# Thread: How do we know that black holes form?

1. What if the matter that is going to comprise a black hole were to get very close to the critical density, and then didn't collapse? Like if gravitational time dilation prevented it from actually collapsing? (A local observer would be perpetually a billionth or so away from seeing the collapse, but because they are unable to see the future, they wouldn't know time has stopped.)

How would know if that were not the case? If time dilation causes all light that escapes the collapsing black hole to be red shifted to an absurdly low frequency, then we would have no means by which to detect it. For a sufficiently low frequency, the antenna needed to detect it might be several astronomical units wide.

So really the question is, how do we know that black holes truly form instead of approximately forming? Given that we have no way to observe it?

2. Time dilation is what a distant observer sees. Locally time passes normally.

3. Originally Posted by Kojax
What if the matter that is going to comprise a black hole were to get very close to the critical density, and then didn't collapse? Like if gravitational time dilation prevented it from actually collapsing? (A local observer would be perpetually a billionth or so away from seeing the collapse, but because they are unable to see the future, they wouldn't know time has stopped.)

How would know if that were not the case? If time dilation causes all light that escapes the collapsing black hole to be red shifted to an absurdly low frequency, then we would have no means by which to detect it. For a sufficiently low frequency, the antenna needed to detect it might be several astronomical units wide.

So really the question is, how do we know that black holes truly form instead of approximately forming? Given that we have no way to observe it?
That's why a black hole used to be called a frozen star. We know it in the sense that we know a several things in physics, i.e. by extrapolation. Close your eyes and drop a rock of the side of the grand canyon. How do you really know that the rock will hit the ground if nobody saw it do so? What if it was impossible to observe hitting the ground in principle?

4. Originally Posted by Kojax
What if the matter that is going to comprise a black hole were to get very close to the critical density, and then didn't collapse?
There is no "critical density". Larger blackholes are less dense than smaller blackholes. To form a blackhole, it is sufficient to place enough mass into a small enough space. But because very large blackholes have very low density, the formation of a large blackhole doesn't require the matter to be compressed. One could build a blackhole from bricks if one had enough of them. What this means is that one cannot invoke resistance to prevent blackhole formation.

5. Originally Posted by Physicist
That's why a black hole used to be called a frozen star. We know it in the sense that we know a several things in physics, i.e. by extrapolation. Close your eyes and drop a rock of the side of the grand canyon. How do you really know that the rock will hit the ground if nobody saw it do so? What if it was impossible to observe hitting the ground in principle?
Yeah, but that example assumes there is nothing more to the situation than mere observation. The effects of General and Special relativity are not mind tricks, or mistakes of observation.

Originally Posted by mathman
Time dilation is what a distant observer sees. Locally time passes normally.
And local time is also what a local observer sees. It's not like one observation is "real" and the other is "false" or "illusory".

The local observer does not observe time to have stopped, but they can only see the present. They can't look into the future. So they have no way of knowing whether the "next tick" of their clock is going to actually happen or not.

Hence, a local observer cannot observe that time stops, nor can they observe that it proceeds normally. They are inherently blind in that respect.

6. Originally Posted by Kojax
Yeah, but that example assumes there is nothing more to the situation than mere observation. The effects of General and Special relativity are not mind tricks, or mistakes of observation.
That was not meant to be an example. It was meant to be an analogy. I.e. how we use physics to make predictions.

Originally Posted by Kojax
Local time is what a local observer sees.
That's not quite right. Before we go on, please let me ask you a question. Do you know what the term local means physically and mathematically? Do you understand how observations are made in relativity? You used the term "c" which makes me a bit nervous as far as your understanding between various kinds of observers and coordinate systems. It's not like one observation is "real" and the other is "false" or "illusory".

Originally Posted by Kojax
The local observer observes time not to have stopped, but they can only see the present.
I'm not certain which observer that you're referring to but it appears to be those sitting on or within the event horizon. However its quite literally impossible for such observers to relay their observations to outside observers or communicate with them in any way.

By the way, upon closer reading this part
and then didn't collapse? Like if gravitational time dilation prevented it from actually collapsing? (A local observer would be perpetually a billionth or so away from seeing the collapse, but because they are unable to see the future, they wouldn't know time has stopped.)
Doesn't make any sense to me. Dilation can't prevent a collapse from happening. Please explain this in a different way. Perhaps I'll understand it if you rephrase if for me. Thanks.

7. Originally Posted by KJW
There is no "critical density". Larger blackholes are less dense than smaller blackholes. To form a blackhole, it is sufficient to place enough mass into a small enough space. But because very large blackholes have very low density, the formation of a large blackhole doesn't require the matter to be compressed. One could build a blackhole from bricks if one had enough of them. What this means is that one cannot invoke resistance to prevent blackhole formation.

Ok. "Density" might be in incorrect term for it. The mass to volume relationship doesn't scale linearly, and it is possible to have an object with lower density than the minimum requirement for black hole, but not become a black hole, if the mass is distributed in certain ways - and so perhaps I should just say the correct term "within the Schwarzchild radius for that mass." Or something along those lines.

Originally Posted by Physicist
That's not quite right. Before we go on, please let me ask you a question. Do you know what the term local means physically and mathematically? Do you understand how observations are made in relativity? You used the term "c" which makes me a bit nervous as far as your understanding between various kinds of observers and coordinate systems. It's not like one observation is "real" and the other is "false" or "illusory".
I've been reading back through my other 2 posts on this thread and I'm pretty sure I didn't use the term "c".

However my understanding of "local" observations is that they are observations made of events occurring within the same coordinate system and frame of motion... etc. .. as the observer.

I was saying that a local observer cannot know whether time has stopped in their frame or not. They would have to know the future. Even right now, how do you know that time hasn't stopped in your frame? You can't know it. You know that the last several seconds have passed, but you can't know whether the next several seconds will pass. Balance of probabilities would suggest that the pattern of time progressing that has been in place from before will continue, but that is only an educated guess.

Therefore, an observer can observe time to "continue normally" even if time stops.

I'm not certain which observer that you're referring to but it appears to be those sitting on or within the event horizon. However its quite literally impossible for such observers to relay their observations to outside observers or communicate with them in any way.
Depends on whether we're talking about an object that has fallen in already or an object that is about to fall in.

If the black hole hasn't quite formed yet, then information is certainly capable of getting out. It will be horribly red shifted, and the EM will be at extremely low frequency, and so you'll need an incredibly big antenna to detect it, but it can get out.

By the way, upon closer reading this part

Doesn't make any sense to me. Dilation can't prevent a collapse from happening. Please explain this in a different way. Perhaps I'll understand it if you rephrase if for me. Thanks.
Of course it can.

What would happen would be a mathematical construct very much like the mathematical explanation for escape velocity. If there were only one very massive object in the whole universe, and you were moving away from that object at escape velocity then you would never turn around fall back in, even though you are perpetually being accelerated toward the object. The reason is because you are being accelerated by smaller and smaller amounts which are never great enough to allow you to slow all the way to zero.

A similar principle would apply to falling into a (nearly formed) black hole. Time dilation would cause the object to fall inward by smaller and smaller amounts in such a manner so it never actually arrives even though it never stops moving.

8. Originally Posted by Kojax
Originally Posted by KJW
There is no "critical density". Larger blackholes are less dense than smaller blackholes. To form a blackhole, it is sufficient to place enough mass into a small enough space. But because very large blackholes have very low density, the formation of a large blackhole doesn't require the matter to be compressed. One could build a blackhole from bricks if one had enough of them. What this means is that one cannot invoke resistance to prevent blackhole formation.
Ok. "Density" might be in incorrect term for it. The mass to volume relationship doesn't scale linearly, and it is possible to have an object with lower density than the minimum requirement for black hole, but not become a black hole, if the mass is distributed in certain ways - and so perhaps I should just say the correct term "within the Schwarzchild radius for that mass." Or something along those lines.
I think you may have misunderstood my point. The mass of a blackhole increases linearly with the Schwarzschild radius. This is in contrast to ordinary objects for which the mass would increase with the cube of the radius. Therefore, large blackholes are less dense than small blackholes, and if the blackhole were large enough, it could be formed from matter of ordinary density without compressing it. Because large pressures are not required to form very large blackholes, one can't invoke resistance to pressure to prevent blackhole formation.

9. Originally Posted by Kojax
… and so perhaps I should just say the correct term "within the Schwarzchild radius for that mass."
That won’t work because if all the matter of an object were contained within a sphere whose radius is the Schwarzschild radius for that mass then that object will be a black hole.
Originally Posted by Kojax
I've been reading back through my other 2 posts on this thread and I'm pretty sure I didn't use the term "c".
You’re quite right, Kojax. I don’t know why I said that. My sincere apologies, Kojax.
Originally Posted by Kojax
However my understanding of "local" observations is that they are observations made of events occurring within the same coordinate system and frame of motion... etc. .. as the observer.
You have to keep something in mind about observers. As Wikipedia describes it (and this sentence holds for GR too) Observer (special relativity) - Wikipedia, the free encyclopedia
Reference frames are inherently nonlocal constructs, covering all of space and time or a nontrivial part of it; thus it does not make sense to speak of an observer (in the special relativistic sense) having a location.
Going back now: I asked if you know what local means because you wrote Local time is what a local observer sees. What do you mean by “sees.” We don’t always use optics for measurements so I don’t know what you mean by “see.” Even when we do use light in measurements we might have to compensate in some way when we record our measurements.
In any case you didn’t explain which observer you were talking about.
Originally Posted by Kojax
I was saying that a local observer cannot know whether time has stopped in their frame or not. They would have to know the future.
How and why would someone have to know the future? Nobody knows the future.
Originally Posted by Kojax
Even right now, how do you know that time hasn't stopped in your frame?
Simple. I looked at the clock and its ticking. By definition time has not stopped in my frame.
Originally Posted by Kojax
You know that the last several seconds have passed, but you can't know whether the next several seconds will pass.
If you mean that I don’t know that the universe will stop working differently in the next few seconds then that’s true. But it seems to have been doing so for about 14 billion years so far so I’m comfortable with assuming that it will continue to do so in the near future. What about you?
Originally Posted by Kojax
Balance of probabilities …
What do you mean by that?
Originally Posted by Kojax
…but that is only an educated guess.
What part of physics is not an educated guess to you?
Originally Posted by Kojax
Therefore, an observer can observe time to "continue normally" even if time stops.
How. I can’t see anyway for that to happen in this universe according to our laws of nature.
Originally Posted by Kojax
If the black hole hasn't quite formed yet, then information is certainly capable of getting out.
That’s why I asked you who the observer is that you’re talking about. No observer inside the event horizon can talk to those outside and none can sit on the event horizon.
Originally Posted by Kojax
Of course it can.
?
When a black hole starts to collapse time slows down for it but time for any particle does not stop until it’s inside or on the event horizon. For that reason time dilation can’t prevent something from collapsing. If you assert otherwise then please give a real world example. I believe that you’re confusing cause and effect with events that occur at the same time.
Originally Posted by Kojax
A similar principle would apply to falling into a (nearly formed) black hole. Time dilation would cause the object to fall inward by smaller and smaller amounts in such a manner so it never actually arrives even though it never stops moving.
Those are well known facts but don’t support your assertion that time dilation can prevent a black hole from forming. You wrote Like if gravitational time dilation prevented it from actually collapsing?
The way it works is like this; if there is enough mass for a star to collapse into a black hole then even though there is gravitational time dilation at work here it’s not to the point where it can prevent matter from forming a black hole. And only when a black hole is already formed can an in-falling object not reach the event horizon.

10. Originally Posted by KJW
I think you may have misunderstood my point. The mass of a blackhole increases linearly with the Schwarzschild radius. This is in contrast to ordinary objects for which the mass would increase with the cube of the radius. Therefore, large blackholes are less dense than small blackholes, and if the blackhole were large enough, it could be formed from matter of ordinary density without compressing it. Because large pressures are not required to form very large blackholes, one can't invoke resistance to pressure to prevent blackhole formation.

This certainly adds things to be accounted for, but pressure is not why the black hole isn't forming. Gravitational time dilation is why it doesn't form.

The objects just keep moving more and more slowly toward one another, but never get close enough for the black hole to form.

This would eliminate the information conservation paradox. Communication is never fully cut off, so there is no paradox.

Originally Posted by Physicist
That won’t work because if all the matter of an object were contained within a sphere whose radius is the Schwarzschild radius for that mass then that object will be a black hole.
But what if it is not within the schwarzchild radius yet?

Suppose there is a nearly collapsed black hole sitting somewhere in space, and a massive object approaches it, containing enough mass to cause the Schwarzchild radius to increase if it were to get close enough.

I'm thinking it would never get close enough. As it started to get close enough time around it would slow down and it would progress by ever slower and slower amounts, but never arrive. It would forever be moving, but never move far enough.

Those of us located outside the nearly formed black hole are the ones who observe the time dilation, of course. Local observers don't realize time is stopping around them.

You have to keep something in mind about observers. As Wikipedia describes it (and this sentence holds for GR too) Observer (special relativity) - Wikipedia, the free encyclopedia

Going back now: I asked if you know what local means because you wrote Local time is what a local observer sees. What do you mean by “sees.” We don’t always use optics for measurements so I don’t know what you mean by “see.” Even when we do use light in measurements we might have to compensate in some way when we record our measurements.
In any case you didn’t explain which observer you were talking about.
I'm not sure it needs to be specified to a very high level of precision. There are essentially two observers who matter.

1) - An observer in the nearly collapsed black hole, as close as possible to the center.

2) - An observer far away from the black hole.

Observer #1 perceives time to continue normally, and remembers the last several seconds of time were passing normally, but because they can't see the future, they don't realize the next second (or fraction thereof) isn't arriving.

Observer #2 is sending and receiving signals from the vicinity of observer #1, but because the signals that are returning from vicinity of observer #1 are increasingly red shifted, the signals coming back are more and more difficult to detect.

How and why would someone have to know the future? Nobody knows the future.
Yes. And that is why it is impossible to know if time stops.

Simple. I looked at the clock and its ticking. By definition time has not stopped in my frame.
You can't know what the clock is doing right now. Only what it was doing distance/c seconds ago.

So all you really know is that time had not stopped yet distance/c seconds ago.

If you mean that I don’t know that the universe will stop working differently in the next few seconds then that’s true. But it seems to have been doing so for about 14 billion years so far so I’m comfortable with assuming that it will continue to do so in the near future. What about you?
Exactly. Our perception that time is going to keep going is based on the fact is has been going for a long time.

An observer falling into a nearly formed black hole is in the same situation. Time has been flowing up till now, so they expect it will keep flowing.

That’s why I asked you who the observer is that you’re talking about. No observer inside the event horizon can talk to those outside and none can sit on the event horizon.
?
When a black hole starts to collapse time slows down for it but time for any particle does not stop until it’s inside or on the event horizon. For that reason time dilation can’t prevent something from collapsing. If you assert otherwise then please give a real world example. I believe that you’re confusing cause and effect with events that occur at the same time.
That's why I brought up the problem of "escape velocity". And I only brought it up because that mathematical relationship has the essential ingredients as this one.

If you can understand "escape velocity", then you can understand how time dilation can stop a black hole from forming. You don't have to actually stop time to stop the collapse. It is enough to slow it down by increasingly amounts.

If time is slowing as the objects fall inward, in the right relationship, then those objects can continue to fall inward forever and still not make it. The rate at which they fall gets ever closer and closer to zero. So close you would be tempted to round it to zero. But it is not quite zero.

11. I've got a wild idea, in reading KJW's #4 post, a crazy stupid idea came to mind.

Now, it has been surmised before that our universe is a black hole itself.....what if the "expansion" we perceive is in acuality just the black hole's continued collapse? We perceive a spacetime expansion only because time is actually going slower and slower to the "universe" outside of our black hole? But the time dialation should not be noticable to us right, as we would see no difference?

I wonder if the time dialation of a black hole has any correlation to the expansion of space outside of the event horizon?

It sounded right in my head, but typing it out, I can't make it work. What's everyone else's thoughts?

I vote for dumb idea. But I'm still anxious to hear what everyone else thinks.

12. Thinking about your idea, Antonio, I might recommend that instead of comparing time dilation "inside" with time dilation "outside", it might work better to compare time dilation from before, with time dilation after. Clearly we can't see outside of our black hole or universe, or whatever it is we are in.

Light that has traveled a long distance is also light that was emitted a long time ago. If the time dilation effect is increasing, and if it was emitted when the time dilation was less, but time dilation has been increasing while it was en route, that would shift its color. However, one problem remains: it would be a blue shift instead of a red shift.

13. So really the question is, how do we know that black holes truly form instead of approximately forming? Given that we have no way to observe it?
I think black hole is said to be formed, when the escape velocity from the mass exceeds the velocity of light.It has nothing to do with the state of matter inside the blackhole.???

14. If that be so, then we still cannot observe whether they form or not, because we have no way of knowing whether the escape velocity from that mass (and distance) does exceed the velocity of light.

Or more importantly, we have no way of knowing whether the mass is located inside its own event horizon. Even the Sun, were its mass contained within a small enough region of space, would be a black hole. Just having enough mass isn't enough.

It's a confusing issue. If a (very heat and pressure resistant) object were located at a point within the Sun that lies within the Schwarzchild radius for a mass of the Sun's size, it still would not be inside a black hole. The reason being that, even though the mass of the Sun has that Schwarzchild radius, not enough of the mass is physically located inside the radius.

A nearly formed black hole would have the same problem. Because not enough of its mass is yet located inside the Schwarzchild radius for that amount of mass, even at the very center of that nearly formed black hole, time still has not entirely stopped. If another massive object approaches the nearly formed black hole, thereby increasing the Schwarzchild radius, the new wider Schwarzchild radius is still only valid if the new massive object reaches it. As long as our new massive object remains outside of it, all regions of the nearly formed black hole are still not entirely cut off from the rest of space.

15. Okay, so bear with me, I'm still really new at this.

If I compare time dilation before to time dilation after, I'm looking at the world line of the black hole, correct? The world line extends backward and forward in time. But the time inside the Event Horizon has stopped, correct? Eg. light cannot escape because it has stopped moving, which means the black hole is "moving" forward in time (from an outside perspective), but from within the Event Horizon, the perspective seems normal, time flows naturally.

To your second paragraph, I think that the time dilation is not increasing, it is slowing isn't it? With the universe's expansion, that tells me the dilation effect is less and less (faster expansion means less time dilation?), which would be red shifted. I think I'm missing something here. One last question, as I'm sure I'm confusing everyone here... if speed increases slows time just as mass does, and getting closer and closer to the speed of light causes time to slow for the traveler, then at the speed of light, time stands still? So, from the perpective of light itself, it's jouney is instantaneous because it is at the speed of light?

16. Originally Posted by Antonio Medina
Okay, so bear with me, I'm still really new at this.

If I compare time dilation before to time dilation after, I'm looking at the world line of the black hole, correct? The world line extends backward and forward in time. But the time inside the Event Horizon has stopped, correct? Eg. light cannot escape because it has stopped moving, which means the black hole is "moving" forward in time (from an outside perspective), but from within the Event Horizon, the perspective seems normal, time flows naturally.
Yeah. That is pretty much the mainstream view. In physics, time is just another dimension. The 4th dimension to an otherwise 3 dimensional universe. Up, down, forward, backward, left, right...... and then time forward and time backward. At the event horizon of a black hole, time would appear to stop for any outside observer, but proceed normally for any local observer.

On this thread I've been proposing a possible alternative view - which would be that time nearly stops but doesn't stop completely, because the black hole never completely forms.

To your second paragraph, I think that the time dilation is not increasing, it is slowing isn't it? With the universe's expansion, that tells me the dilation effect is less and less (faster expansion means less time dilation?), which would be red shifted. I think I'm missing something here.
Wow. Yeah. I guess it depends on which level of effect we are looking at.

Time itself is our first order effect. Time is gradually getting slower, which is our second order effect.

However it is getting slower at a slower and slower rate. Which gives us a third order effect. I'm not sure what the third order effect should be expected to do.

It's like if you look at a car that is accelerating at 10 meters/second/second right now, but in a moment that acceleration will drop down to 5 meters/second/second, and then 1.25 meters/second/second.

It is a deceleration of an acceleration. Pretty confusing to try and figure out what that does.

One last question, as I'm sure I'm confusing everyone here... if speed increases slows time just as mass does, and getting closer and closer to the speed of light causes time to slow for the traveler, then at the speed of light, time stands still? So, from the perpective of light itself, it's jouney is instantaneous because it is at the speed of light?
In special relativity it is easier to figure out what should happen. For an object moving very near the speed of light from Earth to a faraway planet, Lorentz contraction causes the apparent distance to be very nearly zero.

So yes. Arrival appears to be nearly instant, but the reason is because, from your perspective the distance was very nearly zero.

17. So really the question is, how do we know that black holes truly form instead of approximately forming? Given that we have no way to observe it?
You know by explicitly calculating what happens when a spherical body gravitationally collapses; this is called the interior Schwarzschild metric ( as opposed to the usual exterior vacuum one ). It is quite a bit more complicated than the usual textbook exterior metric, but it can nonetheless be treated in analytically closed form. The main result is that the gravitational collapse of any point within that body happens in finite proper time, so the black hole forms completely.

On an empirical level, what you can do is not look at the event horizon ( which is not possible as you rightly point out ), but rather examine the photon sphere. This is the lower bound of any gravitationally stable orbit around a black hole, and is located half a Schwarzschild radius outside the event horizon; if you can emit a photon there in such a way that it describes a stable circular orbit, you know you are dealing with a Schwarzschild black hole.

18. Sorry, forgot - the reason is that any free fall orbit that crosses the photon sphere from the outside inevitably also crosses the event horizon, so if you find a photon sphere outside your central body, you know that the collapse must have been complete. Please note though that all of this is true only classically - if you consider quantum effects during the collapse, the story may or may not be different.

I wish I could force my mind to accept such outre' concepts as the "dilation of time" and "tick-rates define reality"...just think, a couple of "thought experiments" proposed over a century ago

have now become "true facts" that "experiments prove".

......

"Imaginary" defines reality! I am beginning to understand the Leary philosophy now..."drop out and stay stoned as long as possible"...this would possibly enable me to see the "how modern

theory can defy reality". (this is a dedicated area of "astrophysics & cosmology?" In which new "Universe?" There's so many, I'm lost!)

20. Originally Posted by Markus Hanke
You know by explicitly calculating what happens when a spherical body gravitationally collapses; this is called the interior Schwarzschild metric ( as opposed to the usual exterior vacuum one ). It is quite a bit more complicated than the usual textbook exterior metric, but it can nonetheless be treated in analytically closed form. The main result is that the gravitational collapse of any point within that body happens in finite proper time, so the black hole forms completely.

On an empirical level, what you can do is not look at the event horizon ( which is not possible as you rightly point out ), but rather examine the photon sphere. This is the lower bound of any gravitationally stable orbit around a black hole, and is located half a Schwarzschild radius outside the event horizon; if you can emit a photon there in such a way that it describes a stable circular orbit, you know you are dealing with a Schwarzschild black hole.

The main issue with this is, we have to get a singularity before there can be an event horizon or a photon sphere. If the singularity hasn't formed yet, then no part of the region counts as an event horizo" or a photon sphere.

Until a sufficient amount of mass is located within the Schwarzchild radius, even the very center is not yet inside an event horizon or photon sphere, and is still in communication with the outside universe.

So, since the entire mass is still in communication with the outside universe, we can look at it all from the perspective of an observer far away from the mass. What do we see? We see time getting slower and slower. We see the mass getting more and more compact, but at a slower and slower rate. But, no matter how long we watch, the rate of progress will never drop all the way to zero. And no matter how long we watch, the mass will never quite arrive within its own Schwarzchild radius.

Now, here we must note a difference between theory and practice. In theory, communication is not cut off no matter how close the matter gets to its own Schwarzchild radius (and it will never fully arrive). In practice, the red shifting of the EM that does escape from that region might be so extreme that the frequency becomes something like 1 cycle per century. So, in practice we may not be able to observe the EM that is emitted, but that is a practical concern, not a theoretical one.

21. The main issue with this is, we have to get a singularity before there can be an event horizon or a photon sphere. If the singularity hasn't formed yet, then no part of the region counts as an event horizo" or a photon sphere.
I don't think that's true, but I don't immediately have any references to support my opinion. Just from my understanding of the geometry of this, the event horizon would form before the collapse is complete, namely at exactly the time when all the matter gets compressed to a radius equalling the event horizon radius, not the other way around. This is actually an interesting question, and I will need to do some further research into it.

Thanks for bringing this up

22. Originally Posted by Markus Hanke
I don't think that's true, but I don't immediately have any references to support my opinion. Just from my understanding of the geometry of this, the event horizon would form before the collapse is complete, namely at exactly the time when all the matter gets compressed to a radius equalling the event horizon radius, not the other way around. This is actually an interesting question, and I will need to do some further research into it.

Thanks for bringing this up
As a result of the density of small blackholes being larger than the density of large blackholes, if the density remains uniform throughout the collapse, then the event horizon must form when the entire ball of mass falls inside its Schwarzschild radius because no smaller part of the ball has its mass inside the corresponding Schwarzschild radius.

23. then the event horizon must form when the entire ball of mass falls inside its Schwarzschild radius
That seems to support my understanding that the horizon actually forms before the collapse is complete.

24. "Singularity" was probably not the right term for me to use. "Singularity" is like a point in space containing all the mass. I'm thinking we never actually get a singularity at all.

But I don't want to say "critical density" either, because that is also incorrect. What I mean to say is that we have to have a black hole before we can have an event horizon or a photon sphere.

There is always a Schwarzchild radius. The Sun has a Schwarzchild radius. However until enough of an object's mass finds its way inside the Schwarzchild radius, there is no event horizon or photon sphere (even the very center of the mass is not inside the event horizon until an event horizon forms, although it is inside the Schwarzchild radius). But if time is slowing down in a continuous progression as the matter falls toward its own Schwarzchild radius, then not enough of the matter would ever make it inside that radius for a true black hole to form.

25. Originally Posted by Kojax
What I mean to say is that we have to have a black hole before we can have an event horizon or a photon sphere.
By definition, we have a blackhole when we have the event horizon. But we do not need to have a blackhole (event horizon) to have a photon sphere. The photon sphere lies definitely outside the Schwarzschild radius, and we only need the mass to be inside the photon sphere for the photon sphere to exist.

26. The photon sphere lies definitely outside the Schwarzschild radius, and we only need the mass to be inside the photon sphere for the photon sphere to exist.
Ok, so the question becomes - is there any quantum effect that can stop a collapse which has progressed past the photon sphere, before all the mass crosses the horizon radius ( Schwarzschild or otherwise ) ? Is neutron degeneracy pressure sufficient for this, and if so, under what circumstances ?

27. But if time is slowing down in a continuous progression as the matter falls toward its own Schwarzchild radius, then not enough of the matter would ever make it inside that radius for a true black hole to form.
That isn't the case though - the proper time it takes for something to fall towards and through the horizon is finite and well defined, even for a fully formed black hole. The coordinate time of a far-away stationary observer ( which does diverge ) is physically meaningless here, because he is not located where the collapse happens. You cannot use a clock that is located far away to determine local physics, because, due to curvature, measurements of space and time are purely local. So yes, a far-away observer never "sees" the black hole form, simply because he is not there but far away, but someone falling together with the collapsing star most certainly does. That isn't a paradox, but a logical consequence of relativity.

28. I'm thinking we never actually get a singularity at all.
So long as we stay within the confines of classical gravitation and deal only with ordinary matter and radiation, then the presence of an event horizon automatically implies a region of geodesic incompleteness through the Penrose theorem, i.e. a classical singularity.

29. Originally Posted by Markus Hanke
That isn't the case though - the proper time it takes for something to fall towards and through the horizon is finite and well defined, even for a fully formed black hole.

You mean that according to local time, right? Or "proper" time of an infalling observer.

The coordinate time of a far-away stationary observer ( which does diverge ) is physically meaningless here, because he is not located where the collapse happens. You cannot use a clock that is located far away to determine local physics, because, due to curvature, measurements of space and time are purely local.
I think you assign too much importance to the question of what observer we choose. There are no privileged observers.

Absolutely all observer (no exceptions) are equally valid for absolutely all observations, provided they are able to observe at all. That's a foundation stone of relativity. There isn't one observer who is "right" and another who is "wrong".

A distant observer with the ability to detect EM of all frequencies (even extremely low frequencies) watching a black hole in the final stages of its formation would see the matter get closer and closer to reaching the Schwarzchild radius, but continually slow down more and more.

They would never stop seeing EM come from that region. They could watch for a billion years and there would still be EM coming from there. It would be extremely low frequency, lower and lower, and therefore harder to detect (but that is a practical problem, not a theoretical problem.)

So yes, a far-away observer never "sees" the black hole form, simply because he is not there but far away, but someone falling together with the collapsing star most certainly does. That isn't a paradox, but a logical consequence of relativity.
A local observer can't know whether time has stopped. If they can't know it, then of course they don't observe it to happen. They don't observe it not to happen either. They simply don't observe anything about that particular topic.

In order to know time has stopped, you would have to know the future before it happens.

Originally Posted by KJW
By definition, we have a blackhole when we have the event horizon. But we do not need to have a blackhole (event horizon) to have a photon sphere. The photon sphere lies definitely outside the Schwarzschild radius, and we only need the mass to be inside the photon sphere for the photon sphere to exist.
I wasn't aware of that difference. Now I looked it up. Seems the photon sphere is 1.5 times the Schwarzschild radius.

So it seems that within the photon sphere, the angle of departing light determines whether it can escape at all or not. So some parts of the nearly formed black hole could only be seen from some vantage points, I guess?

30. Another interesting trait of a nearly formed black hole would be that, while EM is able to leave the interior, light from the outside cannot reach the interior unless an equal amount of energy/mass first leaves the region to make room for it. And probably whenever energy/mass leaves the interior, the nearly formed black hole simply becomes more compact rather than allow any EM to enter from the outside.

Point is, while communication would remain possible, it would only be one way communication, since new EM can't get into the interior.

Hard to say, though. A single photon doesn't have much mass/energy. Maybe it can get in without quite pushing the density (for want of a better term) over the limit.

31. Originally Posted by Kojax
Another interesting trait of a nearly formed black hole would be that, while EM is able to leave the interior, light from the outside cannot reach the interior unless an equal amount of energy/mass first leaves the region to make room for it.
I see no reason for that to be true unless one is getting into extreme cases such as the requirement through Pauli's principle that no two fermions cannot be in the same quantum state. A degeneracy pressure forms preventing density from increasing such as one finds in neutron stars. However that doesn't mean that the star can't contain bosons which aren't subject to Pauli's principle.

Originally Posted by Kojax
Point is, while communication would remain possible, it would only be one way communication, since new EM can't get into the interior.
I haven't been following this thread. Please explain what communication has to do with this subject?

Originally Posted by Kojax
A single photon doesn't have much mass/energy.
The mass-energy density of a photon is a delta function since the photon has no size whatsoever. The volume of a photon is zero.

32. You mean that according to local time, right? Or "proper" time of an infalling observer.
The proper time of an in-falling observer is local to his free fall frame, since it is what his clock physically measures.

I think you assign too much importance to the question of what observer we choose.
No, I don't think I do. The thing is that an external Schwarzschild observer never sees the horizon form, while a free-fall observer ( Gullstrand-Painleve observer ) just falls through it in finite time, so for him the horizon does indeed form. If we assume for a minute that a central singularity exists, then he will reach it also in finite proper time. So yes, the question of what observer we choose is important, because these observers do not agree on the question of horizon formation.

There isn't one observer who is "right" and another who is "wrong".
Indeed, but the point is that observers in a curved space-time are local. The horizon not forming is true for an external Schwarzschild observer, but only locally in his frame; falling through it in finite time is true for a Gullstrand-Painleve observer, but again only locally in his own frame. That is what I meant when I said that space and time are purely local notions in GR - everyone is right, but only in their own local frames. All of these frames are related by simple coordinate transformations, since all observers share the same space-time.

A local observer can't know whether time has stopped.
He can, by falling through the horizon and hitting something that is located below it, or simply by measuring the tidal forces he is subjected to ( they are a function of r ). If he measures himself to have fallen through the horizon in a finite amount of his own time, then that means time couldn't have stopped anywhere along his world line. So mathematically the question becomes whether or not the event horizon is a region of geodesic incompleteness - it is trivially easy to show that it is not. On an empirical level though the problem is of course that once he has fallen through he can't tell us about it

33. Originally Posted by Markus Hanke
The proper time of an in-falling observer is local to his free fall frame, since it is what his clock physically measures.
And what does that matter? The infalling observer does not disagree that time has stopped.

He/she can't disagree because it is absolutely impossible for a local observer to have any opinion at all as to whether time has stopped in their own frame of reference. Why? Because no observer ever observes time to slow down in their own frame, even in cases where it is obvious that time is moving slower in their own frame (such as humans on Earth observing a GPS satellite's clock.)

It takes two clocks to observe time dilation. If they use the distant observer's clock as their second clock then when they hold their own clock up and compare it against the distant observer's clock, they note that the light coming from the distant observer's clock is insanely blue shifted - indicating that either

1) - There is a large Doppler effect.

or

2)- There is a strong time dilation effect between the clocks, with the distant clock being the fast one.

#1 can be ruled out because the distance between the two observers is not growing. (It may even be shrinking if the distant observer is located on the opposite side of the black hole, or at some angle.)

#2 is a known effect, confirmed by the Pound-Rebka experiment.

Pound

If we assume for a minute that a central singularity exists, then he will reach it also in finite proper time.
But that is exactly what we cannot assume. The moment before a black hole forms, no central singularity exists.

Even at the very core of the massive object, time is still moving.

Indeed, but the point is that observers in a curved space-time are local. The horizon not forming is true for an external Schwarzschild observer, but only locally in his frame; falling through it in finite time is true for a Gullstrand-Painleve observer, but again only locally in his own frame. That is what I meant when I said that space and time are purely local notions in GR - everyone is right, but only in their own local frames. All of these frames are related by simple coordinate transformations, since all observers share the same space-time.

An observer that enters the nearly formed black hole and leaves again would disagree with you. (Since entering and leaving again are still theoretically possible).

They would return to see that lots and lots and lots of time had passed while they were gone.

He can, by falling through the horizon and hitting something that is located below it, or simply by measuring the tidal forces he is subjected to ( they are a function of r ). If he measures himself to have fallen through the horizon in a finite amount of his own time, then that means time couldn't have stopped anywhere along his world line. So mathematically the question becomes whether or not the event horizon is a region of geodesic incompleteness - it is trivially easy to show that it is not. On an empirical level though the problem is of course that once he has fallen through he can't tell us about it
Only if we get a singularity in the first place. If we never get a singularity then he can always tell us what is happening.

And you're right he doesn't observe himself to stop. But that's just because local observers can't observe that. Instead he observes time outside the black hole to be moving very fast.

Originally Posted by Physicist
I

I haven't been following this thread. Please explain what communication has to do with this subject?
If communication is not fully cut off then we don't get to just wave our hands and say one or the other observer is "wrong", and the other is "right".

As long as it is possible to unite the two observers' perspectives via a transformation of some kind, they're both right.

The mass-energy density of a photon is zero since the photon has no size whatsoever. Therefore the volume of a photon is zero.
Despite having zero mass, a photon still has energy, and energy is all it takes to create a gravitational effect. Mass and energy are truly interchangeable in General Relativity.

34. The infalling observer does not disagree that time has stopped.
Yes he does. His world line extends smoothly and continuously between any two radial points, while always being of finite proper length. This precludes both geodesic incompleteness and the notion of time somehow "stopping" anywhere.

Instead he observes time outside the black hole to be moving very fast.
Time ( very nearly ) stopping for him would mean that he sees the entire history of the universe playing out before him. That does not happen, as is immediately obvious once we draw his world line onto a Kruskal-Szekeres diagram. See also : general relativity - Does someone falling into a black hole see the end of the universe? - Physics Stack Exchange

35. I'll add to your sagacity a more simple case to analyse, because much less things are moving.
It is the black-sea examples.<-- image.

I kind of think a pure water at normal density the size of a galaxy would be near its own Schwarzschild radius.
I call it the black sea, it is very cold, because I fear the pressure at the center would be so great, cold fusion may start just to ruin my experiment.

It misses a few tons of water to increase the mass above the limit. A mighty cosmic pirate, with a black-beard, lands its boat on the black-sea. G force may be strong, but tidal force bearable. Critical mass is still not reached, but the dock is ready to be fill with water. There is a great mast, not just for pride, but to escape way above the Schwarzschild radius.
I imagine G force to be so strong that the whole picture would already be very very red-shifted.

Now a cosmic bucket will drop the missing water into the boat (but no enough to sink it).

A black hole form. Not instantly but instantly. Does the pirate flag is severed form the event horizon ?

Seriously though, I think the whole experiment is doable, but reality will start to "conspire" to avoid any paradox.
Like:
-no boat strong enough could be build anymore,
-or observation would already be quite impossible to do before the horizon form (making it a smooth transition) ...
-not only pressure, but space-time curvature at the center will already have ripped apart the water molecule (electron getting lost, quark going berserk)

Shoot away !

36. The way I see it is quite simply like so - GR is a purely classical theory. On the other hand, a gravitational collapse is only approximately classical, and even then only in the initial stages. Furthermore, quantum field theories in curved space-times are nowhere near mathematically rigorous, or even completely understood in all aspects. Combine all of this, and it is doubtful that classical GR on its own is much more than a broad approximation to what really happens during gravitational collapse; it is quite possible that even outside and near the horizon space-time is no longer classical in any meaningful sense of the world. To make a long story short - it is quite possible that true event horizons as we understand them from classical GR just don't form in the real universe; this would in fact help to resolve a number of conceptual difficulties, at least as long as singularities don't exist either ( which I am pretty sure of ).

Let's just remember that GR is purely classical, and a lot of questions are unresolved when it comes to quantum gravity, that's all I'm saying

37. Originally Posted by Markus Hanke
Yes he does. His world line extends smoothly and continuously between any two radial points, while always being of finite proper length. This precludes both geodesic incompleteness and the notion of time somehow "stopping" anywhere.
.

Time dilation has to be defined as something you determine using two or more clocks in different frames, never just one. Or else the whole idea becomes utterly meaningless.

Once we've defined it as requiring two clocks, any arguments about what just one observer sees when that one observer is only using clocks in one frame, are meaningless also.

Time ( very nearly ) stopping for him would mean that he sees the entire history of the universe playing out before him. That does not happen, as is immediately obvious once we draw his world line onto a Kruskal-Szekeres diagram. See also : general relativity - Does someone falling into a black hole see the end of the universe? - Physics Stack Exchange
There is a glaring flaw in that argument.

Unless the infalling observer is moving at exactly C, then an infalling ray of light will certainly catch them.

The reason is because:

1) - The infalling observer never reaches the Schwarzchild radius for their mass+the rest of the mass.

2) - Until the infalling ray of light reaches the observer, the infalling ray of light is always experiencing an amount of time dilation equal to or less than the infalling observer.

The beam of light will never cross the Schwarzchild radius, but it will certainly catch the observer (who will also never cross the Schwarzchild radius).

Originally Posted by Boing3000
I'll add to your sagacity a more simple case to analyse, because much less things are moving.
It is the black-sea examples.<-- image.

I kind of think a pure water at normal density the size of a galaxy would be near its own Schwarzschild radius.
I call it the black sea, it is very cold, because I fear the pressure at the center would be so great, cold fusion may start just to ruin my experiment.

It misses a few tons of water to increase the mass above the limit. A mighty cosmic pirate, with a black-beard, lands its boat on the black-sea. G force may be strong, but tidal force bearable. Critical mass is still not reached, but the dock is ready to be fill with water. There is a great mast, not just for pride, but to escape way above the Schwarzschild radius.
I imagine G force to be so strong that the whole picture would already be very very red-shifted.

Now a cosmic bucket will drop the missing water into the boat (but no enough to sink it).

A black hole form. Not instantly but instantly. Does the pirate flag is severed form the event horizon ?
Except, before the bucket of water can get close enough to cause a black hole to form, time stops. Or well, it slows down, and keeps slowing down by just enough to keep the bucket of water from ever reaching the Schwarzchild radius (for itself + the rest of the matter).

The trouble with black hole theories that involve an event horizon forming is, they don't treat the system as a dynamic system. They treat it like things are going to change one at a time. First the bucket of water gets close, then a gravitational time dilation effect follows. But if we treat it as a dynamic system, then both the bucket of water approaching and the gravitational time dilation effect getting stronger are happening simultaneously.

When they happen simultaneously, the bucket of water never has the chance to get close enough.

38. Originally Posted by Kojax
Except, before the bucket of water can get close enough to cause a black hole to form, time stops.
Why ? time is nowhere stopped, and that is why I bring that different experiment. The more depth you are in the sea, and the more observer time is slowing, but nothing is stopped anywhere. The water is falling to fill the boat, and the schwarzschild radius is moving up toward sea level.

Originally Posted by Kojax
The trouble with black hole theories that involve an event horizon forming is, they don't treat the system as a dynamic system. They treat it like things are going to change one at a time
I don't understand why you say that. Balck hole formation theories are very dynamic things.

Originally Posted by Kojax
When they happen simultaneously, the bucket of water never has the chance to get close enough.
But they happen simultaneously, that's the whole point. Drop the water one droplet at a time if you want. Nothing change in the picture, and nowhere time "appears" to stop.

More specifically what is observed is a smooth closing of the horizon, because even before the Schwarzschild has reach sea level the picture has allready turned very black and distant. This is an relative EVENT horizon. Not a barrier of any kind for objects, the choice of another coordinate system like Kruskal-Szekeres helps you realize that. Only event reporting (more specifically, exporting observation) is "impossible". Events are still happening there.

And the dynamic of the process imply that when the last drop of water close in, the gravitational bending accompanying it cannot reach the center of the black sea before thousands of year (proper time) and the other side in another thousands.

Because the Schwarzschild formula is encompassing a hole swath a space does not imply instantaneously the other side is aware that the schwarzschild is 'now upper'.

And it does not change anything. The boat mast is "in reality" already broken, and the black-sea already black, and you would already require nearly infinite amount of energy to "remove" water from its pull, before the last drop comes in. The black-sea was already a black-hole, it is just a more "gentle" formation than those resulting from a suppernova collapse

39. Time dilation has to be defined as something you determine using two or more clocks in different frames, never just one. Or else the whole idea becomes utterly meaningless.
Yes, that's right. Time dilation is just a geometric relationship between frames.

Once we've defined it as requiring two clocks, any arguments about what just one observer sees when that one observer is only using clocks in one frame, are meaningless also.
I thought the discussion was about whether or not free fall through an event horizon can happen, or not. You don't need two clocks to determine that; a far-away stationary observer plays no role in this at all. All you need to consider is the proper length of the world line of the falling observer and whether or not it is smooth and continuous everywhere. The relationship of that world line to some far away observer is simply not relevant.

The beam of light will never cross the Schwarzchild radius, but it will certainly catch the observer (who will also never cross the Schwarzchild radius).
No, that is trivially wrong, as seen by a simple plot on a Kruskal-Szekeres diagram. Both light and massive objects cross the event horizon; the proper length of a world line ( falling from rest at infinity ) between two radial points is

in geometrised units, which is finite for every pair of radial points, both outside and inside the event horizon, and certainly also across it. This is because space-time ( and hence world lines in it ) is everywhere smooth and continuous, including at the horizon, since all curvature invariants remain regular everywhere; consider for example the Kretschman invariant :

There's no second frame required for any of this, so time dilation doesn't come into it; it's just simply a matter of space-time being smooth and regular at the horizon, so geodesics extends smoothly across it while remaining of finite length. So yes, both light and massive objects cross the horizon - the article I referenced is quite correct, which is supported by all standard textbooks, most notably Taylor/Wheeler "Exploring Black Holes" ( chapter 3 ) and MTW "Gravitation" ( chapter 31 ). If you believe this to be not the case, please give a proper mathematical argument to show that event horizons are geodesically incomplete, or that the proper length of a geodesic from any arbitrary external point to the horizon is infinite.

The trouble with black hole theories that involve an event horizon forming is, they don't treat the system as a dynamic system.
That is incorrect also. The appropriate metric to use for a gravitational collapse is the Vaidya-Bonnet metric ( not Schwarzschild ), which is a "dynamic system" in the sense that the metric coefficients explicitly depend on time. The result about finite world lines extending across the horizon is the same there also.

40. Originally Posted by Markus Hanke
Yes, that's right. Time dilation is just a geometric relationship between frames.

I thought the discussion was about whether or not free fall through an event horizon can happen, or not. You don't need two clocks to determine that; a far-away stationary observer plays no role in this at all. All you need to consider is the proper length of the world line of the falling observer and whether or not it is smooth and continuous everywhere. The relationship of that world line to some far away observer is simply not relevant.
I'm going to quit calling them "Nearly Formed Black Hole" and just call them NFBH. It's much shorter to type.

The issue here is, there is no such thing as an absolute "speed of time". Saying local time is "fast" or "slow" depends entirely on what you feel like calling it.

If an observer near the NFBH can see the universe outside moving very very fast (almost infinitely fast), then they must agree that the relative difference in the rate at which time flows between their own frame and the outside universe's frame is very very big.

Whether that is because time is "flowing normally" at the NFBH and "very very fast" outside the NFBH, or "flowing very fast" inside the NFBH but "even faster" outside the NFBH - That is just a "glass half full" or "glass half empty" type of question.

No, that is trivially wrong, as seen by a simple plot on a Kruskal-Szekeres diagram. Both light and massive objects cross the event horizon; the proper length of a world line ( falling from rest at infinity ) between two radial points is

in geometrised units, which is finite for every pair of radial points, both outside and inside the event horizon, and certainly also across it. This is because space-time ( and hence world lines in it ) is everywhere smooth and continuous, including at the horizon, since all curvature invariants remain regular everywhere; consider for example the Kretschman invariant :

No deductive logical construct can ever exceed the truth value of its assumptions.

This all assumes an event horizon is already present. If an event horizon were already present before you start, then an infalling object would eventually reach it. I can concede that is probably true.

However if event horizons never form, and don't exist because they don't ever form, then what I've said above remains the most likely. An infalling object

1) - Does get overtaken by beams of light entering from the outside. (Allowing it to witness the end of the universe before it reaches the Schwarzchild radius).

2) - Never gets close enough to cause a complete black hole to form.

What other conditions must the argument meet?

There's no second frame required for any of this, so time dilation doesn't come into it; it's just simply a matter of space-time being smooth and regular at the horizon, so geodesics extends smoothly across it while remaining of finite length. So yes, both light and massive objects cross the horizon - the article I referenced is quite correct, which is supported by all standard textbooks, most notably Taylor/Wheeler "Exploring Black Holes" ( chapter 3 ) and MTW "Gravitation" ( chapter 31 ). If you believe this to be not the case, please give a proper mathematical argument to show that event horizons are geodesically incomplete, or that the proper length of a geodesic from any arbitrary external point to the horizon is infinite.
Smoothness is a red herring. It determines whether you can use calculus, but otherwise has no effect on anything.

Just because a change is incremental doesn't mean it won't have cumulative consequences.

That is incorrect also. The appropriate metric to use for a gravitational collapse is the Vaidya-Bonnet metric ( not Schwarzschild ), which is a "dynamic system" in the sense that the metric coefficients explicitly depend on time. The result about finite world lines extending across the horizon is the same there also.
Alright. I'm not familiar with Vaidya- Bonnet. From the wiki page it looks like its main difference is that it uses M(u) instead of M as its basis for determining the collapse radius. But I'm not sure what is meant by M(u), and I think it would take literally days of time I don't have to pour over the math looking for it.

Vaidya metric - Wikipedia, the free encyclopedia

Why does Vaidya-Bonnet allow collapse while Schwarzchild's metric does not?

I'm concerned that this discussion is going to degenerate into a big vocabulary game of throwing out new terms, and then stalling until I can learn each new term, and analyse all of its consequences on my own. (Only to discover in most cases that the new term is simply a retelling of an old one, with the same consequences as the old one.)

I understand that sometimes it can be interesting to cast a question from a new perspective, or redo the math using a new and different coordinate system. But it is only productive and useful if the new coordinate system adds insight not found in the previous system. (And even then it should still be possible to describe it in the old system if they are truly equivalent systems. Less elegant, but still possible.)

41. The issue here is, there is no such thing as an absolute "speed of time". Saying local time is "fast" or "slow" depends entirely on what you feel like calling it.
Yes, there's no such thing. Local time is never "fast" or "slow".

If an observer near the NFBH can see the universe outside moving very very fast (almost infinitely fast)
He doesn't.

This all assumes an event horizon is already present.
No, not at all. It assumes only spherical symmetry, absence of angular momentum, absence of other sources of energy-momentum, and the space-time being static and stationary. Given that, our space-time is necessarily smooth and regular everywhere other than at r=0, irrespective of whether or not there is an event horizon. But if there is, that region is also smooth and regular.

Smoothness is a red herring. It determines whether you can use calculus, but otherwise has no effect on anything.
If space-time was to cease being smooth, than classical GR as a theory no longer applies ( that's what happens at r=0 ), and the discussion becomes moot. As it stands, the geodesic structure of Schwarzschild space-time is smooth and continuous everywhere outside r=0, which can be seen by the fact that the curvature invariants formed from the Riemann tensor are all regular in that region. That is certainly not a red herring, but just means that locally Minkowskian physics apply everywhere.

Why does Vaidya-Bonnet allow collapse while Schwarzchild's metric does not?
They both allow collapse, but the difference is that the Schwarzschild metric assumes that space-time outside the central mass is a perfect vacuum, whereas the Vaidya-Bonnet metric accounts for in falling and outgoing sources of energy-momentum, as would be found in a real collapse type situation. Mathematically, in the Schwarzschild metric the mass parameter M is constant, whereas in the Vaidya solution it is a function of time, so it is allowed to vary. Hence you can model the evolution of a black hole, as opposed to just a static state as in Schwarzschild.

I understand that sometimes it can be interesting to cast a question from a new perspective, or redo the math using a new and different coordinate system. But it is only productive and useful if the new coordinate system adds insight not found in the previous system. (And even then it should still be possible to describe it in the old system if they are truly equivalent systems. Less elegant, but still possible.)
The Vaidya-Bonnet metric is a different space-time from Schwarzschild, with a different geometry; it is not the same solution to the field equations, so it is more than just a different choice of coordinates. Unfortunately it is also much more complicated. The Schwarzschild metric is simple only because it is highly restrictive and symmetric in its assumptions, and the price to pay is the fact that in the real universe we would never find such black holes. One has to use more realistic descriptions to see what is actually happening; Schwarzschild is a good approximation in many cases though.

I'm concerned that this discussion is going to degenerate into a big vocabulary game of throwing out new terms
I'm sorry that you feel that way, but the geometry of Einstein manifolds involves more than just Schwarzschild coordinates ( which are arbitrary and not in any way privileged ). As such, it is important to go beyond the description in a specific set of coordinates and understand the geometry of the space-time in question, as opposed to just an arbitrary coordinate choice. That is what I am attempting to do - we all know that Schwarzschild coordinates become singular at the event horizon, but that singularity is a feature of the coordinate system, not the physics of the space-time itself; choose a different set of coordinates for the same space-time, and the singularity at the horizon vanishes, i.e. Kruskal-Szekeres, Gullstrand-Painleve, Novikov, Eddington-Finkelstein etc etc. The only singularity that is physical is at r=0, because all coordinate systems are undefined there. To distinguish between coordinate singularities and physical singularities, one usually employs the various invariants of the curvature tensor, such as the Kretschmann scalar which I quoted.

Basically, I am attempting to point out that coordinate systems are arbitrary choices, but the underlying geometry of the space-time is not; we should hence focus on quantities that do not depend on the choice of coordinate system ( i.e. quantities on which everyone agrees ) when analysing this situation.

Does that make sense ?

42. Originally Posted by Markus Hanke
Yes, there's no such thing. Local time is never "fast" or "slow".

He doesn't.

No, not at all. It assumes only spherical symmetry, absence of angular momentum, absence of other sources of energy-momentum, and the space-time being static and stationary. Given that, our space-time is necessarily smooth and regular everywhere other than at r=0, irrespective of whether or not there is an event horizon. But if there is, that region is also smooth and regular.

If space-time was to cease being smooth, than classical GR as a theory no longer applies ( that's what happens at r=0 ), and the discussion becomes moot. As it stands, the geodesic structure of Schwarzschild space-time is smooth and continuous everywhere outside r=0, which can be seen by the fact that the curvature invariants formed from the Riemann tensor are all regular in that region. That is certainly not a red herring, but just means that locally Minkowskian physics apply everywhere.

They both allow collapse, but the difference is that the Schwarzschild metric assumes that space-time outside the central mass is a perfect vacuum, whereas the Vaidya-Bonnet metric accounts for in falling and outgoing sources of energy-momentum, as would be found in a real collapse type situation. Mathematically, in the Schwarzschild metric the mass parameter M is constant, whereas in the Vaidya solution it is a function of time, so it is allowed to vary. Hence you can model the evolution of a black hole, as opposed to just a static state as in Schwarzschild.

The Vaidya-Bonnet metric is a different space-time from Schwarzschild, with a different geometry; it is not the same solution to the field equations, so it is more than just a different choice of coordinates. Unfortunately it is also much more complicated. The Schwarzschild metric is simple only because it is highly restrictive and symmetric in its assumptions, and the price to pay is the fact that in the real universe we would never find such black holes. One has to use more realistic descriptions to see what is actually happening; Schwarzschild is a good approximation in many cases though.
The assumptions of Schwarzchild are things like no-spin, and no-charge, right? So very much like when your highschool physics teacher has you calculate the trajectory of a baseball, but lets you leave out air resistance?

I don't see a problem here. Almost all objects in the universe have some charge and some spin, but most of them don't have very much of either. If an object had a lot of spin temporarily, it would begin emitting gravity waves until it slowed down again. If it had a lot of charge, it would start to attract charged particles of the opposite charge and repelling charged particles of the same charge - leading to a gradual accumulation of opposite charge to balance it out.

Small effects are important if you want perfect accuracy, but it seems very unlikely that they would make the difference between an object crossing the Schwarzchild radius or not crossing.

I'm sorry that you feel that way, but the geometry of Einstein manifolds involves more than just Schwarzschild coordinates ( which are arbitrary and not in any way privileged ). As such, it is important to go beyond the description in a specific set of coordinates and understand the geometry of the space-time in question, as opposed to just an arbitrary coordinate choice. That is what I am attempting to do - we all know that Schwarzschild coordinates become singular at the event horizon, but that singularity is a feature of the coordinate system, not the physics of the space-time itself; choose a different set of coordinates for the same space-time, and the singularity at the horizon vanishes, i.e. Kruskal-Szekeres, Gullstrand-Painleve, Novikov, Eddington-Finkelstein etc etc. The only singularity that is physical is at r=0, because all coordinate systems are undefined there. To distinguish between coordinate singularities and physical singularities, one usually employs the various invariants of the curvature tensor, such as the Kretschmann scalar which I quoted.

Basically, I am attempting to point out that coordinate systems are arbitrary choices, but the underlying geometry of the space-time is not; we should hence focus on quantities that do not depend on the choice of coordinate system ( i.e. quantities on which everyone agrees ) when analysing this situation.

Does that make sense ?

My issue with using lots of coordinate systems goes back to Occam's razor. If you don't need a new coordinate system to describe what you are observing, then why introduce one? As near as I can tell, there is nothing requiring a new coordinate system that has actually been observed. GR predicts some interesting things, but as far as I am aware, the only things that are actually backed by observation are those predictions that don't require a new coordinate system. The blue shifting of light, or change in apparent location of stars behind the sun, or even the clocks on GPS satellites - none of those require a new coordinate system.

Also there is simply potential to confuse the issues.

The Kruzkel-Szekeres argument from before, for example. It's an interesting way to solve the problem of seeing the coordinate system break down at the event horizon, but it still appears to be using only a local t. (Or really V and U, which relate to T indirectly.) So it would have nothing to say about the relative observations of observers near and far away from the NFBH.

Going off of Wiki, the formulas that relate R and T to V and U in the exterior region appear to be the following. (And the exterior region is the only one I'm concerned with, since I'm discussing a nearly formed black hole, rather than a black hole that has collapsed inside its own Schwarzchild radius.)

Kruskal

Sinh of any variable X translates as :

Cosh of any variable X translates as:

The e constant is approximately 2.71828.

We can see that as T gets very large, both V and U will tend toward infinity. From looking at it, it appears to me that while the Kruskel Szekeres model does a good job of making sense of the theory of event horizons, it in no way confirms, even slightly, that these event horizons can actually form.

As for world lines, nothing prevents us from being able to draw a world line for an event that will never happen. So the ability to draw a world line that leads to the interior of a black hole also tells us nothing.

The temptation of creating a lot of coordinate systems is that it is easy to delude ourselves into thinking that just because we can imagine something, that it must therefore be real by virtue of our having imagined it.

43. I am a simple minded geologist and my interest in astronomy focuses on planetary formation and evolution, not on black holes. However, from this simple minded position my answer to Kojax's original question is itself simple. We have made calculations as to how a black hole should behave. We have made observations of what may be black holes. We find that these observations provide a close match to expectation. We find no alternative explanation that provides a superior match. So, although we do not know that black holes form, the evidence points very strongly to the fact that they do.

There are two advantages to this simple minded approach: it produces a shorter post that is easier to write and to read; it avoids having to learn all the maths; (And thirdly, it is the correct answer).

44. The assumptions of Schwarzchild are things like no-spin, and no-charge, right?
Yes - the assumptions are that the body is static and stationary, is perfectly spherically symmetric, and has zero angular momentum and zero net charge.

I don't see a problem here. Almost all objects in the universe have some charge and some spin, but most of them don't have very much of either.
No, but these quantities are conserved, so a slowly spinning large star will become an extremely fast spinning small neutron star or black hole - with all associated relativistic effects.

If you don't need a new coordinate system to describe what you are observing, then why introduce one?
Because in GR unlike in Newtonian mechanics, observers and coordinate systems are physically significant only locally, as I have tried to explain before. So, if you want to know what happens locally during a gravitational collapse, you have to use a coordinate system that is local to the collapsing matter, not Schwarzschild coordinates that are local to a stationary observer very far away from the collapsing object. So yes, a new coordinate system is indeed needed.

So it would have nothing to say about the relative observations of observers near and far away from the NFBH.
You are right, it doesn't. But then, that is not the question at stake - we already know that a far-away observer never observes the horizon forming ( that is not in question ), but the question was whether or not that means that locally in the region around the Schwarzschild radius an event horizon is actually present or not. In other words, the question is whether or not a particle - massive or otherwise - is able to fall into and/or escape from the region beyond the Schwarzschild radius. And in order to examine this, we need to examine local physics, no what a distant observer may or may not observe.

The temptation of creating a lot of coordinate systems is that it is easy to delude ourselves into thinking that just because we can imagine something, that it must therefore be real by virtue of our having imagined it.
Coordinates, including Schwarzschild ones, are not real - that's the whole point here. The only thing that is real in any sense of the word is the relationship between events, i.e. the geometry of the space-time, not the way we label the events. Hence you cannot conclude from Schwarzschild coordinates becoming singular at the horizon that the horizon itself never forms, since this is simply an anomaly in the way your distant observer assigns labels to events, it has no physical relevance; instead, we need to examine the geodesic structure of the space-time ( which is independent from any specific set of coordinates ), which is what I was attempting to show. To cut a long story short - the geometry and structure of space-time - independent from any coordinate system - is perfectly regular in the region around the Schwarzschild radius.

So let me ask you this again ( you seem to have ignored this bit ) - can you give a mathematically rigorous argument that world lines do not extend across the Schwarzschild radius, and instead become infinitely long ?

45. Originally Posted by Ophiolite
We have made observations of what may be black holes. We find that these observations provide a close match to expectation. We find no alternative explanation that provides a superior match. So, although we do not know that black holes form, the evidence points very strongly to the fact that they do.
That's the issue. The evidence points equally in both directions. A black hole that nearly forms, but never quite gets there would emit EM but it would be so low in frequency that no detector that has ever yet been built on Earth would be capable of detecting it.

So, when we see a region of space that appears to be emitting no light at all, but which is affecting the region around it by its gravity, in a way that suggests there should be a star there, either theory will equally describe it.

Originally Posted by Markus Hanke

You are right, it doesn't. But then, that is not the question at stake - we already know that a far-away observer never observes the horizon forming ( that is not in question ), but the question was whether or not that means that locally in the region around the Schwarzschild radius an event horizon is actually present or not. In other words, the question is whether or not a particle - massive or otherwise - is able to fall into and/or escape from the region beyond the Schwarzschild radius. And in order to examine this, we need to examine local physics, no what a distant observer may or may not observe.

Using purely local physics only makes sense in a closed system. However a NFBH is not a closed system. It is still in contact with the outside universe.

It is possible that, in the last (locally measured) microsecond before collapse, some event in the outside universe could prevent it. It's not likely in practice, but it's certainly possible in theory.

So the fact an infinite amount of time will pass in the outside universe before that locally measured microsecond is over is significant. Even if the local observer doesn't see anything going on locally during that microsecond.

Coordinates, including Schwarzschild ones, are not real - that's the whole point here. The only thing that is real in any sense of the word is the relationship between events, i.e. the geometry of the space-time, not the way we label the events. Hence you cannot conclude from Schwarzschild coordinates becoming singular at the horizon that the horizon itself never forms, since this is simply an anomaly in the way your distant observer assigns labels to events, it has no physical relevance; instead, we need to examine the geodesic structure of the space-time ( which is independent from any specific set of coordinates ), which is what I was attempting to show. To cut a long story short - the geometry and structure of space-time - independent from any coordinate system - is perfectly regular in the region around the Schwarzschild radius.

What we are really disagreeing about is Relativity of Simultaneity. It's a "when?" question. In relativity, an event can occur at two different times for two different observers (and it's not an illusion that it is happening at that time.) A far away observer perceives that an infalling object only reaches the Schwarzchild radius after infinity time has passed. From their perspective "when it happened" is infinity years in the future.

You appear to be arguing that, even though a faraway observer observes the object to reach its destination infinity years in the future, the object is in fact arriving right now. Indeed, you appear to be suggesting that the faraway observer experiences this to be true, even though they cannot see it.

This gets even more confusing if we go one step further, and start considering Quantum Mechanics, wherein an event becoming "real" is determined by wave function collapse, and wave function collapse doesn't occur until an event is observed. So now the wave function associated with the infalling object passing the Schwarzchild radius has to somehow collapse without being observed?

Isn't everything just a lot simpler if we take things at face value? Why pump so much theory into something if there is no evidence to suggest that it is necessary for us to do so?

So let me ask you this again ( you seem to have ignored this bit ) - can you give a mathematically rigorous argument that world lines do not extend across the Schwarzschild radius, and instead become infinitely long ?
I can't. Because I'm pretty sure world lines cross it and they don't become infinitely long when they are measured locally. Only when they are measured from far away (an issue you don't appear to disagree on.)

The question is: who is drawing the world line? What frame of reference are they in?

Clearly we don't disagree that a distant observer perceives an infinitely long line. What we appear to disagree is whether the infalling observer near the NFBH perceives a world line of 0 for events transpiring outside the NFBH. (Or actually very nearly zero, because our observer has not fallen in yet.)

46. Originally Posted by Markus Hanke
So, if you want to know what happens locally during a gravitational collapse, you have to use a coordinate system that is local to the collapsing matter, not Schwarzschild coordinates that are local to a stationary observer very far away from the collapsing object. So yes, a new coordinate system is indeed needed.
Don't you mean that a coordinate system is needed to describe it, not to know it?

Originally Posted by Markus Hanke
Coordinates, including Schwarzschild ones, are not real - that's the whole point here.
Why not? I've always found it troublesome to talk about things being real/not real in physics. It always seems rather arbitrary to me. E.g. it's quite literally true to say that energy isn't real but it's hardly a useful thing to say since it's so useful and corresponds to so many real things, just like time.

A coordinate system in relativity is defined as a collection of clocks and rods and since those are real then so too is the coordinate system. I thought that there was an illustration of this in MTW but I can't seem to find it. Do you recall where it is, Markus?

47. Don't you mean that a coordinate system is needed to describe it, not to know it?
Yes, that's what I mean.

Why not?
Because they are just arbitrary labels given to events in space-time. It is the events that are physically real, not the labels we give them, which is why the laws of physics are diffeomorphism invariant; the coordinate basis we choose to describe a system has no intrinsic physical meaning. Of course rods and clocks are real, but the labelling on the rods and the clock faces are arbitrary choices, as are the labels we give each of those devices themselves in the grid. That is what I meant.

A coordinate system in relativity is defined as a collection of clocks and rods and since those are real then so too is the coordinate system.
I don't agree with this. While the clocks and rods are certainly real, the labels we give each one of them are arbitrary choices - to me the coordinate system signifies those labels, rather than the clocks and rods themselves, which I identify with events. For example, a stationary far-away observer in Schwarzschild space-time may use Schwarzschild coordinates to label the events in that space-time, or he may use Kruskal-Szekeres coordinates, or any other system he chooses. These are all just different ways to label the same events, so in that sense the coordinates you use are arbitrary and without physical reality, whereas the actual geometry ( i.e. the relationship between events ) is not.

Am I making sense ?

I thought that there was an illustration of this in MTW but I can't seem to find it. Do you recall where it is, Markus?
I am not sure which one you are referring to. The arbitrariness of coordinate systems is dealt with on pages 6, 7 and 8, right at the beginning of the book - it is this arbitrariness of event labelling which I mean when I speak of specific coordinate systems to not posses intrinsic physical reality, unlike the events they describe.

48. Originally Posted by Markus Hanke
I don't agree with this. While the clocks and rods are certainly real, the labels we give each one of them are arbitrary choices - to me the coordinate system signifies those labels, rather than the clocks and rods themselves, which I identify with events.
That is a very important distinction. Infinite confusion -- especially among those with non-mainstream views (often deriving from a lack of understanding the literature) -- results if one conflates coordinates with physical reality. As you -- and MTW -- point out, coordinates are arbitrary labels that we may choose independently, and thus cannot be regarded as reality in any operationally useful sense.

49. Originally Posted by tk421
Originally Posted by Markus Hanke
I don't agree with this. While the clocks and rods are certainly real, the labels we give each one of them are arbitrary choices - to me the coordinate system signifies those labels, rather than the clocks and rods themselves, which I identify with events.
That is a very important distinction. Infinite confusion -- especially among those with non-mainstream views (often deriving from a lack of understanding the literature) -- results if one conflates coordinates with physical reality. As you -- and MTW -- point out, coordinates are arbitrary labels that we may choose independently, and thus cannot be regarded as reality in any operationally useful sense.
Also, we must admit every possible coordinate system.

50. Originally Posted by KJW
Also, we must admit every possible coordinate system.
Subtle and profound, as always, KJW. Glad that you're here!

51. Originally Posted by Markus Hanke
Yes, that's what I mean.

Because they are just arbitrary labels given to events in space-time. It is the events that are physically real, not the labels we give them, which is why the laws of physics are diffeomorphism invariant; the coordinate basis we choose to describe a system has no intrinsic physical meaning. Of course rods and clocks are real, but the labelling on the rods and the clock faces are arbitrary choices, as are the labels we give each of those devices themselves in the grid. That is what I meant.
A coordinate system is just a way of organizing stuff. Suppose you are in a lawyer's office looking through evidence for a case, and you see that he has chosen to organize the files alphabetically. Clearly that is an arbitrary decision. He could have used reverse alphabetic order, or he could have organized them by chronology in the order he encountered them, or maybe just thrown the files into the cabinet randomly.

But the fact a given organization system is being used has no bearing or relevance to the evidence itself.

Indeed, there is no reason why the result you get using one organizational system should differ from the result you get using any other organizational system, unless one of the systems is causing you to make a logical error in your assessment.

But what logical error is created by using Schwarzchild's coordinate system?

I don't agree with this. While the clocks and rods are certainly real, the labels we give each one of them are arbitrary choices - to me the coordinate system signifies those labels, rather than the clocks and rods themselves, which I identify with events. For example, a stationary far-away observer in Schwarzschild space-time may use Schwarzschild coordinates to label the events in that space-time, or he may use Kruskal-Szekeres coordinates, or any other system he chooses. These are all just different ways to label the same events, so in that sense the coordinates you use are arbitrary and without physical reality, whereas the actual geometry ( i.e. the relationship between events ) is not.

Am I making sense ?

You're making sense, I think.

We just totally disagree.

As for Kruskal-Szekeres, I've been studying this matter further..... and it really really looks like Kruskal-Szekeres system of coordinates yields the same result as Schwarzchild coordinates in this matter anyway.

The wiki article states this:

The location of the event horizon (r = 2GM) in these coordinates is given by ,. Note that the metric is perfectly well defined and non-singular at the event horizon.
Kruskal

But if you look at the equations that describe V and U, you'll see that the only difference between them is the term at the end, where t is either operated upon by the Sinh or Cosh function.

t= infinity is the only condition under which those two equations would yield the same outcome.

And the closer t gets to infinity, the closer they get to yielding the same outcome, so if t gets very near infinity, the two equations get very near to yielding the same outcome. (That is important because it suggests that an approximately infinite T is sufficient to match observation.)

52. I should do the math before I say that.

So for cosh x = sinh x we get. Multiply both sides of the equation by 2 and we have .

Then subtract from both sides and we're left with . That can also be written as As x approaches infinity, both terms approach zero, making them equal in spite of having opposite signs. I see no other solution for the equation. Do you?

edit:

I had previously been thinking maybe there was another solution, since the final values are + or - of each other, but since they are added to other values in the equation, that still doesn't make the whole equation meet the + or - requirement.

53. Ok, so the point is, even in Kruskel-Szekeres, we face the same mechanical problem. The singularity is also a theoretical problem, but coordinate changes such as Kruskel-Szekeres resove the theoretical problem. However they don't resolve the mechanical problem.

The mechanical problem is that, mechanically, an object can't reach the Schwarzchild radius in finite time (as measured from the clock of a distant observer.)

And because communication is not entirely cut off, it is always possible to confirm that it hasn't reached the Schwarzchild radius yet. If the infalling object had a very fast computer, and were capable of sending and receiving signals at very high frequencies, then you could send a very low frequency signal to it (which becomes high frequency on the way), and wait for it to send a signal back (which would become extremely low frequency by the time you get it). That would allow you to confirm that, at the time you sent your signal, the object had still not reached the Schwarzchild radius.

That works as a thought experiment at least. Practically it would be unlikely we could ever construct an infalling object that was capable of sending and receiving signals quickly enough (by the standard of its local time).

54. an object can't reach the Schwarzchild radius in finite time (as measured from the clock of a distant observer.)
Of course not, this is a mathematical fact, and not in dispute.

And because communication is not entirely cut off, it is always possible to confirm that it hasn't reached the Schwarzchild radius yet
I don't understand the point in this. We already know that from the far-away observer's point of view the horizon is never reached. There is no dispute over this. At the same time, it completely misses the original point.

and wait for it to send a signal back
Those signals travel along null geodesics, which become increasingly complicated as the emitter approaches the horizon. From just above the horizon, the signal would take a very long time to get back out, but that tells you nothing about how the emitter keeps freely falling, it only gives you an indication of the geometry of the outgoing null geodesics along which your signal travels. Do you see the distinction ? The outgoing signal does not travel along straight radial lines.

We just totally disagree.
The disagreement is over the notion of "time". In terms of far-away Schwarzschild time, the horizon infall time is infinite, so the horizon is never reached by anything. However, what I am trying to point out is that this particular concept of time is valid only in your far-away frame - if you apply it to other remote frames, it ceases to be physically meaningful, because that other frame will not share the same concept of time. So, while Schwarzschild infall time is infinite, proper infall time for a test particle in free fall is not, simply because they don't share the same concept of "time", which, in curved space-times, is a purely local notion and not globally valid. Yes, in the frame of your stationary distant observer the horizon is never reached, but that does not mean that locally at the Schwarzschild radius no horizon forms, and no particle falls through it. It's just that, in order to cross the horizon ( and hence find out whether or not it has formed ), you need to actually be there ( i.e. fall freely into it ), as opposed to just watch while at rest somewhere far away.

Indeed, there is no reason why the result you get using one organizational system should differ from the result you get using any other organizational system
The reason is that in curved space-times, coordinate measurements coincide with physical measurements ( i.e. proper quantities ) only locally. Due to the presence of curvature, two remote frames separated by extended regions of curved space-time will not necessarily agree on measurements; they agree only on invariant quantities. For example, all observers agree on the length of a world line, since this is an invariant - even distant Schwarzschild observers agree that the world line of an in-falling particle is of finite length, since they are just two observers in the same space-time.

and it really really looks like Kruskal-Szekeres system of coordinates yields the same result as Schwarzchild coordinates in this matter anyway.
No. If you plot the trajectory of an infalling particle onto a Kruskal-Szekeres embedding diagram, it will remain of finite length and crosses the horizon - unlike in Schwarzschild coordinates, where you get an infinitely long curve asymptotically approaching the horizon. So they clearly don't yield the same result. And as KJW pointed out, you need to admit every possible system of coordinates, which means that Schwarzschild coordinates with their infinite in-fall time are in no way physically privileged.

t= infinity is the only condition under which those two equations would yield the same outcome.
Yes of course, that's the point. Locally at the far away observer's frame these notions of time coincide, but if you go away from that observer and towards the black hole, they do not. That is why the in-fall time in Kruskal-Szekeres coordinates is not the same as in Schwarzschild coordinates. If you want to actually calculate the coordinate in-fall time in the KS system itself, you have to perform the usual integration of radial coordinate velocity over radial position, and see what that yields. Or, as suggested above, just draw it onto an embedding diagram and see what happens.

55. It seems that our core disagreement is not about what happens, but what it means.

Let's look at a similar question in special relativity. Suppose an object with mass, like a space ship, were moving at the full speed of light from our perspective. Not 99.999999999999999999% of C, but actually 100% of C. How much deceleration sustained over how long a time would it take that object to slow down to the point where time no longer appeared to be standing still inside the space ship (from our perspective)?

The answer? Infinity. It would require infinity acceleration to get it down from 100% to 99.999999999999999999%

Because of that, it is impossible to make a meaningful transformation between those two perspectives. It's safe to say that all communication would be cut off forever between ourselves and that space ship. Since time appears to stand still on that space ship, it can't even emit EM signals to prove to us that it exists. We would be unable to see it.

Now, do you see how this relates to black holes? There is no way to make a meaningful transformation between the frames of the infalling observer and the outside observer after the infalling object reaches the Schwarzchild radius because infinity time will have passed on the outside.

We don't need to worry about making that transformation. And so long as we are only worried about making transformations across a period of time that can be meaningful to both observers (the infalling observer, and the outside observer), we only need to concern ourselves with the period of time within which the black hole has not yet formed.

The moment after the infalling object reaches the Schwarzchild radius may be meaningful to the infalling object, but it's a nonsense time to the outside observer (infinity plus 1?)

Originally Posted by Markus Hanke
Those signals travel along null geodesics, which become increasingly complicated as the emitter approaches the horizon. From just above the horizon, the signal would take a very long time to get back out, but that tells you nothing about how the emitter keeps freely falling, it only gives you an indication of the geometry of the outgoing null geodesics along which your signal travels. Do you see the distinction ? The outgoing signal does not travel along straight radial lines.
The point is it allows an outside observer to fix a date and time to the moment the object disappears forever (which it never does). Two observers in different frames may disagree both about duration, and simultaneity, however they both must agree that signals are received after they are emitted.

So if at some date/time "t" a distant observer sends a signal to the infalling object and the infalling object sends a signal back, then both the distant observer and the infalling object agree the signal was received after "t". Therefore both observers agree that the infalling object had not yet reached the Schwarzchild radius at" t".

Now, no matter how far in the outside observer's future you decide to set "t", this handshake will still be possible to create.

Yes of course, that's the point. Locally at the far away observer's frame these notions of time coincide, but if you go away from that observer and towards the black hole, they do not. That is why the in-fall time in Kruskal-Szekeres coordinates is not the same as in Schwarzschild coordinates. If you want to actually calculate the coordinate in-fall time in the KS system itself, you have to perform the usual integration of radial coordinate velocity over radial position, and see what that yields. Or, as suggested above, just draw it onto an embedding diagram and see what happens.
How can that be possible? If the event horizon is located at a point where t = infinity, then how would you get a result that t doesn't equal infinity?

Do you mean that you can get a finite falling time by continually updating your frame of reference as you go (using purely local time for the infalling object? That is not in dispute. Clearly an infalling observer observes only a short duration for the fall.

56. How much deceleration sustained over how long a time would it take that object to slow down to the point where time no longer appeared to be standing still inside the space ship (from our perspective)?
You see, this question is physically meaningless because a photon isn't a valid frame of reference at all. You cannot transform time and space measurements to or from the "frame of light", because no such frame exists.

Because of that, it is impossible to make a meaningful transformation between those two perspectives.
Indeed

There is no way to make a meaningful transformation between the frames of the infalling observer and the outside observer after the infalling object reaches the Schwarzchild radius because infinity time will have passed on the outside.
No, that's not right - going from Schwarzschild coordinates ( stationary far away ) to Gullstrand-Painleve coordinates ( freely falling into the black hole ) is a simple coordinate transformation. The reason is that both describe the same space-time, just from different perspectives. This is very different from the relationship between photons and massive particles.

And so long as we are only worried about making transformations across a period of time that can be meaningful to both observers (the infalling observer, and the outside observer)
The only concept of "time" that is meaningful to both observers - and on which they both agree - is the time that a clock which is co-moving with a shell of gravitationally collapsing matter ( or that is freely falling into the black hole later ) physically measures. All observers agree on this, because it is local proper time at the place the event takes place. That is the point I was trying to make all along.

The moment after the infalling object reaches the Schwarzchild radius may be meaningful to the infalling object, but it's a nonsense time to the outside observer (infinity plus 1?)
Exactly, Kojax. That is all I have been saying all along Both observers are right, but only in their own local frames. If we can agree on this - as it seems we do now - then all is fine.

Therefore both observers agree that the infalling object had not yet reached the Schwarzchild radius at" t".
That is true, but it doesn't tell you anything about whether or not the object has subsequently continued to fall through the horizon. In both the "frozen star" as well as standard scenario, all signals will eventually cease. A far-away observer cannot distinguish between these scenarios, but the in-falling observer can.

Now, no matter how far in the outside observer's future you decide to set "t", this handshake will still be possible to create.
Yes, of course ( practical challenges of actually doing this aside for now ), but because your outside frame shares no concept of simultaneity with the freely falling frame, this does not allow you to conclude that the black hole hasn't formed, in the sense that there's no true horizon. In fact one could argue the exact opposite - precisely because your signals behave in this way physically means that there must be an event horizon present at the Schwarzschild radius

How can that be possible? If the event horizon is located at a point where t = infinity, then how would you get a result that t doesn't equal infinity?
The same way as the poles on earth are coordinate singularities in long/lat coordinates, yet if you calculate the length of a geodesic crossing a pole, the result will be finite and well defined. The same thing in Schwarzschild coordinates - the horizon is a coordinate singularity, but if you calculate the length of a world line extending across it, the result is finite and well defined. This is possible because in both cases the local space is smooth and regular at the coordinate singularity.

Do you mean that you can get a finite falling time by continually updating your frame of reference as you go (using purely local time for the infalling object?
I am not really sure what you mean by "updating" - a free fall frame is simply one where there is no proper acceleration. This is true everywhere along the in-fall trajectory, in fact this is what defines the free fall geodesic. Such a geodesic is parametrised by the proper ( i.e. local ) time of the object, and every segment of the geodesic you pick out - regardless of whether that is above, below, or across the horizon - is of finite length.

57. Originally Posted by Markus Hanke

The only concept of "time" that is meaningful to both observers - and on which they both agree - is the time that a clock which is co-moving with a shell of gravitationally collapsing matter ( or that is freely falling into the black hole later ) physically measures. All observers agree on this, because it is local proper time at the place the event takes place. That is the point I was trying to make all along.
If you use comoving clocks in special or general relativity, then there is no need to bother making any transformations at all. Time dilation only applies to any situation when you use a non-local clock.

But that doesn't allow us to deal with situations where two objects that have two different clocks may interact. Those interactions require a transformation because it is not possible to use a comoving clock to complete them.

Exactly, Kojax. That is all I have been saying all along Both observers are right, but only in their own local frames. If we can agree on this - as it seems we do now - then all is fine.

What useful information do we gain by taking the infalling observer's perspective? Until they reach the interior of the Schwarzchild radius, it is not impossible to retrieve them from their present location. In the course of infinite time, all possibilities become certainties. Sooner or later, after an infinite number of die rolls, nature will eventually roll a "1", and retrieve the object before it can cross the Schwarzchild radius.

The existence of the outside universe, and possibility (however improbable) of being retrieved, guarantees that no object will ever get the chance to fall all the way in.

That is true, but it doesn't tell you anything about whether or not the object has subsequently continued to fall through the horizon. In both the "frozen star" as well as standard scenario, all signals will eventually cease. A far-away observer cannot distinguish between these scenarios, but the in-falling observer can.
The faraway observer can simply repeat the experiment, if they want to know. It is true that they will only know what was happening D/C seconds ago, but well...... that is true of absolutely all observations of absolutely all systems.

Look up in the sky. You're not seeing where the Sun is now. You're seeing where it was just over 8 minutes ago.

Yes, of course ( practical challenges of actually doing this aside for now ), but because your outside frame shares no concept of simultaneity with the freely falling frame, this does not allow you to conclude that the black hole hasn't formed, in the sense that there's no true horizon. In fact one could argue the exact opposite - precisely because your signals behave in this way physically means that there must be an event horizon present at the Schwarzschild radius
It shares the same concept of simultaneity that all systems share. So long as a beam of EM radiation can make the trip to, and then return from, a distant object in finite time, relativity of simultaneity applies.

The same way as the poles on earth are coordinate singularities in long/lat coordinates, yet if you calculate the length of a geodesic crossing a pole, the result will be finite and well defined. The same thing in Schwarzschild coordinates - the horizon is a coordinate singularity, but if you calculate the length of a world line extending across it, the result is finite and well defined. This is possible because in both cases the local space is smooth and regular at the coordinate singularity.
It applies because we are including a self reference in our measurements. We scale our notion of "time" to match the very thing we wish to measure.

That's no different than scaling the length of your meter stick to measure a long distance, such as the distance between planet Earth and Proxima Centauri. If you lengthen your meter stick so that it's .04 light years long, you can argue that Proxima Centauri is only 100 meters away.

And in reality, an object traveling very near the speed of light might (due to Lorentz contraction) find the distance to be just that short. We could then tell ourselves that, because the object perceives itself (according to its local clock) to travel the distance in 3 millionths of a second, that therefore Faster than Light travel is possible!!!

It only takes 4 years to get there if you use a clock on Earth to measure it. A clock on board the space ship measure 3 millionths of a second (to travel 100 meters at very nearly light speed). Do you see where this kind of logic leads?

58. If you use comoving clocks in special or general relativity, then there is no need to bother making any transformations at all. Time dilation only applies to any situation when you use a non-local clock.
Exactly.

But that doesn't allow us to deal with situations where two objects that have two different clocks may interact.
You can only meaningfully compare clocks which are together at rest in the same frame. What you could do is let a clock fall along a trajectory which brings it very close to the event horizon, and slingshots it back out to re-unite with the far away clock; this type of orbit exists.

What useful information do we gain by taking the infalling observer's perspective?
We are able to locally distinguish between standard GR and the "frozen star" idea, but only in the in-falling frame. Whether this information is deemed "useful" or not is a different matter, since it cannot be communicated.
I don't really understand the rest of the comment, to be honest.

The faraway observer can simply repeat the experiment, if they want to know.
Such on observer will just find that the return signal takes longer and longer to reach him, but he can't distinguish the reasons for that, because that happens the exact same way in both scenarios.

So long as a beam of EM radiation can make the trip to, and then return from, a distant object in finite time, relativity of simultaneity applies.
No, relativity of simultaneity applies only to inertial frames in flat Minkowski space, not extended regions of curved space-time.

It applies because we are including a self reference in our measurements.
I'm afraid I fail to see the connection between your comment and the original post, so I don't understand what you are trying to say here.

59. Originally Posted by Markus Hanke

You can only meaningfully compare clocks which are together at rest in the same frame. What you could do is let a clock fall along a trajectory which brings it very close to the event horizon, and slingshots it back out to re-unite with the far away clock; this type of orbit exists.
That's false.

Comparing clocks between two objects in different frames can be very useful, and meaningful.

The GPS satellites are in a different frame from the consumers on Earth who use GPS, but we still communicate with them and specifically communicate with their clocks. Those clocks are basically the core of the whole system. But they are in a different frame, so we use a transformation.

Just because something is in a different frame doesn't mean it is in a different physical system altogether. Even in a closed system, there could be two or more objects in different frames, interacting as part of that system.

An infalling observer near a black hole can still be thought of as a part of the physical system of a distant observer. So long as there is interaction, there is meaning.

Such on observer will just find that the return signal takes longer and longer to reach him, but he can't distinguish the reasons for that, because that happens the exact same way in both scenarios.
Actually no. The infalling observer won't find the signal taking longer and longer to reach him. He will in instead find the times to be shorter and shorter.

As his frame slows down, all other frames appear to speed up. The overall distance that the EM must travel is only increasing by trivially small amounts. However for most (or actually, all) of the distance between the infalling observer and the distant observer, the EM will be in a frame where time is faster than in the infalling observer's frame. The round trips will appear to take less and less time.

The distant observer, on the other hand, will see the EM taking longer and longer times to go back and forth. But that's ok, because the distant observer has literally infinity years to wait, before the infalling observer will become unavailable to communicate with.

No, relativity of simultaneity applies only to inertial frames in flat Minkowski space, not extended regions of curved space-time.
What reason is there not not apply it to those situations? Either way you still cannot become aware of something until an EM signal makes you aware.

Nearly all interaction is based on EM, so the whole basis to describe how any two objects interact has to be based mostly on how they trade EM signals.

60. Comparing clocks between two objects in different frames can be very useful, and meaningful.
Ok, I grant you that my choice of terminology ( "meaningfully" ) was unfortunate, so you are right with your GPS example.
What I was trying to point out is merely the fact that if two clocks are distant, they do not share the same concept of simultaneity - so in effect, you don't really "compare clocks", but rather you compare a clock against a signal you received from another distant clock. In a curved space-time this becomes a tricky endeavour since the signal itself propagates at finite speed along non-trivial geodesics, so you need to take this into account. This is why GPS clocks need to be carefully calibrated even prior to launch. To eliminate the effects of non-Euclidean null geodesics, you would have to bring the clocks sufficiently close together so that they share a notion of simultaneity.

Do you know what I mean by this distinction ?

The infalling observer won't find the signal taking longer and longer to reach him. He will in instead find the times to be shorter and shorter.
I was referring to the distant Schwarzschild observer, who is the receiver of the return signals sent by the in-falling one.

What reason is there not not apply it to those situations?
It applies only to Minkowski space-time.

Either way you still cannot become aware of something until an EM signal makes you aware.
Yes. Just remember that said signal is propagating along null geodesics, which are increasingly non-trivial as you approach the event horizon.

61. I'm trying to figure out what the impasse is in our two perspectives.

My best guess right now is that it comes down to our definition of interaction and what a system of interactions is. If two objects are in very different frames, but still interacting, then I would define them as being part of the same system of interactions.

But how does that system change if we include time dilation? Or if the EM they are using in order to complete those interactions begins traveling down curved paths or geodesics? I don't think the fact EM is traveling a non-linear path means we can treat them as two separate systems. They still have to be treated as one system because of the interaction. An interaction that crosses multiple frames must still be analyzed as a single interaction.

From an interaction standpoint, the fact a black hole is never observed to form means it doesn't form. So long as it is possible to send a signal to the NFBH, and get a signal back, then we are talking about complete interactions here. Not partial interactions. We're not just watching the black hole and seeing it not form. We are truly observing that it has not formed.

Originally Posted by Markus Hanke
Ok, I grant you that my choice of terminology ( "meaningfully" ) was unfortunate, so you are right with your GPS example.
What I was trying to point out is merely the fact that if two clocks are distant, they do not share the same concept of simultaneity - so in effect, you don't really "compare clocks", but rather you compare a clock against a signal you received from another distant clock. In a curved space-time this becomes a tricky endeavour since the signal itself propagates at finite speed along non-trivial geodesics, so you need to take this into account. This is why GPS clocks need to be carefully calibrated even prior to launch. To eliminate the effects of non-Euclidean null geodesics, you would have to bring the clocks sufficiently close together so that they share a notion of simultaneity.

Do you know what I mean by this distinction ?
Well,.... I kind of understand what you mean.

But an infalling object near a NFBH and the distant observer were originally in the same frame before the infalling observer approached the black hold (or presumably not in very different frames.)

If we look at a less extreme example, perhaps an observer in a near orbit around a large Neutron star, and a distant observer, we could imagine that the two were constantly exchanging messages, one after another. The curvature of space time between them doesn't stop the messages from getting through, and time is still progressing in both frames (at different speeds).

I was referring to the distant Schwarzschild observer, who is the receiver of the return signals sent by the in-falling one.
Yeah. For the distant observer, the round trips appear to take longer and longer, but never an infinite amount of time.

Although in all practicality the EM coming back will get more and more redshifted, and harder to detect. And also the infalling object needs to be able to receive and respond to the message quickly (because when measured from its own frame, these messages are coming very fast.) But these are practical concerns, not theoretical concerns.

It applies only to Minkowski space-time.
But the basis for relativity of simultaneity still applies, and still applies perfectly. The basis being that objects primarily interact by exchange of EM signals. So for the purposes of interacting, "now" is whenever the EM arrives.

Describing how objects perceive time outside of the limits of an interaction is pointless. "Time" is only a meaningful concept if you are interacting.

If you were cut off all by yourself, then it wouldn't matter how long anything takes, because there is nothing else to synchronize with. But as long as any two objects are still interacting, time still matters. In the case of the NFBH and distant observer, the transformation is clearly not linear. Instead of time passing at one linear rate for one and time passing at another linear rate for the other, time is passing at an accelerated rate (or decelerated, depending on which side of the interaction you prefer to observe the interaction from.)

---------------------------------

At the risk of over-reaching, I'd like to analyze another aspect of the NFBH concept. The narrowing of the angle of approach and departure for light. As our infalling object approaches the Schwarzchild radius, getting further and further inside of the Photon Sphere, it seems its view outward into space becomes narrower and narrower. In order to continue to communicate with it, our distant observer needs to be located within the "field of view" of the infalling observer.

So there is a mechanism to compensate against the possibility of the infalling observer perceiving itself to get bombarded by an infinite amount of EM as it falls through the last bit of its descent. It may witness the "end of the universe", but it will only witness a very small area of that universe. When its view of the progression of time outside becomes infinity fast, its view of space outside will become infinity small.

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