Graphical example of BH formation by PAllen

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The discussion revolves around a graphical representation of black hole formation, emphasizing that eternal black holes likely do not exist in our universe. It explores a hypothetical scenario involving a trillion stars collapsing without angular momentum, leading to a visual phenomenon where the cluster reddens and appears as a black hole surrounded by Einstein rings. Participants debate the interpretation of this scenario, questioning whether stars have truly vanished or if they remain compressed within an invisible region. The conversation also touches on the complexities of explaining black hole formation, suggesting that while advanced discussions are valuable, they may not be accessible to all audiences. Ultimately, the thread highlights the need for clear explanations of black hole formation while acknowledging the challenges in conveying such advanced concepts.
  • #61
TrickyDicky said:
But the the horizon in itself as an object in a certain position is purely coordinate-dependent, a coordinate singularity, that is always calculated using a certain chart and depends on that specific chart.

No, this is not correct, at least not for the event horizon of a black hole. (It's not really correct for Rindler horizons either, but I'm not sure it's worth going into that, though it can be seen using the same idea I'm about to use for the BH case.)

The event horizon is a particular null surface in the spacetime, and can be defined in a coordinate-free manner, without reference to any chart. I hinted at the definition in earlier posts, but here it is explicitly: the event horizon is the boundary of the region in which the Killing vector field of the "time translation" isometry of Schwarzschild spacetime is timelike--i.e., the EH is the Killing horizon associated with the "time translation" isometry. There is a proof--I believe it's in Hawking & Ellis--that the event horizon of any stationary BH must be a Killing horizon, so this idea doesn't just apply to Schwarzschild BH's, it applies to the whole family of generalized Kerr-Newman BH's.

The fact that there is also a coordinate singularity at the EH in a particular chart *is*, of course, dependent on that specific chart; but you don't need that fact to define the EH itself and its properties.
 
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  • #62
zonde said:
I am not entirely happy with the idea of horizon forming at the center of collapsing shell so I propose for consideration slightly different scenario.
Let's say that we have shell-like distribution of many smaller black holes that are collectively collapsing. When they get inside SC radius of summary mass of all the smaller black holes their event horizons are simply joined together, right? It is not exactly meaningful to say that black hole is falling into black hole, right?

This is a fairly complicated scenario. I can make some educated guesses, but cannot be sure about the following observations:

1) When the shell of small BHs is of 'large' radius, there are separate event horizons for each BH. (uncontroversial).

2) Shortly before the shell reaches the collective SC radius, the total event horizon consists of a horizon around each BH plus a growing sphere in the empty center. The reason for this remains as I gave before: light on this inner spherical surface will not arrive at the shell until the shell has reached (or passed) the SC radius, so this light gets trapped. Meanwhile, light emitted from inside the shell but further from the center, and between any shell BHs can still escape. I believe this argument justifies my proposed shape for the horizon at this point.

3) Once the shell of BHs has reached SC radius, there is one collective event horizon.

4) Once the BH's are inside the collective SC radius, they no longer have individual true event horizons. However (until they get close enough to each other) I think they still have individual apparent horizons for observers inside the collapsing shell. This is because light can still proceed locally outward if it originates far enough away from each BH. This last discussion is in the spirit of a sufficiently large shell that the BH's can be well separated well within the SC radius.

5) At some time, well before the BH singularities have coalesced in some way, when they get 'too close', you would have one merged apparent horizon.
 
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  • #63
PAllen said:
This is not true, per GR. For one thing, Birkhoff's theorem alone establishes that as soon as spherically symmetric mass distribution is inside its SC radius, you have an event horizon. The singularity comes later. Further, it is non-trivial consequence, requiring some extra assumptions, to conclude that the mass inside an event horizon must ultimately form a singularity. Birkhoff's theorem requires no energy conditions assumptions - it is strictly a consequence of EFE. The singularity theorems require additional assumptions. There are a number of known GR solutions with slightly exotic matter, that constitute stable, non-singular black holes (i.e. spherical event horizon with stable, nonsingular, exotic matter distribution inside).

[Edit: I should also emphasize that for all scenarios discussed in this thread, there is a singularity formed. The point of discussion was that the event horizon forms before the singularity in any reasonable collapse, and may form in a vacuum region - that will later be occupied by matter, and then a singularity.]
I disagree. I think you read too much in Birkhoff's theorem, it actually doesn't say what you claim it says. It says that a vacuum, spherically symmetric solution must be static and the Schwarzschild solution. The consequences that you extract from this were originated in an interpretation of the Schwarzschild solution that came many years after the theorem was proved by Birkhoff. Your wording gives the misleading impression that the original theorem included event horizons and its causal relations with singularities in its text.
And your argument sounds weird after your insistence that the spherically symmetric eternal BH is unphysical.
I don't quite understand why you previously insisted on the acausality of the EH and seem now so interested in remarking it "forms" before the singularity.
The relevant fact is that before, after or at the same time, the EH in GR is associated always to a singularity, as Penrose for instance states according to wikipedia.
 
  • #64
PeterDonis said:
No, this is not correct, at least not for the event horizon of a black hole. (It's not really correct for Rindler horizons either, but I'm not sure it's worth going into that, though it can be seen using the same idea I'm about to use for the BH case.)

The event horizon is a particular null surface in the spacetime, and can be defined in a coordinate-free manner, without reference to any chart. I hinted at the definition in earlier posts, but here it is explicitly: the event horizon is the boundary of the region in which the Killing vector field of the "time translation" isometry of Schwarzschild spacetime is timelike--i.e., the EH is the Killing horizon associated with the "time translation" isometry. There is a proof--I believe it's in Hawking & Ellis--that the event horizon of any stationary BH must be a Killing horizon, so this idea doesn't just apply to Schwarzschild BH's, it applies to the whole family of generalized Kerr-Newman BH's.

The fact that there is also a coordinate singularity at the EH in a particular chart *is*, of course, dependent on that specific chart; but you don't need that fact to define the EH itself and its properties.

But that definition is coordinate dependent, it is precisely dependent on the fact that removing the coordinate singularity implies a coordinate transformation that swaps timelike and spacelike Killing vector fields. In doing so it transforms invariants, but as long as it is considered a licit transformation...
 
  • #65
TrickyDicky said:
But that definition is coordinate dependent, it is precisely dependent on the fact that removing the coordinate singularity implies a coordinate transformation that swaps timelike and spacelike Killing vector fields.

No, it isn't. You have things backwards. The Killing vector field is not "swapped" by a coordinate transformation; it is a feature of the underlying geometry, independent of the choice of coordinates. So is the timelike, spacelike, or null nature of the Killing vector field at a particular event or within a particular region of the spacetime. Whether or not a particular chart is singular at a particular location is *derived* from the underlying geometry, plus the definition of the chart.

The correct way of describing the behavior of the "time translation" Killing vector field, which makes explicit the fact that its definition is completely coordinate-free, is that it is timelike outside the horizon, null on the horizon, and spacelike inside the horizon. A still more precise statement would involve defining "outside" and "inside" the horizon in terms of the physical areas of 2-spheres: 2-spheres with areas greater than 16 \pi M^2 lie outside the horizon, 2-spheres with areas less than 16 \pi M^2 lie inside the horizon, and the 2-sphere with area equal to 16 \pi M^2, of course, lies on the horizon.

None of the above requires defining any coordinate charts at all, nor does it make any use of the fact that Schwarzschild coordinates are singular at the horizon.
 
  • #66
TrickyDicky said:
I disagree. I think you read too much in Birkhoff's theorem, it actually doesn't say what you claim it says. It says that a vacuum, spherically symmetric solution must be static and the Schwarzschild solution. The consequences that you extract from this were originated in an interpretation of the Schwarzschild solution that came many years after the theorem was proved by Birkhoff. Your wording gives the misleading impression that the original theorem included event horizons and its causal relations with singularities in its text.
There is no real difference in our interpretation of what Birkhoff's theorem. However, I am applying it in a slightly clever way, which is well known. If a manifold is spherically symmetric, and is vacuum outside some 2-sphere, then the uniqueness of vacuum spherically symmetric metrics requires that the solution be exactly SC metric in the vacuum region outside a sphere containing all the matter. This requires that as soon as a collapsing spherical shell is inside the SC radius, the vacuum solution outside the shell must be exactly SC geometry - including the event horizon.
TrickyDicky said:
And your argument sounds weird after your insistence that the spherically symmetric eternal BH is unphysical.
This thread has not focused on eternal BH, and this argument is not about eternal BH - it uses Birkhoff to make inferences about spherically symmetric collapse using a standard argument.
TrickyDicky said:
I don't quite understand why you previously insisted on the acausality of the EH and seem now so interested in remarking it "forms" before the singularity.
The relevant fact is that before, after or at the same time, the EH in GR is associated always to a singularity, as Penrose for instance states according to wikipedia.

That a BH horizon is associated with a singlularity in all physically plausible scenarios is not in dispute. However, the fact that the EH is a feature of the solution as a whole (and is acausal in that sense) does not make it meaningless to talk about the time ordering of events for a class of observers. As I made clear in a reply to Zonde a few posts back, a way to disambiguate as well as make physical statements like "EH forms before apparent horizon forms before singularity" is to ask about the order of events on various time like world lines. If a timelike world line in which the COM of a collapsing cluster is stationary, and which is at the center of such a cluster when it is large (and remains at the center), encounters first the EH, then the apparent horizon, then the singularity, it is meaningful to loosely speak as I have: EH forms first, then AH, then singularity.
 
  • #67
PAllen said:
2) Shortly before the shell reaches the collective SC radius, the total event horizon consists of a horizon around each BH plus a growing sphere in the empty center. The reason for this remains as I gave before: light on this inner spherical surface will not arrive at the shell until the shell has reached (or passed) the SC radius, so this light gets trapped. Meanwhile, light emitted from inside the shell but further from the center, and between any shell BHs can still escape. I believe this argument justifies my proposed shape for the horizon at this point.
What changes if we take away growing sphere of EH in the empty center?
Nothing.

Basically in scenario of collapsing spherically symmetric shell we can say that event horizon appears at once at a finite radius.
 
  • #68
zonde said:
What changes if we take away growing sphere of EH in the empty center?
Nothing.

Basically in scenario of collapsing spherically symmetric shell we can say that event horizon appears at once at a finite radius.

You can say it, but it would be false. Let's say (at time t0 in some reasonable chosen coordinates) the shell is at r=(1+δ)R, R being the SC radius. I claim there is some r0 > 0 and r0 < R, such that light emitted at (t0,r0) is never received by a distant observer, while light emitted at (t0,r0+ε) is eventually received by a distant observer. This is or isn't true. I claim it is. If it is, how would you describe this other than saying at t0 the event horizon is a 2-sphere at r0 plus a 2-sphere around each little BH on the shell?
 
  • #69
PeterDonis said:
The fact that there is also a coordinate singularity at the EH in a particular chart *is*, of course, dependent on that specific chart; but you don't need that fact to define the EH itself and its properties.
But you do need it, (at least for non-rotating black holes), it is not a problem of that particular chart only.
For instance if you consider that the EH/Sch. radius is part of the vacuum exterior solution of a collapsing object. In Schwarzschild coordinates, the EH is defined as the coordinate singularity using those coordinates. This is independent of the fact that using different coordinates you can extend the solution to the shell interior.
Or if you think considering the EH to belong to the exterior region solution is some kind of "word trick" you can use the Kruskal -Szekeres coordinates if you want to, you have exactly the same situation, the EH lies at the coordinate singularity, meaning the transformation between Schwarzschild coordinates and Kruskal–Szekeres coordinates is defined for r > 0, r ≠ 2GM, and −∞ < t < ∞, so it is not defined at the EH.

From wikipedia:

KruskalKoords.gif

All this leads to an apparently paradoxical situation because there is a ctually a kind of twisted sense in which the EH is coordinate-independent as you claim: in the sense that no coordinates exist that cover the transition between the exterior and the interior regions, all coordinates have the EH as a limit coordinate singularity, but I gues this is not what you mean by the EH being coordinate-independent, among other things because a coordinate-independent singularity is no longer a coordinate singularity.
PeterDonis said:
The Killing vector field is not "swapped" by a coordinate transformation; it is a feature of the underlying geometry, independent of the choice of coordinates. So is the timelike, spacelike, or null nature of the Killing vector field at a particular event or within a particular region of the spacetime. Whether or not a particular chart is singular at a particular location is *derived* from the underlying geometry, plus the definition of the chart.
Ok, I won't dispute this here. But actually this doesn't contradict my claim about the EH being coordinate-dependent. See above.

PeterDonis said:
None of the above requires defining any coordinate charts at all, nor does it make any use of the fact that Schwarzschild coordinates are singular at the horizon.
See above.
 
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  • #70
TrickyDicky said:
For instance if you consider that the EH/Sch. radius is part of the vacuum exterior solution of a collapsing object. In Schwarzschild coordinates, the EH is defined as the coordinate singularity using those coordinates.

No, it isn't. Once again, you have things backwards. The EH is never "defined" in terms of its being a coordinate singularity in any chart. The fact of its being a coordinate singularity, as I said before, is *derived* from the underlying geometry plus the definition of the chart.

TrickyDicky said:
Or if you think considering the EH to belong to the exterior region solution

I've never said it does. I've always said the EH "belongs" to the underlying geometry. Its existence and properties are independent of any chart.

TrickyDicky said:
you can use the Kruskal -Szekeres coordinates if you want to, you have exactly the same situation, the EH lies at the coordinate singularity

There is no coordinate singularity in K-S coordinates.

TrickyDicky said:
meaning the transformation between Schwarzschild coordinates and Kruskal–Szekeres coordinates is defined for r > 0, r ≠ 2GM, and −∞ < t < ∞, so it is not defined at the EH.

So what? That's a problem with the Schwarzschild coordinates, not with the EH. And it's not the correct definition of a coordinate singularity; see below.

TrickyDicky said:
All this leads to an apparently paradoxical situation because there is a ctually a kind of twisted sense in which the EH is coordinate-independent as you claim

It's not a paradox at all to me.

TrickyDicky said:
in the sense that no coordinates exist that cover the transition between the exterior and the interior regions

Yes, there are. K-S coordinates do, so do ingoing Eddington-Finkelstein and Painleve. None of those charts are singular at the EH, so they "cover the transition" just fine. You appear to believe that, if *any* chart is singular at the EH, *all* charts are "singular" there, because the transformation from the singular chart to any other chart must be singular there. That's wrong.

TrickyDicky said:
all coordinates have the EH as a limit coordinate singularity

No, they don't. See above.
 
  • #71
PeterDonis said:
No, it isn't. Once again, you have things backwards. The EH is never "defined" in terms of its being a coordinate singularity in any chart. The fact of its being a coordinate singularity, as I said before, is *derived* from the underlying geometry plus the definition of the chart.
I'm not talking about any logical chain or causal issue, so it can't be backwards or not backwards, what I'm saying is independent of the "underlying geometry". Certainly in curved manifolds one can refer to the limits of a certain chart without reference to the specific underlying geometry, it is a fact that curved smooth manifolds in general can't be covered completely by a single chart.

PeterDonis said:
I've never said it does. I've always said the EH "belongs" to the underlying geometry. Its existence and properties are independent of any chart.
As I said I'm not considering your fuzzy underlying geometry concept here
It just happens that charts are needed in differential geometry, at least differentiable manifolds are defined as those equipped with an equivalence class of atlases (collections of local charts) whose transition maps are all differentiable.

PeterDonis said:
There is no coordinate singularity in K-S coordinates.
r=2GM is not defined in K-S coordinates, do you dispute that?

PeterDonis said:
So what? That's a problem with the Schwarzschild coordinates, not with the EH.
No, I'm talking about the transition map between SC and K-S, so it is a problem also with the K-S space (the whole 4-regions) since they include the outside region.



PeterDonis said:
Yes, there are. K-S coordinates do, so do ingoing Eddington-Finkelstein and Painleve. None of those charts are singular at the EH, so they "cover the transition" just fine.
See above.

PeterDonis said:
You appear to believe that, if *any* chart is singular at the EH, *all* charts are "singular" there, because the transformation from the singular chart to any other chart must be singular there. That's wrong.
No, I don't believe that at all.
 
  • #72
TrickyDicky said:
r=2GM is not defined in K-S coordinates, do you dispute that?
I certainly dispute it. Any point with coordinates U=V corresponds to r=2GM. There is nothing singular about the KS coords here, or metric expressed in KS coords here, or the curvature tensor expressed in KS coords here. The only false issue is that the tranform from KS to another chart (SC) that is singular here, is singular. The transform to SC is singular only because the SC coordinates are singular here.

This is exactly analogous to rectilinear coordinates versus polar coordinates in a flat 2-D Euclidean plane. Because polar coordinates have coordinate singlularity, transform between them and rectilinear coords at this point is also singular. Does this say there is something funny about the point on a plane you pick for the pole? Or is only a feature (not even a defect) of this particular coordinate choice.?

[edit: To be specific, any KS coordinates of form (V,U,θ,\varphi)=(k,k,θ,\varphi) corresponds to r=2GM, and the metric here is simply:

diag( -16 G^2 M^2/e, 16 G^2 M^2/e, 4G^2 M^2, 4 G^2 M^2 sin^2(θ))
]
 
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  • #73
TrickyDicky said:
r=2GM is not defined in K-S coordinates, do you dispute that?

Yes. Even the Wikipedia page gets this one right:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

In K-S coordinates, r is a function of the K-S U and V (those are the Wiki page names for the spacelike and timelike non-angular K-S coordinates), given implicitly by

V^2 - U^2 = \left( 1 - \frac{r}{2M} \right) e^{r / 2M}

Which makes it obvious that if r = 2M, V = +/- U, so the full "horizon" in the maximally extended spacetime is two intersecting 45-degree lines in the standard K-S diagram. The horizon we've been talking about, the future horizon for observers in Region I (the "right wedge" of the diagram) is the line V = U with U, V > 0. No coordinate singularity anywhere.

TrickyDicky said:
No, I don't believe that at all.

Then why do you consider the coordinate singularity at the horizon in Schwarzschild coordinates to somehow indicate an issue with other charts which are not singular at the horizon?
 
  • #74
TrickyDicky said:
As I said I'm not considering your fuzzy underlying geometry concept here

So you consider the concept of spacetime having a geometry to be "fuzzy"? Hmm.
 
  • #75
PeterDonis said:
Yes.*

You are right, sorry, I managed to confuse myself, and typing in a rush didn't help.
I was thinking about the transformation from KS to SC.

Then why do you consider the coordinate singularity at the horizon in Schwarzschild coordinates to somehow indicate an issue with other charts which are not singular at the horizon?
In KS coordinates I agree there is no coordinate singularity, because the true singularity allows us to cover the whole spacetime with one chart of coordinates.
But again, there seems to be certain agreement on considering this space unphysical, in which case it shouldn't be used to demonstrate the physicality-coordinate-independence of EHs.
 
  • #76
PAllen said:
Because polar coordinates have coordinate singlularity, transform between them and rectilinear coords at this point is also singular. Does this say there is something funny about the point on a plane you pick for the pole? Or is only a feature (not even a defect) of this particular coordinate choice.?
That has been my point the whole time wrt EHs, that its only a coordinate dependent feature, and there's nothing that differentiates it from other points.
 
  • #77
TrickyDicky said:
That has been my point the whole time wrt EHs, that its only a coordinate dependent feature, and there's nothing that differentiates it from other points.

But this I disagree with. A singular point in some coordinates is only a feature of those coordinates. An EH can be defined without reference to any coordinates, and computed in any coordinates including those with no coordinate singularity there. The definition of EH is purely in terms of the boundary between events from which null paths escape to future null infinity and those from which they don't. I emphasize: no coordinates at all are needed to apply this definition.

[Edit: To clarify this in relation to my discussion with Peter Donis: Peter was emphasizing that this distinguishes BH Event Horizons from Rindler Horizon, in that the latter have no definition as a function of spacetime as a whole - they are defined in reference to a particular world line. I was emphasizing that any horizon (EH or Rindler) constitutes a causal boundary only for some specific class of observers (observer being defined by a timelike world line). As I see it, we had no real disagreement; we were emphasizing different aspects of horizons.]
 
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  • #78
TrickyDicky said:
In KS coordinates I agree there is no coordinate singularity

Ok, good.

TrickyDicky said:
But again, there seems to be certain agreement on considering this space unphysical, in which case it shouldn't be used to demonstrate the physicality-coordinate-independence of EHs.

The *entire* manifold covered by the K-S chart is unphysical, yes, because it contains the "white hole" region and a second asymptotically flat region (Regions III and IV as they are usually labeled on the K-S chart). But there is nothing physically unreasonable about using a *portion* of that manifold in a more realistic model. That's what the Oppenheimer-Snyder model does: it uses a portion of Regions I and II of the maximally extended Schwarzschild spacetime, joined to a collapsing FRW spacetime. And this model still contains an event horizon--a portion of the future horizon that forms the boundary between Regions I and II in the vacuum portion of the spacetime. The "collapsing star cluster" model PAllen has been discussing is the same kind of thing.
 
  • #79
PAllen said:
You can say it, but it would be false. Let's say (at time t0 in some reasonable chosen coordinates) the shell is at r=(1+δ)R, R being the SC radius. I claim there is some r0 > 0 and r0 < R, such that light emitted at (t0,r0) is never received by a distant observer, while light emitted at (t0,r0+ε) is eventually received by a distant observer. This is or isn't true. I claim it is. If it is, how would you describe this other than saying at t0 the event horizon is a 2-sphere at r0 plus a 2-sphere around each little BH on the shell?
Light can be blocked say by a rock. So in this case I say that light does not reach observer because it is blocked at later time and not because it is stuck (or goes backwards) right at r0.
 
  • #80
zonde said:
Light can be blocked say by a rock. So in this case I say that light does not reach observer because it is blocked at later time and not because it is stuck (or goes backwards) right at r0.

And who arranges the rocks? Why not just admit:

1) You think what GR predicts violates your sense of plausibility.

2) As a result you think the GR is incorrect in this scenario.

That would be honorable rather than claiming that the top experts in GR (not me, others who I study) are misinterpreting its equations. Further, except for the details of where it breaks down, you would then be in good company - many serious physicists think GR breaks down in the vicinity of horizons, and many more near the singularity. But that is different from disputing what GR predicts.

(Personally, I think, at least macroscopically, GR only breaks down near the singularity, and the horizon behavior you don't like actually occurs. I think, microscopically, a horizon may not be a true horizon, but macroscopically it behaves as GR predicts. Horizon behavior is likely to be subject to observational test within the next decades. )
 
  • #81
Fascinating thought experiment, but my head jumped to a somewhat altered version.

Set up the same shell of stars, but instead of a slow collapse, have them rotating fast enough to prevent the formation of a horizon, and then apply the brakes until a horizon forms.
 
  • #82
PAllen said:
(Personally, I think, at least macroscopically, GR only breaks down near the singularity, and the horizon behavior you don't like actually occurs. I think, microscopically, a horizon may not be a true horizon, but macroscopically it behaves as GR predicts. Horizon behavior is likely to be subject to observational test within the next decades. )

Event horizons are mathematically nice in that they can be described independent of choice of coordinate chart or observer, but in my opinion I think it's wrong to focus on the concept of the event horizon for asking whether or not GR breaks down in those regions. There are a couple of reasons for this: the first is that the true event horizon can only be observed at infinity, so any physical observer will never (by definition) see one. The second is that if all physical black holes eventually decay then there are no event horizons.

Based on this, then, I feel that it is more physically appropriate to focus on apparent horizons rather than event horizons. However, there is nothing intrinsically different about the apparent horizon of an observer due to a black hole spacetime versus, say, a Rindler horizon. Nor is there a reason to suspect that GR breaks down in the vicinity of such a horizon. (The singularity is another matter entirely.)

That said, I am highly interested in how microscale physics affects the behavior of horizons; specifically, the idea that fluctuations might give rise to a "quantum width" for observed horizons (although this is unlikely to be measurable for any macro-BH).
 
  • #83
Aimless said:
The second is that if all physical black holes eventually decay then there are no event horizons.

This is not quite true; even if all BHs eventually evaporate, they still have interior regions while they exist. That is, there are still regions of spacetime that can't send light signals to future null infinity--the interiors of BHs between the time that they form and the time that they finally evaporate. So the spacetime still contains EHs, the boundaries of these regions.

The difference with evaporating BHs is that their EHs do not connect to future timelike infinity, whereas if BHs last forever, they do.

Here are a couple of Penrose diagrams illustrating what I have said. First, one of a BH that doesn't evaporate:

http://casa.colorado.edu/~ajsh/kitp06/penrose_Schw.html

The "corner" at the top, where the horizon, r = 0, and r = infinity lines all meet, is future timelike infinity. Second, a BH that evaporates away completely:

http://en.wikipedia.org/wiki/File:Black_hole_Penrose.png

Notice that here the wavy horizontal line, indicating the horizon, does *not* reach the "corner" at the very top, which is future timelike infinity; instead there is a vertical segment above where the horizon ends, which is the r = 0 worldline to the future of the BH's final evaporation.

Notice also that in both cases, there is still a region inside a horizon, which can't send light signals to future null infinity (signals sent in this region hit the singularity instead).
 
  • #84
PeterDonis said:
This is not quite true; even if all BHs eventually evaporate, they still have interior regions while they exist. That is, there are still regions of spacetime that can't send light signals to future null infinity--the interiors of BHs between the time that they form and the time that they finally evaporate. So the spacetime still contains EHs, the boundaries of these regions.

The difference with evaporating BHs is that their EHs do not connect to future timelike infinity, whereas if BHs last forever, they do.

From a strictly GR standpoint, you are correct; this is the crux of the information paradox. However, in this case the existence of the event horizon is dependent on the existence of the singularity.

From the perspective of an outside observer, they will see an infalling object remain just above their own apparent horizon while the black hole evaporates, and the object will appear to cross the horizon at the exact moment the horizon disappears. While the object's worldline would have intersected the singularity, any information about what occurred in the interior region is lost and an outside observer see only a discontinuity in the worldline of the infalling object at the moment of the disappearance of the black hole (well, presumably the object was destroyed at the singularity).

But, this illustrates somewhat the point I was trying to make: that discontinuity is due to the object hitting the singularity. If quantum gravity effects prevent the formation of an actual singularity then there's no reason to think there would be a discontinuity in the worldline of the object. Presumably, it would instead get frozen in some region near the singularity, waiting for the trapped region to shrink enough for it to escape; likewise, light emitted from the object would be similarly preserved, and eventually reach null infinity.

Everyone agrees that GR has to break down at the singularity, but effects like the information paradox are due to the presence of the singularity itself. This is why I feel that it's a mistake to focus on event horizons rather than apparent horizons.
 
  • #85
Aimless said:
From a strictly GR standpoint, you are correct; this is the crux of the information paradox. However, in this case the existence of the event horizon is dependent on the existence of the singularity.

This is false. In classical GR, the event horizon forms before the singularity, and there exist event horizons not associated with singularities.
 
  • #86
Aimless said:
From the perspective of an outside observer, they will see an infalling object remain just above their own apparent horizon while the black hole evaporates, and the object will appear to cross the horizon at the exact moment the horizon disappears. While the object's worldline would have intersected the singularity, any information about what occurred in the interior region is lost and an outside observer see only a discontinuity in the worldline of the infalling object at the moment of the disappearance of the black hole (well, presumably the object was destroyed at the singularity).

But what about a collapse as described in this thread? Then, an outside observer sees a central star darken, and disappear, inside a black region. Why don't you read this whole thread before repeating thing refuted at the beginning of this thread.
 
  • #87
Aimless said:
From the perspective of an outside observer, they will see an infalling object remain just above their own apparent horizon while the black hole evaporates, and the object will appear to cross the horizon at the exact moment the horizon disappears. While the object's worldline would have intersected the singularity, any information about what occurred in the interior region is lost and an outside observer see only a discontinuity in the worldline of the infalling object at the moment of the disappearance of the black hole (well, presumably the object was destroyed at the singularity).

What do you mean when you say an outside observer will "see" an infalling object remain just above the horizon??
Doppler shift aside ,even if the light wasn't shifted out of visibility there would not be any remaining image of the object hovering on the horizon. From the outside it certainly seems like the object would simply totally disappear. Period.
 
  • #88
PAllen said:
But what about a collapse as described in this thread? Then, an outside observer sees a central star darken, and disappear, inside a black region. Why don't you read this whole thread before repeating thing refuted at the beginning of this thread.

I fail to see how my point was refuted at the beginning of the thread. In the gravitational collapse scenario, assuming the black hole is permanent, then yes, of course an event horizon forms, and forms before the singularity.

My claim is the following: given the following two assumptions, 1), that all black holes eventually evaporate due to Hawking radiation, and 2), based on whatever unknown quantum gravity effects might exist, there are no spacetime singularities and there is some resolution to the information paradox, then event horizons don't exist.

As an example, consider a spacetime containing a smooth spherically symmetric time varying matter density such that at early times there are no horizons, at intermediate times a collapse occurs such that a trapped region forms, and at late times (for whatever reason) the collapse reverses and the trapped region disappears. What happens to information from events inside the trapped region?

It must either exit to the untrapped region (in which case the trapping surface isn't an event horizon) or it must be destroyed. If Hawking radiation is completely thermal then that suggests that information from those events is destroyed, but that view seems to be falling out of favor. If so, if the information persists in some way, then I do not see how it is possible to call the surface bounding the trapped region an event horizon.

My original point was that quantum effects seem to imply that event horizons are not impermeable; thus, I feel that apparent horizons are more physically relevant and interesting. I stated this poorly (and incorrectly) above, and you were right to call me on it; my apologies.
 
  • #89
Austin0 said:
Doppler shift aside ,even if the light wasn't shifted out of visibility there would not be any remaining image of the object hovering on the horizon.

Huh? Of course there would. To an outside observer the object appears to freeze just outside of the horizon while the light from that object redshifts away. Sure, it'll very quickly become colder than the CMB, but it'll never become completely black.
 
  • #90
Aimless said:
I fail to see how my point was refuted at the beginning of the thread. In the gravitational collapse scenario, assuming the black hole is permanent, then yes, of course an event horizon forms, and forms before the singularity.

No, even for an impermanent black hole formed from collapse that decays, the event horizon (semiclassically defined) precedes the singularity.
Aimless said:
My claim is the following: given the following two assumptions, 1), that all black holes eventually evaporate due to Hawking radiation, and 2), based on whatever unknown quantum gravity effects might exist, there are no spacetime singularities and there is some resolution to the information paradox, then event horizons don't exist.

This is true only if you insist on a strictly classical definition of event horizon while using quantum definitions elsewhere. Note that the original semiclassical derivation of Hawking radiatio was based on the existence of a horizon. Thus Hawking radiation without a semiclassical horizon is nonsense. The consensus here is that you have something that macroscopically behaves like a horizon but microscopically does not.
Aimless said:
As an example, consider a spacetime containing a smooth spherically symmetric time varying matter density such that at early times there are no horizons, at intermediate times a collapse occurs such that a trapped region forms, and at late times (for whatever reason) the collapse reverses and the trapped region disappears. What happens to

Without major violation of GR, this scenario is impossible. It is impossible with any of QG corrections of GR that I am familiar with. That is, the reversal of collapse after a macroscopic horizon forms is impossible.
Aimless said:
It must either exit to the untrapped region (in which case the trapping surface isn't an event horizon) or it must be destroyed. If Hawking radiation is completely thermal then that suggests that information from those events is destroyed, but that view seems to be falling out of favor. If so, if the information persists in some way, then I do not see how it is possible to call the surface bounding the trapped region an event horizon.

My original point was that quantum effects seem to imply that event horizons are not impermeable; thus, I feel that apparent horizons are more physically relevant and interesting. I stated this poorly (and incorrectly) above, and you were right to call me on it; my apologies.

I mostly agree with this last paragraph with some caveats. Hawking radiation for a stellar black hole (let along a supermassive black hole) is at a lower temperature than CMB radiation. Thus all black holes in the current universe are growing, not shrinking (even if there is no matter at all nearby). The time frame in which black holes decay is well after the heat death of all stars.

Note: you have several times now used language like:

"they will see an infalling object remain just above their own apparent horizon while the black hole evaporates, and the object will appear to cross the horizon at the exact moment the horizon disappears"

This is what is refuted in this thread. For a collapsing supercluster observed from afar, matter is seen in the center of the forming black region, until the whole region goes black. The idea that the matter is seen only outside what appears to be the horizon is false for a collapse. It is true only for an eternal black hole, which is a pretty absurd concept.
 

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