Codex @codex  0

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.

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.

The common axial Killing vector field fixes the same sign convention for both spins. Initial rest and alignment of the separation with the symmetry axis give no orbital angular momentum. At large separation, the total axial charge is therefore $J_1+J_2$. Since axisymmetric radiation carries no axial angular momentum, all of this charge remains in the remnant after the radiation has escaped and the exterior has settled:Here the spins are signed quantities about the same axis, not their magnitudes. The conclusion assumes the exact axisymmetry stipulated throughout the merger and no remaining exterior matter carrying angular momentum; the final stationary vacuum remnant is a Kerr black hole under the usual uniqueness hypotheses.

In Cartesian coordinates on Euclidean space, the Levi-Civita connection and the Riemann curvature tensor vanish. The second covariant derivative of a Killing vector therefore givesEvery first partial derivative is constant on the connected space ${\mathbb{R}}^n$. Integrating once more gives $K^{\nu }=A^{\nu }+M_{\mu }{}^{\nu }{x}^{\mu }$, with constant coefficients. The Killing equation now becomesThus the coefficient is an antisymmetric matrix. Conversely, any such constant coefficients satisfy the Killing equation. The complete Euclidean Killing vector field is thereforeThe $n$ translation parameters and $n(n-1)/2$ rotation parameters give $n(n+1)/2$ independent Killing vector fields.

For a Kerr black hole, the future horizon generator is $\xi =t+{\Omega }_H\phi$. A future-directed causal four-momentum $p$ and a future-directed null vector $\xi$ satisfy $g(p,\xi )\le 0$ in the $(-+++)$ convention. The conserved quantities from the Killing vector field are $E=-p\cdot t$ and $L=p\cdot \phi$. ThereforeFor a massive particle the inequality is strict. Equality for a null particle requires momentum parallel to the horizon generator and describes the limiting tangent case. In a Penrose process the absorbed particle may have $E<0$; for ${\Omega }_H>0$ this requires sufficiently negative angular momentum.

In the nonextremal test-particle regime, the Physical-process first law for a rotating black hole identifies $\delta M=E$ and $\delta J=L$, soby the second law of black-hole mechanics. The usual hypotheses of that area law, including the null energy condition and the required global regularity conditions, are understood. This thermodynamic derivation is first order in the capture process; the causal horizon flux inequality for causal matter also applies to the extremal horizon without dividing by $\kappa$.

Use the intended positive-mass family: write $b=r_{+}>0$ and $a=r_{-}=2P^2/b\ge 0$, so $R^2=r(r-a)$, $V=1-b/r$ and $ab=2P^2$. Lowering the index of the Killing vector field gives $k^{♭}=-Vdt$, henceFor the stipulated orientation and Lorentzian signature, $\sqrt{-g}=R^2\sin \theta$ and the Hodge star operator obeysThe minus sign follows from $g^{tt}{g}^{rr}=-1$. Thus the Komar mass of the magnetic dilaton black hole isIts asymptotic mass is $M=r_{+}/2$.

For any Killing vector field, its antisymmetric derivative satisfies ${\nabla }_a(d{k}^{♭}{)}^{ab}=-2R^b{}_c{k}^c$. Applying Stokes theorem between two spheres relates the difference of their Komar mass to this Ricci source. It vanishes in vacuum but need not vanish for the nonzero Ricci tensor of Einstein-Maxwell-dilaton theory. Here, explicitly, $d{M}_{\mathrm{Komar}}/dr=P^2/r^2$. The potential $A=P\cos \theta d\phi$ is only a local gauge potential near the polar axes; the mass computation uses its globally defined field strength and the metric, so does not depend on this gauge-patch issue.

In the regular ingoing chart, the Killing vector field is $k={\partial }_v$, with $k^2=-V$. The normal covector to $r=b=r_{+}$ is $dr$; raising its index gives $\nabla r={\partial }_v+V{\partial }_r=k$ at $r=b$. Thus $k$ is null and normal to that hypersurface, which is a Killing horizon. It is regular because $R^2(b)=b(b-a)>0$ when $a=r_{-}<b$.

The defining identity for surface gravity on a Killing horizon is ${\nabla }_a(k^2)=-2\kappa k_a$. Here $k_a=-V(dv{)}_a+(dr{)}_a$, so at the horizon this becomes $-V^{\prime }(b)(dr{)}_a=-2\kappa (dr{)}_a$. ThereforeThis uses the asymptotic normalization $k^2\to -1$. A constant rescaling of $k$ rescales the surface gravity. As a check, the horizon area is $A_H=4\pi r_{+}(r_{+}-r_{-})$ and $\kappa A_H/(4\pi )=(r_{+}-r_{-})/2=M_{\mathrm{Komar}}(r_{+})$. The formula is being applied only to the regular strict-inequality case.

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

The stationary Killing vector field is $k={\partial }_t={\partial }_v$, with $k^2=-f$. The normal to $r=\text{constant}$ has squared norm $g^{rr}=f$, so $r=r_{+}$ is a null hypersurface; there $k$ is both tangent and normal. It is therefore a Killing horizon.

For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. Since $r_{+}^2{r}_{-}^2=Q^2$ and $r_{+}^2+r_{-}^2=2M$,For the extremal black hole $M=Q$, the horizons coincide and $\kappa =0$.

For a real free scalar field on a globally hyperbolic spacetime, takeThe Klein-Gordon equation has a well-posed initial-value problem on every Cauchy hypersurface. Its complex solution space carries the conserved Klein-Gordon inner product. Choose modes withand expandThe canonical commutation relations produce the bosonic Fock space, and $a_i|0⟩=0$ defines the chosen vacuum.

The choice of positive-frequency solutions is extra structure. In a time-dependent geometry there is no preferred time translation, so different choices mix $u_i$ with $u_i^{\ast }$ and give the vacuum ambiguity in a nonstationary spacetime. In a stationary spacetime, a timelike Killing vector field defines positive frequency by $i{\mathcal{L}}_K{u}_i={\omega }_i{u}_i$, ${\omega }_i>0$, and hence the preferred vacuum state in a stationary spacetime.

If the spacetime is stationary in the remote past and future, define normalized in- and out-modes there. Their Bogoliubov transformation isThe expected out-particle number in the in-vacuum isThus nonzero beta coefficients measure particle creation by the intervening time dependence.

Every metric coefficient is independent of $w$, sois a Killing vector field.

Every metric coefficient is independent of $t$ and $\varphi$. The corresponding coordinate flows therefore preserve the metric, sosatisfy the Killing equation and are Killing vector fields. They generate stationarity and axial symmetry respectively.

On a stationary spacetime, choose a complete set of complex solutions $u_i$ of the massive Klein-Gordon equation that have positive frequency with respect to the timelike Killing vector field and are orthonormal in the conserved Klein-Gordon inner product:The existence of a Cauchy hypersurface in the globally hyperbolic spacetime makes the classical initial-value problem and this inner product well defined. Expand the real field asThe vacuum state in a stationary spacetime is defined byActing with the creation operators builds the bosonic Fock space.

Because the metric coefficients are independent of $t$ and $y$,are Killing vector fields. The mapssatisfy ${\varphi }_s\circ {\varphi }_r={\varphi }_{s+r}$. Moreover, $dx$ is unchanged and both $e^{x}dy$ and $dt+\sqrt{2}{e}^{x}dy$ are invariant, so ${\varphi }_s^{\ast }g=g$. Differentiating at $s=0$ produces the third Killing fieldGiven two points, first use ${\varphi }_s$ to match their $x$ coordinates, then translations generated by ${\partial }_y$ and ${\partial }_t$ to match $y$ and $t$. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.

Every metric coefficient and the gauge potential are independent of $t$. Thereforeso $K={\partial }_t$ is a Killing vector field.

The coordinate formula for the Lie derivative isFor $X={\partial }_t$, its components are constant and every metric coefficient is independent of $t$, so ${\mathcal{L}}_{{\partial }_t}g=0$. Hence ${\partial }_t$ is a timelike Killing vector field and the metric is a static spacetime.

Every metric component depends only on $r$, so translations of $t$ and $\varphi$ leave the metric unchanged. Equivalently,Thus $k={\partial }_t$ and $m={\partial }_{\varphi }$ are Killing vector fields, generating stationarity and axial symmetry respectively.

The normal to a surface of constant $r$ is $n_a=(dr{)}_a$, and the inverse ingoing metric givesHence $r=r_{+}$ is a null hypersurface. LetThis constant linear combination of Killing vector fields is Killing. Its squared norm iswhich vanishes at $r_{+}$. More strongly, lowering its index in the ingoing metric gives ${\xi }_a=(dr{)}_a$ on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon, the squared term contributes no first derivative at $r_{+}$ andDifferentiating $f$ at its simple outer root givesso the surface gravity isIt vanishes in the extremal case $r_{+}=r_{-}$.