-energy tensor is $T_{ab}={\nabla }_a\varphi {\nabla }_b\varphi -g_{ab}[\frac{1}{2}(\nabla \varphi {)}^2+V]$. The scalar stress-energy divergence identity explains its conservation on the field equation.

For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

A four-vector is a collection of four components $A^{\mu }=(A^0,\mathbf{A})$ that transforms by the same Lorentz transformation as $(ct,\mathbf{x})$. Use the metric signature $(-,+,+,+)$. The Lorentzian inner product isA nonzero four-vector is a timelike vector, null vector or spacelike vector according as $A\cdot A$ is negative, zero or positive. Reversing the metric signature reverses the signs used to name these three classes, without changing their geometric meaning.

For a frame moving at speed $v$ along the positive $x$ axis, write $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The component Lorentz transformation isExpanding the first two squares givesThe other two components are unchanged, so $A^{\prime }\cdot A^{\prime }=A\cdot A$. Consequently a timelike vector remains timelike under a Lorentz transformation.

For a nonzero null vector, $|\mathbf{A}|=|A^0|$ and $A^0\ne 0$. Set $a=A^0$ and $\hat{\mathbf{n}}=\mathbf{A}/a$. Then $|\hat{\mathbf{n}}|=1$ and $A=a(1,\hat{\mathbf{n}})$. If the zero four-vector is included among null vectors, take $a=0$ and any unit vector.

For two future-pointing null vectors, write $A=a(1,\hat{\mathbf{n}})$ and $B=b(1,\hat{\mathbf{m}})$ with $a,b>0$. The sum of future-pointing null vectors satisfiesbecause the ordinary inner product of two unit vectors is at most one. The sum is null if their spatial directions coincide, and timelike otherwise. Its positive time component makes the sum nonzero and future-pointing.

Use geometrized units and metric signature $(-+++)$, with $M>0$. In the exterior of Schwarzschild spacetime, put $f=1-2M/r$. The Schwarzschild tortoise coordinate satisfiesThe retarded and advanced null coordinates $u=t-r_{\ast }$ and $v=t+r_{\ast }$ then give $d{s}^2=-fdudv+r^{2}d{\Omega }^2$. The logarithmic divergence of $r_{\ast }$ suggests exponentiating these null coordinates. In the right exterior define the Kruskal–Szekeres coordinatesTheir product eliminates $t$:Since $du=-4MdU/U$ and $dv=4MdV/V$, the Schwarzschild metric becomesHere $r$ is an implicitly defined function of $UV$. The derivative of the right side with respect to $r$ is $-r{e}^{r/(2M)}/(4M^2)$, which is nonzero at $r=2M$. The inverse function theorem therefore makes $r(UV)$ smooth across that surface, and the coefficient of $dUdV$ tends to $-16M^2/e$. Thus the Schwarzschild event horizon is a coordinate singularity of the original chart, while this Lorentzian metric remains regular there.

Extend the Kruskal–Szekeres coordinates to all real $U,V$ with $UV<1$. The signs give two exterior regions, $U<0<V$ and $V<0<U$, a future black hole region $U,V>0$, and a past white hole region $U,V<0$. The event horizons are $U=0$ or $V=0$, intersecting at the bifurcation surface. The boundary $UV=1$ has $r=0$ and is a genuine Schwarzschild singularity, as the Kretschmann scalar $48M^2/r^6$ diverges there.

Finally, $T=(V+U)/2$ and $X=(V-U)/2$ give $UV=T^2-X^2$ and a radial metric proportional to $-d{T}^2+d{X}^2$. Hence radial null geodesics have slopes $\pm 1$, the event horizons are $T=\pm X$, and the singular boundaries are $T^2-X^2=1$. This constructs the maximal Kruskal extension; a black hole produced by collapse need not contain the second exterior or the white hole of that eternal extension.

Fix a classical globally hyperbolic spacetime with metric signature $(-+++)$; in this solution take $\hslash =1$. A free real Klein-Gordon field can be specified bywhere $\mu$ is its mass, $\xi$ its curvature coupling and $\mathcal{R}$ the scalar curvature. Global hyperbolicity ensures a well-posed initial-value problem on a Cauchy hypersurface $\Sigma$ and the existence of retarded and advanced propagators. Thus compactly supported field and normal-derivative data determine a classical solution. This fixes the dynamics, but not a Fock vacuum.

For real solutions with suitable support or falloff, the symplectic form on scalar-field solutions iswith $n$ the future unit normal. The field equation makes the current conserved, so this symplectic form is independent of $\Sigma$ when boundary flux vanishes. Quantize the initial data by the canonical commutation relation: with $\pi =n^a{\nabla }_a\varphi$ and the delta function defined relative to $d\Sigma$, $[\hat{\varphi }(x),\hat{\pi }(y)]=i{\delta }_{\Sigma }(x,y)$ and the two equal-field commutators vanish. Equivalently, construct the field algebra using the causal propagator. A state on that algebra is additional input.

To construct a particle representation, complexify the classical solution space. Its conserved Klein-Gordon inner product isIt is indefinite on all complex solutions. Choose a complete positive-norm subspace and an orthonormal mode basis $u_i$ with $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Such a choice is encoded by a compatible complex structure on the Klein-Gordon solution space. Its positive subspace gives the one-particle Hilbert space, and the associated bosonic Fock space contains symmetrized many-particle states. The field expansion isThe creation operator $a_i^{†}$ adds a particle in mode $i$, the annihilation operator removes one, and the number operator is $N_i=a_i^{†}{a}_i$. For continuous mode labels the sums and Kronecker deltas become integrals and delta functions, or one can work with normalized wave packets.

The ambiguity is precisely that the field equation and global hyperbolicity do not select that positive subspace. A different normalized basis may mix $u_j$ and $u_j^{\ast }$ by a Bogoliubov transformation, and then its annihilation operators mix $a_j$ and $a_j^{†}$. Its Fock vacuum and number operators differ. The Hadamard condition constrains physically acceptable short-distance singularities and allows local renormalization, but it still leaves many states. Hence there is generally no observer-independent particle count on an arbitrary dynamical geometry.

In a stable strictly stationary spacetime, a globally future timelike Killing vector field $K$ gives a preferred time translation. Fix its normalization and suitable boundary conditions, and choose positive-frequency solutions satisfying $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. The corresponding positive spectral subspace gives the preferred vacuum state in a stationary spacetime. Unitary changes of basis within it leave the vacuum and the particle notion unchanged. This construction assumes a well-defined positive stationary generator; stationarity by itself is insufficient if $K$ becomes spacelike, as in a Kerr ergoregion, or if unstable or zero modes obstruct the ground-state construction. It selects a preferred ground state under the stated assumptions, not a unique state among all thermal and excited states.

If the geometry is suitably stationary in the asymptotic past and future, choose those preferred mode spaces separately, giving the in-vacuum and out-vacuum. Propagate the past modes through the intervening region using the field equation and compare them to the future modes using the conserved Klein-Gordon inner product. Adopt the conventionThe canonical identities for a bosonic Bogoliubov transformation read $\alpha {\alpha }^{†}-\beta {\beta }^{†}=I$ and $\alpha {\beta }^T=\beta {\alpha }^T$. Extracting the future annihilation operator with the same inner product givesIn the in-vacuum, only $⟨a_j{a}_k^{†}⟩={\delta }_{jk}$ contributes to $⟨b_i^{†}{b}_i⟩$. Therefore the particle number from Bogoliubov coefficients isNonzero $\beta$ is the production of future particles from the past vacuum. Summing over future modes gives the total expected particle number when that sum is finite. For infinitely many modes, a Hilbert-Schmidt operator $\beta$ is the condition for unitary implementability between these pure bosonic Fock space representations; finite-volume or wave-packet calculations must respect the relevant measures and convergence. Particle production is determined by the negative-frequency mixing, rather than by identifying a single instantaneous vacuum throughout the time-dependent region.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

Use geometrized units $G=c=1$ and metric signature $(-,+,+,+)$. Put $f=1-2M/r$ in the Schwarzschild metric. The Schwarzschild tortoise coordinate satisfies $d{r}_{\ast }/dr=f^{-1}$, soSubstituting $dt=dv-f^{-1}dr$ gives the Ingoing Eddington-Finkelstein coordinates:The radial metric tensor has determinant $-1$ and inverse components $g^{vv}=0$, $g^{vr}=1$, $g^{rr}=f$. Thus it is nondegenerate and analytic at $r=2M$. The same expression defines a Lorentzian metric for every $r>0$, extending the exterior across the future Schwarzschild event horizon into the black hole. It does not include the other exterior or the white hole of the full Kruskal spacetime. At $r=0$, the Kretschmann scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges, so this is a curvature singularity, not a removable coordinate singularity.

The dominant energy condition says that, for every future timelike vector $u^a$, the energy current $J^a=-T^a{}_b{u}^b$ is future causal or zero. Choose normal coordinates whose future unit time vector at $p$ is $u^a=(1,0,0,0)$. With metric signature $(-,+,+,+)$ and symmetric stress-energy tensor,A vector with these components is future causal or zero exactly whenThis proves necessity. Conversely, every future unit timelike vector can be made the time axis of an orthonormal frame by a proper orthochronous Lorentz transformation, and that frame can be extended to normal coordinates at $p$. Requiring the displayed inequality in every such chart therefore gives the dominant energy condition for every future timelike vector; positive rescaling handles nonunit vectors. Future null vectors follow by continuity if they are included in the definition. The condition must hold in every local Lorentz frame; one chart alone is insufficient.

Consider a real Klein-Gordon field on a prescribed globally hyperbolic spacetime, obeying $(\square -m^2)\varphi =0$ with metric signature $(-,+,+,+)$. A curvature coupling can be included as $m^2\mapsto m^2+\xi R$. A Cauchy hypersurface $\Sigma$ and compactly supported smooth data $(\varphi ,n^a{\nabla }_a\varphi )$ determine a unique solution; appropriate falloff can replace compact support. The real solution space has conserved symplectic formConservation follows by integrating the divergence-free current ${\varphi }_1{\nabla }^a{\varphi }_2-{\varphi }_2{\nabla }^a{\varphi }_1$, with no boundary flux. On complex solutions the conserved Klein-Gordon inner product isIt is indefinite on the full complex solution space.

Choose a complete positive-norm mode subspace, with modes $u_j$ satisfyingEquivalently choose a compatible complex structure on the Klein-Gordon solution space. The mode labels may be continuous, in which case sums and Kronecker symbols become integrals and Dirac delta functions. Construct the one-particle Hilbert space from these modes and its bosonic Fock space. Promote the field to the operator-valued distributionFor a foliation with spatial metric determinant $h$, the conjugate momentum density is $\hat{\pi }=\sqrt{h}{n}^a{\nabla }_a\hat{\varphi }$. Mode completeness gives the equal-time canonical commutation relationsThe Fock vacuum obeys $a_j|0⟩=0$, and $a_j^{†}{a}_j$ counts particles in the chosen mode. For local products and a renormalized stress-energy tensor, physically admissible states are further restricted by the Hadamard condition. The field algebra exists without a preferred Fock vacuum.

A different admissible mode splitting can mix positive and negative norms:Orthonormality imposes the canonical identities for a bosonic Bogoliubov transformationThe same field then hasWithout a preferred notion of positive frequency, a nonstationary spacetime supplies no distinguished mode splitting: particle number and vacuum depend on the choice of modes, although the field equation and field algebra do not. With infinitely many modes the Bogoliubov transformation need not be unitarily implementable; finite total mixing requires a Hilbert-Schmidt operator $\beta$.

For a stable strictly stationary spacetime, a chosen future globally timelike Killing vector field $K$ gives a preferred time translation. Choose modes withWhen the corresponding conserved Killing energy is positive and the spectral problem has suitable boundary conditions and no problematic zero modes, this gives the preferred vacuum state in a stationary spacetime and particles relative to $K$. Positive-frequency mode mixing within that same subspace leaves the Fock vacuum unchanged. Rescaling $K$ by a positive constant changes the frequency units but not their sign.

Stationarity alone, if it only means a Killing field timelike near infinity, is insufficient for a global unique particle interpretation. In the Kerr ergoregion, ${\partial }_t$ is spacelike, so positive frequency relative to $t$ does not automatically select a positive-norm subspace throughout the geometry; superradiance illustrates the difficulty. Additional vacuum and boundary choices remain necessary. The customary stationary answer therefore assumes a suitable timelike stationary flow and a stable positive-energy quantization; it does not assert that every stationary black-hole extension has one globally preferred vacuum.

Use $G=c=1$, the Einstein field equations with zero cosmological constant, and metric signature $(-,+,+,+)$. Write $m=m_{a}d{x}^a$ for the differential one-form dual to the axial Killing vector field. To fix the sign of the volume formula, use the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$ and $(\star F{)}_{ab}=\frac{1}{2}{ϵ}_{abcd}{F}^{cd}$. The Killing equation gives $(dm{)}_{ab}=2{\nabla }_a{m}_b$, and its contracted curvature identity givesTo see the curvature step, tracing the Killing equation gives ${\nabla }_b{m}^b=0$. The second covariant derivative of a Killing vector then gives ${\nabla }^b{\nabla }_b{m}_a=-R_{ab}{m}^b$, while commuting the covariant derivatives gives ${\nabla }^b{\nabla }_a{m}_b=R_{ab}{m}^b$. Subtracting these two terms is precisely ${\nabla }^b(dm{)}_{ab}$ above. The component Hodge star operator converts this divergence to $d\star dm$ with the displayed minus sign.

In a vacuum spacetime region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing spacelike submanifolds in the same homology class bounding a vacuum three-dimensional region $W$, Stokes theorem yieldsThus the Komar angular momentum is independent of the enclosing vacuum spacelike two-manifold, provided the surfaces have the same orientation and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial Killing vector field suffices.

For a regular filling hypersurface $\Sigma$ with $\partial \Sigma =S$, the Einstein field equations giveThe pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. ThereforeThis is the stress-energy current from a Killing vector integrated over the slice, with the signs fixed by the stated Hodge star operator convention.

There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial \Sigma =S-{\bigcup }_i{S}_i$. The actual Komar angular momentum with inner boundaries identity isFor a vacuum Kerr black hole, $T_{ab}=0$ on the exterior slice but $J\ne 0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The Komar angular momentum independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.

Evaluate the metric inner products of the proposed orthonormal tetrad to first order in the vector cosmological perturbation:The orthonormal tetrad therefore has metric signature $(-,+,+,+)$ to the required order. Put $q=ϵ/a^2$ and $b=\mathbf{e}\cdot \mathbf{B}$. The photon momentum components areThey satisfy the null vector condition to first order, with $|\mathbf{e}|=1$.

Use $\eta$ as parameter in the time component of the geodesic equation:The derivative of the direction is first order, since the background photon moves on a straight comoving line; $B_{i}d{e}^i/d\eta$ is consequently second order. With $d\mathbf{x}/d\eta =\mathbf{e}+O(B)$,The supplied Levi-Civita connection coefficients giveThe spatial gradients and all background expansion terms cancel. Since ${ϵ}^{\prime }/ϵ$ vanishes in the background, its product with $b$ is also second order. The photon energy redshift from a vector metric perturbation is thereforeAn independent check uses the covariant component $p_0=g_{0\mu }{p}^{\mu }=-ϵ+O(B^2)$. The covariant geodesic equation gives $d{p}_0/d\eta =(2p^0{)}^{-1}{\partial }_{\eta }{g}_{\alpha \beta }{p}^{\alpha }{p}^{\beta }=ϵe^i{\dot{B}}_i$; the conformal expansion term vanishes by the null vector condition and reproduces the same result.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

In metric signature $(-,+,\dots ,+)$, a spacelike submanifold of a Lorentzian manifold has an induced metric tensor defining a positive-definite quadratic form. With the overall reversed metric signature, its induced metric tensor instead has only negative eigenvalues. The definition applies to any dimension: enclosing charge-integration surfaces in four dimensions are two-dimensional examples, whereas a spacelike Cauchy hypersurface in four dimensions is three-dimensional.

With metric signature $(-,+,+,+)$, a nonzero vector $v$ is spacelike when its Lorentzian inner product with itself satisfies $g(v,v)>0$. In Minkowski spacetime, this means its spatial components have larger squared Euclidean length than its time component. The classification is preserved by Lorentz transformations. A timelike vector has negative squared Lorentzian inner product, and a null vector has zero squared Lorentzian inner product.