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The Equivalence principle permits Riemann normal coordinates at any event, in which the first derivatives of the metric tensor, and hence the Christoffel symbols, vanish. A putative universal local gravitational stress-energy tensor constructed from the metric and its first derivatives would vanish in that freely falling description. If it were a genuine tensor, it would then vanish in every description, even where other coordinates assign nonzero gravitational energy density. The gravitational field therefore does not have a local stress-energy tensor analogous to that of matter, with coordinate-independent energy density and the usual energy-balance interpretation.

Gravitational energy is not represented by a universal local covariant stress-energy tensor. This does not assert that all local curvature tensors vanish or are forbidden: the Riemann curvature tensor measures tidal gravity, and higher-curvature tensors can be constructed. They do not provide the sought ordinary local gravitational energy density. Boundary charges, quantities associated with a background, and suitable averaged wave energies have different definitions and evade that local requirement. This is the absence of a local gravitational stress-energy tensor.

Extract the Riemann curvature tensor from ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ using the curvature 2-forms just computed. Contraction $R_{bd}=R^a{}_{bad}$ gives the diagonal Ricci tensor in the orthonormal coframe:For example, the $00$ component receives $a^2$ from each of the $1$ and $2$ directions and $q$ from direction $3$. The signs in the spatial components reflect the same Lorentzian index lowering used for the connection 1-forms.

For a null vector $k$ in this frame, $(k^0{)}^2=(k^1{)}^2+(k^2{)}^2+(k^3{)}^2$. ThereforeThe Einstein field equations imply $8\pi T_{ab}{k}^a{k}^b=R_{ab}{k}^a{k}^b$, since the metric term vanishes for a null vector; an included cosmological-constant term would also vanish. Choose $k^a=(1,0,0,1)$ and apply the null energy condition. It follows thatConversely, this inequality makes the same contraction nonnegative for every null vector, so it is exactly the null energy condition for a planar warped spacetime when its matter stress-energy tensor is defined by the Einstein field equations. Null vectors tangent to the planar slices saturate the condition. As checks, constant $A$ gives flat spacetime and $A=e^{bz}$ gives constant negative sectional curvature with $a^{\prime }=0$, also saturating it.

Use signature $(-,+,+,+)$ and write $h_{ij}$ for the spatial induced metric, $D_i$ for its spatial covariant derivative, and $q^2=h^{ij}{D}_i\varphi D_j\varphi$. The negative shift vector convention gives $n^{\mu }=N^{-1}(1,N^i)$ and $\Pi =n^{\mu }{\partial }_{\mu }\varphi$. Decomposing the scalar field derivative givesVarying the scalar field matter action with respect to the inverse metric tensor, including the variation of $\sqrt{-g}$, gives the stress-energy tensorRaising both indices gives the requested $T^{\mu \nu }$. Contracting with the unit normal twice, or once with the spatial projection tensor, givesProjecting both indices of the stress-energy tensor onto the spatial hypersurface givesTwo printed identifications require correction. The canonical momentum of the displayed density is ${\pi }_{\varphi }=\sqrt{h}\Pi$, not $\Pi$: explicitly ${\mathcal{L}}_m=N\sqrt{h}[{\Pi }^2/2-q^2/2-V]$ and differentiation with respect to $\dot{\varphi }$ supplies $\sqrt{h}$. Thus $\Pi$ is the normal scalar-field momentum. Also, the printed trace-free stress has an extra factor $1/2$ and uses ${\delta }_{ij}$ where the general spatial induced metric requires $h_{ij}$. In an orthonormal spatial frame the latter becomes ${\delta }_{ij}$, but the coefficient is still one. For a gradient $(q,0,0)$ in that frame, direct projection gives ${\tilde{S}}_{11}=2q^2/3$, rather than the printed $q^2/3$. The PDF's energy-density gradient is spatial ${\partial }_i\varphi {\partial }^i\varphi$; the TeX aid's spacetime index there is an OCR defect.

A symmetry preserves the action, possibly up to a boundary term. In a classical field theory, this connects transformations of fields to conserved quantities. In a quantum field theory, it also constrains states, observables and scattering amplitudes. Continuous symmetries have Noether currents; discrete symmetries can be equally important without supplying a current through the same theorem.

For a local density depending on fields ${\Phi }_a$ and their first derivatives, define ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\Phi }_a)$ and ${\mathcal{E}}_a=\partial \mathcal{L}/\partial {\Phi }_a-{\partial }_{\mu }{\Pi }_a^{\mu }$. Suppose a one-parameter transformation has fixed-coordinate variation $\delta {\Phi }_a=\epsilon R_a$ and satisfies $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$. The Noether's first theorem asserts thatFor Grassmann-valued fields the derivatives and ordering are chosen consistently; the same integration-by-parts argument applies. To prove the theorem, expand the density variation and integrate its derivative term once algebraically:Comparison with the assumed divergence gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{\mathcal{E}}_a{R}_a$. On shell every Euler-Lagrange expression vanishes. Integrating this continuity equation over space gives the Noether charge $Q=\int j^0{d}^{3}x$ and $dQ/dt=-{\int }_{\partial {\mathbb{R}}^3}\mathbf{j}\cdot d\mathbf{S}$. Thus $Q$ is conserved whenever the boundary flux vanishes. Boundary conditions are part of the conservation statement; the local identity alone does not discard nonzero flux.

Translations illustrate the theorem in every relativistic field theory. For $R_a=a^{\nu }{\partial }_{\nu }{\Phi }_a$ and $K^{\mu }=a^{\mu }\mathcal{L}$, the current is $a^{\nu }{T}^{\mu }{}_{\nu }$, whereThe four charges are energy and momentum. For the Klein-Gordon field they reduce to the expressions derived in Question 1. Lorentz invariance gives angular momentum and boost currents. With a symmetric improved stress-energy tensor, these are $M^{\lambda \mu \nu }=x^{\mu }{T}^{\lambda \nu }-x^{\nu }{T}^{\lambda \mu }$. The canonical Dirac expression also carries an intrinsic spin current; improvement incorporates it into the symmetric tensor. Together translations and Lorentz transformations form the Poincare group.

Internal symmetries act on field components without moving spacetime points. The global phase symmetry of a complex scalar field, $\delta \varphi =-i\epsilon \varphi$, hasIts charge distinguishes particles and antiparticles. A Dirac phase transformation $\delta \psi =-i\epsilon \psi$ gives $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$. This is preserved by the real-scalar Yukawa interaction, because the phases of $\bar{\psi }$ and $\psi$ cancel; consequently fermion number is conserved even while scalar and fermion particles interact. A multiplet of real scalars with a potential depending only on ${\sum }_a{\varphi }_a^2$ has orthogonal internal rotations. Their antisymmetric currents can be written $j_{ab}^{\mu }={\varphi }_a{\partial }^{\mu }{\varphi }_b-{\varphi }_b{\partial }^{\mu }{\varphi }_a$.

Discrete examples include $\varphi \mapsto -\varphi$ for an even real-scalar potential, parity symmetry in quantum field theory, and charge conjugation, which interchanges a complex scalar with its conjugate and reverses the electromagnetic potential. Discrete transformations are not generated by an infinitesimal continuous parameter, so Noether theorem does not attach a local conserved current to each of them. Nevertheless they forbid interaction terms and relate physical processes.

A global symmetry in field theory uses parameters constant across spacetime and can act nontrivially on physical states. A gauge redundancy allows arbitrary spacetime-dependent parameters and relates descriptions of the same physical configuration. In scalar electrodynamics, replacing the scalar's global phase by a local phase requires the compensating gauge transformation of $A_{\mu }$ and the gauge covariant derivative; the invariant contraction uses the opposite representation for ${\varphi }^{\ast }$. In Yang-Mills theory, matter transforms in a representation of a non-Abelian gauge group and the gauge potential transforms so that covariant derivatives and field strengths transform covariantly. Gauge fixing, such as Coulomb gauge, chooses representatives of these descriptions rather than changing physical predictions. The current-contracted propagator identity of Question 3 illustrates why a noncovariant gauge choice leaves scattering Lorentz invariant.

The arbitrary local parameter produces a stronger statement than one independent conserved charge for each function. For the abelian scalar transformation, vary the action with an arbitrary compactly supported $\alpha (x)$ and integrate its derivative term:ThereforeThis consequence of Noether second theorem, the Noether identity for abelian scalar gauge symmetry, holds off shell: gauge invariance makes the field equations dependent. In Hamiltonian language the same redundancy is reflected by the Gauss law constraint in gauge theory and the elimination of unphysical gauge degrees of freedom. Gauge transformations that vanish at the boundary are redundancies; transformations with nontrivial boundary behaviour can instead carry physical surface charges. This is why identifying every gauge transformation with a zero-charge operation would be too strong.

Finally, symmetry of the action need not imply symmetry of a chosen vacuum. For a complex scalar with a symmetry-preserving potential whose minima occur at nonzero $|\varphi |$, different constant phases label degenerate vacua. Choosing one breaks the global phase symmetry of the state; the Goldstone theorem supplies a massless mode under its usual relativistic assumptions. When the same phase is gauged, the Higgs mechanism describes the phase degree of freedom becoming the longitudinal polarization of a massive gauge boson; the local redundancy is still present in the underlying description. At the quantum level a classical symmetry can also fail through a quantum anomaly, so a classical Noether derivation alone does not guarantee an exact quantum Ward identity. Conserved charges, physical global transformations and local gauge redundancy are related but distinct consequences of symmetry.

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

One standard form of the Penrose singularity theorem assumes a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector $k$, and a closed future trapped surface. It concludes future null geodesic incompleteness: some future-inextendible null geodesic has a finite upper endpoint of its affine parameter.

The focusing mechanism is the Null Raychaudhuri equation. The normal generators have zero null twist, soAn initial ${\theta }_0<0$ therefore gives a conjugate point to a spacelike surface within affine distance at most $2/|{\theta }_0|$, assuming the generator can be continued that far. Such a generator ceases to lie on the achronal boundary after its first focal point. Compactness of the trapped surface supplies a uniform bound for all normalized initial null normals. Future completeness would consequently make its future boundary compact. Projection along timelike curves to a connected Cauchy hypersurface is injective on the achronal boundary and has open image. Compactness makes the image closed as well; the nonempty image must therefore be the entire hypersurface, contradicting its noncompactness. This explains why the global assumptions supplement local focusing.

For Reissner-Nordstrom spacetime, the Maxwell stress-energy tensor satisfies the null energy condition; the Einstein field equations imply the required null convergence condition. The spheres in the band just found are closed and trapped. Apply the theorem to a maximal Cauchy development with a noncompact Cauchy hypersurface and containing one such sphere. That globally hyperbolic spacetime must be future null-geodesically incomplete.

The Penrose theorem at a Cauchy horizon needs care: the full maximal analytic Reissner-Nordstrom spacetime has inner Cauchy horizons and is not globally hyperbolic. It does not satisfy every hypothesis of the displayed theorem. In the exact solution some incomplete geodesics of the globally hyperbolic development reach a smoothly extendible Cauchy horizon in finite affine parameter. The theorem asserts incompleteness of the development, not that every such endpoint is a curvature singularity. The separate curvature singularity at $r=0$ does not justify silently dropping the theorem's global hypothesis.

The laws of black-hole mechanics initially relate geometric quantities in a way resembling thermodynamics. Hawking radiation supplies a physical temperature: in units $c=1$, while displaying $G,\hslash ,k_B$,The Hawking temperature $T_H$ is measured with the stationary time normalized at infinity. The radiation's thermal occupation factor is physically observable; propagation to infinity also introduces greybody factors, so the distant spectrum need not be a perfect blackbody spectrum at every frequency.

The Zeroth law of black-hole mechanics says that surface gravity $\kappa$ is constant on a stationary Killing horizon under its standard hypotheses. Through the Hawking temperature, this becomes uniform equilibrium temperature. The First law of black-hole mechanics isUsing $T_H$ identifies its area term as $T_H\delta S_{\mathrm{BH}}$. Integrating $\delta S_{\mathrm{BH}}=k_B\delta A/(4G\hslash )$ gives the Bekenstein-Hawking entropy, up to an additive constant. Thus $M$ is the energy, the angular and charge terms are work terms, and the geometrical law is the ordinary thermodynamic first law with a fixed entropy normalization.

The second law of black-hole mechanics, or Hawking's area theorem, gives nondecreasing area in the classical setting with the requisite energy and predictability assumptions. It corresponds to increasing Bekenstein-Hawking entropy. Semiclassical black-hole evaporation can decrease the area: the classical null energy condition need not hold for the quantum expectation of the stress-energy tensor. The appropriate extension is the generalized second law, that $S_{\mathrm{BH}}+S_{\mathrm{outside}}$ does not decrease, with the exterior entropy and its renormalization treated consistently. Hawking's temperature identification motivates this law; thermality alone does not prove every form of it.

The third law of black-hole mechanics is the unattainability, by an admissible finite physical process, of zero surface gravity. With Hawking temperature it becomes unattainability of absolute zero. It is not the assertion that an extremal black hole has vanishing entropy: its area can remain nonzero when $\kappa =0$.

For a Schwarzschild black hole, $\kappa =1/(4GM)$ and $A=16\pi G^2{M}^2$, giving $T_H=\hslash /(8\pi k_{B}GM)$ and $S_{\mathrm{BH}}=4\pi k_{B}G{M}^2/\hslash$. Their product satisfies $T_{H}d{S}_{\mathrm{BH}}=dM$, explicitly checking the first law. Quantum radiation turns the temperature and entropy in the mechanical analogy into physical thermodynamic quantities.

A small direction-dependent temperature change $T=\bar{T}(1+\Theta )$ shifts a thermal spectrum according to $\bar{f}(ϵ/(1+\Theta ))=\bar{f}(ϵ)-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$ at first order. Hence $\boxed{\Theta =\delta T/\bar{T}}$ is the angular photon temperature perturbation. For a general spectrum the same ansatz describes an energy-independent brightness dilation; a literal thermodynamic temperature interpretation additionally assumes a thermal shape without spectral distortions.

Use $E=ϵ/a$ in the tetrad stress-energy tensor. The background photon energy density isIntegration by parts gives $-\int dϵ{ϵ}^4{\bar{f}}^{\prime }=4\int dϵ{ϵ}^3\bar{f}$, assuming the boundary term vanishes. Thus each directional energy perturbation is four times its temperature perturbation. Writing $⟨\cdot {⟩}_{\Omega }=(4\pi {)}^{-1}\int d\Omega$, we haveIn the paper's convention for photon angular temperature moments, there is no $(2\ell +1)$ factor multiplying ${\Theta }_{\ell }$. Legendre polynomial orthogonality gives $⟨\Theta {⟩}_{\Omega }={\Theta }_0$ and $⟨e^i\Theta {⟩}_{\Omega }=-i{\hat{k}}^i{\Theta }_1/3$. Comparing with the stated density and velocity conventions yieldsFor completeness, $⟨e^i{e}^j{P}_2(\hat{\mathbf{k}}\cdot \mathbf{e}){⟩}_{\Omega }=({\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3)/5$. The $(-i{)}^2$ quadrupole phase then gives $\boxed{{\Pi }_{\gamma }=-3{\Theta }_2/5}$. These moment relations are photon angular temperature moments and keep the anisotropic stress convention consistent with the subsequent hierarchy.

The one-particle distribution function is a scalar function on the future mass shell. Choose its normalization so that, in a local orthonormal tetrad, $f{d}^{3}x{d}^{3}p$ is the expected photon number in a phase-space cell on a constant-time slice. Spin degeneracy and any phase-space normalization factors are included in $f$; with an occupation-number convention they would instead appear explicitly in the measure.

More covariantly, the photon number crossing a spacelike element is $f{p}^{\mu }d{\Sigma }_{\mu }{d}^{3}p/E$. The factor $p^{\mu }d{\Sigma }_{\mu }$ reduces to $E{d}^{3}x$ on a local constant-time slice, recovering the cell interpretation. The measure $d^{3}p/E$ is Lorentz invariant on the mass shell.

A photon carries energy $E$ and momentum $p^i$, while its velocity is $p^i/E$. Energy density, momentum density and momentum flux are thereforeThey assemble into the stress-energy tensorThe tetrad momenta are related to coordinate components by the tetrad basis. For an isotropic distribution, angular averaging gives $T^{\hat{i}\hat{j}}={\rho }_{\gamma }{\delta }^{ij}/3$ and $T^{\hat{0}\hat{i}}=0$. Thus the photon gas has pressure ${\rho }_{\gamma }/3$, consistent with its massless equation of state.

Differentiate the electromagnetic stress-energy tensor. The current term isThe other two terms cancel by the homogeneous Maxwell equations. To see the contraction explicitly, multiply ${\partial }_{\mu }{F}_{\beta \alpha }+{\partial }_{\beta }{F}_{\alpha \mu }+{\partial }_{\alpha }{F}_{\mu \beta }=0$ by $F^{\mu \alpha }$. The first and third contracted terms are equal after swapping indices and using antisymmetry; the middle is $-\frac{1}{2}{\partial }_{\beta }(F^{\mu \alpha }{F}_{\mu \alpha })$. ThereforeIt cancels differentiation of the trace term, leavingThe sign is tied to the printed convention ${\partial }_{\mu }{F}^{\mu \nu }=-{\mu }_0{J}^{\nu }$ and the displayed field-strength tensor.

Use $x^0=ct$. The stress-energy tensor component $T^{00}$ is energy per unit volume; $c{T}^{i0}$ is energy flux in direction $i$; $T^{0j}/c$ is momentum density in direction $j$; and $T^{ij}$ is the flux of momentum component $j$ through a surface normal to direction $i$. In a symmetric tensor, energy flux equals $c^2$ times momentum density. Thus ${\partial }_{\mu }{T}^{\mu 0}=0$ is energy conservation and ${\partial }_{\mu }{T}^{\mu j}=0$ is momentum conservation.

Take the oscillation to be $z(t)=L\cos \omega t$; a different initial phase does not affect a period average. In the mass-density normalization of the quadrupole formula, the second mass moment tensor of the point mass has only $I_{33}=m{z}^2$ nonzero. Its trace-free mass quadrupole moment is thereforeThe constant part does not radiate. Since $d^3({\cos }^2\omega t)/d{t}^3=4{\omega }^3\sin 2\omega t$, the tensor contraction isUsing $⟨{\sin }^2(2\omega t)⟩=1/2$ in the quadrupole formula gives the averaged positive radiated powerThe mass quadrupole moment oscillates at twice the source's angular frequency, and the source energy loss has the opposite sign. The expression uses the leading slow-motion weak-field approximation, $\omega L\ll c$. With SI stress-energy tensor components, the mass-density integrand is $T_{00}/c^2$; the displayed $c^{-5}$ quadrupole formula uses that mass normalization. This is the point-mass contribution requested by the model: an apparatus maintaining an accelerating mass would also contribute its own mass quadrupole moment.

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.

For a closed universe, the spatial Cauchy hypersurface is compact without boundary. There is consequently no asymptotic surface term in the gravitational Hamiltonian. On solutions, the Hamiltonian constraint and momentum constraint giveThis is the vanishing canonical Hamiltonian of a closed universe: evolution generated by lapse and shift is a gauge transformation, and there is no asymptotic time-translation charge. It does not mean the local stress-energy tensor or Riemann curvature tensor is zero. In an asymptotically flat universe the same bulk constraint equations in field theory hold, but a boundary Arnowitt-Deser-Misner energy can be nonzero.

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.

Take $\lambda >0$, $m>0$, use metric $(+,-)$ and set $\kappa =\sqrt{2\lambda }m$. The two scalar-field vacua are $\varphi =\pm m$; the positive kink joins $-m$ at the left end to $m$ at the right. The square completion for a one-dimensional kink givesThus its Bogomolny bound is $E\ge 4\sqrt{2\lambda }{m}^3/3$. Equality requires ${\varphi }_x=\sqrt{2\lambda }(m^2-{\varphi }^2)$; separating variables, or differentiating the resulting hyperbolic tangent, gives$X$ is the translational collective coordinate. The antikink is $-{\varphi }_K$ for the centered odd profile. Differentiating the first-order equation gives the static Euler-Lagrange field equation ${\varphi }_{xx}=4\lambda \varphi ({\varphi }^2-m^2)$. Hence the Bogomolny equation really does solve the second-order theory. The two distinct vacua and the above kink require the positive parameters; the degenerate case $m=0$, or the unstable negative-$\lambda$ potential, does not have this topological kink sector.

For a slowly moving kink, put $\varphi (x,t)={\varphi }_K(x-X(t))$. The saturated first-order equation implies $V({\varphi }_K)={\varphi }_K^{\prime 2}/2$, soThe stress-energy tensor has $T^{00}={\varphi }_t^2/2+{\varphi }_x^2/2+V$ and $T^{01}=-{\varphi }_t{\varphi }_x$. Using ${\varphi }_t=-v{\varphi }_K^{\prime }$ in the approximate translate therefore gives $E=M+M{v}^2/2$ and $P=Mv$. Equivalently, substitution into the action gives the collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+M{\dot{X}}^2/2$. These are the translational dynamics of a phi-four kink, withThe displayed remainder orders refer to the exact constant-velocity solution. The uncontracted translated profile is only an approximation: its error as a field starts at $O(v^2)$, but the static profile is an energy stationary point, so the corresponding static-energy error starts at $O(v^4)$ and does not change the displayed $v^2$ coefficient.

Since the relativistic action is invariant under Lorentz transformations, a Lorentz boost gives the exact moving kink, for $|v|<1$,Its conserved energy and momentum are $E={\gamma }_{v}M$ and $P={\gamma }_{v}Mv$, whose expansions agree with the preceding calculation. The width contraction is essential to solving the exact time-dependent equation.

Use the centered profile $K(x)=m\tanh (\kappa x)$. An appropriate approximate separated-lump initial condition isIt tends to $-m$ as $x\to -\infty$ and $m$ as $x\to +\infty$. Near $-a$ and $a$ it is a positive kink, and near zero it is an antikink; the other two tails cancel to exponentially small accuracy. The condition $\kappa a\gg 1$ makes these interpretations accurate. This sum is legitimate smooth initial data, not an exact static multi-kink solution. Its topological charge isand its energy is close to $3M$, with exponentially small interactions.

Adjacent kink–antikink pairs attract: the kink–antikink attraction from the stress tensor gives a negative midpoint pressure and hence an inward force on each outside kink. Direct overlap of the two outer kink tails is exponentially smaller than each adjacent kink–antikink overlap. The initial field is odd and its velocity zero, so the equation's reflection-plus-sign symmetry preserves $\varphi (-x,t)=-\varphi (x,t)$. The central zero remains at $x=0$ and the outer cores move in symmetrically. They collide with the central antikink. Excess energy can excite localized kink oscillations and outgoing radiation; the usual relaxational outcome is a remaining centered kink plus radiation after transient collisions or oscillations. On an infinite line radiation can carry energy away from the core even though total energy is conserved. Conservation of $Q=1$ prevents complete disappearance into vacuum, but it does not determine every bounce or exclude a temporarily re-emerging kink–antikink pair. The nonintegrable collision should not be described as exact elastic passage of all three solitons.

A gravitational wave can propagate through a region with vanishing stress-energy tensor; its gravitational field is not a matter current $T_{ab}{m}^b$. Nevertheless, exact axisymmetry gives an axial Killing vector field, and the vacuum identity $d\star dm=0$ makes the Komar angular momentum equal on enclosing spacelike submanifolds in the same homology class. Applying the same identity to the spacetime tube between surrounding spacelike submanifolds at different times gives zero net flux of this charge through the tube.

Consequently gravitational waves in exact axisymmetry carry no net angular momentum about the symmetry axis. They can still carry energy and decrease the total mass. In a mode description, rotationally invariant radiative data have zero azimuthal mode number; exact axisymmetry therefore excludes the axial angular momentum flux present in nonaxisymmetric merger radiation. This does not mean there are no gravitational waves, or that radiation cannot transport angular momentum when the exact symmetry is absent. The relevant conserved charge includes any horizon contributions identified in the preceding Komar angular momentum with inner boundaries identity.