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For an axial Killing vector $\xi$ and zero cosmological constant, set $F^{ab}={\nabla }^b{\xi }^a$. The Killing equation and Ricci identity give ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. In a vacuum region this vanishes, so Stokes theorem makes the Komar angular momentum equal on homologous enclosing surfaces. The region may contain gravitational waves with nonzero Weyl curvature: Ricci flatness, rather than complete flatness, is what removes the flux. This proves absence of axial angular-momentum transport by exactly axisymmetric gravitational radiation, without assuming that a radiating spacetime is stationary.

An exact axial Killing vector field makes Komar angular momentum independent of the surrounding vacuum surface. Therefore gravitational waves in a configuration with exact axisymmetry carry no net angular momentum about that symmetry axis, although they can carry energy. This is specific to axial angular momentum; it is not an absence of radiation.

Komar angular momentum is the Komar charge associated with an axial Killing vector field in four-dimensional general relativity. In geometrized units its normalization is $(16\pi {)}^{-1}{\int }_S\star dm$, where $m$ is the metric-dual differential one-form. With zero cosmological constant, vacuum Einstein field equations make $d\star dm=0$, so homologous surrounding surfaces in the vacuum region give the same charge.

Use metric compatibility and a torsion-free connection. The coordinate formula for the Lie derivative of the metric tensor isSubstituting ${\partial }_a{g}_{bc}={\Gamma }^d{}_{ab}{g}_{dc}+{\Gamma }^d{}_{ac}{g}_{bd}$ and collecting connection terms gives $({\mathcal{L}}_{\xi }g{)}_{ab}={\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a$. Thus a Killing vector obeys ${\nabla }_{(a}{\xi }_{b)}=0$, and conversely this Killing equation makes its flow preserve the metric. Its contraction gives ${\nabla }_a{\xi }^a=0$.

With the curvature convention fixed in Q1, contract the vector Ricci identity to obtainThe second term vanishes because the Killing divergence is zero. Metric compatibility converts the first term to ${\nabla }^b{\nabla }_a{\xi }_b$, and the Killing equation changes this to $-{\nabla }^b{\nabla }_b{\xi }_a$. Therefore the second covariant derivative of a Killing vector satisfiesThis derivation fixes the sign by the stated Ricci identity rather than by an unspecified convention for curvature.

For the axial Killing vector set $F^{ab}={\nabla }^b{\xi }^a$. It is antisymmetric and its divergence is ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. Work in geometrized units $G=c=1$ and with zero cosmological constant, as required by the asymptotically flat setting. If a spatial shell $\Omega$ has oriented boundary $S_2-S_1$, the given form of Stokes theorem applied to the Komar angular momentum givesThe vacuum Einstein field equations give $R_{ab}=0$ everywhere in the shell outside the matter source. The shell can still contain gravitational waves, since those have nonzero Weyl curvature rather than a nonzero Ricci tensor. HenceThis is angular momentum conservation in an axisymmetric vacuum region. Moving the outer surface through the radiative region adds no angular momentum. In an exactly axisymmetric spacetime there is no outward transport of the axial angular momentum by the waves. Axisymmetry does not imply stationarity, so this conclusion does not exclude radiation of energy.

For completeness the Einstein equations express the bulk current asOn an axisymmetric spacelike slice, $\xi$ is tangent to the slice and its contraction with $d^3{\Sigma }_a$ vanishes; there the trace contribution to the angular-momentum volume integral drops out. One must retain any inner surface charges if the integration region has inner boundaries. The vacuum shell argument itself does not require a matter-only representation of the total charge.

The current $Y$ is conserved on arbitrary slices, not just those to which $\xi$ is tangent. Indeed symmetry of the Ricci tensor and antisymmetry of ${\nabla }_a{\xi }_b$ giveThe last equality follows because a metric-preserving Killing flow also preserves the scalar curvature. Equivalently, the stress-energy conservation law and invariance of the stress trace give the same conserved current.

Let two spacelike hypersurfaces terminate on large enclosing spheres $S_1,S_2$ at the same asymptotic end. Join the spheres by a timelike tube lying wholly in the vacuum exterior. On this tube ${\nabla }_b{F}^{ab}=0$, so Stokes theorem gives zero difference of their Komar surface integrals, independently of the tube's shape. Alternatively, use the given Gauss theorem on a spacetime volume between the slices: ${\nabla }_a{Y}^a=0$, and the lateral flux is zero because $Y=0$ in the exterior vacuum tube. Taking the enclosing spheres to infinity with the usual asymptotic falloff therefore provesAll surface orientations in these equations use the same convention as the stated Komar integral. The argument concerns enclosing surfaces at the same asymptotic end, so it does not discard additional charges from other ends or inner boundaries.

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.