A BBH is a system consisting of two black holes in close orbit around each other.

The last stable orbit or innermost stable circular orbit (ISCO) is the innermost complete orbit before the transition from inspiral to merger.


If these BHs are likely EM charged (either Reissner–Nordström or Kerr–Newman) then the EM repulsion could be able to withstand the gravitational attraction.

The two types of charged black holes are Reissner–Nordström black holes (without spin) and Kerr–Newman black holes (with spin).


Now the EM force is said to be 40 orders of magnitude stronger then gravity, and this might give rise to a stable bound orbiting binary system, where the gravitational attraction is balanced by the EM repulsion.

It might just come down to the distance (orbiting distance), where the two forces (gravity and EM) could balance out.

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But this won't be possible because increasing the charge of a Black Hole (for a given mass) beyond the value of charge at which it becomes extremal is not possible. Well, General Relativity doesn't forbid doing so per se but that would make the singularity solution a Naked Singularity solution instead of a Black Hole solution and according to the Cosmic Censorship Conjecture, it is forbidden.

If they somehow would have the amount of charge what is needed for the electrostatic force to be stronger than the gravitational pull they could repell each other I think but I am not 100% sure about this answer.

Would two Reissner-Nordstrom black holes with like charges repel or attract?

The weak cosmic censorship hypothesis asserts there can be no singularity visible from future null infinity. In other words, singularities need to be hidden from an observer at infinity by the event horizon of a black hole. Mathematically, the conjecture states that, for generic initial data, the maximal Cauchy development possesses a complete future null infinity. The strong cosmic censorship hypothesis asserts that, generically, general relativity is a deterministic theory, in the same sense that classical mechanics is a deterministic theory. In other words, the classical fate of all observers should be predictable from the initial data. Mathematically, the conjecture states that the maximal Cauchy development of generic compact or asymptotically flat initial data is locally inextendible as a regular Lorentzian manifold.


Now I do not clearly see how this would disallow for two likely charged black holes (at certain distance) to create a stable orbit where gravity and the EM force would balance out.


  1. Is there any distance of orbit where two likely charged BHs could create a stable orbit?
  • $\begingroup$ Opposite EM charges attract each other, rather than repel. So, I am not quite sure what you are trying to ask. $\endgroup$
    – mmeent
    Dec 16 '19 at 23:12
  • $\begingroup$ @mmeent thank you edited. I meant likely charged. $\endgroup$ Dec 16 '19 at 23:18
  • $\begingroup$ Around a charged black hole, charged particles can have periodic orbits inside and outside the horizon, across different worlds. See all possible orbits with nice graphics here: arxiv.org/pdf/1011.5399.pdf $\endgroup$
    – safesphere
    Dec 17 '19 at 7:40
  • $\begingroup$ A pro martial artist in Hollywood? Wow! Why physics? $\endgroup$
    – safesphere
    Dec 18 '19 at 7:53
  • $\begingroup$ @safesphere :) I actually went to University for math, and had to do exams in physics together with the physics students. $\endgroup$ Dec 18 '19 at 17:16

Two (non-maximally) charged Reissner-Nordstöm black holes will always attract each other, i.e. the gravitational force will always be stronger than any repulsion from like chargers. (The turnover point is exactly when both black holes are maximally charged.)

If the charges have the same sign, the net attraction is, of course, smaller than it would have been had the black holes not been charged. Since uncharged black holes can orbit each other stably, so can two similarly charged black holes.

Note however, that this binary would emit EM radiation in addition to gravitational waves, which will cause the binary's orbit to decay over time.

  • $\begingroup$ The turnover point is exactly when both black holes are maximally charged. How is this known? There is no 2-hole exact solution, is there? $\endgroup$
    – G. Smith
    Dec 17 '19 at 0:25
  • $\begingroup$ @G.Smith: The reason this is known, is because you can find exact solutions for 2 (and in fact any number) extremal RN black holes. $\endgroup$
    – mmeent
    Dec 17 '19 at 0:32
  • $\begingroup$ How interesting. I didn’t know that. Are the extremal holes stationary, with their gravitational attraction and electrostatic repulsion balanced? $\endgroup$
    – G. Smith
    Dec 17 '19 at 0:37
  • $\begingroup$ @G.Smith Yes they are. $\endgroup$
    – mmeent
    Dec 17 '19 at 0:42
  • $\begingroup$ Here is a link to a paper by Hartle and Hawking that discusses this solution (originally derived by Majumdar and Papapetrou) : projecteuclid.org/download/pdf_1/euclid.cmp/1103858037 $\endgroup$
    – mmeent
    Dec 17 '19 at 0:48

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