= Solution
For $M>0$ and $Q^2>2M^2$, the physical exterior starts at $r=b>2M$. Throughout it, $f>0$, so the putative zero of $f$ is not in the exterior and no regular horizon shields the <curvature singularity>. The normal to $r=b$ has positive limiting squared norm $g^{rr}=f(b)>0$, identifying a timelike singular boundary. Outgoing radial <null geodesics> obey $dr/dt=f(r)>0$ and escape from arbitrarily close to that boundary. Hence
$$
\boxed{Q^2>2M^2\quad\text{gives a timelike naked singularity}.}
$$
At $Q^2=2M^2$, the singular boundary coincides with $f=0$ and is null. There is no regular shielding horizon of positive area; this marginal singular extremal case must be separated from both the regular <black hole> and the supercritical timelike naked singularity. If “naked” is used simply to mean unshielded by a regular horizon, equality belongs on that side of the distinction, but its causal boundary is different.
The usual positive-mass restriction matters. For $M<0$, $f(r)>0$ at every positive radius and there is no positive-radius horizon; the central zero-area boundary at $r=0$ is singular and naked. For $Q=0$ its angular sectional curvature is $2M/r^3$, which diverges there. For $Q\ne0$, $C=r(r-b)$ with $b<0$, and the angular sectional curvature $[1-f(C')^2/(4C)]/C$ also diverges as $r\downarrow0$. The displayed family is undefined at $M=0$ with nonzero $Q$; its uncharged zero-mass limit is flat.
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