Say there are 2 stellar black holes (equal mass and like charged), can they repel each other? I believe when people say things like charged black hole it is not the same as magnetar where it generates extremely powerful magnetic field right?


1 Answer 1


A real black hole is unlikely to have any significant electric charge, because a distribution of matter that has enough charge to overcome the gravitational attraction would — for that very reason — not collapse to form a black hole.

(Magnetic fields are a different story. Spinning black holes with accretion disks are thought to generate very strong magnetic fields, which are presumably responsible for the astrophysical jets that have been observed in several systems [1].)

For a more interesting answer, suppose we ignore the obstacles to forming such a highly charged object. The Reissner-Nordström solution describing a non-rotating charged black hole with mass $M$ and charge $Q$ is $$ ds^2=-A(r)dt^2+\frac{dr^2}{A(r)}+r^2d\Omega^2 \tag{1} $$ with $$ A(r)\equiv 1-\frac{2GM}{r}+\frac{GQ^2}{r^2} \tag{2} $$ in units with $c=1$ and in which Coulomb's law would have the simple form $F=Q^2/r^2$. Equations (1)-(2) are from equation (11.15) in [2]. There are three cases:

  • The usual case is $Q<M$, which describes a black hole in the usual sense. The special case $Q=0$ is the Schwarzschild black hole. Two objects of this type attract each other, because the repulsive electric interaction is not strong enough to overcome the attractive gravitational interaction.

  • The opposite case is $Q>M$, which doesn't describe a black hole at all, because there is no horizon in this case. This solution has a naked singularity instead. Two objects of this type would repel each other, but they're not black holes, so this doesn't satisfy what the question is asking for.

  • The extremal case is $Q=M$. This case is remarkable, because two objects of this type (with same-sign charges) don't attract each other or repel each other. The class of exact solutions called the Majumdar-Papapetrou solutions describes any finite number of these objects in static equilibrium with each other, as explained in section 1.2 in [3]. These are black holes (they do have event horizons), but they don't quite repel each other, so this also doesn't satisfy what the question is asking for.

Based on these examples, the answer to the question seems to be "no": if the objects are black holes, then they don't repel each other; and if they do repel each other, then they're not black holes. But these are admittedly only examples, not a general proof.

By the way, although the question asked about electrical repulsion, the paper [4] claims that two black holes can repel each other for a different reason: angular momentum. Here's an excerpt:

the remarkable feature of the newly found binary black-hole configurations is that the black holes in them can repel each other due to prevailing of the spin-spin repulsion over gravitational attraction, which prevents the two black holes from merging into a single black hole.

I don't know if this claim has been independently checked yet, but it's an intriguing claim.


[1] https://en.wikipedia.org/wiki/Astrophysical_jet

[2] Marolf, 2012. "Chapter 11: Black holes and branes in supergravity," pages 273-324 in Black Holes in Higher Dimensions, edited by Horowitz (Cambridge University Press)

[3] Horowitz, 2012. “Chapter 1: Black holes in four dimensions," pages 3-25 in the same book

[4] Manko, Ruiz, "'Black hole-naked singularity' dualism and the repulsion of two Kerr black holes due to spin-spin interaction," https://arxiv.org/abs/1803.03301

  • $\begingroup$ I would like to clarify, what happens to the Q=M black hole if we keep firing charged into it with an electron canon? It seems like the singularity will not disintegrate, so prevents us from forming a Q > M black hole that way? $\endgroup$ Oct 18, 2019 at 13:56
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    $\begingroup$ @VladimirAkopyan Two things to consider: first, an electron has $Q>M$ already. Second, an electron cannon would need to propel the electrons with enough energy to overcome the immense electrostatic repulsion, and we need to account for this energy when calculating the final total mass. $\endgroup$ Oct 18, 2019 at 15:13

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