Why the black hole density is called the infinite density.

yuiop

I looked into this a long time ago and by studying the interior (internal?) Schwarzschild solution for the non vacuum case with constant density, it certainly appears that the event horizon (where gravitational time dilation goes to zero) initially forms at the centre and then moves out towards the Schwarzschild radius as the collapse continues. The situation is basically unchanged for a sphere of matter with density increasing proportionally to 1/r towards the centre.

PeterDonis

This is not quite correct. What forms at the center (r = 0) and moves outward until it reaches the Schwarzschild radius is the absolute horizon, not the apparent horizon.

The apparent horizon is "where gravitational time dilation goes to zero", though I would prefer to describe it as the location where outgoing light rays stay at the same radius, since "time dilation" is observer dependent but the paths of light rays are not. That is, the apparent horizon is locally defined; you can tell it's there just by looking at the paths of light rays locally. The apparent horizon forms when the surface of the star is just falling inward past the Schwarzschild radius corresponding to the star's total mass; before that, there is *no* apparent horizon anywhere.

The absolute horizon is the boundary between the region of spacetime that cannot send signals "to infinity" and the rest of the spacetime. That is, the absolute horizon is globally defined; you have to know the entire future paths of light rays to know where it is. This boundary coincides with the apparent horizon once it forms (for the idealized case of a black hole that is static and "eternal" once formed), but the absolute horizon forms at r = 0 before that and expands outward at "the speed of light" (which is a bit of a misleading way to state it, since as more of the star's matter passes inward the outgoing null surface moves outward more and more slowly) until it meets the surface of the star at just the point where the apparent horizon forms.

dustinthewind

So let me know if I read you wrong. But it looks like you are suggesting then that the event horizon stops moving as soon as it reaches the outer surface of the object its engulfing? I would guess as this event horzon is expanding there is some reduction in the radius of the object? But it sounds as if this horizon and the surface meet up and then find equilibrum. This would suggest to me some kind of maximum density maybe.

Does the mass per volume remain constant with the expansion of the event horizon?

What is interesting is that earths gravity reduces from its maximum when inside the earth linearly. So inside this object we can have the gravitational force reduce by r/a*GravHole? However an object reaching the speed of light at the surface would require infinite force. Yet gravity is with respect to the center of the object as it decreases by 1/r^2. I wonder how its possible for an object to experience infinite force from the gravitational law when it reaches the event horizon?

I suppose for this to be possible it would require for the density at the surface to be infinite but the gravitational law only takes into accound mass. Would this suggest that there is infinite mass located at the edge of the event horizon? Or maybe that the distance to the center of the black hole becomes zero due to lorentz contraction.

And I believe it would be certainly in error to assume a black hole to have infinite mass. And to assume it has infinite density is the same as saying it has infinite mass. But it certainly has limited mass other wise there would be no limit to its gravity.

I guess I like the best the concept that on approach of the black hole the lorentz contraction closes you distance to the center of the black hole to zero.

PeterDonis

"Moving" is not a precise term as it stands, and I should have clarified how I was using it. The EH "stops moving" in the sense that its radial coordinate no longer changes.

None of what I said has anything to do with the density of the collapsing object, or indeed with its radius. The EH is not a "thing" that can affect other things; it is just a "marker" that we place in the spacetime to pick out the boundary of the region that cannot send light signals to infinity. Saying that "the EH moves outward" just means that the radius of the point where we place the "marker" increases with time.

Again, the EH is not a "thing"; it has no mass and "mass per volume" doesn't mean anything with respect to it. You seem to be confusing the trajectory of the EH with the trajectory of the actual matter of the collapsing star. They are not the same.

No. Gravity decreases inside the Earth because more and more of the Earth's mass is above you instead of below you. Once the collapsing object has formed a singularity at the center, r = 0, there is no place anywhere where any of the mass of that object is above you. It's all been crushed into the singularity and ceased to exist.

This formula is incorrect; it doesn't include the relativistic correction factor. The correct formula is (in units where G = c = 1):

[tex]a = \frac{M}{r^{2} \sqrt{1 - \frac{2M}{r}}}[/tex]

This obviously increases without bound as r -> 2M, i.e., as the horizon is approached. However, this formula only applies to an observer who is "hovering" at a constant radius; but it is impossible for any massive object (i.e., anything except light or some other massless radiation) to "hover" exactly at the horizon, because it would have to move at the speed of light to do so.

No, both of these are incorrect. See above.

"Gravity" is not a term with a single meaning. The curvature of spacetime, as shown by tidal gravity, is finite at the horizon. The "acceleration due to gravity" for a "hovering" observer goes to infinity, but as noted above, this is simply because the horizon is a null surface, so anything "hovering" at the horizon would have to move at the speed of light, which would indeed require "infinite acceleration", just as it would in special relativity. This does not in any way require "infinite gravity".

This is also wrong. A freely falling observer sees a finite distance to the center (r = 0) when he is at the horizon (r = 2M). "Lorentz contraction" does affect the distances he sees, compared to those that a "hovering" observer sees, while he is outside the horizon; but there is no "hovering" observer at the horizon (or inside it).

TEjedi

OK, so a black hole forms when the gravity of a set amount of mass has expelled enough energy that it can no longer support itself. When it collapses it forms a singularity. The singularity is infintely dense and infinetly small. How did a set amount of mass form something infinite? The gravitational signature exist of something with a set mass but if we examine the process in reverse wouldnt we end up with something infinetly large at the limit of density approaching 0? I know that reversing the process is impossible.

PeterDonis

I'm not exactly sure what you mean by this, so it's hard to say whether or not it's correct.

Because all of the individual pieces of the set amount of mass came together at a single point at the center, so, speaking somewhat loosely, you have a finite amount of mass in a zero volume, which equates to infinite density.

However, that's not the whole story, because as previous posts in this thread have pointed out, once the singularity is formed at r = 0, the matter in the object that originally collapsed to form it ceases to exist. All that remains is the infinite spacetime curvature at r = 0. The reason that curvature persists is that once it forms, it is self-perpetuating; the vacuum Einstein Field Equation (vacuum because, as I said just now, there is no matter left from the original collapsing object) says that that configuration of spacetime curvature will remain the same for all time.

The time reverse of a black hole singularity forming from a collapsing sphere of matter, would be a "white hole" singularity suddenly exploding into an expanding sphere of matter. Whether or not the matter would expand indefinitely, i.e., whether it would continue to expand forever to unbounded values of the radius (and therefore continue to decrease in density to arbitrarily small values) would depend on how the explosion happened; there are infinitely many ways such an explosion could happen, just as there are infinitely many ways that an object can collapse into a black hole.

yuiop

Consider a sphere of dust of even density and radius R. When the sphere collapses to a radius of 9/8 Rs where Rs is the Schwarzschild radius equal to 2M/c^2, a clock at the centre of the mass stops ticking relative to Schwarzschild coordinate time. This is the where the absolute horizon forms and as the dust sphere radius collapses towards R=Rs this absolute horizon moves outwards until the absolute horizon, the apparent horizon, the Schwarschild radius, the event horizon and the surface of the dust sphere are all at the same radius. After that I am not saying anything specific about where all the dust ends up or how gravity acts when proper time reverses or becomes imaginary relative to Schwarzschild coordinate time.
Just for fun here is the interior Schwarzschild solution expressed as a ratio of proper time (d[itex]\tau[/itex]) to coordinate time (dt):

[tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{R_s}{R}}-\frac{1}{2}\sqrt{1-\frac{R_s r^{2}}{R^{3}}}[/tex]

where r is the location of the clock whose proper time we are interested in and R is the radius of the surface of the dust sphere.

Now when R = 9/8 Rs and the clock is at the centre (r=0):

[tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{8}{9}}-\frac{1}{2}\sqrt{1-\frac{0}{R^{3}}} = 0[/tex]

while for clock at the surface of the same dust sphere, R = 9/8 Rs = r, so:

[tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{8}{9}}-\frac{1}{2}\sqrt{1-\frac{R_s (9/8 R_s)^2}{(9/8R_s)^{3}}} = \frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{8}{9}} = \frac{1}{3}[/tex]

Now let's say the dust sphere continues to collapse so that R = 16/15 Rs then the relative proper time of the clock at the centre (r=0) is:

[tex] \frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{15}{16}}-\frac{1}{2}\sqrt{1-\frac{0}{R^{3}}} = \frac{3}{8}-\frac{1}{2} = - \frac{1}{8} [/tex]

so now, the clock at the centre is running backwards relative to clocks outside the dust sphere! How this affects how the dust continues to collapse is anyones guess.


No, because as mentioned above the volume of the dust sphere is getting smaller as the absolute horizon moves outwards.

Yes it would be an error to assume the black hole has infinite mass. Finite mass in zero volume implies infinite density but does not imply infinite mass.

PeterDonis

I think this formula only applies to a static solution, not to a collapsing solution. The only derivation of it I'm familiar with, the one in MTW, certainly restricts it to the static case. The limiting condition you are deriving, R = 9/8 R_s, is the limit at which the pressure gradient required to sustain static equilibrium goes to infinity; that is what dtau/dt going to zero means when taken in proper context, that the "acceleration due to gravity" at the center, which is what the pressure gradient has to work against, is now infinite.

sshai45

I'm curious about this. Does this mean that the curved space just exists of its own and doesn't require any energy (mass) present (unlike the curvature caused by "ordinary" objects) to sustain it? Does this also imply the destruction of energy (mass), and so that "conservation of energy" is invalid here? Is it also possible that if energy can be annihilated, then that perhaps there is also some other process out there that can create it out of nothingness, and this could explain the big bang? Or am I just all wet here? (e.g. if there's nothing "before" the Big Bang, then you can't really say energy got "created out of nowhere" since there was no time "before" when there was "no energy") And if there is no mass there, then what does it mean to say a black hole has a given mass?

Passionflower

Careful, the t-coordinate does not represent time (time on a clock at infinity) over the whole spacetime.

PeterDonis

It means that this particular configuration of curved spacetime (not just space), once it is formed, is static--it stays the same forever. (This is obviously an idealization, which assumes that nothing else ever falls into the black hole, and also that we can ignore quantum effects like Hawking radiation.) But the configuration still has to form in the first place; some massive object has to collapse to form it.

(Technically, there is a mathematical solution, the "maximally extended Schwarzschild spacetime", which describes a black hole that is not formed by a collapsing object; but that's a mathematical solution only. As far as I know, nobody believes that it is physically reasonable.)

Good question. There are two answers:

(1) From the standpoint of someone far away from the collapsing object, and from the black hole that forms from it, there is still "mass" present; it does not disappear when the collapsing object vanishes into the singularity. By "mass" I mean that, for example, if you were orbiting the original object before it collapsed (i.e., you were far enough away that you were outside the original object to begin with), your orbit would not change during the collapse and the formation of the black hole (this is in the idealized case of a spherically symmetric collapse with no radiation emitted by the collapsing matter); from the orbital parameters alone you would not even be able to tell that the collapse had happened. So from a global perspective the whole process does "conserve energy".

(2) Locally, however, there is indeed an issue with energy conservation when the collapsing matter vanishes into the singularity. However, strictly speaking, physicists do not think that this prediction of GR is what actually happens; rather, they take this prediction of GR, that a singularity of infinite density and infinite spacetime curvature forms, to be an indication that this situation is beyond GR's domain of validity. That is why you will see physicists saying that GR "breaks down" at the singularity, and that we need some new theory, presumably a theory of quantum gravity, to tell us what actually happens to the collapsed object. We would then want that new theory to explain how local energy conservation is maintained.

The "initial singularity" of the big bang, with the universe starting at infinite density, is another prediction of GR that, strictly speaking, physicists do not think describes what actually happens; again, physicists take this prediction as an indication that GR is being pushed beyond its domain of validity. There are various theories about how the early universe evolved to a point where the standard "hot big bang" theory, which includes the standard GR model of the universe's expansion, can take over. Some of these theories involve the creation of the matter and energy we observe out of something different, but as far as I know, none of them involve creation "out of nothing"--there is always *something* there, even if it's just a quantum field.

It means that you can put objects in orbit about the black hole, measure their orbital parameters, and compute a "mass" for the hole using Kepler's Third Law. This mass, M, is what appears in the Schwarzschild metric that describes the curvature of spacetime around the hole.

sshai45

@PeterDonis: Thanks for the detailed response. I wonder about one more thing: if there's no energy and no "mass" in the hole, then from where does the energy that powers Hawking Radiation originate?

PeterDonis

There is no *matter* in the hole (once the object that collapses to form it has vanished into the singularity), but there is "energy" and "mass" there. One way to tell is, as I said in my previous post, to put an object in orbit about the hole. Another, as you note, is to observe that a real black hole can radiate energy.

As far as "where the energy comes from", the best simple answer is probably that it comes from the curvature of spacetime; as the hole radiates, its mass slowly decreases (for example, if you were in orbit about the hole, you would see your orbital parameters slowly changing to reflect a slowly decreasing mass), so the spacetime around the hole slowly becomes less curved. However, there is a *lot* more lurking here, enough for multiple threads (and there have been plenty on these forums already).

sshai45

Yet then where is this energy, anyways, when you just said the equations describing the situation are the "vacuum" Einstein Field Equations and the "stress energy tensor" is "zero" everywhere?

yuiop

What does it represent then?

<EDIT> Having given it yet more thought, the t coordinate in the interior solution is the same as the t coordinate in the exterior solution (ie the reference coordinate clock at infinity) because the 2 solutions join smoothly at the boundary.

For example, when the dust sphere has a radius of 9/8 Rs and the test clock is at the surface of the dust sphere so that r = R = 9/8 Rs the proper time rate according to the interior solution is:

[tex]\frac{d\tau }{dt} = \frac{3}{2}\sqrt{1-\frac{8}{9}}-\frac{1}{2}\sqrt{1-\frac{ (9/8)^2}{(9/8)^{3}}} = \frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{8}{9}} = \frac{1}{3}[/tex]

while the proper time rate according to the exterior solution is:

[tex]\frac{d\tau }{dt} = \sqrt{1-\frac{8}{9}} = \frac{1}{3}[/tex]

The solutions are in agreement, so the proper times are relative to the reference Schwarzschild coordinate clock at infinity in both cases. Perhaps I misunderstand what you are getting at?

Passionflower

Inside it represents space and the r coordinate represents time.

That does not, as some often do, imply, that space becomes time and vice versa.
It is only the coordinates that change their mapping.

PeterDonis

That's part of the "a lot more lurking here". But the short answer is that a black hole that is radiating is *not* a solution of the vacuum Einstein Field Equations. Those equations are classical and don't include quantum effects. Hawking radiation is a quantum effect. That's why I was careful to state that the particular vacuum solution that describes a "static" black hole only applies if we ignore quantum effects. If you want a longer answer, you should probably start a separate thread since this is really a separate question from the question in the OP.