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Let's say a charged particle is falling into a black hole.

What happens immediately after the particle crosses the horizon? It seems to me that one of these two situations will take place.

  1. The charge is instantaneously distributed on the surface of the horizon. It looks like this scenario breaks the local conservation of electric charge. Not only this, but charge density moving instantly from one end of the horizon to another would generate an infinite electromagnetic field.

  2. The charge is not instantaneously distributed on the surface of the horizon. But this would violate the no-hair theorem, because the black hole would have a non-uniform charge distribution.

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    $\begingroup$ The no hair theorem applies to a stationary black hole i.e. you need to give the charge time to reach the singularity and the field to equilibrate. For the external observer this will take an infinite time. $\endgroup$
    – John Rennie
    Commented Jul 13, 2018 at 5:17

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Neither options OP described are realized. Instead the black hole would considerably disturb the electromagnetic field of a charge even before it crosses the black hole horizon, so that distant observer would see the electric field as originating on the surface of horizon with a minor, rapidly decaying disturbance by the falling charge.

First let us consider the charge of magnitude $q$ that is slowly lowered into an initially neutral black hole, so that electromagnetic field could be described as (quasi)static. The reference for this is:

  • Hanni, R. S., & Ruffini, R. (1973). Lines of force of a point charge near a Schwarzschild black hole. Physical Review D, 8(10), 3259, doi:10.1103/PhysRevD.8.3259.

As the charge is moved closer and closer to the horizon more and more of its field lines enter the insides of the black hole. But, since the total flux of the electric field through the horizon must remain zero, in addition to "ingoing" field lines there must also be "outgoing" field lines. This situation is somewhat similar to a charge near a conducting sphere: there are induced surface charges, with the charges of the opposite sign localized on the surface part closest to the external charge but the net total charge of the conductor is zero. In case of black hole the effect is, of course, purely geometric in origin, nevertheless we can integrate the total flux of ingoing field lines and find the effective induced charge as well as the region of its localization. As the separation of charge and the black hole horizon decreases, the magnitude of induced charge approaches $-q$ and it is localized in the region closest to the external charge, while the rest of the black hole horizon carries the opposite induced charge of a nearly uniform density.

Field lines for the charge at $r=4M$ and $r=2.2M$: Image from Hanni & Ruffini

By the time the charge is almost touching the horizon the overall electric field of the system could be approximated by the black hole carrying the charge $q$ and a dipole ($-q$,$q$) situated where the charge touches the horizon. As the charge crosses the horizon, dipole component drops to zero while the monopole part, describing the charged black hole remains.

If we consider dynamics of the charge falling into the black hole the qualitative picture would remain the same except that there would be delays in the responses from induced effective charges at the surface of the horizon as well as dissipation in the form of electromagnetic radiation leaving the system (or absorbed by the horizon). Another way to look at this is that the falling charge excites various quasinormal modes of electromagnetic (and gravitational) field. They would be decaying exponentially with the decay constant of order $r_s/c$. And while for the test charge it would take infinitely long time by the clock of an outside observer to cross the horizon, the electromagnetic field of the system would be rapidly stabilizing, and from the outside it would be impossible to distinguish any charge distributions after several $r_s/c$ intervals.

Final note: the no-hair theorem does not really apply here, since it describes the end-point of the evolution of infalling matter.

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    $\begingroup$ Thanks ! So if the no-hair theorem only describes the end-point of the evolution and that for an external observer this takes an infinite time to reach, is this theorem irrelevant for external observers (and therefore all of us)? $\endgroup$
    – user140255
    Commented Jul 13, 2018 at 21:48
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    $\begingroup$ @Undead: Not at all, the theorem tells us which characteristics of black hole would remain at the end and which are radiated away (or fall into bh). For example for the situation we have considered, the charge would remain, while dipole moment (and quadrupole moment, etc.) would asymptotically (and quite rapidly) approach zero as the system stabilizes. $\endgroup$
    – A.V.S.
    Commented Jul 14, 2018 at 5:44
  • $\begingroup$ "As the charge crosses the horizon, dipole component drops to zero" - This is never observed from outside so classically the dipole is viewed forever. However, the separation of charges drops to the Planck length in a fraction of a second. Does such a Planck dipole have any physical meaning or can we say that the dipole becomes principally unmeasurable and effectively disappears in a fraction of a second? +1 $\endgroup$
    – safesphere
    Commented Dec 25, 2020 at 14:23
  • $\begingroup$ Does such a Planck dipole have any physical meaning Not in classical GR or in models of quantum gravity where “nothing special” happens at the horizon crossing. However there are plenty of speculative black hole modifications/alternatives that could produce (potentually measurable by outside observer) “echoes” from near horizon interactions. $\endgroup$
    – A.V.S.
    Commented Dec 25, 2020 at 15:20

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