Charged black hole

Question

It is known that Einstein's equations admit solutions for charged black holes. The Reissner–Nordström metric in case of a non-rotating charged black hole and for rotating charged black holes there is the Kerr–Newman metric.
In Reissner–Nordström metric I can calculate electric field, it has the following form
$$
F_{0r}=-F_{r0}=frac{Q}{r^2}
$$
But I can not understand the following thing. In my point of view when two charged particles are interacting, they exchange photons between each other. If I apply this argumentation in the case of the interaction between a charged black hole and a probe charge, I will get zero force because the black hole cannot emit photons and absorbs all photons. Can we explain how it works at a quantum level point of view? (I mean QED and not Quantum gravity!!).

Void · Answer

The issue is with your picture of the electromagnetic (and generally any) interaction as arising because of the exchange of real particles. However, the electrostatic interactions $sim 1/r$ arise thanks to the "exchange of virtual particles". Virtual particles are not particles, they are called "particles" because they appear in a similar context as actual quantum particles in the mathematical expressions. However, "virtual particles" can never be observed, they are quantum-field excitations that are invariably attached to their source and cannot be particle-like.
One defining property of "virtual particles" is that they can "travel" at a speed larger than light. More specifically, their momentum can be space-like and their mass formally imaginary. In other words, Coulombic-type interactions $sim Q_1 Q_2/r$ are "mediated by photons traveling faster than the speed of light". This means that such a virtual particle can without any issue "tunnel" outside of a black hole and mediate interactions.

Charged black hole

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