<blockquote>
  <p>... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?</p>
</blockquote>

<p>I'm not completion sure what distinction you're drawing between <em>concentration</em> and <em>density</em>, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.</p>

<p>For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is <a href="https://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29" rel="nofollow noreferrer">Birkhoff's theorem</a>: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">shell theorem</a>. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.</p>

<p>So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be <a href="https://physics.stackexchange.com/questions/45963/do-black-holes-have-infinite-areas-and-volumes">careful</a> about how you define 'volume' when comparing overall densities.</p>

<p>What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what <em>will</em> fall into the black hole in the future.</p>

<p>This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.</p>

<p>Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the <a href="https://en.wikipedia.org/wiki/No-hair_theorem" rel="nofollow noreferrer">no-hair theorem</a>. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the <a href="https://en.wikipedia.org/wiki/Penrose%E2%80%93Hawking_singularity_theorems" rel="nofollow noreferrer">singularity theorems</a> guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.</p>

CC BY-SA 3.0
2016-12-06T06:32:40.253Z
Also check out https://en.wikipedia.org/wiki/Hoop_Conjecture
Rag
CC BY-SA 3.0
2014-11-09T23:57:08.580Z
2014-11-09T23:57:08.580Z
2017-04-13T12:40:15.377Z
Stan Liou
<p>"Concentration of matter" and "density" are equivalent terms, i.e. density is measured in mass per unit volume, so yes, it is exactly the concentration of matter within a given volume that determines whether or not an object will form a black hole.</p>

<p>Any body has a so-called Schwarzchild radius - the radius within which, should all its matter become concentrated, the body will form a black hole. AFAIK Black Holes do not have a low density of matter within the black hole, in fact the density there is for all intents and purposes infinite.</p>

<p>Further to this, any reliable description of the environment within a black hole is, at best, educated conjecture. Bar Hawking Radiation and some other possible forms of minor 'bleeding', black holes give up no information about their contents. The only indication we have is of their mass (due to their effect on spacetime) which indicates (as far as observations are concerned) that they are very dense indeed.</p>

CC BY-SA 3.0
2014-09-10T18:20:52.947Z
The density is decidedly not infinite over the entire black hole region. This would require infinite energy. Whether or not there is a point of infinite density within the horizon is hardly commonly agreed on, as well.
zibadawa timmy
CC BY-SA 3.0
2014-09-10T18:34:16.357Z
Hence 'for all intents and purposes' ;)
Guy Stimpson
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2014-09-10T23:50:30.690Z
Yet your argument rests on that idea.
HDE 226868
CC BY-SA 3.0
2014-09-11T19:34:00.773Z
In specific, the distinction I'm attempting to get across is based on the homogeneity of density across the volume of the event horizon - ie, is there difference between a diffuse cloud versus a tightly bound sphere of equivalent mass within it's Schwarzchild radius?
Ian Moriarty
CC BY-SA 3.0
2014-09-11T22:03:42.447Z
I think it might be good to mention the term 'homogeneity' in the question, as this wasn't obvious. In that case, I largely retract my answer! My speculation would be that if the cloud and the sphere had the same initial mass that, once the BH had formed, the matter would be distributed in the same way. However, this is only a supposition.
Guy Stimpson
CC BY-SA 3.0
2014-09-10T17:04:56.070Z
2014-09-10T17:04:56.070Z
Guy Stimpson
<p>I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?</p>

<p>Forked from <a href="https://astronomy.stackexchange.com/questions/6305/is-it-possible-that-a-ultra-large-portion-of-the-space-we-live-in-is-already-ins">"Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?"</a></p>

<black-hole><astrophysics>
Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?
Ian Moriarty
# Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?

(Asked by: username_0 on 2014-09-10)

I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

Forked from ["Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?"](https://astronomy.stackexchange.com/questions/6305/is-it-possible-that-a-ultra-large-portion-of-the-space-we-live-in-is-already-ins)

**Tags:** black-hole, astrophysics


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## 2 answers

### Answer by username_1 on 2014-11-09. Score: 3 | Accepted answer

> ... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

I'm not completion sure what distinction you're drawing between *concentration* and *density*, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.

For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is [Birkhoff's theorem](https://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29): outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's [shell theorem](https://en.wikipedia.org/wiki/Shell_theorem). Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.

So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be [careful](https://physics.stackexchange.com/questions/45963/do-black-holes-have-infinite-areas-and-volumes) about how you define 'volume' when comparing overall densities.

What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what *will* fall into the black hole in the future.

This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.

Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the [no-hair theorem](https://en.wikipedia.org/wiki/No-hair_theorem). In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the [singularity theorems](https://en.wikipedia.org/wiki/Penrose%E2%80%93Hawking_singularity_theorems) guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.

> **Comments:**
> **username_2**: Also check out https://en.wikipedia.org/wiki/Hoop_Conjecture  

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### Answer by username_3 on 2014-09-10. Score: 2

"Concentration of matter" and "density" are equivalent terms, i.e. density is measured in mass per unit volume, so yes, it is exactly the concentration of matter within a given volume that determines whether or not an object will form a black hole.

Any body has a so-called Schwarzchild radius - the radius within which, should all its matter become concentrated, the body will form a black hole. AFAIK Black Holes do not have a low density of matter within the black hole, in fact the density there is for all intents and purposes infinite.

Further to this, any reliable description of the environment within a black hole is, at best, educated conjecture. Bar Hawking Radiation and some other possible forms of minor 'bleeding', black holes give up no information about their contents. The only indication we have is of their mass (due to their effect on spacetime) which indicates (as far as observations are concerned) that they are very dense indeed.

> **Comments:**
> **username_4**: The density is decidedly not infinite over the entire black hole region. This would require infinite energy. Whether or not there is a point of infinite density within the horizon is hardly commonly agreed on, as well.  
> **username_3**: Hence 'for all intents and purposes' ;)  
> **username_5**: Yet your argument rests on that idea.  
> **username_0**: In specific, the distinction I'm attempting to get across is based on the homogeneity of density across the volume of the event horizon - ie, is there difference between a diffuse cloud versus a tightly bound sphere of equivalent mass within it's Schwarzchild radius?  
> **username_3**: I think it might be good to mention the term 'homogeneity' in the question, as this wasn't obvious. In that case, I largely retract my answer! My speculation would be that if the cloud and the sphere had the same initial mass that, once the BH had formed, the matter would be distributed in the same way. However, this is only a supposition.  

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