Geometric electrovac equations and flux normalization — Examples

Set $a=0$ and retain the electric parameter $q$. Then $\mathrm{\Sigma }=r^2$, $\mathrm{\Delta }=r^2-2Mr+q^2$, and the KN line element reduces term by term to $-(1-\frac{2M}{r}+\frac{q^2}{r^2})(d{x}^0{)}^2+{(1-\frac{2M}{r}+\frac{q^2}{r^2})}^{-1}d{r}^2+r^{2}d{\mathrm{\Omega }}^2\mathrm{.}$ Its potential becomes $-qd{x}^0\mathrm{/}r$. For every open $J\subset (0,\mathrm{\infty })$ on which $f=\mathrm{\Delta }\mathrm{/}{r}^2\mathrm{\ne }0$, the resulting metric, potential and field on the common pole-free chart are the restrictions of the electric RN datum with $q_e=q,q_m=0$ in [F2]. When $r<0$ is admitted by the KN definition, the same coordinate expressions hold, but that chart is outside the positive-radius RN datum in [F2]. No global-domain equality or uniqueness theorem is used.

Set $q=0$ with $a$ arbitrary. The potential and field vanish, while $\mathrm{\Delta }=r^2-2Mr+a^2$ and the displayed metric is the vacuum Kerr form. Since the KN residual identity of [F1] remains valid at $q=0$, its Ricci tensor vanishes. This is the vacuum rotating limit, not a classification claim.

Set both $a=0$ and $q=0$. On the component $M>0$ and $r>2M$, the metric function is $1-2M\mathrm{/}r$. With $s=2M$ and $m_{\ast }=c^{2}M\mathrm{/}{G}_N$, [F3] identifies it with the Schwarzschild exterior metric form. The conclusion is only the coordinate substitution and local metric identity.

Since $(+\mathrm{\ell }{)}^2=(-\mathrm{\ell }{)}^2$, both lapse functions equal $1-2M\mathrm{/}r+{\mathrm{\ell }}^2\mathrm{/}{r}^2$. Every metric coefficient is consequently identical.

At fixed $x^0$ and admitted radius $r>0$, the sphere map $u\mapsto (x^0,r,u)$ from the oriented unit sphere into the RN product is a smooth embedding; it is a product-factor inclusion, so it is injective with injective differential and has the continuous inverse given by the sphere projection. The source sphere is compact and has no boundary, and the fields are smooth on a neighborhood since $r>0$ and $f(r)\mathrm{\ne }0$. Thus its pulled-back field and dual are smooth compactly supported top forms, and the flux integrals below are the well-defined operation in [F4]. The two fields are ${\mathcal{F}}_{\pm }=\mp \mathrm{\ell }{r}^{-2}d{x}^0\wedge dr$, so ${\mathcal{F}}_{-}=-{\mathcal{F}}_{+}$. If $f>0$, use the positively oriented coframe $e^0=\sqrt{f}d{x}^0$, $e^1=f^{-1\mathrm{/}2}dr$, $e^2=rd\theta$, $e^3=r\sin \theta d\varphi$. If $f<0$, use $e^0=(-f{)}^{-1\mathrm{/}2}dr$, $e^1=-(-f{)}^{1\mathrm{/}2}d{x}^0$, $e^2=rd\theta$, $e^3=r\sin \theta d\varphi$. In both cases $d{x}^0\wedge dr=e^0\wedge e^1$ and $e^2\wedge e^3=r^2{\omega }_{S^2}$, so [F2] gives ${\star }_g(d{x}^0\wedge dr)=-r^2{\omega }_{S^2}$ and ${\star }_g{\mathcal{F}}_{\pm }=\pm \mathrm{\ell }{\omega }_{S^2}$. For the explicit sphere parametrization $s(\theta ,\varphi )=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$, differentiation gives $s\cdot (s_{\theta }\times s_{\varphi })=\sin \theta$, so the outward form pulls back to $\sin \theta d\theta \wedge d\varphi$, as in [F1]. The sine density is continuous, hence integrable by [F3]; its primitive is $-\cos \theta$. Its integral on $\delta \le \theta \le \pi -\delta$, $0\le \varphi \le 2\pi$ is $4\pi \cos \delta$, and the two omitted caps have total area $4\pi (1-\cos \delta )\to 0$ by the same bounded polar density; the seam has zero width in the azimuthal integral. Thus ${\int }_{S^2}{\omega }_{S^2}=2\pi [-\cos \theta {]}_0^{\pi }=4\pi$. The normalized oriented electric sphere fluxes are therefore $(4\pi {)}^{-1}{\int }_{S_r}{\star }_g{\mathcal{F}}_{\pm }=\pm \mathrm{\ell }$. Thus the metric does not determine the sign of the electric flux parameter.

At $r=2\mathrm{\ell }$, $Q^2={\mathrm{\ell }}^2$ and ${\mathrm{Ric}}_{22}=Q^2\mathrm{/}{r}^2=1\mathrm{/}4\mathrm{\ne }0$, while [F1] gives $S=0$. By continuity the same nonvanishing holds on a sufficiently small open neighborhood of that radius.

The hypotheses of the refuted statement hold on this smooth RN chart but its conclusion fails because the displayed Ricci component is nonzero. Hence the universal implication is false.