Stationary Axisymmetric Vacuum Reduction

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Stationarity and axisymmetry reduce the four-dimensional vacuum equations to an elliptic boundary-value problem on a two-dimensional half-plane. The dynamical variables can be packaged either as a complex Ernst potential or as a symmetric-space matrix current. The remaining metric coefficient is then obtained by quadrature. What determines a regular black-hole exterior is not arbitrary Dirichlet data: asymptotic charges, axis regularity, a horizon rod and its null generator, topology, and compatibility at rod endpoints must all be imposed together.

Required background. Analysis for Mathematical Relativity supplies elliptic and harmonic-map tools, while Spacetime Symmetries and Killing Fields supplies commuting Killing fields, twists, and orbit spaces.

Helpful background. Einstein Equations provides the vacuum system, Causality and Global Structure supplies the simply connected exterior used for global potentials, Partial Differential Equations and Green Functions supports the half-plane problem, and Black-Hole Geometry and Mechanics supplies horizon generators and angular velocity.

Circularity and Weyl coordinates

Let $T$ generate stationary time translations and let $Φ$ generate a period- $2\pi$ axial action, with $[T,\Phi]=0$ . In vacuum the twist identities, together with regularity on the axis and asymptotic flatness, imply

T\wedge\Phi\wedge dT=0, \qquad T\wedge\Phi\wedge d\Phi=0.

These circularity conditions make the two-planes orthogonal to the group orbits integrable. They are field-equation consequences in vacuum, not part of the definition of two commuting symmetries Carter 1969, Theorem 2. In wedge products, $T$ and $\Phi$ denote their metric-dual one-forms. Their derivation and the matter terms that can obstruct them are given in Chruściel, Costa, and Heusler 2012, §8.1, pp. 41–42.

Write $G_{AB}=g(K_A,K_B)$ for $K_A=(T,\Phi)$ and define

\rho^2=-\det G =g(T,\Phi)^2-g(T,T)g(\Phi,\Phi).

On the orbit space of a regular domain of outer communications, $\rho$ is a nonnegative harmonic function. Simple connectedness permits a global harmonic conjugate $z$ . The orbit metric is conformally flat in $(\rho,z)$ , and its boundary $\rho=0$ decomposes into axis rods and horizon rods. This topological step is indispensable: a merely local harmonic conjugate does not give one global half-plane problem.

The vacuum Ernst system

On a region where $f=-g(T,T)>0$ , use the Weyl–Papapetrou form

g=-f(dt-A\,d\phi)^2 +f^{-1}\left[e^{2k}(d\rho^2+dz^2)+\rho^2d\phi^2\right].

Take $dt\wedge d\rho\wedge dz\wedge d\phi$ positive, with the Hodge star defined by $\alpha\wedge*\beta=\langle\alpha,\beta\rangle_g\,d\mathrm{vol}_g$ , and define the twist potential $\chi$ by $d\chi=*(T^\flat\wedge dT^\flat)$ . Vacuum gives $d*(T\wedge dT)=0$ , so $χ$ exists globally after simple connectedness is used. With this orientation, the first-order relations are

\partial_\rho\chi=\frac{f^2}{\rho}\partial_zA, \qquad \partial_z\chi=-\frac{f^2}{\rho}\partial_\rho A.

The signs follow directly from $T^\flat=-f(dt-A\,d\phi)$ and $T^\flat\wedge dT^\flat=-f^2dt\wedge dA\wedge d\phi$ . In a positively oriented orthonormal coframe, $*(e^0\wedge e^1\wedge e^3)=e^2$ and $*(e^0\wedge e^2\wedge e^3)=-e^1$ . Reversing the spacetime orientation, or defining the twist with an extra minus sign, reverses $χ$ and both displayed signs without changing the metric. This translation must be made before comparing Ernst potentials between sources. Introduce $\mathcal E=f+i\chi$ and

\Delta_3=\partial_\rho^2+\rho^{-1}\partial_\rho+\partial_z^2.

The reduced vacuum equations are the Ernst equation

(\operatorname{Re}\mathcal E)\Delta_3\mathcal E =(\partial_\rho\mathcal E)^2+(\partial_z\mathcal E)^2.

Once $f$ and $χ$ are known, $A$ follows from the twist relations and $k$ from

\begin{aligned} \partial_\rho k &=\frac{\rho}{4f^2} \left[f_\rho^2-f_z^2+\chi_\rho^2-\chi_z^2\right],\\ \partial_z k &=\frac{\rho}{2f^2} \left[f_\rho f_z+\chi_\rho\chi_z\right]. \end{aligned}

The Ernst equation is exactly the integrability condition for the last two relations. The reduction to a harmonic map and its target geometry are derived in Chruściel, Costa, and Heusler 2012, §6.2, pp. 31–32 and §§8.2–8.3, pp. 42–45.

For rotating black holes, $T$ becomes spacelike in the ergoregion, so the stationary-potential chart is not the best global uniqueness variable. One instead reduces with the axial field, whose norm is positive off the axis, and packages its norm and normalized twist potential into a positive matrix $\mathbf\Phi$ . Its current $J_a=\mathbf\Phi^{-1}\partial_a\mathbf\Phi$ satisfies

\partial_\rho(\rho J_\rho)+\partial_z(\rho J_z)=0.

This is the globally useful symmetric-space form. It is equivalent to the Ernst description where both are regular, but it avoids pretending that the stationary field stays timelike throughout Kerr.

Axis, horizon, and infinity data

At infinity, normalize $f\to1$ , $A\to0$ , $χ\to0$ , and $k\to0$ . The next terms encode ADM mass and Komar angular momentum. On each regular axis rod, $Φ=0$ and elementary flatness fixes the limiting circumference-to-radius ratio; in the displayed gauge this sets the regular-axis constant of $k$ to zero and excludes a helical time identification. For the axial twist potential, $dY=*(\Phi\wedge d\Phi)$ vanishes tangentially on an axis rod, so $Y$ is constant there. Its north–south jump is proportional to $GJ$ , with the sign fixed by the chosen orientation and Komar normalization.

On a horizon rod, the Gram matrix has a one-dimensional null kernel generated by

\chi_H=T+\Omega_H\Phi.

Regularity fixes the limiting behavior of the potentials at the rod and at both axis–horizon corners. The rod length supplies additional geometric data; charges alone do not specify a general disconnected rod structure. For a connected, nondegenerate, asymptotically flat vacuum black hole, the Kerr data match these conditions and the divergence or distance identity proves that no second regular map has the same data. A conical defect on an axis rod is a failure of the boundary conditions, even if the formal Ernst equation holds.

Check the static reduction

Set $χ=0$ and take $A=0$ . Show that the real Ernst equation becomes the axisymmetric Laplace equation for $U=\tfrac12\log f$ . What does this check—and what does it not prove?

Solution

The real equation is $f\Delta_3f=|\nabla f|^2$ . Therefore

\Delta_3\log f =f^{-1}\Delta_3f-f^{-2}|\nabla f|^2=0,

and hence $\Delta_3U=0$ away from the axis and horizon rod. This checks the sign and normalization of the nonlinear term: a wrong sign would not reduce to the Weyl harmonic potential. It does not prove Schwarzschild uniqueness, because many formal harmonic functions exist. Horizon regularity, absence of conical defects, asymptotic flatness, completeness, and the global rod structure are still needed to select the regular black-hole exterior.

References

Carter, Brandon. “Killing Horizons and Orthogonally Transitive Groups in Space-Time.” Journal of Mathematical Physics 10 (1969): 70–81. DOI.
Chruściel, Piotr T., João Lopes Costa, and Markus Heusler. “Stationary Black Holes: Uniqueness and Beyond.” Living Reviews in Relativity 15 (2012): 7. DOI. Open PDF.