Is there any argument against this situation: two black holes, one within the event horizon of the other, and the system is stable.

It is interesting to me because that if this works, we could have the system's event horizon not in the shape of a perfect sphere. (The event horizon would probably be moving with the orbit)

  • $\begingroup$ No clue, but boy that would be interesting to think about. Wouldn't any assymetry in the size of the two black holes cause the other to consume the smaller one? $\endgroup$
    – geoffc
    Sep 18, 2015 at 19:52
  • $\begingroup$ A very interesting question, but shouldn't the size of black holes be taken into account? $\endgroup$
    – Aneek
    Sep 20, 2015 at 10:07

4 Answers 4


Inside the event horizon is a weird place to be. Inside a black-hole space-time is flowing towards the singularity faster than the speed of light (as measured from outside the black-hole. The effect is that any thing inside a black hole will, in a finite (and usually short) amount of time end up at the singularity.

Models of merging black holes exist such as video from nasa. It requires super-computers to solve Einstein's equations numerically.

As I noted in a comment, no stable orbit can exist within 1.5 times the radius of the event horizon. The orbital velocity is c at that distance (1.5 Schwarzschild); it's also called the "photon sphere". Within that radius all orbits are unstable. And the radiation of gravitation waves means that as two black holes orbit each other, they will lose energy, and so their orbits will decay. As the black holes approach, the event horizons are distorted, and merge into an droplet shape


two black holes, one within the event horizon of the other

The premise is wrong. There can be no such thing. What you and me, external observers, understand to be a "black hole" is actually the whole volume inside the event horizon. It is a chunk of spacetime causally disconnected from where we are now.

When two black holes get close enough together, the event horizons bulge towards each other. When they touch and become a single event horizon, the merger process has begun in force (sometimes this is called "collision"). The only possible outcome at that point is that the two BHs will merge and become a single, bigger black hole.

See this video for an example:


(note: the two red spheres at the end of the simulation have no physical reality, please ignore them)

  • $\begingroup$ It's a technical point, but your discussion of "event horizons" here isn't precise. The event horizon is a global property of the spacetime. You can choose a sequence of spacelike surfaces (as was done in the movie) where the intersection of the event horizon on those slices will appear to merge when taken in sequence, but that's not the same. $\endgroup$
    – Brick
    Sep 19, 2015 at 0:25

The event horizon is a global property of the spacetime. There is only one event horizon in this case. For there to be "two" black holes, we usually mean that their apparent horizons are separate at some time, and the ultimately merge.

This technicality aside, there's a ton of research in this are because black hole mergers are a likely source of gravitational radiation, and there are multiple experiments around the world attempting to directly detect such radiation, including LIGO in the US. There are also several space-based experiments proposed but not currently funded.

A lot of the theoretical work in this are uses large numerical simulations on supercomputers to compute important quantities, including event and apparent horizons. Due to the emission of gravitational radiation, the black holes that you describe will slowly in-spiral and will eventual merge completely. (The gravitational radiation carries away energy, so their orbital radius will shrink.) Eventually they will settle (asymptotically) into either a Kerr or a Schwarzschild solution.


It is an unlikely scenario.

The event horizon is the point where orbital velocity exceeds C (the speed of light).

Matter is incapable of equalling or exceeding C within its frame. Therefore, a stable orbit is implausible, since the second black hole would need to be moving faster than C. Thus, the orbit is inherently unstable, and decaying.

We should be able to detect a velocity shift in the orbits of objects outside the event horizon in the accretion disk, and the event horizon should also have a bulge, but depending upon the current spot in orbit, that may become undetectably small as the smaller black hole spirals in closer and closer.

Note that I'm not up on all the details of frame dragging and other black hole related weirdnesses, so this is only a first order examination of the principles, hence implausible rather than impossible.

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    $\begingroup$ The orbital velocity is c at 1.5 Schwarzschild, its called the "photon sphere". Within that radius all orbits are unstable. $\endgroup$
    – James K
    Sep 18, 2015 at 20:39
  • 1
    $\begingroup$ There's a lot wrong with this answer, and there's a lot of speculation. The science of BH merger is well understood - no need to speculate. $\endgroup$ Sep 19, 2015 at 0:18
  • $\begingroup$ @FlorinAndrei And the progenitor of much of it now thinks it mostly wrong... Hawking has changed his views. $\endgroup$
    – aramis
    Sep 19, 2015 at 2:55

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