In an electrically charged black hole, such as the one described by the Reissner-Nordström metric (i.e. with no angular momentum), where would the electric charge be situated (neglecting any charged particles falling into the black hole)?
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2 Answers
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Depending on how you feel philosophically about these things, either on the horizon or at the central singularity.
answered Jul 29, 2014 at 17:19
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$\begingroup$ I appreciate your response. What about in the case of a black hole described by the Kerr-Newman metric (i.e. with a ring singularity)? $\endgroup$ Jul 29, 2014 at 17:22
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$\begingroup$ Same thing. The Kerr solution is an electrovac solution (and reissner-nordstrom is a special case of the Kerr solution) $\endgroup$ Jul 29, 2014 at 17:47
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1$\begingroup$ Of course, that makes a lot of sense. Thanks for the insight. $\endgroup$ Jul 29, 2014 at 17:55
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There is no charge anywhere in the charged black hole solutions (the divergence of the electric field is zero everywhere). Gauss's law applied to any surface enclosing the singularity will tell you that all of the charge is inside that surface. However the singularity itself is not part of the spacetime. There are two ways of looking at this. You can say that the charge is located at the singularity, even though technically the singularity isn't a place. Or you can say that when a spacetime is topologically nontrivial, you can have charge (in the sense of Gauss's law) without any source. This is appealing because it's simpler, in the sense that you don't need a fundamental notion of charge any more, just the source-free Maxwell equations. It's conceivable that all charge in the real world is source-free in this way.
