Zero-rotation and zero-charge metric substitutions

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.