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It is said that nothing escapes from black holes, not even light. All particles are now thought to be excitation of different fields (electric field, electromagnetic field, photon field, etc).

Does it follow that there are no fields (of whatever kind) inside the event horizon of a black hole? IFFY (if and only if) there are fields, are particles created inside the black hole?

Qmechanic
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Roy Closa
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3 Answers3

9

Sure, the reisner-Nordström solution is given by:

$$ds^{2} = -f(r)dt^{2} + \frac{1}{f(r)}dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\phi^{2}$$

where

$$f = 1 -\frac{2M}{r} + \frac{q^{2}}{r^{2}}$$

This is a solution to the Einstein-Maxwell equations with 4-vector potential:

$$A_{a}ds^{a} = \frac{q}{r}dt$$

So, this solution has an electric field that is nonzero everywhere, including inside the horizon.

4

We don't know for certain what is inside the event horizon of a black hole, but we expect there to be fields inside the event horizon that are just extensions of the fields outside of the event horizon. The only restriction is that events that happen inside the event horizon cannot be causally linked to effects outside the event horizon.

However, a complete description of the event horizon at the very smallest scales requires a theory of quantum gravity, which we do not yet have.

gandalf61
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Besides the electric field, as Jerry Schirmer mentions in his answer, that is non-zero everywhere one can have a scalar field that is exists inside the horizon. See for example the (2+1) dimensional solution reported here Conformally dressed black hole in 2+1 dimensions.

The metric function is found to be:

$$f(r) = \cfrac{(r+B)^2(r-2B)}{rl^2} $$

and the scalar field:

$$\Psi = \sqrt{\cfrac{8B}{\kappa(r+B)}}$$

where $B>0$. The black hole horizon is located at $r_h=2B$ and the scalar field remains finite for all $r>0$.

Noone
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