Past exam of the general relativity II course of the University of Oxford / 2015 / 2 / b

For the three cases, distinguish a curvature singularity from a coordinate singularity. The Reissner-Nordstrom Kretschmann scalar isIt diverges at $r=0$ when either $M$ or $Q$ is nonzero. Any positive root of $F$ instead admits Eddington-Finkelstein coordinates: set $v=t+\int dr/F$ to obtainThe determinant of its $(v,r)$ block is $-1$, including at $F=0$. Hence positive roots of $F$ are removable coordinate singularities of the static chart. The apparent degeneracy at the poles of the angular chart is also a coordinate singularity.

If $M^2<Q^2$, thenThere is no positive root and no horizon hiding $r=0$ in the extended Reissner-Nordstrom spacetime. Its Kretschmann scalar diverges there, so $r=0$ is a naked singularity. It is a timelike curvature singularity, as the normal to nearby $r=\text{constant}$ hypersurfaces has positive squared norm $g^{rr}=F$.

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.

For $0<Q^2<M^2$ there are two distinct positive roots:Both roots are coordinate singularities of the static chart. In the standard extension, $r_{+}$ is the outer event horizon and $r_{-}$ is an inner Cauchy horizon. The curvature singularity is $r=0$. Between the horizons $F<0$, so the radial direction is timelike; for $0<r<r_{-}$, $F>0$ again and the central curvature singularity is timelike. These are the three radial regions of the global structure of a charged spherical black hole.^[ref]

For completeness, if $Q=0$ and $M>0$, the Schwarzschild spacetime limit has just the positive horizon $r=2M$ and a spacelike curvature singularity at $r=0$; the formal root $r_{-}=0$ of $r^{2}F$ is not a regular inner horizon. If $M=Q=0$, the metric tensor is flat and the radial origin is regular in Cartesian coordinates.