A few related side questions:

  • If the two oppositely-charged Reissner Nordstrom black holes collide would their charge be negated and result in a regular Schwarzschild black hole?
  • Would a positively-charged Reissner Nordstrom black hole repel or attract a proton or other positively-charged particle?
  • Could a positively-charged particle formed inside such a black hole escape?
  • Maybe a bit further off-topic, but would the merger of two Kerr black holes with exactly opposite angular momentum evaporate with a massive release of Hawking radiation? Or would just form a more massive Schwarzschild black hole? Or would revert to a less massive state (such as a neutron star)? Or would something entirely different occur?

** I'm sorry for adding so many extra questions, I just didn't want to mass-post a bunch of highly related queries. Hope that's ok!**


2 Answers 2


Two likely charged Black Holes will always attract each other.

As you can appreciate, if the charges are vanishingly small (say, as compared to the mass) then they must attract as the effect of the gravitational attraction would be overwhelmingly greater than the effect of the electrodynamic repulsion between the Black Holes. Now, let's start cranking up the charges of the Black Holes (keeping the mass invariant) and it might happen that at some point they stop either attracting or repelling each other. Then, if we further crank up the charges beyond this limit then they would start repelling each other.

The crack is that when we solve as to at what values of charges (for given values of masses) the Black Holes would neither repel nor attract, it turns out that this will happen when both the Black Holes are extremal (i.e., have zero temperature). Now, if you want the Black Holes to repel then you would have to further crank up the charge (keeping the mass constant) of any one of the Black Holes. 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 two Black Holes with exactly opposite values of charges collide then the result would be a Kerr (Schwarzschild, if it is not rotating after the collision) Black Hole.

As already mentioned in a previous answer, the conservation of charge demands so to happen.

Whether a "non-Black Holic" charged object would be repelled or attracted by a charged Black Hole depends on the mass to charge ratio of the particle (as well as that of the Black Hole) and where the particle is.

See, for large enough distances from the Black Hole, the electric force, and the gravitational force both go as $\dfrac{qQ}{r^2}$ and $\dfrac{mM}{r^2}$ - where $q,m$ are charge and mass of the particle and $Q, M$ are charge and mass of the Black Hole respectively. So, clearly, whether the particle would be attracted or repelled depends on the ratio $\dfrac{q}{m}\dfrac{Q}{M}$.

For distances that are not far from the Black Hole (and thus, one must use proper GR to determine whether the particle should be attracted or repelled), whether the particle is attracted or repelled depends on its position as well as the charges and the masses of the Black Hole and the particle. One can work out the exact conditions is one uses the full relativistic machinery.


Any particle in the interior can never escape irrespective of the value of its charge.

The simple point is that the reason a particle can never escape from the interior is that the timelike direction inside the Black Hole is along the radial coordinate. Thus, just like a particle always travels towards its future, a particle inside a Black Hole always travels along the decreasing radial coordinate.

Although the result of a collision of Black Holes is a complex phenomenon, the generically valid rule is that the result will always be a Black Hole with a surface area greater than the sum of the surface areas of the previous Black Holes.

First of all, the release of energy during the Black Hole collision would not be due to the Hawking mechanism. It would be due to classical gravitational waves that are emitted during the collision. But this collision will always follow the rule that $A_f > \displaystyle\Sigma_{i} A_i$, the famous law of Black Hole mechanics due to Hawking (where $A_f$ denotes the surface area of the resulting Black Hole and $A_i$s are the surface areas of the colliding Black Holes).


My answer for the main question : 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.

First question : Two oppositely charged black holes would collide they would form a schwarzschild black hole according to the conservation of the charge.

Second question : It depends on how massive is the black hole and how much charge it has . If the force which is repelling the particle is stronger than the gravitational pull it would repellent positivelyrics charged particles, but if the black hole is too massive or it doesn't have enough charge or the particle is going fast in the direction of the black hole it would fall towards it.

Third question : No , it would not because it would still have to go faster than the speed of light to escape.

Fourth question : If their angular momenta are equal and opposite they would form a schwarzschild black hole by using the law of angular momentum conservation.

  • $\begingroup$ Whether 2 opposite $J$ black holes will form a $J=0$ Schwarzschild black hole probably cannot be reasoned based on two-body conservation laws: Gravitational radiation can carry angular momentum, c.f.: the observed "kick" in black hole collection, which is an active area of research. Here, there is an apparent massive violation of momentum conservation (for 2 bodies), as the radiation carries away large amounts of momentum. $\endgroup$
    – JEB
    Commented Sep 4, 2017 at 14:36
  • $\begingroup$ Thanks for your comment , I forgotten it somehow. I will edit my answer according to your comment . $\endgroup$
    – L.Gyula
    Commented Sep 4, 2017 at 22:02

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