Let's assume that you had a black hole binary system and everything said here is possible. Their large masses would lead to a large emission of gravitational waves. The loss of orbital energy and angular momentum to this gravitational radiation should ultimately cause the two black holes to merge into a single black hole containing the combined mass of the merged black holes.

With this said, does a greater combined mass of the newly formed black hole mean that the radius of the new event horizon is greater than the radius of the event horizons of the singular black holes? I don't know if there is a formula connecting mass to radius, or if this is essentially a conceptually hypothetical concept, and the answer is simply that the larger the mass of the black hole, the larger the radius of its event horizon


The "radius" (there is no physical surface) of the event horizon of a rotating black hole, depends on both it's mass $M$ and angular momentum $J$, and is given by the equation $$ r = \frac{GM}{c^2} +\sqrt{\left(\frac{GM}{c^2}\right)^2- \left(\frac{J}{Mc}\right)^2}.$$

It is thus difficult (for me anyway) to give a straightforward answer to your question. When two black holes merge, they will each have their own masses and angular momenta, plus there will be angular momentum in the orbit. The gravitational waves emitted during the merger can take away mass from the system as a whole (e.g. the final mass of the first observed black hole merger was three solar masses less than the summed masses of the merging components).

So, in general yes, for non-spinning black holes the event horizon grows as the total mass. But if the merger results in a black hole with maximal spin, where $J = GM^2/c$, then the final event horizon could be half the size given by the Schwarzschild radius ($2GM/c^2$), even though the total mass is larger than the black holes that contributed to it.

However, I think what is certainly true, is that it is not possible to add mass (even in the form of another black hole) to a given black hole and for the event horizon to get smaller, whatever the angular momentum contributed by that mass (section 4.2 of "Black Holes", by Raine & Thomas, 2015, Imperial College Press).

  • $\begingroup$ A detail, but the gravitational waves can also carry away angular momentum. $\endgroup$ Mar 20 '18 at 20:56

Yes, there is a simple formula connecting mass with radius. It is called the Schwarzschild radius. It is really well explained in Wikipedia, with examples.


The Schwarzschild radius is the radius of a sphere such that, if all the mass of an object were to be compressed within that sphere, the escape velocity from the surface of the sphere would equal the speed of light.

Formula: $$r_s = \frac{2MG}{c^2}$$ As you can see, with exception of radius and mass itself, all other are constants (c = speed of light, G = gravitational constant). So the double the mass, the double the radius.

Therefore yes, after merging the size would be greater (I think this is your original question), althought not sure if it is the exact sum due to some kind of energy loss during the merge. As a side note, there is no need of having a binary system of black holes colliding in order to determine or change the size of a black hole, it can grow just by eating surrounding particles, gas, etc.

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    $\begingroup$ Except that a merged black hole binary would not be a Schwarzschild black hole. Your answer needs further consideration/complexity, since the event horizon of a Kerr black hole is smaller. $\endgroup$
    – ProfRob
    Nov 11 '17 at 12:45
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    $\begingroup$ You are right. I was considering the simplest example to show that is not a "conceptually hypothetical concept", but your answer is more accurate. From that I wonder, if spin contracts the event horizon and makes the singularity ring-shaped, if there could be a spin so fast it contracts the event horizon inside the singularity radius, or it is impossible because some concept I miss. $\endgroup$
    – gopejavi
    Nov 11 '17 at 16:18
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    $\begingroup$ Material cannot be accreted such that it increases $J/M$ to $>1$. $\endgroup$
    – ProfRob
    Nov 15 '17 at 10:19

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