The positive mass theorem for multiple rotating charged black holes

Abstract

In this paper a lower bound for the ADM mass is given in terms of the angular momenta and charges of black holes present in axisymmetric initial data sets for the Einstein–Maxwell equations. This generalizes the mass-angular momentum-charge inequality obtained by Chrusciel and Costa to the case of multiple black holes. We also weaken the hypotheses used in the proof of this result for single black holes, and establish the associated rigidity statement.

Appendices

Revisiting the heuristic arguments

The heuristic physical arguments which motivate (1.1) go back to Penrose’s original derivation of the Penrose inequality [21]. Typically in such arguments, it is assumed that the end state of gravitational collapse is a single Kerr–Newman black hole. However, a more appropriate assumption for the end state is a finite number of mutually distant Kerr–Newman black holes moving apart with asymptotically constant velocity. This should be the result, if for instance, two distant black holes were initially moving away from each other sufficiently fast. We will now describe the heuristic arguments for the mass-angular momentum-charge inequality in this setting. It appears that this has not been previously considered in the literature.

Let $m_{i}$, $\mathcal {J}_i$, $q_{i}$ denote the ADM masses, angular momenta, and total charges of the end state black holes. Then the total (ADM) mass, angular momentum, and charge of the end state is $m=\sum m_i$, $\mathcal {J}=\sum \mathcal {J}_{i}$, $q=\sum q_i$. In a Kerr–Newman black hole these quantities satisfy the Eq. [14]

$$\begin{aligned} m_{i}^{2}=\frac{A_{i}}{16\pi }+\frac{q_{i}^2}{2}+\frac{\pi (q_{i}^{4}+4\mathcal {J}_{i}^{2})}{A_{i}}, \end{aligned}$$

(5.1)

where $A_{i}$ denotes horizon area. Moreover, as a function of $A_{i}$ (keeping $\mathcal {J}_{i}$ and $q_{i}$ fixed), the right-hand side is nondecreasing precisely when

$$\begin{aligned} A_{i}\ge 4\pi \sqrt{q_{i}^{4}+4\mathcal {J}_{i}^{2}}, \end{aligned}$$

(5.2)

and this inequality is always satisfied with equality only for extreme black holes. Thus, computing the minimum value of the right-hand side of (5.1) yields

$$\begin{aligned} m_{i}^2\ge \frac{q_{i}^{2}+ \sqrt{q_{i}^4 + 4\mathcal {J}_{i}^2}}{2}, \end{aligned}$$

(5.3)

with equality only for extreme black holes. Let $m_{0}$, $\mathcal {J}_{0}$, $q_{0}$ denote the ADM mass, angular momentum, and total charge of an initial state. Under appropriate hypotheses, such as axisymmetry and the existence of a twist potential, angular momentum is conserved $\mathcal {J}_{0}=\mathcal {J}=\sum \mathcal {J}_{i}$. Moreover, by assuming that no charged matter is present, the total charge is conserved $q_0=q=\sum q_i$, and since gravitational waves may only carry away positive energy $m_0 \ge m = \sum m_i$.

Lemma 5.1

Let $a_i, b_i\in \mathbb {R}$ and let $a=\sum a_i$, $b=\sum b_i$. Then

$$\begin{aligned} \left( a^4+b^2\right) ^{1/4} \le \sum \left( a_i^4 + b_i^2\right) ^{1/4}. \end{aligned}$$

(5.4)

Proof

Let $c_i=|b_i|^{1/2}$ and $c=\sum c_i$, then

$$\begin{aligned} |b|^{1/2} \le \left( \sum |b_i|\right) ^{1/2}=\left( \sum c_i^2\right) ^{1/2} \le \sum c_i = c. \end{aligned}$$

(5.5)

Hence $b^2\le c^4$. We conclude that

$$\begin{aligned} \left( a^4 + b^2\right) ^{1/4} \le \left( a^4+c^4\right) ^{1/4} \le \sum \left( a_i^4+c_i^4\right) ^{1/4} = \sum \left( a_i^4 + b_i^2\right) ^{1/4}. \end{aligned}$$

(5.6)

$\square$

Lemma 5.2

Let $a_i, b_i\in \mathbb {R}$ and let $a=\sum a_i$, $b=\sum b_i$. Then

$$\begin{aligned} \sqrt{a^2 + \sqrt{a^4+b^2}} \le \sum \sqrt{a_i^2 + \sqrt{a_i^4+b_i^2}}. \end{aligned}$$

(5.7)

Proof

By Lemma 5.1

$$\begin{aligned} \left( a^4+b^2\right) ^{1/2} \le \left( \sum \left( a_i^4+b_i^2\right) ^{1/4} \right) ^2. \end{aligned}$$

(5.8)

Thus, it follows that

$$\begin{aligned} \sqrt{a^2 + \sqrt{a^4+b^2}} \le \sqrt{\left( \sum a_i\right) ^2 +\left( \sum \left( a_i^4+b_i^2\right) ^{1/4} \right) ^2} \le \sum \sqrt{a_i^2 + \sqrt{a_i^4+b_i^2}} \end{aligned}$$

(5.9)

$\square$

Now, let $a_i=q_i$, and $b_i=2\mathcal {J}_i$, then we get

$$\begin{aligned} \sqrt{2} m = \sqrt{2} \sum m_i \ge \sum \sqrt{q_i^2 + \sqrt{q_i^4+4\mathcal {J}_i^2}} \ge \sqrt{q^2 + \sqrt{q^4+4\mathcal {J}^2}}. \end{aligned}$$

(5.10)

Squaring both sides yields the desired result (1.1). We conclude that the heuristic arguments are sufficiently robust to support the mass-angular momentum-charge inequality, even for spacetimes with multiple black holes moving apart from one another at high velocities.

The extreme Kerr–Newman and Majumdar–Papapetrou harmonic maps

First we record formulas for the extreme Kerr–Newman harmonic map. Recall that in Boyer-Lindquist coordinates the Kerr–Newmann metric takes the form

$$\begin{aligned} -\frac{\Delta - a^{2}\sin ^{2}\theta }{\Sigma } dt^{2}+ & {} \frac{2a\sin ^{2}\theta }{\Sigma }\left( \widetilde{r}^{2}+a^{2}-\Delta \right) dtd\phi \nonumber \\+ & {} \frac{(\widetilde{r}^{2}+a^{2})^{2} - \Delta a^{2}\sin ^{2}\theta }{\Sigma } \sin ^{2}\theta d\phi ^{2} + \frac{\Sigma }{\Delta } d\widetilde{r}^{2} + \Sigma d\theta ^{2} \end{aligned}$$

(6.1)

where

$$\begin{aligned} \Delta = \widetilde{r}^{2} + a^{2} +q^{2} -2m\widetilde{r}, \quad \Sigma = \widetilde{r}^{2} + a^{2}\cos ^{2}\theta , \end{aligned}$$

(6.2)

and the electromagnetic 4-potential is given by

$$\begin{aligned} \mathbf {A} = -\frac{q_{e}\widetilde{r}}{\Sigma } \left( dt+ a\sin ^{2}\theta d\phi \right) - \frac{q_{b}\cos \theta }{\Sigma } \left( a dt + (\widetilde{r}^{2}+a^{2}) d \phi \right) , \end{aligned}$$

(6.3)

The event horizon is located at the larger of the two solutions to the quadratic equation $\Delta =0$, namely $\widetilde{r}_{+}=m+\sqrt{m^{2}-a^{2}-q^{2}}$, where the angular momentum is given by $\mathcal {J} = ma$. For $\widetilde{r}>\widetilde{r}_{+}$ it holds that $\Delta >0$, so that a new radial coordinate may be defined by

$$\begin{aligned} r=\frac{1}{2}(\widetilde{r}-m+\sqrt{\Delta }), \end{aligned}$$

(6.4)

or rather

$$\begin{aligned}&\widetilde{r} = r + m + \frac{m^{2}-a^{2}-q^{2}}{4r}, \quad m^{2} \ne a^{2}+q^{2} \nonumber \\&\widetilde{r} = r + m, \quad m^{2}=a^{2}+q^{2}. \end{aligned}$$

(6.5)

Note that the new coordinate is defined for $r>0$, and a critical point for the right-hand side of (6.5) ($m^{2}\ne a^{2} + q^{2}$) occurs at the horizon, so that two isometric copies of the outer region are encoded on this interval. The coordinates $(r,\theta ,\phi )$ then form a (polar) Brill coordinate system, which is related to the (cylindrical) Brill coordinates via the usual transformation $\rho =r\sin \theta$, $z=r\cos \theta$. Finally, the harmonic map $(u_\mathrm{KN},v_\mathrm{KN},\chi _\mathrm{KN},\psi _\mathrm{KN}):\mathbb {R}^{3}{\setminus }\Gamma \rightarrow \mathbb {H}^{2}_{\mathbb {C}}$, $U_\mathrm{KN}=u_\mathrm{KN}+\log \rho$, which determines the extreme Kerr–Newman solution is given by

$$\begin{aligned}&u_\mathrm{KN}=-\frac{1}{2}\log \left[ \left( \widetilde{r}^{2}+a^{2}+\frac{a^{2}\sin ^{2}\theta (2m\widetilde{r}-q^{2})}{\Sigma }\right) \sin ^{2}\theta \right] ,\nonumber \\&v_\mathrm{KN}=ma\cos \theta (3-\cos ^{2}\theta )-\frac{a(q^{2}\widetilde{r}-ma^{2}\sin ^{2}\theta )\cos \theta \sin ^{2}\theta }{\Sigma },\nonumber \\&\chi _\mathrm{KN}=-\frac{qa\widetilde{r}\sin ^{2}\theta }{\Sigma },\nonumber \\&\psi _\mathrm{KN}=\frac{q(\widetilde{r}^{2}+a^{2})\cos \theta }{\Sigma }. \end{aligned}$$

(6.6)

The Euler-Lagrange equations satisfied by this and any other harmonic map $\Psi :\mathbb {R}^{3}\rightarrow \mathbb {H}^{2}_{\mathbb {C}}$ are given by

$$\begin{aligned} \Delta u-2e^{4u}|\nabla v+\chi \nabla \psi -\psi \nabla \chi |^{2} -e^{2u}(|\nabla \chi |^{2}+|\nabla \psi |^{2})&=0,\nonumber \\ {\text {div}}\left[ e^{4u}(\nabla v+\chi \nabla \psi -\psi \nabla \chi )\right]&=0,\nonumber \\ {\text {div}}(e^{2u}\nabla \chi )-2e^{4u}\nabla \chi \cdot (\nabla v+\chi \nabla \psi -\psi \nabla \chi )&=0,\nonumber \\ {\text {div}}(e^{2u}\nabla \psi )+2e^{4u}\nabla \psi \cdot (\nabla v+\chi \nabla \psi -\psi \nabla \chi )&=0. \end{aligned}$$

(6.7)

Consider now the Majumdar–Papapetrou spacetime $\left( \mathbb {R}\times (\mathbb {R}^{3} {\setminus }\cup _{n=1}^{N}p_{n}),ds^2\right)$ with

$$\begin{aligned} ds^2=-f^{-2}dt^2+f^{2}\delta , \quad f=1+\sum _{n=1}^{N}\frac{m_{n}}{r_{n}}, \end{aligned}$$

(6.8)

where $m_{n}=\sqrt{(q^{e}_{n})^{2}+(q^{b}_{n})^{2}}$ represents the mass and total electromagnetic charge of each black hole, $\delta$ is the Euclidean metric, and $r_{n}$ is the Euclidean distance to each puncture. Axisymmetry may be imposed by choosing the punctures $p_{n}$ to lie on the z-axis. Cylindrical coordinates $(\rho ,z,\phi )$ in 3-space give rise to Brill coordinates with $U_\mathrm{MP}=-\log f$, and the 4-potential is given by

$$\begin{aligned} \mathbf {A}=\kappa fdt+\sqrt{1-\kappa ^2}\sum _{n=1}^{N}\frac{m_{n}(z-z_{n})}{r_{n}}d\phi , \quad 0\le \kappa \le 1. \end{aligned}$$

(6.9)

The constant $\kappa$ relates the electric and magnetic charges to the mass by $q_{n}^{e}=\kappa m_{n}$ and $q_{n}^{b}=\sqrt{1-\kappa ^{2}}m_{n}$. Typically the Majumdar–Papapetrou spacetime is stated without magnetic charges, however through a duality rotation

$$\begin{aligned} E=(\cos \vartheta )\tilde{E}-(\sin \vartheta )\tilde{B}, \quad B=(\sin \vartheta )\tilde{E}+(\cos \vartheta )\tilde{B}, \end{aligned}$$

(6.10)

magnetic charge may be introduced so that $\kappa =\cos \vartheta$. Since $E=\kappa \nabla \log f$ and $B=\sqrt{1-\kappa ^{2}}\nabla \log f$, the electromagnetic potentials are obtained from (2.7)

$$\begin{aligned} d\chi _\mathrm{MP}=\kappa \rho (\partial _{z}f d\rho -\partial _{\rho }f dz), \quad d\psi _\mathrm{MP}=\sqrt{1-\kappa ^{2}}\rho (\partial _{z}f d\rho -\partial _{\rho }f dz), \end{aligned}$$

(6.11)

so that

$$\begin{aligned} \chi _\mathrm{MP}=\kappa \sum _{n=1}^{N}\frac{m_{n}(z-z_{n})}{r_{n}}, \quad \chi _\mathrm{MP}=\sqrt{1-\kappa ^{2}}\sum _{n=1}^{N}\frac{m_{n}(z-z_{n})}{r_{n}}. \end{aligned}$$

(6.12)

Lastly, since this spacetime is static there is no angular momentum, and hence $v_\mathrm{MP}=0$. This, combined with the fact that $\chi _\mathrm{MP}$ and $\psi _\mathrm{MP}$ are proportional leads to a harmonic map with a 2-dimensional target that is isometric to hyperbolic space, namely $(u_\mathrm{MP},v_\mathrm{MP},\chi _\mathrm{MP},\psi _\mathrm{MP}):\mathbb {R}^{3}{\setminus }\Gamma \rightarrow \mathbb {H}^{2}\subset \mathbb {H}_{\mathbb {C}}^{2}$ where $U_\mathrm{MP}=u_\mathrm{MP}+\log \rho$.