A Penrose-type inequality with angular momentum and charge for axisymmetric initial data

Abstract

A lower bound for the ADM mass is established in terms of angular momentum, charge, and horizon area in the context of maximal, axisymmetric initial data for the Einstein–Maxwell equations which satisfy the weak energy condition. If, on the horizon, the given data agree to a certain extent with the associated model Kerr–Newman data, then the inequality reduces to the conjectured Penrose inequality with angular momentum and charge. In addition, a rigidity statement is also proven whereby equality is achieved if and only if the data set arises from the canonical slice of a Kerr–Newman spacetime.

Appendices

Appendix A. Weyl coordinates

Here we prove Lemma 2.1. In [9] the existence of Weyl coordinates was established by first constructing so called pseudospherical coordinates $(\rho _{s},z_{s},\phi )$, in which the initial data boundary $\partial M$ is represented by a semi-circle of radius $\frac{m_0}{2}$ about the origin in the $\rho _s z_s$-plane. This contrasts with Weyl coordinates in which the boundary takes the form of an interval on the z-axis in the orbit space. Pseudospherical coordinates are valid on the planar region ${\mathbb {C}}_{+}\setminus D_{m_0/2}=\{\rho _s+iz_s\mid \rho _{s}>0, r_s>m_0/2\}$, where $r_s^2=\rho _s^2+z_s^2$. In these coordinates the metric takes the standard ‘Brill’ form

$$\begin{aligned} g=e^{-2U_{s}+2\alpha _{s}}(d\rho _{s}^{2}+dz_{s}^{2})+\rho _s^{2}e^{-2U_{s}} (d\phi +A_{\rho _s} d\rho _{s}+A_{z_s} dz_{s})^{2}. \end{aligned}$$

(A.1)

This structure for the metric is preserved under any coordinate change of the plane which yields a conformal transformation, and Weyl coordinates are a particular example of this. The metric coefficients are axisymmetric, smooth up to the boundary in ${\mathbb {C}}_{+}\setminus D_{m_0/2}$ with $\alpha _s=0$ on the $z_s$-axis, and satisfy the fall-off

$$\begin{aligned} U_{s}=O_{1}(r_{s}^{-1/2-\epsilon }),\quad \alpha _{s}=O_{1}(r_{s}^{-1/2-\epsilon }), \quad A_{\rho _s}=O_{1}(r_{s}^{-3/2-\epsilon }),\quad A_{z_s}=O_{1}(r_{s}^{-3/2-\epsilon }).\nonumber \\ \end{aligned}$$

(A.2)

Weyl coordinates $(\rho ,z,\phi )$ are constructed from pseudospherical coordinates as follows. Define complex coordinates $\zeta _{s}=\rho _{s}+iz_{s}$ and $\zeta =\rho +iz$ and consider the holomorphic diffeomorphism $f:{\mathbb {C}}_+\setminus D_{m_0/2}\rightarrow {\mathbb {C}}_+$ given by

$$\begin{aligned} \zeta =f(\zeta _{s})=\zeta _{s}-\frac{m_0^{2}}{4\zeta _{s}}\quad \Rightarrow \quad \rho =\frac{\rho _{s}(r_{s}^{2}-\frac{m_0^{2}}{4})}{r_{s}^{2}},\quad z=\frac{z_{s}(r_{s}^{2}+\frac{m_0^{2}}{4})}{r_{s}^{2}}.\nonumber \\ \end{aligned}$$

(A.3)

Observe that

$$\begin{aligned} \frac{\partial {\zeta }}{\partial {\zeta _s}}=1+\frac{m_0^2}{4\zeta _s^2}, \end{aligned}$$

(A.4)

which is smooth up to the boundary of ${\mathbb {C}}_+\setminus D_{m_0/2}$ and is nonzero except at the points $\zeta _s=\pm \frac{m_0}{2}i$. Thus by the inverse function theorem, the inverse transformation is holomorphic and has bounded derivatives away from the poles $\zeta =\pm m_0i$ of the horizon. Near these points we have

$$\begin{aligned} \left| \frac{\partial {\zeta }}{\partial {\zeta _s}}\right| \ge C^{-1}|\zeta _s \mp \frac{m_0}{2}i|\quad \Rightarrow \quad \left| \frac{\partial {\zeta _s}}{\partial {\zeta }}\right| \le \frac{C}{|\zeta _s \mp \frac{m_0}{2}i|}. \end{aligned}$$

(A.5)

In particular, all first derivatives of the real and imaginary parts admit the bound

$$\begin{aligned} \left| \frac{\partial {\rho _s}}{\partial {\rho }}\right| + \left| \frac{\partial {\rho _s}}{\partial {z}}\right| + \left| \frac{\partial {z_s}}{\partial {\rho }}\right| + \left| \frac{\partial {z_s}}{\partial {z}}\right| \le \frac{C}{|\zeta _s \mp \frac{m_0}{2}i|} \end{aligned}$$

(A.6)

near the poles.

The relationship between U, $\alpha$ of Weyl coordinates and $U_s$, $\alpha _s$ of pseudospherical coordinates is given by [9]

$$\begin{aligned} U(\rho ,z)=U_{s}(\rho _{s},z_{s})-\log \frac{\rho _{s}}{\rho },\quad \quad \alpha (\rho ,z)=\alpha _{s}(\rho _s,z_s) +\log \frac{|\zeta _{s}|^{2}-\frac{m_0^{2}}{4}}{|\zeta _{s}^{2}+\frac{m_0^{2}}{4}|}.\nonumber \\ \end{aligned}$$

(A.7)

Note that the second term on the right-hand side of both expressions depends only on the coordinate transformation. For the Schwarzschild solution

$$\begin{aligned} U_{s,0}(\rho _{s},z_{s})=-2\log \frac{2r_{s}+m_0}{2r_{s}},\quad \quad \alpha _{s,0}(\rho _s,z_s)=0, \end{aligned}$$

(A.8)

and the expressions for the Schwarzschild data $U_0$ and $\alpha _0$ in Weyl coordinates may then be obtained from the above formulas. We may then write $U=U_{0}+{\overline{U}}$ and $\alpha =\alpha _{0}+{\overline{\alpha }}$ where

$$\begin{aligned} {\overline{U}}(\rho ,z):=U(\rho ,z)-U_{0}(\rho ,z) =U_{s}(\rho _{s},z_{s})-U_{s,0}(\rho _{s},z_{s}), \end{aligned}$$

(A.9)

and

$$\begin{aligned} {\overline{\alpha }}(\rho ,z):=\alpha (\rho ,z)-\alpha _{0}(\rho ,z) =\alpha _{s}(\rho _{s},z_{s}). \end{aligned}$$

(A.10)

It immediately follows that ${\overline{U}}$ and ${\overline{\alpha }}$ are uniformly bounded and satisfy the desired decay at infinity. Furthermore since $U_s$, $U_{s,0}$, and $\alpha _s$ are smooth, the regularity properties of ${\overline{U}}$ and ${\overline{\alpha }}$ depend on the coordinate transformation $f^{-1}$, and the only possible issues arise at the poles.

Consider the partial derivative

$$\begin{aligned} \frac{\partial {\overline{U}}}{\partial \rho }=\left( \frac{\partial U_{s}}{\partial \rho _{s}}-\frac{\partial U_{s,0}}{\partial \rho _{s}}\right) \frac{\partial {\rho _{s}}}{\partial {\rho }} +\left( \frac{\partial U_{s}}{\partial {z_{s}}}-\frac{\partial U_{s,0}}{\partial {z_{s}}}\right) \frac{\partial {z_{s}}}{\partial {\rho }}. \end{aligned}$$

(A.11)

Since the horizon is a minimal surface

$$\begin{aligned} \frac{\partial }{\partial r_{s}}(U_{s}-\frac{1}{2}\alpha _{s}) =\frac{2}{m_0}=\frac{\partial U_{s,0}}{\partial r_{s}}\quad \quad \text { when }r_s=\frac{m_0}{2}. \end{aligned}$$

(A.12)

In particular this holds at $(\rho _{s},z_{s})=(0,\pm m_0/2)$. Moreover, since $\alpha _{s}=0$ on the axis and $\partial _{r_s}$ coincides with $\pm \partial _{z_{s}}$ there, we have

$$\begin{aligned} \left( \frac{\partial U_{s}}{\partial z_{s}}-\frac{\partial U_{s,0}}{\partial z_{s}}\right) \left( 0,\pm \frac{m_0}{2}\right) =0. \end{aligned}$$

(A.13)

Next, use the fact that all functions are axisymmetric to find

$$\begin{aligned} \frac{\partial U_{s}}{\partial \rho _{s}}\left( 0,\pm \frac{m_0}{2}\right) =\frac{\partial U_{s,0}}{\partial \rho _{s}}\left( 0,\pm \frac{m_0}{2}\right) =0. \end{aligned}$$

(A.14)

Therefore the first derivatives of $U_{s}-U_{s,0}$ vanish at the poles. This, combined with the smoothness of this function up to the boundary, shows that even though $\partial _{\rho }\rho _s$ and $\partial _{\rho }z_s$ may blow-up at these points in a manner controlled by (A.6), the full expression (A.11) remains bounded. Similar considerations may be used to treat the $\partial _z {\overline{U}}$ and the derivatives of ${\overline{\alpha }}$.

Appendix B. Relation of ${\overline{\beta }}$ to surface gravity

Here we compute ${\overline{\beta }}={\overline{\alpha }}-2{\overline{U}}$ on the horizon rod for the Kerr black hole. Let us recall the constant time slice Kerr metric $g_{kerr}$ in Weyl coordinates [25]. We will denote the mass and angular momentum of the Kerr metric by m and ${\mathcal {J}}=ma$, while the notation for half the horizon rod length will be $m_0$. Then

$$\begin{aligned} g_{kerr}=e^{-2U_{kerr}+2\alpha _{kerr}}(d\rho ^{2}+dz^{2})+\rho ^{2}e^{-2U_{kerr}}d\phi ^{2}, \end{aligned}$$

(B.1)

where

$$\begin{aligned}&e^{-2U_{kerr}+2\alpha _{kerr}} =\frac{m_0^{2}\left( r_{+}+r_{-}+2m\right) ^{2}+a^{2}(r_{+}-r_{-})^{2}}{4m_0^{2}r_{+}r_{-}}, \end{aligned}$$

(B.2)

$$\begin{aligned}&\rho ^{2}e^{-2U_{kerr}} \nonumber \\&\quad = \frac{m_0^{2}\left( r_{+}+r_{-}+2m\right) ^{2}+a^{2}(r_{+}-r_{-})^{2}}{m_0^{2}\left( (r_{+}+r_{-})^{2}-4m^{2}\right) +a^{2}(r_{+}-r_{-})^{2}}\rho ^{2} \nonumber \\&\qquad -\frac{\left[ am(r_{+}+r_{-}+2m)(4m_0^{2}-(r_{+}-r_{-})^{2})\right] ^{2}}{\left[ m_0^{2}\left( (r_{+}+r_{-})^{2}-4m^{2}\right) +a^{2}(r_{+}-r_{-})^{2}\right] \left[ m_0^{2}\left( r_{+}+r_{-}+2m\right) ^{2}+a^{2}(r_{+}-r_{-})^{2}\right] },\nonumber \\ \end{aligned}$$

(B.3)

with $r_{\pm }=\sqrt{\rho ^{2}+(z\pm m_0)^{2}}$. Write $U_{kerr}=U_0+{\overline{U}}_{kerr}$ and $\alpha _{kerr}=\alpha _0+{\overline{\alpha }}_{kerr}$, where $U_0$ and $\alpha _0$ are the corresponding Schwarzschild functions. It follows that for $|z|<m_0$ we have

$$\begin{aligned} {\overline{U}}_{kerr}(0,z)=-\frac{1}{2} \log \left( \frac{m^{2}(m+m_{0})^{2}}{m_0^{2}(m+m_{0})^{2}+a^{2}z^{2}}\right) , \end{aligned}$$

(B.4)

and

$$\begin{aligned} {\overline{\alpha }}_{kerr}(0,z)=\frac{1}{2}\log \frac{\left[ m_0^{2}(m+m_{0})^{2}+a^{2}z^{2}\right] ^{2}}{4m_0^{4}m^{2}(m+m_{0})^{2}}. \end{aligned}$$

(B.5)

Notice that ${\mathcal {J}}=0$ implies that ${\overline{U}}_{kerr}(0,z)={\overline{\alpha }}_{kerr}(0,z)=0$ as expected, since half the horizon rod length is given by

$$\begin{aligned} m_0=\sqrt{m^{2}-a^{2}} =\sqrt{m^2-\frac{{\mathcal {J}}^{2}}{m^{2}}}. \end{aligned}$$

(B.6)

We now have that on the horizon rod

$$\begin{aligned} {\overline{\beta }}_{kerr}(0,z)={\overline{\alpha }}_{kerr}(0,z)-2{\overline{U}}_{kerr}(0,z) =\log \frac{m(m+m_{0})}{2m_0^{2}}\ge 0. \end{aligned}$$

(B.7)

Consider now the surface gravity of the Kerr black hole

$$\begin{aligned} \kappa =\frac{\sqrt{m^4-{\mathcal {J}}^2}}{2\left( m^3+m\sqrt{m^4-{\mathcal {J}}^2}\right) }. \end{aligned}$$

(B.8)

Comparing the two formulas produces

$$\begin{aligned} {\overline{\beta }}_{kerr}(0,z)=-\log (4m_0\kappa ). \end{aligned}$$

(B.9)