Opposite electric charges give the same RN metric

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.