# Why the black hole density is called the infinite density.

1. Apr 5, 2012

### 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.

2. Apr 5, 2012

### Passionflower

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.

3. Apr 5, 2012

### jnorman

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

4. Apr 5, 2012

### 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.

5. Apr 5, 2012

### Staff: Mentor

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.

6. Apr 5, 2012

### Passionflower

No that is simply not correct.

Correct because the object's wordline simply stops.

7. Apr 5, 2012

### phyzguy

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?

8. Apr 5, 2012

### Passionflower

No it doesn't.

9. Apr 5, 2012

### Staff: Mentor

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?

10. Apr 5, 2012

### 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).

11. Apr 5, 2012

### Passionflower

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.

12. Apr 5, 2012

### Staff: Mentor

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.)

13. Apr 5, 2012

### Passionflower

I do not disagree with that.

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.

14. Apr 5, 2012

### Staff: Mentor

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.

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".

15. Apr 7, 2012

### 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 $\Delta$x$\Delta$p$\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.

16. Apr 7, 2012

### pervect

Staff Emeritus

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.

17. Apr 7, 2012

### dustinthewind

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.

18. Apr 8, 2012

### 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.

19. Apr 8, 2012

### Staff: Mentor

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.

20. Apr 8, 2012

### 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.

Last edited: Apr 8, 2012
21. Apr 8, 2012

### Staff: Mentor

"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):

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

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.

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).

22. Apr 8, 2012

### 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.

23. Apr 8, 2012

### Staff: Mentor

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.

24. Apr 9, 2012

### 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$\tau$) 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.

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.

25. Apr 9, 2012

### Staff: Mentor

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.