Imagine a sphere of black holes surrounding a piece of space. Will this piece be separated from the rest of normal spacetime (at least for some time, till these black holes finally attracted themselves).

So, seen from the outside, we have a black hole, but with a non-singular interior.

Is this possible?

  • $\begingroup$ Do you mean a collection of black holes arranged on the surface of a sphere and allowed to move inwards until their horizons overlap? $\endgroup$ Aug 23, 2014 at 7:08
  • $\begingroup$ @JohnRennie correct... $\endgroup$
    – draks ...
    Aug 23, 2014 at 7:18
  • 3
    $\begingroup$ Good luck working out the equations of motion. Describing the merger of two black holes is still an ongoing research effort. Describing the merger of a sphere of them is way out of reach. $\endgroup$ Aug 23, 2014 at 7:30
  • 1
    $\begingroup$ @JohnRennie a qualitative answer would be sufficient... $\endgroup$
    – draks ...
    Aug 23, 2014 at 8:39

3 Answers 3


The radius of the event horizon of a black hole of mass $m$ is given by:

$$ r_s = \frac{2GM}{c^2} \tag{1} $$

Let's consider your idea of taking $n$ black holes of mass $M$ and arranging them into a sphere. The total mass is $nM$, and the radius of the event horizon corresponding to this mass is:

$$ R_s = n\frac{2GM}{c^2} \tag{2} $$

Now let's see how closely we have to pack our black holes to get them to form a spherical surface with their event horizons overlapping. The cross sectional area of a single black hole is $\pi r_s^2$, and since we have $n$ of them their total cross sectional area is just $n \pi r_s^2$. the surface area of a sphere of radius $R$ is $4\pi R^2$, and we can get a rough idea of $R$ by just setting the areas equal:

$$ 4\pi R^2 = n \pi r_s^2 $$

Giving us:

$$ R = \frac{\sqrt{n}}{2}r_s $$

Use equation (1) to substitute for $r_s$ and we find that the radius of our sphere of packed black holes is:

$$ R = \frac{\sqrt{n}}{2}\frac{2GM}{c^2} \tag{3} $$

Bit if you compare equations (2) and (3) you find that $R < R_s$ because $\sqrt{n}/2 < n$. That means when you try and construct the sphere of black holes that you imagine you won't be able to do it. An event horizon will form before you can get the individual black holes to overlap. You won't be able to construct the black shell that you want and it's impossible to trap a normal bit of space inside a shell of black holes.

However there is a situation a bit like the one you're thinking about, and it's called the Reissner–Nordström metric. A normal black hole has just the single event horizon, but if you electrically charge the black hole you get a geometry with two event horizons, and inner one and an outer one. When you cross the outer horizon you enter a region of spacetime where time and space are switched round, just as in an uncharged black hole, and you can't resist falling inwards towards the second horizon. However when you cross the second horizon you're back into normal space. You can choose a trajectory that misses the singularity and travels outwards through both horizons so you re-emerge from the black hole. If you're interested I go into this further in my answer to Entering a black hole, jumping into another universe---with questions.

As for what the spacetime inside the second horizon looks like, well it's just spacetime. It's highly curved spacetime, but there's nothing extraordinary about it.

  • $\begingroup$ It should be noted that as is the example with two black holes (sorry I do not know about any public access) there will probably be "support" in the form of a conic singularity. I.e., the space-time would be in fact unstable and a direct simulation would show an immediate merger. Or differently, there is no reason for the black holes not to fall to the center and the dynamically consistent Einstein equations would not allow such a space-time in any plausible sense. $\endgroup$
    – Void
    Aug 23, 2014 at 12:04
  • $\begingroup$ +1 for knowing the premiss was incorrect $\endgroup$
    – user16438
    Aug 23, 2014 at 12:55
  • $\begingroup$ How is the outer event horizon relevant? It's hidden by the inner amalgamated event horizon anyway. The problem seems to be constructed so that the pulls from the various black holes cancel out symmetrically, as with any massive shell. The answer is essentially "business as usual, until an inner event horizon hits you, which will happen soon." $\endgroup$ Aug 24, 2014 at 4:02
  • $\begingroup$ @Blackbody The actual definition of "event horizon" is the boundary you cannot cross to get to the rest of the universe, ever. And it has no effect other than this - you can't know you are crossing an event horizon from any local experiment, even in principle. Thus in a very real sense there is only ever an "outer" event horizon. The "inner" ones you're picturing don't demarcate anything of importance, and in fact wouldn't exist according to the definition by the time the outer one formed. $\endgroup$
    – user10851
    Aug 24, 2014 at 5:33
  • $\begingroup$ If you arrange some black holes isotropically far enough from each others and you let them fall towards the center of the distribution maybe you could realize at some instant the configuration the OP proposes. Sure it is not a static solution but at some point you could have a horizon containing a spherical singularity containing another horizon containing empty space. Maybe the best way to realize this would be taking a spherical shell of gas with stellar density in a way that it starts collapsing locally the same way stars usually do. Thus you should produce a spherical singularity.. $\endgroup$
    – AoZora
    May 24, 2019 at 8:45

It is in principle possible, at least for some time, to have a collection of black holes gravitating along the surface of a sphere such that one can still escape from the inside of that sphere. In other words, the inside of the sphere need not be hidden behind a gravitational horizon.

However, as soon as the density of black holes exceeds a critical value, a black holes merger is unavoidable and the whole collection of black holes will collapse into one giant black hole. For a large number $N$ of black holes this critical density is very small. General relativity tells us that a black hole merger is unavoidable as soon as the collection of $N$ black holes, each with horizon circumference $l$, fits within a sphere of circumference $N l$. This means that even if you would have no more than a ring of black holes that are close to touching (I.e. their horizons being close to overlapping) you would not be able to prevent a giant black hole merger.


There is a known example where it looks like a black hole on the outside but is flat spacetime on the inside. Imagine the funnel shaped exterior of a black hole and take the entire part outside of the event horizon, which is a spherical shell of surface area $4\pi r^2$. Then take a spherical ball of Minkowski space of radius $r$ and sew the two together spaces together along the common spherical shell of surface area $4\pi r^2$. Thinking back about the funnel picture of the black hole it is like replacing the lower part of the funnel with a flat disc.

There will be a curvature discontinuity where you sewed the two vacuum solutions together so you need a concentrated stress-energy at that surface between the two vacuum solutions. This can correspond to a spherical shell of particles or fluid with no rest mass moving away from the center point at the speed of light, but placed exactly on the event horizon so it doesn't get any farther from the center.

So: black hole solution on the outside, flat space on the inside, just like you asked about. However, it is very unstable, any tiny change and the inside will collapses to a singularity on the inside. Plus we don;t know of anything that can move at the speed of light yet also be confined to an infinitesimally thin spherical shell. So it's more a mathematical example than a practical one, but it does show that just because something looks like a black hole on the outside, it doesn't have to be anything other than flat on the inside if we are OK with exotic things confined to infinitesimally thin regions.

I first learned about this solution from the American Journal of Physics, I'll edit this sentence when I find the reference.


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