 Solution

ID: past-exam-of-the-general-relativity-ii-course-of-the-university-of-oxford/2015/2/b/i/solution

M^{2} < Q^{2}

, then

r^{2} F = (r - M)^{2} + (Q^{2} - M^{2}) > 0.

There 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 = constant

hypersurfaces has positive squared norm

g^{rr} = F

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