Why the black hole density is called the infinite density.

[manvirsingh]

d=m/v, d= density, m= mass, v= volume

If volume is 0 than the density is infinite.But if black hole has radius it must has a volume.So how its density can be infinite.

 

[Passionflower]

d=m/v, d= density, m= mass, v= volume

If volume is 0 than the density is infinite.But if black hole has radius it must has a volume.So how its density can be infinite.

Who says a black hole has a radius?
Perhaps you allude to the event horizon and sometimes in ill-written articles the event horizon has a radius but in reality the event horizon has no radius at all.

 

[jnorman]

the event horizon does, indeed, have a radius. it is the singularity which has zero volume and infinite density.

 

[Naty1]

The fact that the 'density of a black hole singularity is infinite' is almost certainly a sign that neither quantum mechanics nor general relativity is accurate. Our mathematics probably doesn't work there.

It's kind of an analog to the electric charge of a classical electron...a point particle, no volume, with a finite charge...means 'infinite' charge density at 'zero radius' ...but of course it isn't; physicsts know that such math doesn't work. It's all an approximation.

 

[PeterDonis]

Strictly speaking, what is infinite at the "center", r = 0, of a black hole spacetime is spacetime curvature, not density. A black hole spacetime, if by that we mean the strict mathematical solution to the Einstein Field Equation that describes a black hole, is a vacuum everywhere, so there is no "mass" in the sense in which "mass" is used to define density; density requires a non-zero stress-energy tensor, and the SET of a black hole is zero everywhere.

However, this strict mathematical solution (the maximally extended Schwarzschild spacetime) is not really applicable to a real black hole. A real black hole would have to be formed by the collapse of a massive object (i.e., an object with nonzero SET), so the spacetime would not be vacuum everywhere; inside the object, there would indeed be a well-defined nonzero "density".

But once the object collapses to a singularity at r = 0, the object, including its nonzero SET, is no longer there, so there is no longer a nonzero density there either. At most, the density could be thought of as being "infinite" for a single instant, when the entire mass of the object reaches r = 0 and forms the singularity. After that, what I said above still applies: spacetime curvature is infinite at r = 0, but density is not, because there is no nonzero SET there--the object that formed the hole is gone.

 

[Passionflower]

the event horizon does, indeed, have a radius. it is the singularity which has zero volume and infinite density.

No that is simply not correct.

the object that formed the hole is gone.

Correct because the object's wordline simply stops.

 

[phyzguy]

No that is simply not correct.

Passionflower - I'm trying to understand why you say that the event horizon does not have a radius. I'm not necessarily disagreeing, just trying to understand on what you base the statement. Doesn't a black hole have an event horizon with a well-defined radius?

 

[Passionflower]

Doesn't a black hole have an event horizon with a well-defined radius?

No it doesn't.

 

[PeterDonis]

Who says a black hole has a radius?
Perhaps you allude to the event horizon and sometimes in ill-written articles the event horizon has a radius but in reality the event horizon has no radius at all.

No it doesn't.

Can you elaborate on these statements? I think you're using an interpretation of the word "radius" which is stricter than the rest of the posters are using--not that yours is incorrect, just that it probably should be explained. To see what I'm getting at, would you also say that the black hole's horizon doesn't have a well-defined area? Or would you say that it *does* have a well-defined physical area, but *not* a well-defined physical radius?

 

[PAllen]

I think what Passionflower is referring to is that the r=<constant> 'surface' in SC is not a physical radius in any meaningful sense. Why? The r coordinate is timelike when r< 'horizon', spacelike above, and r = <horizon> is lightlike sphere. To any frame (in the GR technical sense) near the horizon, the horizon is a light surface moving at c, like any other.

On the other hand, a distant observer sees a literal 'black hole in space' of a certain size ( assuming there is no nearby matter falling in).

 

[Passionflower]

Can you elaborate on these statements? I think you're using an interpretation of the word "radius" which is stricter than the rest of the posters are using--not that yours is incorrect, just that it probably should be explained. To see what I'm getting at, would you also say that the black hole's horizon doesn't have a well-defined area? Or would you say that it *does* have a well-defined physical area, but *not* a well-defined physical radius?

Well even that question is not simple.

The class of observers that are stationary in a Schwarzschild spacetime can certainly chart the event horizon as a sphere with a certain distance removed from them but not a sphere with a distance removed from a center, because there is no center.

 

[PeterDonis]

The class of observers that are stationary in a Schwarzschild spacetime can certainly chart the event horizon as a sphere with a certain distance removed from them but not a sphere with a distance removed from a center, because there is no center.

The definition of the physical area of the horizon doesn't depend on the horizon being a certain physical distance from the center. It depends only on the horizon being a null hypersurface whose intersection with any spacelike hypersurface of "constant time" is a spacelike 2-sphere. The physical area of that 2-sphere then turns out to be invariant; it doesn't depend on which particular spacelike hypersurface is defined as the "surface of constant time". The area is always 16 pi M^2 in geometric units.

Also, strictly speaking, saying "there is no center" is coordinate-dependent. More precisely, saying that a spacelike hypersurface of "constant time" doesn't have a "center" at r = 0 is coordinate dependent. Surfaces of constant Schwarzschild time don't; but surfaces of constant Painleve time, for example, do. (Or, if you object that the singularity itself can't really be a physical "center" since it's a singularity, there are still spacelike geodesics lying in surfaces of constant Painleve time that can approach r = 0 arbitrarily closely, whereas this is not true for surfaces of constant Schwarzschild time.)

 

[Passionflower]

The definition of the physical area of the horizon doesn't depend on the horizon being a certain physical distance from the center. It depends only on the horizon being a null hypersurface whose intersection with any spacelike hypersurface of "constant time" is a spacelike 2-sphere. The physical area of that 2-sphere then turns out to be invariant; it doesn't depend on which particular spacelike hypersurface is defined as the "surface of constant time". The area is always 16 pi M^2 in geometric units.

I do not disagree with that.

Also, strictly speaking, saying "there is no center" is coordinate-dependent. More precisely, saying that a spacelike hypersurface of "constant time" doesn't have a "center" at r = 0 is coordinate dependent. Surfaces of constant Schwarzschild time don't; but surfaces of constant Painleve time, for example, do. (Or, if you object that the singularity itself can't really be a physical "center" since it's a singularity, there are still spacelike geodesics lying in surfaces of constant Painleve time that can approach r = 0 arbitrarily closely, whereas this is not true for surfaces of constant Schwarzschild time.)

Free falling observers outside of the event horizon do not even observe the event horizon as a sphere and inside there is no center. I suggest you study the nature of space-like singularities.

 

[PeterDonis]

Free falling observers outside of the event horizon do not even observe the event horizon as a sphere

They can't "observe" it at all outside the horizon since light from it can't reach them. But a "network" of freely falling observers, falling through the horizon at different angular coordinates, could in principle exchange measurements that showed that the horizon was a 2-sphere with area 16 pi M^2.

and inside there is no center. I suggest you study the nature of space-like singularities.

I understand the nature of spacelike singularities, but the fact that the r = 0 singularity is spacelike does not in itself prevent it from being a "center" with respect to an appropriately chosen family of spacelike hypersurfaces. If there were *no* family of spacelike hypersurfaces that all intersected the spacelike line at r = 0, then you would be right that there was no "center". But there is such a family--the family of surfaces of constant Painleve time, which are spacelike all the way down to r = 0 (unlike the surfaces of constant Schwarzschild "time", which are timelike inside r = 2M). With respect to those spacelike surfaces, the singularity at r = 0 *is* a "center".

 

[dustinthewind]

If you think of it in terms of space time contraction as the matter reaches the speed of light it becomes flat. However no matter can be infinitly flat as if we know its speed exactly we are uncertain about its location by \Deltax\Deltap\geq\frac{\hbar}{2}. This alone would give the in falling matter a thickness. You could possibly think of the infalling matter becoming the event horizon possibly as the mater would flatten radially around the surface of the black hole. But then the question is would this matter actuall comprise the event horizon or not. I am not sure about that.

I have wondered if when a black hole forms it forms on the inside first and then grows rapidly consuming matter in layers and there by forming layers that form the black hole but that is just speculation and I will have to look into that some time if I ever get the time.

 

[pervect]

If you think of it in terms of space time contraction as the matter reaches the speed of light it becomes flat. However no matter can be infinitly flat as if we know its speed exactly we are uncertain about its location by \Deltax\Deltap\geq\frac{\hbar}{2}. This alone would give the in falling matter a thickness. You could possibly think of the infalling matter becoming the event horizon possibly as the mater would flatten radially around the surface of the black hole. But then the question is would this matter actuall comprise the event horizon or not. I am not sure about that.

I have wondered if when a black hole forms it forms on the inside first and then grows rapidly consuming matter in layers and there by forming layers that form the black hole but that is just speculation and I will have to look into that some time if I ever get the time.

The event horizon of a black hole is actually lightlike. This follows from it being a null surface, and you can even think of the event horizon as being "trapped light".

Thus, while it is true that the relative velocity between the event horizon of the black hole, and infalling matter is always "c", the only valid frame of reference here is the frame of reference of the infalling manner. The event horizon of the black hole does not have a "frame of reference", in any standard sense of the term, any more than light does.

And in the frame of reference of the infalling matter, it's not flattened at all, nor does its density go up, as the event horizon approaches it at the speed of light.

If you consider some stationary observer close to (but outside) the event horizon, valid frames for the staionary observer do exist. In this frame, the matter does appear to be compressed as it falls by the stationary observer, at nearly (but lower than) the speed of light. From the viewpoint of the matter itself, it still has normal density, of course.

The quantum mechanical relation is true, but not needed for GR, which is a classical theory. ANd most of the real issues here are classical ones.

 

[dustinthewind]

The event horizon of a black hole is actually lightlike. This follows from it being a null surface, and you can even think of the event horizon as being "trapped light".

Thus, while it is true that the relative velocity between the event horizon of the black hole, and infalling matter is always "c", the only valid frame of reference here is the frame of reference of the infalling manner. The event horizon of the black hole does not have a "frame of reference", in any standard sense of the term, any more than light does.

And in the frame of reference of the infalling matter, it's not flattened at all, nor does its density go up, as the event horizon approaches it at the speed of light.

If you consider some stationary observer close to (but outside) the event horizon, valid frames for the staionary observer do exist. In this frame, the matter does appear to be compressed as it falls by the stationary observer, at nearly (but lower than) the speed of light. From the viewpoint of the matter itself, it still has normal density, of course.

The quantum mechanical relation is true, but not needed for GR, which is a classical theory. ANd most of the real issues here are classical ones.

Ok yeah I was referring to an outside observer referring to the black hole in a frame equivilant to the frame of the black hole's motion if that is appropriate. I guess it is unusual to combine quantum mechanics and general relativity. And its just speculation on my part but it didn't seem right that an observer watching matter fall in would observe infinitly flat matter.

 

[yuiop]

I have wondered if when a black hole forms it forms on the inside first and then grows rapidly consuming matter in layers and there by forming layers that form the black hole but that is just speculation and I will have to look into that some time if I ever get the time.

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]

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.

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]

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.

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 Earth's 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]

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?

"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.

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.

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.

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

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.

What is interesting is that Earth's 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?

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.

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.

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

a = \frac{M}{r^{2} \sqrt{1 - \frac{2M}{r}}}

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.

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.

No, both of these are incorrect. See above.

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.

"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".

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.

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 infinitely 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 wouldn't we end up with something infinetly large at the limit of density approaching 0? I know that reversing the process is impossible.

 

[PeterDonis]

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.

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

When it collapses it forms a singularity. The singularity is infinitely dense and infinetly small. How did a set amount of mass form something infinite?

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 gravitational signature exist of something with a set mass but if we examine the process in reverse wouldn't we end up with something infinetly large at the limit of density approaching 0?

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]

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.

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 Schwarzschild 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\tau) to coordinate time (dt):

\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}}}

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):

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

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

\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}

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:

\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}

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.

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

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

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.

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]

Just for fun here is the interior Schwarzschild solution expressed as a ratio of proper time (d\tau) to coordinate time (dt)

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]

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.

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]

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.

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

 

[PeterDonis]

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?

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.)

Does this also imply the destruction of energy (mass), and so that "conservation of energy" is invalid here?

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.

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?

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.

And if there is no mass there, then what does it mean to say a black hole has a given mass?

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]

@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?

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]

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).

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]

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

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:

\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}

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

\frac{d\tau }{dt} = \sqrt{1-\frac{8}{9}} = \frac{1}{3}

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]

What does it represent then?

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]

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?

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.

 

[Austin0]

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.

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.

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??
Thanks

 

[PeterDonis]

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

First, please note my later post in response to yuiop; his specific formulas only apply (I believe) to the case of a static equilibrium, *not* to the case of a collapsing object.

That said, your general intuition here is correct. As the object collapses, the density at its center increases, and the "time dilation" at its center becomes more extreme, relative to an observer far away (another way to look at it would be to say that the "potential well" inside the object becomes deeper). See next comment.

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??

Because there is a "potential well" there, and it gets deeper as the object collapses. Please note that the "potential", or "time dilation" factor, which is what we're discussing here, is different from the "acceleration due to gravity", which is what gets smaller as you go deeper inside a gravitating body (like the Earth) because more of it is above you, as I said in that previous post you quoted.

Also note that, strictly speaking, "acceleration due to gravity" only applies to a *static* observer, one who is "hovering" at a constant radial coordinate r. Once the object has collapsed inside its Schwarzschild radius, there are no such observers in the interior vacuum region inside the event horizon (i.e., outside the surface of the collapsing object but still below the horizon). (I'm not sure if there can still be static observers inside the collapsing object once it has collapsed inside the horizon; I think not, but I have not looked at the math in detail.)

 

[yuiop]

Would this mean that the collapsing dust cloud [above] as it was approaching the Schwarzschild radius would be collapsing more rapidly on the surface with an increasing density and dilation gradient from there to a minimum at the center?

That seems a reasonable assumption. However it more likely that the dust cloud would be denser nearer the centre during the initial stages before the time dilation effects become significant and start evening out or even reversing the density gradient. The equation I used assumes an even density gradient which is unlikely in practice. However, it can be shown that if the density increases towards the centre inversely proportional to radius of the dust sphere, that the results are the same. Also, as Peter pointed out the interior solution is a static solution and in reality a dust sphere so close to the Schwarzschild density is almost certainly going to be collapsing rapidly and not static. It is a pity we do not have a solution for the dynamic case.

Prior to the actual collapse and formation of the singularity at the center why would the clocks there be more dilated than at the surface??

If you had a very deep well that went to the centre of the Earth, the acceleration of gravity reduces as you get nearer the centre, but the gravitational time dilation increases. Gravitational time dilation is a function of gravitational potential which can be expressed as -GM/r. As r gets smaller, -GM/r gets larger (and more negative).

NOTE: I am assuming an idealised non rotating Earth or a well dug from a pole. Rotation of the Earth will cause increased time dilation nearer the surface due to velocity induced time dilation of SR.

 

[PeterDonis]

It is a pity we do not have a solution for the dynamic case.

If you mean an analytical solution, we do for the idealized case of a perfectly spherically symmetric collapse. It's the Oppenheimer-Snyder solution, from their 1939 paper. The abstract is here:

http://prola.aps.org/abstract/PR/v56/i5/p455_1

Unfortunately the paper itself is behind a paywall.

The interior of that solution is basically the time reverse of the Big Bang: a contracting FRW metric which is matched at the boundary to the exterior vacuum Schwarzschild metric. The density inside the collapsing object, in this model, increases as the object collapses, ultimately going to infinity as the radius of the object goes to zero (i.e., as it forms the singularity at r = 0).

There have been plenty of numerical simulations of less idealized collapses; unfortunately I don't have any good general references handy.

 

[yuiop]

Unfortunately the paper itself is behind a paywall.

Any equations? I would be interested in how they avoid negative or imaginary proper time during the collapse.

 

[PeterDonis]

Any equations? I would be interested in how they avoid negative or imaginary proper time during the collapse.

The OH solution is treated in most GR textbooks; in MTW it's in section 32.4. I can post some more details from there if desired, but I don't have time at the moment. But the fact that the proper time behaves itself in the interior should be obvious just from the fact that it's a contracting FRW metric; proper time, which is just FRW coordinate time, obviously increases monotonically during the collapse, just from looking at the FRW line element.

The only technical point is that the most natural time coordinate in the FRW interior, the one in the standard FRW metric, doesn't match up with the most natural time coordinate in the Schwarzschild exterior, which is the Schwarzschild one; but as the exercises in MTW show, one can "fix" this, if desired, by finding a single time coordinate that covers both regions.

 

[Austin0]

That said, your general intuition here is correct. As the object collapses, the density at its center increases, and the "time dilation" at its center becomes more extreme, relative to an observer far away (another way to look at it would be to say that the "potential well" inside the object becomes deeper). See next comment.
Because there is a "potential well" there, and it gets deeper as the object collapses. Please note that the "potential", or "time dilation" factor, which is what we're discussing here, is different from the "acceleration due to gravity", which is what gets smaller as you go deeper inside a gravitating body (like the Earth) because more of it is above you, as I said in that previous post you quoted.

That seems a reasonable assumption. However it more likely that the dust cloud would be denser nearer the centre during the initial stages before the time dilation effects become significant and start evening out or even reversing the density gradient. The equation I used assumes an even density gradient which is unlikely in practice. However, it can be shown that if the density increases towards the centre inversely proportional to radius of the dust sphere, that the results are the same.
If you had a very deep well that went to the centre of the Earth, the acceleration of gravity reduces as you get nearer the centre, but the gravitational time dilation increases. Gravitational time dilation is a function of gravitational potential which can be expressed as -GM/r. As r gets smaller, -GM/r gets larger (and more negative).

SO consider a homogeneous cloud well outside R_s. R=4 R_s
Would, in this case, the contraction acceleration and density occur with a differential decreasing towards the center?

Thank you both for clearing up my confusion regarding potential and acceleration factor.
This actually answered a question I have posted before without answer ; What is the dilation factor mid point between two equally massive bodies compared to on the surface of one. From the above I would now assume it would be greater?

PS IS the time running backwards at the center after collapse derived from Rindler or is it intrinsic in the Schwarzschild equations? Thanks

 

[sshai45]

The interior of that solution is basically the time reverse of the Big Bang: a contracting FRW metric which is matched at the boundary to the exterior vacuum Schwarzschild metric. The density inside the collapsing object, in this model, increases as the object collapses, ultimately going to infinity as the radius of the object goes to zero (i.e., as it forms the singularity at r = 0).

So does this mean that if you could stand inside the star as it collapses to a BH, you'd find that the matter and what not around you was changing in a way that looked like a little mini "Big Crunch"?

 

[PeterDonis]

SO consider a homogeneous cloud well outside R_s. R=4 R_s
Would, in this case, the contraction acceleration and density occur with a differential decreasing towards the center?

For the case of a perfectly *homogeneous* dust cloud, the density would be the same everywhere in the cloud (at least if you are using the standard "comoving" coordinates); that's the scenario that Oppenheimer and Snyder analyzed (because it's the only one they could treat analytically). The "contraction acceleration" is not really a good term to use for this scenario, because there is no single point within the cloud that counts as the "center" towards which everything else is collapsing; *every* point within the cloud is the same as every other (just as in our expanding universe, there is no single "center" from which everything is expanding; *every* point in the universe is the same as every other).

What is the dilation factor mid point between two equally massive bodies compared to on the surface of one. From the above I would now assume it would be greater?

There is no analytical solution for the case of two massive bodies; GR is not linear so you can't just superpose two one-body solutions to get a solution for two bodies. However, for the case where the distance between the bodies is large compared to their masses, I would expect the "potential" at the midpoint between them to be greater (i.e., closer to the "potential" at infinity) than the "potential" at the surface of either body.

PS IS the time running backwards at the center after collapse derived from Rindler or is it intrinsic in the Schwarzschild equations? Thanks

Time does not run backwards at the center after collapse. The statement yuiop made to that effect was not correct; his formulas only apply to a static equilibrium, not a collapsing object, and there is no static equilibrium with a radius less than 9/8 the Schwarzschild radius, which was the condition he needed to make it look like time was "running backwards" at the center.

 

[PeterDonis]

So does this mean that if you could stand inside the star as it collapses to a BH, you'd find that the matter and what not around you was changing in a way that looked like a little mini "Big Crunch"?

Yes.

 

[Passionflower]

the most natural time coordinate in the Schwarzschild exterior, which is the Schwarzschild one;

Could you perhaps give a reason why you think there is a natural time coordinate and why you think it is the time coordinate in Schwarzschild coordinates.

 

[Whovian]

Our mathematics probably doesn't work there.

Sincerest apologies for attacking wording, but our mathematics works fine. It's our physics that doesn't work.

 

[PeterDonis]

Could you perhaps give a reason why you think there is a natural time coordinate and why you think it is the time coordinate in Schwarzschild coordinates.

"Natural" because it matches up with the time translation symmetry of the spacetime. But as you have pointed out, the Schwarzschild time coordinate is no longer timelike inside the horizon; so really it is only a natural *time* coordinate outside the horizon; inside the horizon it's still "natural" (because d/dt is still a Killing vector), but it's not a "time" coordinate any more because it's spacelike.

 

[Passionflower]

I think the whole idea which coordinates are more natural a little arbitrary but if I would be forced to pick one it would not be Schwarzschild coordinates.

 

[PeterDonis]

I think the whole idea which coordinates are more natural a little arbitrary

I could have been more precise by saying "coordinates which match up best with a particular symmetry of the spacetime", but that would have been a mouthful.

but if I would be forced to pick one it would not be Schwarzschild coordinates.

Which one would you pick?

 

[Passionflower]

Which one would you pick?

If forced I would go with Gullstrand–Painlevé coordinates.