Full-sphere exterior extension of the Kerr-Newman datum

Let $M,a,q\in \mathbb{R}$ have units of metres, choose $R_0>4\mathrm{\mid }M\mathrm{\mid }$, and set $\mathcal{X}:={\mathbb{R}}_{x^0}\times (R_0,\mathrm{\infty }{)}_r\times S^2,R=r^2+a^2,b=1-Z^2,\mathrm{\Sigma }=r^2+a^2{Z}^2,\mathrm{\Delta }=r^2-2Mr+a^2+q^2,$ where $S^2=\{(X,Y,Z)\in {\mathbb{R}}^3:X^2+Y^2+Z^2=1\}$ has its induced unit round metric $\gamma$ and outward area form $\omega$. Specify the sphere smooth structure by The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation, Definition (b): on $S^2\setminus \{(0,0,\epsilon )\}$ use ${\sigma }_{\epsilon }(X,Y,Z)=(X,Y)\mathrm{/}(1-\epsilon Z)$, $\epsilon =\pm 1$. With $h=u_1^2+u_2^2$, the inverse is $u\mapsto (2u_1,2u_2,\epsilon (h-1))\mathrm{/}(1+h)$, and the transition on $h>0$ is $u\mapsto u\mathrm{/}h$. These maps are smooth on their domains. Give $\mathcal{X}$ the maximal atlas generated by the product charts $(x^0,r,p)\mapsto (x^0,r,{\sigma }_{\epsilon }(p))$; its transitions leave $x^0,r$ fixed and apply the displayed sphere transition. The product is Hausdorff and second countable as a subspace of ${\mathbb{R}}^5$, using rational balls for a countable basis. The inverse parametrizations make the sphere inclusion smooth with injective differential, so ${\gamma }_p(u,v)=u\cdot v$ and ${\omega }_p(u,v)=\det (p,u,v)$ are smooth tensors with the stated outward orientation. Write $\lambda :=XdY-YdX,\bar{g}:=-\frac{\mathrm{\Delta }-a^{2}b}{\mathrm{\Sigma }}(d{x}^0{)}^2+\frac{\mathrm{\Sigma }}{\mathrm{\Delta }}d{r}^2+\mathrm{\Sigma }\gamma +\frac{a^2(R+\mathrm{\Sigma }-\mathrm{\Delta })}{\mathrm{\Sigma }}{\lambda }^2-\frac{2a(R-\mathrm{\Delta })}{\mathrm{\Sigma }}d{x}^0\odot \lambda ,$ $\bar{\mathcal{A}}:=-\frac{qr}{\mathrm{\Sigma }}(d{x}^0-a\lambda ),\bar{\mathcal{F}}:=d\bar{\mathcal{A}}\mathrm{.}$ Here ${\lambda }^2=\lambda \otimes \lambda$ and $u\odot v=(u\otimes v+v\otimes u)\mathrm{/}2$. This is the full-sphere exterior extension of the pole-free datum in Kerr-Newman metric and potential on the exact Boyer-Lindquist domain. The coordinates $x^0,r$ and parameters $M,a,q$ have units of metres; sphere coordinates are dimensionless. The field $\bar{\mathcal{F}}$ is the geometric length-normalized two-form and no SI conversion is part of this definition.

The coordinate functions $X,Y,Z$, their differentials, the induced metric $\gamma$, and $\lambda =XdY-YdX$ are smooth on $S^2$; thus the displayed metric and potential are smooth wherever $\mathrm{\Sigma }>0$. Since $r>R_0>0$, $\mathrm{\Sigma }>0$. Also $\mathrm{\Delta }\mathrm{/}{r}^2=1-2M\mathrm{/}r+(a^2+q^2)\mathrm{/}{r}^2>1\mathrm{/}2,$ so $\mathrm{\Delta }>0$. The scalar $\mathrm{\Delta }-a^{2}b=r^2-2Mr+q^2+a^2{Z}^2$ is positive, hence $\bar{g}({\mathrm{\partial }}_{x^0},{\mathrm{\partial }}_{x^0})<0$; take $N={\mathrm{\partial }}_{x^0}$ as time orientation and $o=d{x}^0\wedge dr\wedge \omega$ as orientation.

On the polar chart away from the poles, put $Z=\cos \theta$ and $b=1-Z^2={\sin }^2\theta$. Then $\lambda =bd\varphi$ and $\gamma =d{\theta }^2+bd{\varphi }^2$. The identity $R^2-{\mathrm{\Sigma }}^2=a^{2}b(R+\mathrm{\Sigma })$ gives $\frac{R^2-a^2\mathrm{\Delta }b}{\mathrm{\Sigma }}-\mathrm{\Sigma }=\frac{a^{2}b(R+\mathrm{\Sigma }-\mathrm{\Delta })}{\mathrm{\Sigma }}\mathrm{.}$ Consequently the angular part $\mathrm{\Sigma }d{\theta }^2+b(R^2-a^2\mathrm{\Delta }b){\mathrm{\Sigma }}^{-1}d{\varphi }^2$ equals $\mathrm{\Sigma }\gamma +a^2(R+\mathrm{\Sigma }-\mathrm{\Delta }){\mathrm{\Sigma }}^{-1}{\lambda }^2$. The time-time and time-angle coefficients are respectively $-(\mathrm{\Delta }-a^{2}b)\mathrm{/}\mathrm{\Sigma }$ and $-a(R-\mathrm{\Delta })b\mathrm{/}\mathrm{\Sigma }$, exactly those obtained by expanding the Boyer-Lindquist formula. Thus $\bar{g}$ and $\bar{\mathcal{A}}$ agree with Kerr-Newman metric and potential on the exact Boyer-Lindquist domain on the overlap.

On the dense pole-free region, Kerr-Newman metric and potential on the exact Boyer-Lindquist domain gives $\det (g_{ij})=-{\mathrm{\Sigma }}^2$ in coordinates $(x^0,r,\cos \theta ,\varphi )$ and signature $(-+++)$ because $\mathrm{\Delta }>0$. At either pole $\lambda =0$ and $b=0$, so $\bar{g}=-\mathrm{\Delta }{\mathrm{\Sigma }}^{-1}(d{x}^0{)}^2+\mathrm{\Sigma }{\mathrm{\Delta }}^{-1}d{r}^2+\mathrm{\Sigma }\gamma$ at that point; these four directions have one negative and three positive squares. Hence $(\mathcal{X},\bar{g},N,o,\bar{\mathcal{F}})$ is a smooth time-oriented Lorentzian datum on the full sphere exterior. The smooth global potential also gives $d\bar{\mathcal{F}}=0$. No extension across a horizon, through $r=0$, or to a maximal spacetime is asserted.