Transition from normal matter into black hole

  • #1
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Thought experiment: single use of "magic"

Setup:
Let's assume we have a giant ball of water in space.
Magic: Lets assume the water does not compress its center under its own gravity. (Constant density of 1 g/cm^3)

Basic stuff:
The mass of this ball of water, (since it does not compress) is directly proportional to its Volume, which is proportional to the CUBE of ball's radius(v = 4/3Pi r^3). (mass aprox= r^3 or CubedRoot(mass) aprox= r)

But, the Schwarzschild radius of this ball is DIRECTLY proportional to its mass. (mass aprox= r)

So, there comes a point, when the ball of water is massive enough, that it's "normal" radius will be LESS than the Schwarzschild radius.

Question:

What does it look like to the various observers, while we add sufficient water to take it under the Schwarzschild radius?

For an outside observer, do the stars that shine through the water ball: fade away as we approach the critical mass, "slide" towards the edge, just "snap off" when we reach it, or some combination?

What about for an observer scuba-diving a few meters beneath the surface that is looking towards the center, and one looking towards the surface?


p.s.
If this particular use of magic renders the entire thought experiment useless, perhaps you see what I'm looking to visualize and can suggest a better way to express the scenario. ( I was initially thinking neutronium would be better than water since it's less compressible and won't undergo fusion, but I assume it's NOT transparent. Also, scuba diving in it would be difficult, at best.)
 

Answers and Replies

  • #2
Ibix
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As I understand it, the pressure at the centre of a static sphere goes to infinity when the ball is 9/8 of a Schwarzschild radius and then it has to collapse. It simply isn't possible to describe what you want to do past that point.
 
  • #3
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As I understand it, the pressure at the centre of a static sphere goes to infinity when the ball is 9/8 of a Schwarzschild radius and then it has to collapse. It simply isn't possible to describe what you want to do past that point.
9/8 > 1
I assume you meant 8/9?
Hmm, could we see the transition if we added the magic-water "really fast"? (This is a HUGE ball of water (multiple AU), so we could expect it to take some time to all "fall in" anyway.)
Edit: Question- if THAT is when we get infinite pressure, isn't that also when we get an infinite density? If so, why is that 8/9th not the defined Schwartzchild radius?
 
  • #4
Ibix
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9/8 > 1
I assume you meant 8/9?
No, I meant 9/8. There is no way to describe a non-collapsing ball of matter smaller than 1.125 times its Schwarzschild radius. This is Buchdahl's Theorem.
Edit: Question- if THAT is when we get infinite pressure, isn't that also when we get an infinite density? If so, why is that 8/9th not the defined Schwartzchild radius?
It hasn't collapsed into a black hole at 9/8 of its Schwarzschild radius. It's just that at this point its internal forces cannot be sufficient to resist its collapse, not even in theory.
 
  • #5
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No, I meant 9/8. There is no way to describe a non-collapsing ball of matter smaller than 1.125 times its Schwarzschild radius. This is Buchdahl's Theorem.
It hasn't collapsed into a black hole at 9/8 of its Schwarzschild radius. It's just that at this point its internal forces cannot be sufficient to resist its collapse, not even in theory.
9/8...oh, of course- brain fart.
I guess I don't see why the situation of "internal forces cannot be sufficient to resist its collapse" wouldn't automatically turn it into a black hole. What else would that mass collapse down to? (checking out Buchdahl's Theorem now... thank you)
 
  • #6
Ibix
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The fact that something must inevitably collapse into a black hole doesn't mean that it is a black hole. Just that it must become one.

You could still escape the surface in a sufficiently powerful rocket. It's just that the pressure needed to maintain the matter in its static distribution diverges as the ball approaches the 9/8 limit.
 
  • #7
pervect
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As I understand it, the pressure at the centre of a static sphere goes to infinity when the ball is 9/8 of a Schwarzschild radius and then it has to collapse. It simply isn't possible to describe what you want to do past that point.

I think the 9/8 number is for the uniform density case? Which does appear to be what the OP is interested in, stating it as "magic".
 
  • #8
Ibix
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I think the 9/8 number is for the uniform density case? Which does appear to be what the OP is interested in, stating it as "magic".
9/4, according to Schutz, for a constant density. Either the 9/8 is an invention of my memory or it's a more general limit.
 
  • #9
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" Just that it must become one."
Does this mean there COULD be an opportunity to observe the transition? If we add water at a faster rate than water collapses into the singularity, would our observers be able to see the surface of the water-ball transition into a black hole?
 
  • #10
PeroK
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" Just that it must become one."
Does this mean there COULD be an opportunity to observe the transition? If we add water at a faster rate than water collapses into the singularity, would our observers be able to see the surface of the water-ball transition into a black hole?

Why water?
 
  • #11
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Why water?
Arbitrary, it's transparent, and an observer can scuba-dive in it. Please see OP for why we want these features for the thought experiment.
 
  • #12
PeterDonis
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The mass of this ball of water, (since it does not compress) is directly proportional to its Volume, which is proportional to the CUBE of ball's radius(v = 4/3Pi r^3).

No, it isn't. It would be if space were Euclidean, but in the presence of a gravitating object like this ball of water, space is not Euclidean. The actual proper volume of the ball will be larger than your calculation. That means its mass will be larger than a fixed proportionality to the cube of its radius would indicate.
 
  • #13
PeterDonis
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I think the 9/8 number is for the uniform density case

No. It applies to any spherically symmetric spacetime.

9/4, according to Schutz, for a constant density. Either the 9/8 is an invention of my memory or it's a more general limit

It's 9/4 of ##M##, or 9/8 of the Schwarzschild radius ##2M##. Unfortunately different sources don't always make it clear what the fraction is actually a fraction of.
 
  • #14
PeterDonis
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You could still escape the surface in a sufficiently powerful rocket.

More precisely, you can do this until the surface has collapsed to the horizon radius, ##2M##. After that point you can't escape from the surface no matter what you do.

Does this mean there COULD be an opportunity to observe the transition?

If you have a static object that is just marginally above the minimum radius, and you add some mass to it and thereby trigger a collapse, you will be able to observe its initial stage, yes--until the object's surface has shrunk to the same area as the area of the horizon of a black hole with the same mass, which will be ##16 \pi M^2##, where ##M## is the mass in geometric units (i.e., ##c = G = 1##). Once the surface has shrunk to the horizon size, not even light can escape from it, so you won't see from the outside anything that happens after that point.

Also, as the object collapses towards the horizon size, if you're watching from far away, the collapse will seem to slow down; in the limit, it will appear to take an infinite amount of time by your distant clock for the object to even reach the horizon size. (In any actual case, the light emitted from the object's surface will be redshifted to wavelengths too long to detect in a fairly short time; we are talking about an idealized case where you can detect radiation of any wavelength, no matter how long.) This fact has misled a lot of people into thinking that the object never actually forms a black hole; it does. The delay is in the light getting out to you far away, because of the curvature of spacetime; it's not in the collapse process itself.
 
  • #15
PeroK
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Arbitrary, it's transparent, and an observer can scuba-dive in it.

I imagine the water pressure would be a bit high for scuba-diving.
 
  • #16
Ibix
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It's 9/4 of ##M##, or 9/8 of the Schwarzschild radius ##2M##. Unfortunately different sources don't always make it clear what the fraction is actually a fraction of.
Schutz is actually perfectly clear. I just apparently forgot that ##M\neq R_S##. Friday was not a good day...
 

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