How certain are we about black hole mass density?

[Dale]

Also, “evenly distributed” is inherently frame variant for dynamic scenarios

 

[PeterDonis]

“evenly distributed” is inherently frame variant for dynamic scenarios

Yes, agreed. In post #35 I was implicitly adopting what amounts to FRW coordinates on the dust region.

 

[sweet springs]

The best way to view it is to consider the "mass" of the hole as a global property of the spacetime geometry, not something that is localized to any particular region. So of your choices, (e) is probably the best.

Thanks. I would like this view be in accordance with the view of observers who are far away from BH so whose space time is almost Minkowsy, that rest energy or mass of BH is localized at BH or at the region nearby BH at least.

 

[PeterDonis]

I would like this view be in accordance with the view of observers who are far away from BH so whose space time is almost Minkowsy, that rest energy or mass of BH is localized at BH or at the region nearby BH at least.

Far away observers can certainly take this view, since the spacetime is vacuum in their region so there is no stress-energy anywhere near them. But they should not expect to be able to move close to the BH and find any localized stress-energy anywhere. They won't. Even if they fall into the hole, they won't see anything but vacuum until they hit the singularity.

 

[Boing3000]

There is no plausible scenario where galaxies just happen to "superpose" so that a superdupermassive black hole gets created "in an instant" as they all reach just the right positions.

No, certainly not plausible, but not that far fetched either. Incidentally the wiki article give a lot of referenced of scale. It even bother to compute the Schwarzschild radius for the "observable universe" that is (probably not coincidentally) 13.7 Billion ...quite like the age of the universe (at light speed without inflation).

Anyway some rough computation got me that you'll only need +- 500 identical Milky-way on top of each other, so they "became" a Black Hole (given that there is enough empty space around).
500 it is not that much, given the average local density is quite low, and local curvature should still be quite flat, or am I missing something ? Anyhow I cannot help wondering what would happens if you were inside such an event. I am guessing not much, but you seem to be quite formal that nothing can exist inside a BH.

It's not a matter of scale, it's a matter of spacetime geometry. The concept of "Schwarzschild radius" only applies to the spacetime geometry of an isolated object or system of objects surrounded by vacuum. It does not apply to the universe as a whole since the universe as a whole is not an isolated system surrounded by vacuum.

Got it.
But now I am wondering if you could "unwind" a BH but changing the curvature outside of it. Would the presence of a lot of matter around a BH (I know that one IS far fetched) change the BH far away apparent Schwarzschild radius ?

 

[martinbn]

It even bother to compute the Schwarzschild radius for the "observable universe" that is (probably not coincidentally) 13.7 Billion ...quite like the age of the universe (at light speed without inflation).

The radius of the observable universe is 46billion (if I remember correctly).

 

[Boing3000]

The radius of the observable universe is 46billion (if I remember correctly).

I find the connection with 13.7 quite bizarre. I don't believe in coincidence. I don't even think black-matter has been factored in.
My computation got me: 8.75 * 10^52 kg for a 13.7 Schwarzschild radius black hole

and this page seems to evaluate the mass of the universe to:
2.072 × 10^53 (without dark matter/energy),

But the wiki page (also 1^53) seems to include "dark content".

I know that at one time the observable universe was smaller than that, and it seems to point to the fact that there was no vacuum "outside" of it (it is now un-observable, but kind of still is indirectly). Or is it that the general expansion modify the curvature enough so that even with "vacuum outside" would not induce a BH geometry ?

 

[PeterDonis]

certainly not plausible, but not that far fetched either

I think you need to take a step back and consider carefully before you make such a claim.

some rough computation got me that you'll only need +- 500 identical Milky-way on top of each other, so they "became" a Black Hole (given that there is enough empty space around).

And you think this is not "far fetched"?

Apart from anything else, you are failing to take into account that 500 Milky Ways, each composed of a few hundred billion stars, plus clouds of gas and dust, would have a very complicated dynamics. The idea that they could somehow just end up "all on top of each other" in an instant and form a black hole at that instant is indeed far-fetched.

now I am wondering if you could "unwind" a BH but changing the curvature outside of it

I have no idea what you mean by this.

 

[PeterDonis]

It even bother to compute the Schwarzschild radius for the "observable universe"

Which is meaningless since the observable universe does not have Schwarzschild spacetime geometry. Wikipedia is not a reliable source.

 

[Boing3000]

And you think this is not "far fetched"?

not that far fetched. And I am carefull, minds you.

Apart from anything else, you are failing to take into account that 500 Milky Ways,

No I didn't, the very reason is that it is *I* that put them in the scenario.

each composed of a few hundred billion stars, plus clouds of gas and dust, would have a very complicated dynamics.

That's, of course, irrelevant. Nobody, especially me, said that it is was likely of even plausible, nor that the dynamic would be "simple"

The idea that they could somehow just end up "all on top of each other" in an instant and form a black hole at that instant is indeed far-fetched.

It is, but not that far fetched. The universe is a big place Peter, we happens to live in a Galaxy that is on a collision course with another. 500 is just 2 order of magnitude away from 2. Beside there are denser clusters than ours. And some cluster (500 ?) are bound enough by gravity so that they should survive the tearing appart from expansion.
Besides on a greater scale than galaxies (cluster) the "volumetric density" required is even lower. So the scenario stands, even if the "instant" comes about after half a million years a building...
So does the scenario of post#16 which is after all a thought experiment. Again: the question is: What does GR predicts for what would happens inside that volume of space (surrounded by as much deep vacuums you need)

I have no idea what you mean by this.

OK, let me explain (beware, that scenario is really far fetched).
1) We found an Earth mass black hole (~1 cm across)
2) We somewhat managed to drop synchronously from from orbits 2 half sphere of great thickness and strength so they meet exactly around the BH (let's say the initial interior radius is (1 meter, 10 meter, 1 kilometer ?). Curvature is strong, but will it systematically destroy the sphere (given it is still only 1G of acceleration somehow)
3) On top of the sphere, we pack as much mass as possible (taking care of momentum so the sphere never "touch" the even horizon) without breaking the sphere.

The question is: Does that added curvature outside the hole, change the radius of the horizon ?(1)

The really really really far fetched thought experiment would be to scale up the experiment to a HUGE proto black hole gaz cloud, kept from collapsing by "external(at outer surface)" mass curvature, that could be suddenly moved outward, creating enough vacuum so the BH form.
What if the mass is suddenly re-set inward, would that affect the BH formation ?(related to (1))

Are such simulation even possible ?

 

[PeterDonis]

not that far fetched

IMO you have not justified this claim at all. But it's a subjective claim, and evidently your standards for what counts as "far-fetched" are different from mine.

What does GR predicts for what would happens inside that volume of space

You haven't specified the scenario precisely enough to tell. You say the dynamics are irrelevant, but unless you know the dynamics you don't know what GR predicts. There is certainly no idealized, highly symmetric model like Schwarzschild spacetime that will let you extract qualitative answers for this case without having to do detailed computations of the dynamics.

Remember that, in the original comments of mine that appear to have spawned this sub-thread, I wasn't saying that a black hole could not form by a process like this. I was only saying that you can't just compute the "density" (using the physically meaningless formula described in the OP of this thread) and expect that to give you a simple rule for exactly when the black hole forms. It doesn't work that way; the dynamics is complicated, whether you want to think about them or not, and that means the criteria for when a black hole is formed are also complicated.

Does that added curvature outside the hole, change the radius of the horizon ?

You haven't specified the scenario precisely enough to tell (because it's possible that the sphere could be close enough to the hole that it would be forced to fall in, increasing the hole's mass). But here is a general statement that might help: if you have a static (meaning it won't collapse), spherically symmetric shell of matter with vacuum inside, the shell has no effect on spacetime curvature in the region inside. That means that whatever is inside is the sole determinant of the curvature inside.

 

[Boing3000]

You haven't specified the scenario precisely enough to tell. You say the dynamics are irrelevant, but unless you know the dynamics you don't know what GR predicts.There is certainly no idealized, highly symmetric model like Schwarzschild spacetime that will let you extract qualitative answers for this case without having to do detailed computations of the dynamics.

Fair enough. I could add that all galaxies are parallel, that half the galaxies are rotating one way, and they come equally from all direction. The goeal is to have an average momentum(and angular momentum) of 0.
Another scenario would be a kugelblitz has described here. The dynamic of interest for me, is what happens INSIDE the BH. I think GR can describe it, but I don't think that any dramatic change should be experienced inside the BH. Well, not faster than light anyway.

I was only saying that you can't just compute the "density" (using the physically meaningless formula described in the OP of this thread) and expect that to give you a simple rule for exactly when the black hole forms. It doesn't work that way; the dynamics is complicated, whether you want to think about them or not, and that means the criteria for when a black hole is formed are also complicated.

Indeed, and I have another scenario in mind. Maybe I'll start another thread with it.

You haven't specified the scenario precisely enough to tell (because it's possible that the sphere could be close enough to the hole that it would be forced to fall in, increasing the hole's mass).

Well, I cannot be more precise. I give some diameter of choices (not too far fetched), I don't see what I can add to that (maybe the Earth BH is non-rotating and uncharged)

But here is a general statement that might help: if you have a static (meaning it won't collapse), spherically symmetric shell of matter with vacuum inside, the shell has no effect on spacetime curvature in the region inside. That means that whatever is inside is the sole determinant of the curvature inside.

But the if is known to be false. I though that Newton shell theorem would not apply anyway (due to strong enough curvature).
I get that global(possibly to infinity ?) geometry is important for BH formation, and that why the universe is not a BH (there is no vacuum outside the universe).

It just get me to wonder if a local measuring device on the interior of the sphere would notice a change in the horizon local radius, when more and more mass was added on the outer shell. I am going to assume no change.

Also, do you have any opinion on the coincidence between the "physically meaningless formula" that nonetheless happens to give 13.7 Bly radius for a pseudo BH with the mass of the observable universe ?

 

[PeterDonis]

the if is known to be false.

It is? Weren't you describing the assembly, from two half-shells, of a spherical shell of matter with vacuum inside? (Remember that a black hole is vacuum.)

I though that Newton shell theorem would not apply anyway

You are wrong. The shell theorem, as I stated it, holds in GR.

a local measuring device on the interior of the sphere would notice a change in the horizon local radius,

There is no such thing as "horizon local radius". The radius of the horizon is a global, not a local, property. There is no way to locally measure where the horizon is or what its radius (more precisely, its area) is.

do you have any opinion on the coincidence between the "physically meaningless formula" that nonetheless happens to give 13.7 Bly radius for a pseudo BH with the mass of the observable universe ?

I already gave my opinion: it's meaningless.

 

[martinbn]

Also, do you have any opinion on the coincidence between the "physically meaningless formula" that nonetheless happens to give 13.7 Bly radius for a pseudo BH with the mass of the observable universe ?

As already pointed out it is meaningless, but what exactly is the coincidence? The radius of the observable universe is a lot more than 13.7bly.

 

[Boing3000]

(Remember that a black hole is vacuum.)

Ho, I thought you were jesting there. OK, you've made up quite a new definition of vacuum (we both know its not, there is at least a singularity, and there is mass anyway). You got me there.

You are wrong. The shell theorem, as I stated it, holds in GR.

Then I'll start the new thread, a new scenario is coming

There is no such thing as "horizon local radius". The radius of the horizon is a global, not a local, property. There is no way to locally measure where the horizon is or what its radius (more precisely, its area) is.

Not as such, but we surely can measure the change of the horizon by shooting a laser and seeing if it hit a detector somewhere or elsewhere (or no more).

I already gave my opinion: it's meaningless.

As allready pointed out it is meaningless, but what exactly is the coincidence? The radius of the observable universe is a lot more than 13.7bly.

OK fine. That's bizarre. We just can compute that all the particle of the universe seems to be at the center of a singularity 13.7 by old. And that's actually confirmed somehow by observation.

I think I heard once Susskind say something like that (kind of: that's like if we are inside a BH) but right now, I cannot find the lecture...

 

[PeterDonis]

OK, you've made up quite a new definition of vacuum (we both now its not, there is at least a singularity, and there is mass anyway)

The singularity is not part of the manifold.

The complete spacetime of a realistic black hole does include a non-vacuum region (occupied by the object that originally collapsed to form the hole). But that does not mean that non-vacuum region is "still inside" the hole forever. I am not changing the definition of "vacuum" at all.

we surely can measure the change of the horizon by shooting a laser and seeing if it hit a detector somewhere or elsewhere

This is not a local measurement. It's a global one.

We just can compute that all the particle of the universe seems to be at the center of a singularity 13.7 by old.

You're missing the point. The spacetime geometry of the universe as a whole is not the Schwarzschild geometry. It's not even close to that geometry. The Schwarzschild geometry is static outside the horizon and asymptotically flat--i.e., the metric goes to Minkowski at infinity. The spacetime geometry of the universe as a whole is static nowhere and has no "infinity" at all, let alone an asymptotically flat one. The observable universe is a portion of the universe as a whole, but also is static nowhere and has no "infinity". So any numerical similarity between some computation you make about the universe and some computation you make about the Schwarzschild geometry is physically meaningless. It's like saying the point on Earth where the prime meridian meets the equator is "somehow the same" as the origin of a Euclidean plane, because they both happen to have coordinates $(0, 0)$.

 

[PeterDonis]

I think I heard once Susskind say something like that (kind of: that's like if we are inside a BH)

Yet another reason for me not to like Susskind.

 

[Boing3000]

This is not a local measurement. It's a global one.

OK. The question become, does this global measurement will detect a change inside that shell ? The answer seems to be no.

You're missing the point. The spacetime geometry of the universe as a whole is not the Schwarzschild geometry. It's not even close to that geometry.

I didn't miss that point because your explanations are perfectly clear.

It's like saying the point on Earth where the prime meridian meets the equator is "somehow the same" as the origin of a Euclidean plane, because they both happen to have coordinates $(0, 0)$.

More like having a formula to compute the surface of the oceans, and obtaining, by coincidence, the age of the ocean. My only point is that is is bizarre.

Actually it is more like having a formula to compute the surface of the oceans in some metric, and obtaining the volume of the ocean in another one. Totally unrelated, not even the same units. But not that bizarre, given the obvious (but missing) link (average depth).

 

[Ibix]

Actually it is more like having a formula to compute the surface of the oceans in some metric, and obtaining the volume of the ocean in another one.

In geometric units, the Schwarzschild radius is 2M. Taking your word for the fact that this is the same as the age of the universe, T, you are saying 2M=T. Either this is just a coincidence (we just happen to be around at the time when T=2M, which happens at some point for a very large range of possible relationships between M and T), or this holds for all times (i.e., the mass of the observable universe is directly proportional to its age). I'm not sure, but I do not believe the latter is correct.

 

[Boing3000]

In geometric units, the Schwarzschild radius is 2M. Taking your word for the fact that this is the same as the age of the universe, T, you are saying 2M=T.

I may very well have made a mistake I used this formula $r_s= \frac{2GM}{c^2}$. That gave me this code that you could check in your browser javascript console (F12)

(function(c,G,yearS,solarMass){ console.log("Mass:"+(((Math.pow(c,2) * ((13.7e+9)*yearS*c)) / (2*G))/1/*solarMass*/).toExponential()+"kg");})(299792458, 6.674e-11, 365 * 24 * 60 * 60, 1.98847e+30)

Either this is just a coincidence (we just happen to be around at the time when T=2M, which happens at some point for a very large range of possible relationships between M and T), or this holds for all times (i.e., the mass of the observable universe is directly proportional to its age). I'm not sure, but I do not believe the latter is correct.

I don't either, that's why I am surprised.
I have heard that we are a a somewhat special time in the universe that make the expansion rate very detectable (it won't be anymore in the far future). Maybe that's the reason.

 

[PeterDonis]

does this global measurement will detect a change inside that shell ?

For the case you described, no, by the shell theorem.