1
$\begingroup$

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!!).

$\endgroup$
  • $\begingroup$ This is not an answer, but: if your logic is correct, then gravitons also cannot escape the black hole, and so a distant charged particle would not be able to feel the force due to either electric repulsion or gravitational attraction (since neither photons nor gravitons should be able to escape the hole). So clearly this way of thinking of virtual particles communicating static potentials has to be modified. $\endgroup$ – Surgical Commander Mar 7 '15 at 5:21
  • $\begingroup$ I don't want to discuss graviton. May be gravity is impossible to quantize. It is difficult question. Let's start with electrodynamic. $\endgroup$ – Peter Mar 7 '15 at 5:48
  • $\begingroup$ Related: physics.stackexchange.com/q/937/2451 , physics.stackexchange.com/q/149581/2451 and links therein. $\endgroup$ – Qmechanic Mar 23 '15 at 1:05
  • $\begingroup$ Well photons and gravitons themselves can not leave the black hole but that does not mean the black hole does not affect the quantum fields around it. Black hole distorts space time and EM creating gravity and electromagnetic waves. This is also the reason gravity and charge and indirectly spin and magnetism get communicated out of a black hole. This is also why hawking radiation happens. $\endgroup$ – Roghan Arun May 11 at 18:01
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.