# Black hole inside of a black hole.... can it be done?

• I
• Grasshopper
No, you can't. The singularity is spacelike; once one is there, it already occupies the future of every event inside the horizon.f

#### Grasshopper

Gold Member
Let's say you have an absolutely giant black hole, so big that items inside of it leisurely approach the singularity, reaching it in about a million years (or whatever time it takes for a black hole to form from matter accumulation). Could matter slowly accumulating somehow form its own black hole on the way to the singularity? Why or why not?

Thanks!

(I won't speculate about machines that humans can use to create small black holes, because until we can do them, I feel that's going too far into science fiction. So no need to worry about a human going into a black hole and then making one on the way down.

Also, if I have the right formula, for proper time to be a million years, then the mass of the black hole would need to be ##M = \frac{1x10^6 y*c^3}{πG} =## 4.06 x 10^48 kg, which I think might be bigger than any black hole we've ever seen by a few orders of magnitude. So it certainly looks doubtful...
)

Algr and russ_watters
Could matter slowly accumulating somehow form its own black hole on the way to the singularity?
No. The definition of a black hole is a region of spacetime that cannot send light signals to infinity. Once you are inside such a region, you're inside it. The idea of having a second such region inside the first doesn't even make sense.

Maybe a better question then is can you form a second singularity (through collapse) inside the horizon of a first singularity. I don’t see any reason why not.

russ_watters and Grasshopper
Maybe a better question then is can you form a second singularity (through collapse) inside the horizon of a first singularity.
No, you can't. The singularity is spacelike; once one is there, it already occupies the future of every event inside the horizon.

To put this another way: the "collapse" that ends in the singularity involves all matter that will ever cross the hole's horizon, not just the "original" collapsing object. All infalling worldlines end on the one singularity.

bhobba, SolarisOne, Grasshopper and 3 others
Maybe a better question
There is a question about multiple "things" inside a black hole horizon that could potentially have the answer "yes", but it's not either of the ones already asked in this thread. I believe it is possible to have multiple "trapped surfaces" (i.e., surfaces at which, locally, radially outgoing light does not move outward) inside a single horizon (and whose futures end in the same singularity). However, even this, AFAIK, requires very precisely tuned initial conditions.

bhobba, geshel, Grasshopper and 1 other person
I agree there is no such thing as a BH inside a BH. I would even ask "how could you tell?"

I am a little curious about the following scenario. You have an observer in orbit far from a BH. Much farther out there is a thin spherical shell of dust moving radially inward, the mass is sufficiently large that when it forms a horizon the horizon radius is larger than that of the observer. Much much farther out there is a second observer. What do the observers see?

I think the outer observer case is simpler. He sees the dust rush in, and when it reaches the Schwarzschild radius, a horizon forms and he has a big BH where there was once a smaller one.

I am not sure what happens to the inner observer. I believe there is a time - before the large horizon forms - where he is unable to escape to infinity. I do not know if the inner observer sees the dust rush past him or not.

vanhees71
No, you can't. The singularity is spacelike; once one is there, it already occupies the future of every event inside the horizon.
When two black holes collide to one, in some time in a merging process are there two singular points within a merged black hole?

PS
In time reverse one black hole spit into two black holes. It seems prohibited so such a BH merging should be an irreversible process.

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vanhees71
I think the outer observer case is simpler. He sees the dust rush in, and when it reaches the Schwarzschild radius, a horizon forms and he has a big BH where there was once a smaller one.
Yes.

I am not sure what happens to the inner observer. I believe there is a time - before the large horizon forms - where he is unable to escape to infinity.
This is correct except that "before the large horizon forms" needs to be changed to "before the large horizon reaches its larger radius". The event on the inner observer's worldline at which he can no longer escape to infinity (or send light signals there) is on the horizon, by definition; that event is where the horizon, as it moves outward from the original smaller radius to the new larger one, passes him.

I do not know if the inner observer sees the dust rush past him or not.
If the inner observer is in free fall, I think he will end up hitting the singularity before the dust passes him (but I have not tried to do the math to check this). However, if the inner observer is close enough to the original hole's horizon for this scenario to be workable, I'm not sure he can be in free fall, since there are no stable free-fall orbits inside ##r = 6M## and no free-fall orbits at all, even unstable ones, inside ##r = 3M##. If the inner observer is not in free fall, it's perfectly possible for him to accelerate in a way that allows him to see the dust fall past him before he hits the singularity.

No, you can't. The singularity is spacelike; once one is there, it already occupies the future of every event inside the horizon.

To put this another way: the "collapse" that ends in the singularity involves all matter that will ever cross the hole's horizon, not just the "original" collapsing object. All infalling worldlines end on the one singularity.
Is this generically true or is this specific to say spherical collapse?

Is this generically true or is this specific to say spherical collapse?
For Reissner-Nordstrom or Kerr spacetime, if you take the maximal analytic extension, it is true that the singularity is timelike, not spacelike, and not all infalling worldlines will end on the singularity. However, in those maximally extended spacetimes, everything from the inner horizon inward (which includes the singularity) is beyond the Cauchy horizon of the exterior--i.e., it is impossible to predict what will actually be inside there from initial conditions specified in the exterior, unless you specify exact vacuum (or electrovacuum) everywhere (so that the maximal analytic extension is the solution)--which eliminates any such black hole that forms from the collapse of an actual material object. So those maximal analytic extensions don't seem trustworthy as a guide to what the interior of an actual charged or spinning black hole formed from the collapse of a charged or spinning object would be like. The only maximal analytic extension that does not have this problem is the Schwarzschild one--there is no Cauchy horizon in that case, so the interior of the maximal extension at least might be trustworthy as a rough guide to generic behavior.

To put this another way, we do not have any exact solutions for the collapse of a charged or spinning body to a Reissner-Nordstrom or Kerr black hole. The only exact solution we have for gravitational collapse of a material body is the Oppenheimer-Snyder model of spherically symmetric collapse to a Schwarzschild hole. We do, however, have numerical simulations of charged and spinning collapses, and as I understand it, they do one of two things: either the simulation blows up as the inner horizon is approached (because of the "infinite blueshift" problem at the inner horizon), or the simulation results in a singularity that is spacelike--i.e., it behaves the way the Schwarzschild black hole's singularity does. So, again, the Schwarzschild interior appears to be the one that is the best rough guide to what generic behavior should be expected in black hole interiors.

Aanta and martinbn
This is excellent stuff. But since @mitochan brought up colliding BHs, I have a related follow up question:

Nothing can travel faster than c, so would this not prevent two colliding BHs from ever instantly becoming one? Because presumably the two singularities are separated by a non-zero distance when the two event horizons meet.

On the other hand, at that point, clearly there is one singular future event for all matter in this composite BH. So would it be more practical to think of a singularity as a time rather than a location in space? (And in that case, would that not sidestep the issues of instantaneous information travel and of the distance separation between the two previous singularities, since by definition, all the matter will eventually reach one, singular fate?)

Nothing can travel faster than c, so would this not prevent two colliding BHs from ever instantly becoming one?
No.

the two singularities
There aren't two singularities. There is only one. The horizon of a black hole merger spacetime, heuristically, is shaped like a pair of trousers instead of a cylinder. The "legs" of the trousers are the two holes before the merger, and the upper part of the trousers is the single hole after the merger. The singularity is at the top of the trousers, to the future of everything inside the horizon.

Aanta and SolarisOne
would it be more practical to think of a singularity as a time rather than a location in space?
Not just more practical--correct, as opposed to wrong.

Grasshopper
There aren't two singularities. There is only one.
The questions regarding a BH merger posed by @mitochan and @Grasshopper are assuming singularities. But what if physics "breaks down" at ##r=0## as most physicists are thinking? What if the mass exists in finite density in the center in a yet unknown state?
Could we still say there aren't two centers?

Dale
There aren't two singularities. There is only one. The horizon of a black hole merger spacetime, heuristically, is shaped like a pair of trousers instead of a cylinder. The "legs" of the trousers are the two holes before the merger, and the upper part of the trousers is the single hole after the merger. The singularity is at the top of the trousers, to the future of everything inside the horizon.
I draw a sketch for better understanding at attached.

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But what if physics "breaks down" at ##r=0## as most physicists are thinking? What if the mass exists in finite density in the center in a yet unknown state?
Could we still say there aren't two centers?
No. The "breakdown" area (roughly, where the spacetime curvature reaches the Planck scale) is still just one area, not two. And the spacetime location of that one area is just where I described, at the top of the trousers.

bhobba and timmdeeg
I draw a sketch for better understanding at attached.
Your sketch is incorrect. The topology of the singularity is a single spacelike line, not one line that branches into two.

mitochan
The topology of the singularity is a single spacelike line, not one line that branches into two.
The primitive deduction,

A BH as well as an ordinary body has a timelike line in our spacetime.
The center of BH has its timelike line also.
The singularity of BH lies in its center.
Thus singularity of BH is on a timelike line

would have one or more errors that I cannot point them out clearly.

A BH as well as an ordinary body has a timelike line in our spacetime.
The center of BH has its timelike line also.
The singularity of BH lies in its center.
Thus singularity of BH is on a timelike line
All four of these statements are false.

bhobba and PAllen
No. The "breakdown" area (roughly, where the spacetime curvature reaches the Planck scale) is still just one area, not two. And the spacetime location of that one area is just where I described, at the top of the trousers.
Ok thanks. Does this mean that this area on Planck scale has not a location in space like the singularity and thus is not part of the manifold? If so then it would be misleading to say it is in the center of the black hole.

You said here
The mass of the eternal black hole is a global property of the spacetime geometry; there is no stress-energy associated with it." we have to distinguish two categories of black holes, the "real" one's which exist in our universe and which have stress-energy and the eternal black holes which haven't.
So it seems that the "global property of the spacetime geometry" of a black hole is represented by the mass regardless whether or not it is associated with stress-energy. On the other side how can something which has no location in space (in case the answer to the question above is yes) have stress-energy?

Does this mean that this area on Planck scale has not a location in space like the singularity and thus is not part of the manifold?
It means we don't know what the correct physics is in that area. The very concept of "part of the manifold" vs. "not part of the manifold" might not make sense.

If so then it would be misleading to say it is in the center of the black hole.
It's already misleading even in the purely classical GR model to say that the singularity is at the center of the black hole, and not because the singularity is strictly speaking not part of the manifold. There are still well-defined limits as ##r \to 0## that define "where the singularity is" in the spacetime. The key thing is that the singularity is a moment in time, not a place in space; it's spacelike, not timelike. A moment in time can't be "at the center" of anything.

how can something which has no location in space (in case the answer to the question above is yes) have stress-energy?
The singularity doesn't have stress-energy, nor does the region within a Planck scale of it. In spacetimes containing black holes formed by the collapse of a massive object, the stress-energy is in the region of spacetime occupied by the massive object.

bhobba, SolarisOne, Vanadium 50 and 1 other person
The singularity doesn't have stress-energy, nor does the region within a Planck scale of it. In spacetimes containing black holes formed by the collapse of a massive object, the stress-energy is in the region of spacetime occupied by the massive object.
Thank you for clarifying this question.

108 The concept of black hole 3: The global view shows diagrams of the head-on merger of two black holes.

Figure 4.18: Spacetime diagram of the event horizon corresponding to the head-on merger of two black holes as computed by Matzner et al. (1995)

Figure 4.19: Spacetime diagram showing the event horizon in the head-on merger of two black holes, as computed by Cohen et al. (2009)

Figure 4.21: Cross-section of the event horizon H of the spiraling merger of two black holes as computed by Cohen et al. (2012)

The first two diagrams show the trousers picture as mentioned by @PeterDonis in this Thread.

I have been searching for Eddington-Finkelstein diagrams which would eventually show the singularities - but not successful. A black hole in these coordinates is depicted in Figure 6.3 (page 136): Radial null geodesics of Schwarzschild spacetime, plotted in terms of ingoing EddingtonFinkelstein coordinates (t, r ˜ ).

Are there possibly principal reasons that one can't calculate the merger in Eddington-Finkelstein coordinates? Nevertheless the upper part of the legs (between them) of the trousers (Fig. 18 - 19) seems to mark the emergence of the new horizon.

Would it be correct that to assume the singularities in Fig. 4.19 within the trouser legs parallel to the horizon? This seems however to lead to the probably misleading notion that the two horizons are "unified" to one instantaneously at that upper part I mentioned above.

EDIT I just reread #12 where @PeterDonis said
There aren't two singularities. There is only one.
Does this mean if there are two Black holes which will collide in future then both have the "same" singularity. Then most I said above would be meaningless.

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Would it be correct that to assume the singularities in Fig. 4.19 within the trouser legs parallel to the horizon?
No. See my previous posts.

Does this mean if there are two Black holes which will collide in future then both have the "same" singularity.
Yes, at the top of the "trousers". See my previous posts.

timmdeeg
However, if the inner observer is close enough to the original hole's horizon for this scenario to be workable, I'm not sure he can be in free fall

My original thought was that at time 1t you had the small black hole and at t2 you had the large black hole and that was that. Once the dust crosses a certain radius you go immediately from the old horizon to the new horizon. I no longer think that.

Here's why. The observer is orbiting at r, r1 < r < 2, where r1 is the initial Schwarzschild radius and r2 is the final one. To escape, he needs to travel from r to r2 before the outer horizon has formed. If he waits too long, he has to travel faster than c to do it, and that means he is now inside the horizon. So it can be said that the horizon (not a physical thing anyway) moves outward.

I was bugged by the shell theorem for a while. How can a spherical shell cause what is effectively an inward force? But the conditions do not apply when spacetime is this curved.

Once the dust crosses a certain radius you go immediately from the old horizon to the new horizon.
That's not correct. The horizon (meaning the event horizon) smoothly increases in size from the smaller to the larger radius. And it does this before the new shell of dust falls in. At the instant when the shell of dust is at the new horizon radius, the horizon has already expanded smoothly to that new radius.

The observer is orbiting at r, r1 < r < 2, where r1 is the initial Schwarzschild radius and r2 is the final one. To escape, he needs to travel from r to r2 before the outer horizon has formed. If he waits too long, he has to travel faster than c to do it, and that means he is now inside the horizon.
Yes. And by keeping track of at what value of ##r## the observer would have to travel faster than ##c## to escape, as a function of time (although you have to be careful of how "time" is defined here), you can track the smooth expansion of the horizon from ##r_1## to ##r_2##.

I was bugged by the shell theorem for a while. How can a spherical shell cause what is effectively an inward force? But the conditions do not apply when spacetime is this curved.
The shell theorem does apply in this case, but all it says is that the shell itself doesn't affect the spacetime curvature inside it. It doesn't say the spacetime inside has to be flat. And in this case, of course, it isn't.

Also, there is no "inward force". The spacetime curvature at a given radius ##r## does not change until the shell reaches that radius ##r##. The reason an observer between ##r_1## and ##r_2## can no longer escape at some point, even though the shell has not yet reached him, is that he would have to travel faster than light, according to the spacetime geometry inside the shell region--i.e., according to the spacetime geometry with the smaller mass. If the shell weren't falling in, he would be able to escape from any ##r## greater than ##2M_1## (the smaller horizon radius). But because the shell is falling in, his global future worldline is different from what it would have been with no shell falling in, even though locally he sees no difference in the spacetime geometry when he starts his escape attempt.

The difficulty here is that the event horizon is teleological--its location depends on the entire future of the spacetime. There is no local counterpart to it in general. For a stationary black hole (i.e., one that nothing ever falls into), there is an apparent horizon co-located with the event horizon, and the apparent horizon can be locally detected (by radially outgoing light rays not moving outward). But in the case of the infalling shell, the apparent horizon stays at the "old" horizon radius ##2M_1## until the instant that the shell reaches the "new" horizon radius ##2M_2 > 2M_1##, and then the apparent horizon discontinuously jumps to the new horizon radius. So the observer with ##2M_1 < r < 2M_2## can't conclude that he is safe because he is outside the apparent horizon. He can't watch for any local change that will tell him the shell is falling in. There is simply no local way for him to know he is safe.

(Strictly speaking, if the shell is made of ordinary matter, the observer inside could see it, for example by ingoing light rays emitted from it, and act on that information. But in the limiting case of an infalling shell of radiation, he would have no possible warning at all, since no information about the shell's coming can get to him faster than light, i.e., faster than the shell itself.)

I believe (a bad way to start a Physics arguement) that there has been recent recordings of black hole mergers. A very fast and energetic merger that generates energy on a galactic scale. So Mergers happen.
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We also know that the black holes in galactic centers are relatively huge with relatively moderate tidal forces at their surface. And these Black holes are huge by any and all measures.
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What happens to a 2-5 solar mass black hole as it wanders into the Galactic Center Black hole of 1^9 Solar masses. Would that not equal the original poster's concept of a BH inside of another BH? Please explain that condition or why it can't happen to this layman?

What happens to a 2-5 solar mass black hole as it wanders into the Galactic Center Black hole of 1^9 Solar masses.
The two holes would merge into one. Since one is so much less massive than the other, the merger would be a much less energetic process than the ones observed by LIGO.

Would that not equal the original poster's concept of a BH inside of another BH?
No, because the two holes just merge into one. In the "pair of trousers" heuristic used earlier in this thread, one "leg" of the trousers would be very, very thin and the other would be very, very thick--but the basic principles of the merge would remain the same.

Also, there is no "inward force". The spacetime curvature at a given radius does not change until the shell reaches that radius .
Hmmmm...

Hmmmm...

On the one hand, I see that this must be the case. Suppose my dust is rocket powered, and at the last moment, the engines turn on and reverse the velocity vectors of the dust particles.

On the other, suppose I am orbiting at a very close radius: 3 (or 6) +ε R. How does spacetime around me know that the dust is causing the horizon to grow? Or not?

As far as the shell theorem, that only works if the force is exactly inverse square. If that were the case, you would get no perihelion advance.

bhobba
suppose I am orbiting at a very close radius: 3 (or 6) +ε R. How does spacetime around me know that the dust is causing the horizon to grow?
It doesn't, until the dust falls past you. As I said, inside the dust shell there is no change in the spacetime geometry. So your orbital parameters will not change until the dust shell is below you.

Note that, as I said, you cannot locally detect the growth of the horizon. It can grow outwards past you and you won't know it (unless you try to escape and fail after that happens), because nothing locally changes when it passes you. The event horizon is not a locally detectable thing.

As far as the shell theorem, that only works if the force is exactly inverse square.
Not in GR. In GR, gravity is not a "force", and the shell theorem is formulated in terms of spacetime geometry, in the form I gave before. If there is a condition in the theorem that corresponds to the "inverse square" in the Newtonian version, it is spherical symmetry. That is a very strong condition, because not just the shell, but the entire spacetime outside the shell, has to be spherically symmetric for the theorem to hold. (If the shell itself is spherically symmetric, and the region outside is vacuum, the region outside will also satisfy spherical symmetry--that is the case we are discussing here.)

"Not a force" is a quibble. If you prefer, call it acceleration.

Let me rephrase. I am orbiting at 3+ε R. The dust gets closer, the horizon moves out,and now I am at 3 - ε R and my orbit is now unstable. How does it know? I know because if I make a small perturbation, it is no longer restored. The restoring force acceleration is now in the opposite direction. But as you say, there's no way of telling that this has happened before light from the dust has reached me.

Since I was talking about orbit stability, replace 3 with 6.

I am orbiting at 3+ε R.
To be clear: you mean just a little outside the innermost stable timelike circular orbit at ##r = 6M## (3 times the Schwarzschild radius), correct?

The dust gets closer, the horizon moves out,and now I am at 3 - ε R and my orbit is now unstable. How does it know?
You will see the change in your orbital parameters when the dust falls past you. That will most likely be way before the horizon moves out. You won't be able to detect any change in the horizon from the behavior of your orbit.

The way your orbit "knows" that its parameters have changed is that the dust fell past you. It has nothing to do with the horizon.

I know because if I make a small perturbation, it is no longer restored.
You know because your orbital parameters have changed and you can now compute that your orbital radius is slightly less than ##6M_2## (where it was slightly greater than ##6 M_1##). Note that, in order to maintain a circular orbit as the shell falls past you so that you can detect the change in your orbital parameters, you have to use rocket thrust. If you don't, you won't be in a circular orbit any more; you'll be in an unstable elliptical orbit (which, I suppose, would be another way of detecting the change as the shell fell past you).

Once you are in your new circular orbit with changed parameters, you can tell that it's unstable by seeing what happens when small perturbations occur, yes.

The restoring force acceleration is now in the opposite direction.
For an inward perturbation, yes. For an outward one, it actually isn't.

SolarisOne
One could imagine a whole galaxy with millions of black holes collapsing inside its Schwarzschild radius.
Although, technically speaking, there is, classically, one global event horizon and a future spacelike singularity (as already mentioned), and r= constant hypersurfaces inside the Schwarzschild radius of this hypothetical galaxy are spacelike, for an observer that's free falling (along with every other star etc ) inside, these individual "black holes" -that have only apparent horizons- are still there, aren't they?
The situation is reminiscent of a closed FLRW spacetime, with the crucial difference that in the FLRW case there is no exterior region, or horizon.!
Also, in the Oppenheimer/ Snyder pressureless dust collapse model, one could imagine that these specs of dust are, themselves, collapsing objects with their apparent horizons, and they all have a common global event horizon and a common future spacelike singularity.
There are also collapsing models where timelike singularities ( with their corresponding Cauchy horizons), are formed before the final spacelike singularity.

By the way, one can also have black holes "inside" other bhs in the cases of Bag of gold or Baby universe spacetimes.
These are not related, of course, with Grasshopper's question, but i think they're worth mentioning.

bhobba