How certain are we about black hole mass density?

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The Wikipedia entry on "White dwarf" has a table listing the densities (in kg/m3) of various objects/substances, and states that an earth-mass black hole has a critical density of 2 x 1030, about 13 orders of magnitude denser than atomic nuclei. Has such an enormous density been calculated, based on a black hole's observed gravitational pull and its observed (Schwarzschild?) radius, or is this a purely theoretical result? Basically, I'm wondering how certain, if at all, we are that such densities are possible.
 

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  • #2
Dale
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an earth-mass black hole has a critical density of 2 x 1030, about 13 orders of magnitude denser than atomic nuclei. Has such an enormous density been calculated, based on a black hole's observed gravitational pull and its observed (Schwarzschild?) radius, or is this a purely theoretical result?
This is purely theoretical. It is also one reason why we believe there are no natural earth-mass black holes. A larger black hole has a lower critical density.
 
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  • #3
Buzz Bloom
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Hi @hkyriazi:

You might want to do a bit of arithmetic. One way to define the density of a black hole is its mass M divided by the volume of a sphere with a radius equal to its Schwarzschild radius rs.

Wikipedia gives a formula for the Schwarzschild radius as
rs = 2.95 M/Msun km.​
https://en.wikipedia.org/wiki/Black_hole#Physical_properties

I was unable to find any usage of he term "critical density" with respect to a black hole except in the table in the article
https://en.wikipedia.org/wiki/White_dwarf .
I do not know the definition of "critical" in this usage.

Regards,
Buzz
 
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  • #4
Nugatory
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It's a calculation, but the gravitational pull doesn't come into it. If all the mass were not inside the event horizon it wouldn't be a black hole, so knowing the mass and the "volume" from the Schwarzschild radius is enough to calculate the "density". (The scare-quotes around the word "volume" are because we're defining the "volume" of the black hole to be the volume of a sphere whose surface area is ##4\pi{R_S}^2##; the actual volume of space inside the event horizon is not well defined.)
 
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  • #5
PeterDonis
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an earth-mass black hole has a critical density of 2 x 1030
This "density" is just a numerical calculation that assumes that the hole's mass is contained within a Euclidean sphere with the surface area of the hole's horizon. Such a calculation has no physical meaning; it's just a number that pop science treatments sometimes throw out for no good reason. A black hole is not a static object with mass contained in a Euclidean 3-sphere with the surface area of the horizon. It's something very different.
 
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  • #6
PeterDonis
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It is also one reason why we believe there are no natural earth-mass black holes.
I don't think this is correct. As I said in my previous post, this "density" calculation has no physical meaning. AFAIK it does not play a role in any hypotheses about what masses of black holes we should expect to observe.
 
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It's a calculation, but the gravitational pull doesn't come into it.
I assume astronomers have looked at the rotation rates of galactic cores, and calculated central "super-massive black hole" masses based on the gravitational pull apparently exerted. I doubt they're able to visualize the black hole event horizon (due to gravitational lensing?) sufficiently to do a pedestrian density calculation.
 
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So, then, is it safe to assume that no pedestrian but respectable, pop science-type calculations have been made of mass densities in black holes, that yield densities much greater than those of neutron star cores?
 
  • #9
PeterDonis
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So, then, is it safe to assume that no pedestrian but respectable, pop science-type calculations have been made of mass densities in black holes
You're missing the point. A black hole is not an ordinary object full of matter. It doesn't have a well-defined density at all.
 
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  • #10
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I assume astronomers have looked at the rotation rates of galactic cores, and calculated central "super-massive black hole" masses based on the gravitational pull apparently exerted. I doubt they're able to visualize the black hole event horizon (due to gravitational lensing?) sufficiently to do a pedestrian density calculation.
It's not necessary to visualize or otherwise measure the size of the event horizon. The Schwarzschild radius is a function only of the mass, so if you know the mass you also know the "volume" and hence the density.

Note also that the "volume" goes up as the cube of the Schwarzschild radius while the Schwarzchild radius is directly proportional to the mass. Thus the "density" goes as the inverse square of the mass; an earth-mass black hole may be 13 orders of magnitude denser than an atomic nucleus, but a supermassive black hole is less dense than ordinary boring room temperature water. Thus, no extraordinary mechanisms are needed to turn a sufficiently large amount of gas into a supermassive black hole; its own ordinary gravitational attraction is enough to compress it to below the Schwarzschild radius and form the black hole.
 
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  • #11
PeterDonis
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no extraordinary mechanisms are needed to turn a sufficiently large amount of gas into a supermassive black hole; its own ordinary gravitational attraction is enough to compress it to below the Schwarzschild radius and form the black hole
I don't think this intuitive reasoning is correct, because, as I've already said, the "density" calculated this way has no physical meaning. Even if we are considering a vast cloud of gas, if it is true that "its own ordinary gravitational attraction" is enough to cause it to become a black hole, then its spacetime geometry is already sufficiently different from flat that its density, even before it forms the hole, can't be calculated by assuming it is occupying a Euclidean volume of space bounded by its surface. So I don't think any intuitions along these lines should be used.
 
  • #12
pervect
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The density (usually of a presureless cloud of "dust") to form a black hole of a given size/mass is well defined though, and I believe is what the popularizations are calculating.

It'd be more interesting (but more complicated) to consider some sort of idealized fluid model that represented matter, and calculate the amount of pressure needed to compress an Earth-sized cloud of this idealized fluid to a low enough volume for it to form a black hole.

The usual model would be degenrate matter, <<wiki link>>

I don't know the experimental bounds on our understanding of neutron degenreate matter (the most reasonable case for a small black hole) would be though. I don't think they're terribly tight, but I'd be surprised if we didn't at least know the correct order of magnitude - at a guess.
 
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PeterDonis
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The density (usually of a presureless cloud of "dust") to form a black hole of a given size/mass is well defined though
Only in the sense that, in an idealized model such as the Oppenheimer-Snyder model of collapse of dust, given an externally measured mass (which is constant throughout the process), and an initial radius for the object when it starts collapsing from rest (which must be at least 9/8 of the Schwarzschild radius for the given mass because of Buchdahl's Theorem, and in practice will be much larger), one can calculate the density of the dust (which is uniform throughout the dust) at the instant when it is just crossing the event horizon of the spacetime as it collapses. But this density is not the same as the density being calculated by these pop science calculations. It is also not "the density of the black hole" in any meaningful sense. Nor is it the density of the dust when it starts collapsing from rest (that will be smaller).

It'd be more interesting (but more complicated) to consider some sort of idealized fluid model that represented matter, and calculate the amount of pressure needed to compress an Earth-sized cloud of this idealized fluid to a low enough volume for it to form a black hole.
This would be relevant for primordial black hole formation (i.e., formation by pressure and density fluctuations in the early universe), but not for formation of black holes by gravitational collapse. Such a collapse does not involve "pushing matter inward" by pressure.

I don't know the experimental bounds on our understanding of neutron degenreate matter
As far as the densities we have a reasonable understanding of? IIRC we have a reasonable approximate model of the equation of state of neutron degenerate matter up to about a hundred times nuclear densities, or about ##10^{18} \text{kg/m^3}##.
 
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  • #14
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A black hole is not a static object with mass contained in a Euclidean 3-sphere with the surface area of the horizon. It's something very different.
May be it is along the line with this that I once calculated the integration of energy momentum pseudo tensor t in region outside Schwartzshild boundary sphere of BH. I got the result equals Mc^2 where M is mass of BH. It could be interpreted that all energy exist outside Schwatzshild sphere, not inside.
 
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  • #15
bobob
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The Wikipedia entry on "White dwarf" has a table listing the densities (in kg/m3) of various objects/substances, and states that an earth-mass black hole has a critical density of 2 x 1030, about 13 orders of magnitude denser than atomic nuclei.
That statement is somewhat misleading. The matter within the black hole has collapsed completely, so aside from any matter falling into a black hole, the interior doesn't contain any matter. All of the matter ends up at the singularity and the interior is just empty space. It would be more accurate to just state the schwarzschild radius of a black hole having the mass of the earth and leave it at that. A better way to understand this is to ask why all matter doesn't end up in black holes and why there are any objects that have the densities like that of white dwarf stars. The answer to that requires more than just general relativity and the relationship you get between a black hole mass and its corresponding schwarzschild radius.

What gives a white dwarf such a large density is what prevents it from collapsing any further and that is due to quantum mechanics and forces other than gravity. In particular, there is a limit to the number of particles you can confine to a given volume due to the exclusion principle. this results in a pressure that resists gravity trying to squeeze those particles into a smaller volume. Finding out the volume at which the pressure balances gravity requires knowing something called the density of states, which depends on things other than just mass and volume. You can look up Chandrasekhar limit to see how this is obtained for a white dwarf and Oppenheimer-Volkoff-Tolman limit for neutron stars. Whether or not further collapse into a more dense, stable body (like a strange star made of uds quarks) is possible is an open question. In any case, the density of matter alone is not sufficient determine whether or not the matter will further collapse. You have to know what is preventing matter from collapsing which requires knowing the number of quantum mechanical states available to the particles confined to that volume.

Has such an enormous density been calculated, based on a black hole's observed gravitational pull and its observed (Schwarzschild?) radius, or is this a purely theoretical result? Basically, I'm wondering how certain, if at all, we are that such densities are possible.
The density of a neutron star (which is much denser than a white dwarf) is certainly possible because the density of a neutron star is not estimated to be that much different than the density of your average heavy nucleus. Lots of experiments have been done to determine nuclear densities, so a density on the order of 10^17 kg/m^3 is not just a theoretical result. One of the reasons nuclear physicists study nuclei is to better understand nuclear matter and refine the equation of state for nuclear matter and consequently for understanding neutron stars and how large they can be without collapsing further.
 
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  • #16
Dale
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I don't think this is correct. As I said in my previous post, this "density" calculation has no physical meaning. AFAIK it does not play a role in any hypotheses about what masses of black holes we should expect to observe.
I disagree. We don’t know of any natural process which will take an isolated earth mass and compress it to a density ~13 times the density of atomic nuclei. We do know of natural processes which will take an isolated million solar mass and compress it to the density of water. Therefore we expect to see million solar mass black holes but not earth mass black holes.

If an amount of matter is uniformly distributed throughout its schwarzschild radius then it is already a black hole fully inside its event horizon. It will of course not remain uniformly distributed, but it is a perfectly valid initial condition that gives an easy plausibility check.

Leonard Susskind uses this type of argument in his GR course.
 
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  • #17
PeterDonis
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I once calculated the integration of energy momentum pseudo tensor t in region outside Schwartzshild boundary sphere of BH
A black hole is vacuum everywhere, outside and inside, so this integral will give zero. If you got M, you did it wrong.
 
  • #18
PeterDonis
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We do know of natural processes which will take an isolated million solar mass and compress it to the density of water.
We know of natural processes which will take a million solar masses worth of highly diffuse matter and have it eventually collapse to a black hole whose "density", if it is calculated using the physically meaningless method described earlier, comes out to the density of water. But, as I've already said several times, this "density" is physically meaningless, and the process is not a process of anything being "compressed". It's gravitational collapse, which is something different.

Also, as I've already said, the physically meaningless "density" calculation has nothing to do with the arguments for why we expect to see million solar mass black holes but not earth mass black holes. Those arguments are about what kinds of objects can undergo gravitational collapse to black holes, and do not depend on the physically meaningless "density" of the hole once it is formed.

If an amount of matter is uniformly distributed throughout its schwarzschild radius then it is already a black hole fully inside its event horizon. It will of course not remain uniformly distributed, but it is a perfectly valid initial condition
Not if the matter is claimed to be at rest. It is impossible for any distribution of matter to be at rest inside a sphere with a surface area having an areal radius less than 9/8 of the Schwarzschild radius corresponding to its externally measured mass. This is a simple consequence of Buchdahl's Theorem.

If the matter is not at rest but collapsing inward, then, while technically this counts as an "initial condition" (since in a deterministic theory you can call any condition an "initial condition" if you like), calling it that does not strike me as a good use of ordinary language, and trying to describe things that way is very likely to confuse people into thinking that a black hole can be viewed as some kind of static object where the matter just happens to be sitting just inside the Schwarzschild radius. Which it isn't.

Leonard Susskind uses this type of argument in his GR course.
Then that just gives me another reason not to be a fan of Susskind. :wink:
 
  • #19
Dale
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But, as I've already said several times, this "density" is physically meaningless, and the process is not a process of anything being "compressed". It's gravitational collapse, which is something different.
I am not talking about gravitational collapse. I am talking about the initial conditions prior to the collapse.

Not if the matter is claimed to be at rest. It is impossible for any distribution of matter to be at rest inside a sphere with a surface area having an areal radius less than 9/8 of the Schwarzschild radius corresponding to its externally measured mass.
I am not making any claims about being at rest, just density. However, my understanding of Buchdahl’s theorem is that it places a limit on static configurations, I.e. momentarily at rest is fine it just won’t remain at rest.

likely to confuse people into thinking that a black hole can be viewed as some kind of static object where the matter just happens to be sitting just inside the Schwarzschild radius. Which it isn't.
I agree.

Then that just gives me another reason not to be a fan of Susskind.
That is fine, but I liked the lecture series. It was not pop sci, but a gentle introduction. The point that he made was a valid point. Basically, the point is that for any given density, there is a finite amount of mass that at that density will form a black hole. So the conditions for formation of a black hole are not at all exotic, all you need is a lot of ordinary matter. Thus we are almost guaranteed that they exist since the conditions for their formation are so ordinary.
 
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  • #20
PeterDonis
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momentarily at rest is fine it just won’t remain at rest.
I don't think this is true either, at least not in any physically reasonable sense.

Mathematically speaking, yes, I think there are solutions that describe a spherically symmetric region of dust with finite extent that starts out with zero surface area, expands to a maximal surface area that could correspond to an areal radius less than 9/8 of the Schwarzschild radius corresponding to its externally measured mass, but still greater than that Schwarzschild radius, is momentarily at rest at that maximal areal radius, and then collapses back to zero radius. But such a solution would have an initial "white hole" singularity as well as a final "black hole" singularity, and is therefore not physically reasonable. (Such a solution would also have a "white hole" past horizon as well as a "black hole" future horizon.)

Also, even mathematically speaking, I don't think there are any valid solutions that have a maximal areal radius that is less than or equal to the Schwarzschild radius corresponding to the externally measured mass. The reason is that such a solution would have to have a single horizon that played the role of both the past and future horizons (since the matter never goes outside the horizon), and that is not possible, because those two horizons must be distinct (for example, on a Kruskal spacetime diagram they are perpendicular to each other--looking at a Kruskal diagram and trying to imagine how the region of matter would look if it never went outside the horizon can also help to see why such a solution is not possible). So if we are talking about the matter occupying a region with surface area less than the horizon area, it is impossible for it to be at rest even momentarily.

the conditions for formation of a black hole are not at all exotic, all you need is a lot of ordinary matter. Thus we are almost guaranteed that they exist since the conditions for their formation are so ordinary.
I agree with this statement as a general statement, but I do not agree with Susskind's way of justifying it.
 
  • #21
PeterDonis
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momentarily at rest is fine
Btw, an even simpler way of seeing that even "momentarily at rest" is impossible inside the horizon is to note that, for the matter to be even momentarily at rest, there must be a point on its worldline at which its 4-velocity points in a direction of constant areal radius (this will be the point at which it is momentarily at rest). But there are no such vectors inside the horizon that are timelike, or even null; all vectors pointing in the direction of constant areal radius inside the horizon are spacelike.
 
  • #22
Dale
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The reason is that such a solution would have to have a single horizon that played the role of both the past and future horizons (since the matter never goes outside the horizon),
Yes, that’s a good point. Anyway, as I said above, I am not making any claims about the initial velocity. But regardless of the velocity the density is a valid thing to include in the characterization of the initial conditions.

I agree with this statement as a general statement, but I do not agree with Susskind's way of justifying it.
It made sense to me.
 
  • #23
PeterDonis
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regardless of the velocity the density is a valid thing to include in the characterization of the initial conditions.
Again, while this is technically true, since in a deterministic system you can call conditions at any time "initial", I don't think referring to the state of dust collapsing inward with a surface area just a bit less than the horizon area as "initial conditions" is a good choice of language. The "initial condition" should be the condition of the original object that started the collapse.

It made sense to me.
I can make sense of it if I interpret him as describing a cloud of dust that is collapsing inward and has a surface area just a bit less than the horizon area for its mass. But for the reasons I have given, I think describing this as an "initial condition" is seriously misleading. Such a cloud will never just occur on its own; it will always be an intermediate state in a process of gravitational collapse from some larger object like a massive star.
 
  • #24
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A black hole is vacuum everywhere, outside and inside, so this integral will give zero. If you got M, you did it wrong.
Honestly I could integrate t only outside Schwartzshild sphere. I did not and even now do not know formula of t inside the shere.
 
  • #25
PeterDonis
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I could integrate t only outside Schwartzshild sphere.
So what? The region is still vacuum and the integral still gives zero.
 

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