Quasi-Local Horizon Angular Momentum and Charge

Angular momentum and electric charge can be assigned to a horizon cross-section using fields at that finite boundary. Electric charge is a Gauss-law flux. Angular momentum pairs an axial vector with the horizon’s normal-bundle connection and, in Einstein–Maxwell theory, includes a gauge- field surface term. The result is quasi-local but not structure free: axial symmetry, its $2\pi$ normalization, orientation, null-normal scaling, and electromagnetic gauge behavior must all be controlled.

Required background. Hypersurface Normals and Tangential Projectors supplies the normal bundle. Helpful background. Exact Spacetimes supplies Kerr–Newman checks; Causality and Global Structure supplies finite inner boundaries; Einstein Equations fixes the Maxwell coupling; Black-Hole Geometry and Mechanics supplies isolated horizons; Variational and Symplectic Geometry supplies Hamiltonian surface charges.

Electric charge from horizon flux

Let $S$ be an oriented spacelike cross-section of an isolated or suitable marginal horizon. With Maxwell action $-\tfrac14\int F_{ab}F^{ab}\sqrt{-g}\,d^4x$, define

$$Q_S=\int_S\star F.$$

The orientation is chosen so that the outward flux of a positively charged Reissner–Nordström black hole is positive. If $S_1$ and $S_2$ are homologous sections joined by a source-free horizon segment, Maxwell’s equation $d\star F=0$ and Stokes’ theorem give $Q_{S_2}=Q_{S_1}$. Charged current crossing the intervening segment changes the charge by its flux.

The integral depends only on $F=dA$, not on the potential gauge. It also does not require axial symmetry. Magnetic charge, when present, is analogously a flux of $F$ with its own normalization; a global potential $A$ may then need gauge patches. Ashtekar and Krishnan give the isolated-horizon charge and its conservation in Ashtekar and Krishnan 2004, §§4.1–4.2.

For Kerr–Newman in the convention used throughout this volume, $F^{tr}\sim Q/(4\pi r^2)$ at large radius, and the sphere integral is exactly $Q$. This check is what pairs the flux convention with the metric term $GQ^2/(4\pi)$ and the potential $\Phi_H=Qr_+/[4\pi(r_+^2+a^2)]$.

Gravitational angular momentum

Now assume $S$ has an axial vector field $\varphi^a$ tangent to $S$, with closed orbits with parameter period $2\pi$ and two fixed points. Let $\ell^a$ be a future null normal and

$$D_a\ell^b=\omega_a\ell^b$$

define the rotation one-form. In vacuum, the horizon angular momentum is

$$J_S^{\mathrm{grav}}[\varphi] =-\frac{1}{8\pi G}\int_S (\varphi^a\omega_a)\,\epsilon,$$

with the displayed sign tied to the cross-section and axial orientation. Reversing $\varphi^a$ reverses $J$; changing its period rescales $J$, which is why the $2\pi$ normalization is physical data.

This expression is invariant under null boosts. If $\ell'=f\ell$ and the other null normal is rescaled inversely, then $\omega'_a=\omega_a+D_a\ln f$. On a closed axisymmetric section,

$$\int_S\varphi^aD_a\ln f\,\epsilon =-\int_S(\ln f)D_a\varphi^a\,\epsilon=0,$$

because an axial Killing field is divergence free. No assumption that $f$ is constant on $S$ is needed for this surface integral. Gourgoulhon and Jaramillo derive the rotation one-form and horizon angular momentum in Gourgoulhon and Jaramillo 2006, §§7.5 and 8.6.1, pp. 222–223 and 237–238.

The electromagnetic angular momentum term

In Einstein–Maxwell theory, the Hamiltonian generating rotations receives a gauge-field contribution. In a gauge adapted to the axial symmetry, $\mathcal L_\varphi A=0$, the total quasi-local charge can be written

$$J_S[\varphi] =-\frac{1}{8\pi G}\int_S (\varphi\mathbin{\cdot}\omega)\,\epsilon -\int_S(\varphi\mathbin{\cdot}A)\star F.$$

The coefficient matches the rationalized Maxwell action and charge used here. In the more common convention with a $1/(4\pi)$ in the Maxwell action and charge, the corresponding term carries that factor; formulas cannot be mixed across the two normalizations.

Under a single-valued symmetry-preserving gauge transformation $A\mapsto A+d\lambda$ with $\mathcal L_\varphi\lambda=0$, the electromagnetic integrand is unchanged. More general large or patchwise gauge transformations require transition data and a consistent Hamiltonian prescription, especially with magnetic charge. The gravitational and electromagnetic pieces separately depend on how the fields are divided, while the total symmetry charge is the object compared with asymptotic angular momentum.

For vacuum Kerr, the electromagnetic term vanishes and direct evaluation of $\omega_a$ gives $J_S=J$. For Reissner–Nordström, axial symmetry is present but both rotational contributions vanish, while the flux integral gives $Q$. For Kerr–Newman, the combined expression—not generally the gravitational piece alone—agrees with the total angular momentum parameter. These cases test vacuum, static charged, and rotating charged limits.

What symmetry is being assumed

An exactly axisymmetric section supplies a canonical axial direction up to orientation once its $2\pi$ orbits are fixed. A generic distorted section has no exact Killing vector, so the displayed $J[\varphi]$ can still be evaluated for a chosen tangent vector but is no longer a unique intrinsic angular momentum. Approximate-Killing-vector procedures are useful in numerical work, yet their optimization, normalization, and error should be reported.

The same warning applies to comparing sections. If different axial fields are chosen independently on successive leaves, a change in $J$ can reflect a change of identification rather than physical flux. A horizon flux law needs a coherent evolution of the symmetry field and the appropriate matter and gravitational contributions.

These quasi-local charges do not assert equality with ADM charges in every spacetime. Equality follows in stationary source-free regions when the Hamiltonian or Komar identity relates the two boundaries. Exterior matter or radiation can carry part of the total charge.

Evaluate a model rotation one-form

On a round section of area radius $R$, take $q=R^2(d\theta^2+\sin^2\theta\,d\phi^2)$, $\varphi=\partial_\phi$, and $\omega=C\sin^2\theta\,d\phi$. With no Maxwell field, compute $J_S$ and find $C$ for a desired positive $J$ in the stated orientation.

Solution

Since $\varphi\mathbin{\cdot}\omega=C\sin^2\theta$ and $\epsilon=R^2\sin\theta\,d\theta\,d\phi$,

$$\int_S(\varphi\mathbin{\cdot}\omega)\epsilon =CR^2(2\pi)\int_0^\pi\sin^3\theta\,d\theta =\frac{8\pi}{3}CR^2.$$

Therefore

$$J_S=-\frac{CR^2}{3G}.$$

A desired positive $J$ requires $C=-3GJ/R^2$ for this axial and surface orientation. Reversing $\varphi$ reverses both the sign of the integral and the reported angular momentum. Adding $D_a\ln f$ to $\omega_a$ changes the integral by zero because $S$ is closed and $D_a\varphi^a=0$.

Isolated-Horizon First Laws asks when these charges combine into an integrable horizon Hamiltonian.

References

• Ashtekar, Abhay, and Badri Krishnan. “Isolated and Dynamical Horizons and Their Applications.” Living Reviews in Relativity 7 (2004): article 10. DOI.
• Gourgoulhon, Éric, and José Luis Jaramillo. “A 3+1 Perspective on Null Hypersurfaces and Isolated Horizons.” Physics Reports 423 (2006): 159–294. DOI.