Black hole event horizon confusion

[mattjbr2]

Leonard Susskind said "everything that ever fell in, to make the black hole, [..] [is] all contained in [...] progressively thinner and thinner shells that approach the horizon asymptotically, never quite getting there" and from the perspective of someone outside the black hole "a shell, called a horizon, inside of which there is nothing that will ever be important to us, because from our perspective nothing ever fell through."
(says it at 10:29 ).

(Let its Schwarzschild radius = Rs) (The following is entirely from the perspective of an outside observer.)

Based on my logic: matter collapses until its radius appears to slow down as it approaches Rs, never actually reaching exactly Rs. Therefore (as time approaches infinity) all the matter will be apparently frozen in a ball with a radius infinitesimally larger than its own Schwarzschild radius and hence no "singularity" (tiny point of infinite density) will ever form anywhere in the universe (from our perspective). So every black hole we ever observe will just be balls of matter with radius= their Rs. Because from our perspective, the ball of matter cannot continue collapsing beyond its own Rs (let alone ever reaching it) because of time dilation.
So, the matter from the supermassive star is still there, it's just now a ball asymptotically approaching a radius equal to its Rs.
But, Prof. Susskind said that from our perspective everything is outside the event horizon with nothing that will ever be important to us inside the event horizon, because from our perspective nothing ever fell through. If we follow his logic, then: as the supermassive star collapses towards its Rs, a sphere of nothingness will grow in the centre of the star, pushing matter away from the centre and eventually out of the event horizon, leaving nothing inside and a shell on the outside. OR: as the supermassive star collapses towards its Rs, everything within its volume just ceases to exist..?
He also says it's something like a hollow sphere with microphysics on the outside (around 35mins in video)
Am I understanding him incorrectly? What's wrong with my picture of the collapse?

 

[.Scott]

The "sphere of nothingness" does not push. Quite the opposite, it pulls to the extreme.
Your "ceasing to exist" is closer, but far from correct.

In the situation where Alice jumps in and Bob views from afar, Alice cruises through the event horizon without issue. Bob sees Alice approach the event horizon, but as she does her time slows down. So Bob never sees Alice cross the event horizon. She simply gets closer and closer without ever crossing while her watch gets closer and closer to the time she saw that she crossed.

Actually, there are technical problems that might keep Bob from being able to track Alice during her approach, so it may be better to say "Bob would see" as oppose to "Bob sees".

That aside, as Alice approaches the event horizon, her time slows so much that the uncertainty principle becomes significant - and Alice's position become less and less certain to Bob. From Bob's perspective, Alice spreads out across the event horizon sphere as a type of interference pattern.

I just watched the video. Susskind never suggests that these "shells" remain intact as additional shells (or "balls") are formed. They do not. This is described by him at 56:30.

If we let Charlie jump in after Alice, he will also form a sphere - but he will also increase the total mass of the black hole - thus causing the event horizon to move outward past Alice's sphere. The result would certainly not be two concentric spheres. It would instead be a single sphere that was some combination of the interference patterns of Alice and Charlie.

 

[Ilythiiri]

The result would certainly not be two concentric spheres. It would instead be a single sphere that was some combination of the interference patterns of Alice and Charlie.

3D object party in a 2D surface? Like a hologram? (:

 

[mattjbr2]

The "sphere of nothingness" does not push. Quite the opposite, it pulls to the extreme.
Your "ceasing to exist" is closer, but far from correct.

In the situation where Alice jumps in and Bob views from afar, Alice cruises through the event horizon without issue. Bob sees Alice approach the event horizon, but as she does her time slows down. So Bob never sees Alice cross the event horizon. She simply gets closer and closer without ever crossing while her watch gets closer and closer to the time she saw that she crossed.

Actually, there are technical problems that might keep Bob from being able to track Alice during her approach, so it may be better to say "Bob would see" as oppose to "Bob sees".

That aside, as Alice approaches the event horizon, her time slows so much that the uncertainty principle becomes significant - and Alice's position become less and less certain to Bob. From Bob's perspective, Alice spreads out across the event horizon sphere as a type of interference pattern.

I just watched the video. Susskind never suggests that these "shells" remain intact as additional shells (or "balls") are formed. They do not. This is described by him at 56:30.

If we let Charlie jump in after Alice, he will also form a sphere - but he will also increase the total mass of the black hole - thus causing the event horizon to move outward past Alice's sphere. The result would certainly not be two concentric spheres. It would instead be a single sphere that was some combination of the interference patterns of Alice and Charlie.

Thanks for taking the time to respond. I don't think you understood my question, as I'm already aware of what you described. I'll rephrase it: as a supermassive star collapses, its warping of spacetime increases as its radius decreases. This leads to time running slower for it, from our perspective. As its radius nears its Schwarzschild radius, its warping of spacetime becomes such that time appears to slow down asymptotically and the star never compresses to its Schwarzschild radius. It just stays frozen for all eternity. So, from our perspective, it's a frozen ball of matter. I quoted Susskind as saying that there's nothing inside a black hole, that everything that ever fell into it to form the black hole is laying in a shell around the event horizon. "everything that ever fell in, to make the black hole, [..] [is] all contained in [...] progressively thinner and thinner shells that approach the horizon asymptotically, never quite getting there" and from the perspective of someone outside the black hole "a shell, called a horizon, inside of which there is nothing that will ever be important to us, because from our perspective nothing ever fell through.". The entire mass of the star must be inside the event horizon, frozen in time. It just has to be, if you slowly simulate the collapse of a supermassive star in your head and think about where each lump of matter is going. Is this compatible with what he's saying? It doesn't sound like it

 

[.Scott]

Once the star collapses, it will approach the event horizon quite rapidly.
The time scale would vary based on the size of the star, but certainly within a few hours it would no longer look like a star. By that time all of its mass that is destined for the event horizon would be frozen in time and within an Angstrom of the event horizon. At that point, it is essentially at the event horizon - and it is no longer recognizable as regular matter. Instead, it has characteristics of a hologram. You can say that it's a frozen ball of matter. As the matter approaches the event horizon, it would "appear" to be a progressively thinner and thinner shell (singular, not shells). But that is only during the initial approach to the horizon (as seen by Alice). In short order, that thin shell will combine with other shells at the event horizon - as its mass causes the black hole to swell and overtake the older "thin shells".

From our perspective, the entire mass of the black hole is at the event horizon - or within a few Planck's distances of it.

 

[mattjbr2]

Once the star collapses, it will approach the event horizon quite rapidly.
The time scale would vary based on the size of the star, but certainly within a few hours it would no longer look like a star. By that time all of its mass that is destined for the event horizon would be frozen in time and within an Angstrom of the event horizon. At that point, it is essentially at the event horizon - and it is no longer recognizable as regular matter. Instead, it has characteristics of a hologram. You can say that it's a frozen ball of matter. As the matter approaches the event horizon, it would "appear" to be a progressively thinner and thinner shell (singular, not shells). But that is only during the initial approach to the horizon (as seen by Alice). In short order, that thin shell will combine with other shells at the event horizon - as its mass causes the black hole to swell and overtake the older "thin shells".

From our perspective, the entire mass of the black hole is at the event horizon - or within a few Planck's distances of it.

I'm still not getting understood. I'll rephrase it again. Imagine a star with a Schwartzschild radius of 50km is collapsing. At the instant where its radius is 55km it still hasn't formed a black hole anywhere as it has yet to reach 50km, so there are still lumps of matter within that star that a sufficiently advanced lifeform would detect. At that instant, imagine a lump of matter that is 25km from the centre of the star. Let that matter = B. Let matter at 55km (surface of star) = A. Keep imagining that lump of matter, B, and what happens to it as we move forward in time in slow motion. As A gets closer to the centre, distortion of spacetime everywhere increases and time slows down even more. As A gets smaller than 55km, B gets smaller than 25km, but time is running much slower at B than it is at A, so B gets smaller to a much mess extent. As A approaches 50km, its time goes slower and slower, while B's time goes even more slower. Eventually time goes so slow that it appears to take infinite time to reach 50km. B is still there slightly below 25km, but still there frozen in time more than A is. What you and Susskind are saying is that as A approaches 50km, B somehow rises from 25km to 50km. I'm saying that B is slightly below 25km. If B was there before the black hole was formed, why would it not be where it was after the black hole forms? Hence all the mass of the star is frozen in a dense ball of matter with a radius equal to slightly larger than 50km and asymptotically reaching that radius. Any infalling matter would take forever to reach A, as A is taking forever to reach 50km itself. And B is still asymptotically getting smaller than 25km but we can't detect it because nothing can escape where A is located. People always only think about the black while after it has formed and never ask what happened moments before it formed and never ask where that matter went (from our perspective). From the perspective of A or B, they violently get squished into a singularity. But we, on the outside never have the time to observe this singularity forming as it takes forever to form. Therefore there are no singularities in the universe as A has never had time to reach 50km (let alone zero radius). So all black holes in the universe are frozen (time, not temperature) shrinking balls of matter with r>Rs. A and B are still there near their original locations. We just can't detect them because of how impossibly deep the gravity well is at 50km. Whats wrong with my logic? Tell me what happens to B, well before A reaches 50km, after it gets close to it, and after it approaches it.

 

[JLowe]

I'm still not getting understood. I'll rephrase it again.

I'm saying that B is slightly below 25km. If B was there before the black hole was formed, why would it not be where it was after the black hole forms? .

I don't understand you either(but I'm no expert so it's all good)

It is my understanding however, and it may be flawed, but the black hole starts at the center where matter is densest, and the event horizon would begin at the core and work it's way outwards as the rest of the star fell into it. At that point, you're back to the story of Alice and Charlie's interference patterns on the horizon immediately outside the core, and the horizon expanding past the point their photons were last seen.

My understanding may be flawed however, because I'm no expert like I said.

 

[Nugatory]

Hence all the mass of the star is frozen in a dense ball of matter with a radius equal to slightly larger than 50km and asymptotically reaching that radius. Any infalling matter would take forever to reach A, as A is taking forever to reach 50km itself. And B is still asymptotically getting smaller than 25km but we can't detect it because nothing can escape where A is located. People always only think about the black while after it has formed and never ask what happened moments before it formed and never ask where that matter went (from our perspective).

You are trying to use the Schwarzschild solution, which describes a black hole that has already formed, to analyze the behavior of the infalling matter before the black hole has formed. That won't work because the black hole hasn't yet formed so the Schwarzschild solution doesn't apply. Instead you have to use something like the Oppenheimer-Snyder solution (google will find some good links) to analyze the collapse - and that will show that the infalling matter does end up inside the horizon when the black hole forms.

 

[PeterDonis]

Leonard Susskind said "everything that ever fell in, to make the black hole, [..] [is] all contained in [...] progressively thinner and thinner shells that approach the horizon asymptotically, never quite getting there" and from the perspective of someone outside the black hole "a shell, called a horizon, inside of which there is nothing that will ever be important to us, because from our perspective nothing ever fell through."

Oh, my goodness. Another example of a physicist using highly misleading ordinary language. What he's describing is first of all coordinate-dependent, and second of all fails to distinguish between coordinate-dependent quantities and invariants. I'm very disappointed to see this in what is supposed to be a video teaching actual science.

Am I understanding him incorrectly?

Yes, but I think it's really his fault for using highly misleading language. See above.

It is my understanding however, and it may be flawed, but the black hole starts at the center where matter is densest

That's not quite how it works. Here is how I would describe the Oppenheimer-Snyder solution that @Nugatory referred to, which is the simplest idealized model we have of gravitational collapse:

An object of uniform density, at some instant of time, is occupying a spherically symmetric region with a finite surface area (surface area is a better way to think of it than radius, because space in these coordinates is non-Euclidean and so the relationship between surface area and radius is not what we are intuitively used to, and surface area is a better measure of our intuitive concept of "how much space the object occupies"), and with all of its parts at rest relative to each other. The interior of the object is homogeneous--that is, every interior point is the same (same density, same local spacetime geometry) at a given instant of time. Here "time" means proper time according to an observer riding along with any small piece of the object--since every spatial point is the same inside the object, all such observers at every interior point can synchronize their clocks, and they will then stay synchronized all through the collapse.

The object then starts to collapse under its own gravity. Each interior observer, riding along with a small piece of the object, sees the density in his vicinity increasing with time. In a finite time after the collapse starts, the density goes to infinity, and it does so at the same instant of time everywhere inside the object. That instant of time is the singularity (which means that the singularity is a moment of time, not a place in space).

To see how the event horizon forms and expands, consider the observer riding along with the piece of matter at the exact center of the collapsing object. Suppose that this observer emits light signals radially outward. At the start of the collapse, these light signals are not affected much by the object's gravity; they reach the surface of the object and escape out to infinity with only a slight delay. But as the collapse proceeds, the object's density increases and the light signals get more and more delayed escaping. Eventually, there will be a light signal that reaches the surface of the object at the exact instant that the surface area is exactly $16 \pi G^2 M^2/ c^4$, where $M$ is the mass of the object (as measured from far away)--i.e., the surface area of the horizon of a black hole with the same mass as the object. This light signal will never escape to infinity; it will remain stuck on a 2-sphere with that same surface area forever. In fact, that light signal lies on the event horizon, and the instant of its reaching the object's surface marks the instant at which the object has collapsed within its own event horizon and the black hole is fully formed. But the event horizon actually starts at the instant when that light signal was emitted from the center of the object; in fact the set of all light signals emitted radially outward, in all possible directions, from that event at the center of the object is the event horizon (each individual light signal is called a "generator" of the horizon).

Note carefully that the horizon does not form because the object is denser at the center; the object is of uniform density throughout. The event horizon does not really "form" at all, in the sense of something locally causing it to form. The definition of the event horizon is global: it is the set of radially outgoing light signals that just barely fail to escape to infinity. But "escape to infinity" is a global concept; to know for sure which light signals just barely fail to escape to infinity, you need to know the entire future of the spacetime. You can't tell for sure, locally, whether a given light signal will escape or not.

 

[mattjbr2]

Oh, my goodness. Another example of a physicist using highly misleading ordinary language. What he's describing is first of all coordinate-dependent, and second of all fails to distinguish between coordinate-dependent quantities and invariants. I'm very disappointed to see this in what is supposed to be a video teaching actual science.
Yes, but I think it's really his fault for using highly misleading language. See above.
That's not quite how it works. Here is how I would describe the Oppenheimer-Snyder solution that @Nugatory referred to, which is the simplest idealized model we have of gravitational collapse:

An object of uniform density, at some instant of time, is occupying a spherically symmetric region with a finite surface area (surface area is a better way to think of it than radius, because space in these coordinates is non-Euclidean and so the relationship between surface area and radius is not what we are intuitively used to, and surface area is a better measure of our intuitive concept of "how much space the object occupies"), and with all of its parts at rest relative to each other. The interior of the object is homogeneous--that is, every interior point is the same (same density, same local spacetime geometry) at a given instant of time. Here "time" means proper time according to an observer riding along with any small piece of the object--since every spatial point is the same inside the object, all such observers at every interior point can synchronize their clocks, and they will then stay synchronized all through the collapse.

The object then starts to collapse under its own gravity. Each interior observer, riding along with a small piece of the object, sees the density in his vicinity increasing with time. In a finite time after the collapse starts, the density goes to infinity, and it does so at the same instant of time everywhere inside the object. That instant of time is the singularity (which means that the singularity is a moment of time, not a place in space).

To see how the event horizon forms and expands, consider the observer riding along with the piece of matter at the exact center of the collapsing object. Suppose that this observer emits light signals radially outward. At the start of the collapse, these light signals are not affected much by the object's gravity; they reach the surface of the object and escape out to infinity with only a slight delay. But as the collapse proceeds, the object's density increases and the light signals get more and more delayed escaping. Eventually, there will be a light signal that reaches the surface of the object at the exact instant that the surface area is exactly $16 \pi G^2 M^2/ c^4$, where $M$ is the mass of the object (as measured from far away)--i.e., the surface area of the horizon of a black hole with the same mass as the object. This light signal will never escape to infinity; it will remain stuck on a 2-sphere with that same surface area forever. In fact, that light signal lies on the event horizon, and the instant of its reaching the object's surface marks the instant at which the object has collapsed within its own event horizon and the black hole is fully formed. But the event horizon actually starts at the instant when that light signal was emitted from the center of the object; in fact the set of all light signals emitted radially outward, in all possible directions, from that event at the center of the object is the event horizon (each individual light signal is called a "generator" of the horizon).

Note carefully that the horizon does not form because the object is denser at the center; the object is of uniform density throughout. The event horizon does not really "form" at all, in the sense of something locally causing it to form. The definition of the event horizon is global: it is the set of radially outgoing light signals that just barely fail to escape to infinity. But "escape to infinity" is a global concept; to know for sure which light signals just barely fail to escape to infinity, you need to know the entire future of the spacetime. You can't tell for sure, locally, whether a given light signal will escape or not.

This is the first time i heard of a "horizon generator". Looking it up lead me to Christopher M Hirata's 2012 Caltech lecture on the nature of horizons. I've seen countless youtube videos on black holes and no one ever talks about this. Thank you!

 

[Dale]

What he's describing is first of all coordinate-dependent, and second of all fails to distinguish between coordinate-dependent quantities and invariants.

I agree. Both the lecture and this thread are placing way too much importance on a set of coordinates which are known to be bad in the region under discussion.

 

[martinbn]

TIve seen countless youtube videos on black holes and no one ever talks about this.

May be the conclusion is that it is better to read reliable texts instead of watching youtube.

 

[JLowe]

May be the conclusion is that it is better to read reliable texts instead of watching youtube.

Can you recommend one?

 

[PeterDonis]

This is the first time i heard of a "horizon generator".

You won't find that term used except in fairly technical discussions, but I think it's a useful way of imagining the horizon.

 

[Nugatory]

Can you recommend one?

For special relativity, it's hard to beat Taylor and Wheeler's "Spacetime Physics".
For general relativity, Sean Carrol's online text: https://www.preposterousuniverse.com/grnotes/
Ben Crowell has some good online stuff.
Our own PeterDonis has a series of insights articles on the Schwarzschild metric.

 

[JLowe]

For special relativity, it's hard to beat Taylor and Wheeler's "Spacetime Physics".
For general relativity, Sean Carrol's online text: https://www.preposterousuniverse.com/grnotes/
Ben Crowell has some good online stuff.
Our own PeterDonis has a series of insights articles on the Schwarzschild metric.

Thanks, I'll check them out.

 

[Yukterez]

But, Prof. Susskind said that from our perspective everything is outside the event horizon with nothing that will ever be important to us inside the event horizon, because from our perspective nothing ever fell through. If we follow his logic, then: as the supermassive star collapses towards its Rs, a sphere of nothingness will grow in the centre of the star, pushing matter away from the centre and eventually out of the event horizon, leaving nothing inside and a shell on the outside.

That's not what Susskind meant. The hollow shell forms if you construct your black hole the way he did, which was the Kugelblitz-Method. If you have a collapsing star you have to add up all the shells of matter, then every shell will always stay larger than it's horizon size, and the radius of the encompassing shell larger than the two shell's combined horizon radius (and so on). You can integrate all the shells (see the Shell theorem), which will give you a density which is always an infinitesimal higher than the critical density. So you are right, in the frame of an external observer the horizon never forms, the radius just converges but never really gets there. The horizon forms only in the frame of an infalling observer.

 

[PeterDonis]

The hollow shell forms if you construct your black hole the way he did, which was the Kugelblitz-Method.

Is there an actual valid reference (textbook or peer-reviewed paper) for this? As has already been commented in this thread, YouTube videos are not good sources.

If you have a collapsing star you have to add up all the shells of matter, then every shell will always stay larger than it's horizon size

This is false in the Oppenheimer-Snyder model, and I don't see how it could be true in any model of collapse to a black hole. I think you, like Susskind, are mistaking coordinate-dependent quantities for invariants.

 

[Yukterez]

I think you, like Susskind, are mistaking coordinate-dependent quantities for invariants.

It's okay for you to think you know more about relativity than me, but do you really think you know more about it than the famous Leonard Susskind himself?

 

[Nugatory]

It's okay for you to think you know more about relativity than me, but do you really think you know more about it than the famous Leonard Susskind himself?

The question isn't whether we know more than Susskind (I don't, and PeterDonis can speak for himself), it is whether we can explain it to non-specialists less misleadingly than Susskind.

There is no doubt that Susskind understands the difference between coordinate-dependent quantities and invariants. But in the video that started this thread he chose to gloss over that distinction, and it just so happens that that distinction is critical to resolving the original poster's question.

 

[Nugatory]

So you are right, in the frame of an external observer the horizon never forms, the radius just converges but never really gets there. The horizon forms only in the frame of an infalling observer.

That cannot be right, because the presence or absence of an event horizon is a property of the spacetime geometry. If it is present in one "frame" it must be present in all of them. A correct statement is that there is no point on the worldline of an external observer than can be said to correspond to the moment when the horizon forms, and that there is a point on the worldine of the infalling matter beyond which no light signal will reach the external observer. But that doesn;t mean it didn't happen, it means that the external observer never sees it happen.

 

[Yukterez]

A correct statement is that there is no point on the worldline of an external observer than can be said to correspond to the moment when the horizon forms

That is true.

and that there is a point on the worldine of the infalling matter beyond which no light signal will reach the external observer. But that doesn;t mean it didn't happen, it means that the external observer never sees it happen.

That sounds contradictory, since it seems to imply that it was only a matter of what one sees (that is a common misconception), while it really is a question of what happens in which frame of reference. In the frame of reference of the external coordinate bookkeeper you not only don't see the infalling observer crossing the horizon, he really never crosses it until an infinite amount of your own coordinate time has passed (which is never), regardless of his own proper time beeing finite.

 

[Yukterez]

I think you, like Susskind, are mistaking coordinate-dependent quantities for invariants.

The coordinate time of the external bookkeeper is also an invariant, since it is the proper time of a field free stationary observer (stationary with respect to the fixed stars). Therefore we can surely say that in his frame of reference the matter never gets compressed beyond it's critical radius.

 

[PeterDonis]

it really is a question of what happens in which frame of reference

Things don't happen "in" frames of reference. Frames of reference are human artifacts constructed to describe what happens, but what happens does not depend on which frame we choose. The fact that a particular frame of reference cannot describe a particular thing that happens, does not mean that the thing does not happen, or does not happen "in" that frame. It just means that frame can't describe all of what happens.

he really never crosses it until an infinite amount of your own coordinate time has passed (which is never)

This is not correct, because there is no invariant way to assign "bookkeeper coordinate times" to events on the infalling observer's worldline. See below.

The coordinate time of the external bookkeeper is also an invariant

The coordinate time of the external bookkeeper is an invariant along his own worldline. It is not an invariant throughout spacetime. In particular, it is not an invariant at events near, at, or below the horizon of the black hole. The only way to assign external bookkeeper coordinate times to any events off the bookkeeper's worldline is to choose a coordinate chart.

 

[PAllen]

Let’s consider any world line remaining very far from a BH that form from collapse of a body. The following are all true, per GR:

1) There is a an event on on the world line marking the transition from the event horizon being all in the causal future to not all being in the causal future. For any event on the world line after this, it is permissible to consider some part of the horizon to be in the present.

2) If one insists on a convention that geodesics 4 orthogonal to a world line define its present, then it is true that for an eternal static world line the horizon is not in its present. However, as soon as you are past the bounding event described in (1), for any event on the stationary world line, there is a colocated observer moving towards the BH such that at that moment, the horizon is in their present by the criterion of 4 orthogonal geodesic.

Unless one rules out the validity of natural simultaneity for distant observers approaching the BH, statements like the horizon is not in the present of distant observers is utter nonsense.

 

[Yukterez]

There is no doubt that Susskind understands the difference between coordinate-dependent quantities and invariants. But in the video that started this thread he chose to gloss over that distinction

That's not true, at t=34m41s he explicity says that it takes a finite proper time for the infalling observer to cross the horizon. That is no contradiction to the fact that in the bookeeper's frame of reference he stays outside of the horizon eternally. Susskind is completelly right with everything he says.

 

[Orodruin]

That is no contradiction to the fact that in the bookeeper's frame of reference he stays outside of the horizon eternally.

Again, without specifying a global coordinate chart - with some simultaneity convention - there is no way to claim this. The ”bookeeper reference frame” you are imagining is only well defined locally for the bookeeper without a global simultaneity convention. You are trying to claim a global extension of a local observable. There is no unique way of doing so. The event horizon is a global phenomenon, you just cannot use a local only description to deal with it. I suggest you read #25 more carefully.

 

[PeterDonis]

That is no contradiction to the fact that in the bookeeper's frame of reference he stays outside of the horizon eternally.

Only if you interpret that claim correctly. The correct interpretation is: the bookkeeper's frame of reference (which really means "the Schwarzschild coordinate chart on the region exterior to the horizon") only covers a restricted portion of the spacetime, and that portion does not contain the horizon or any events on or beneath it.

But that interpretation does not justify the further inferences that the OP was making from Susskind's presentation. In other words, Susskind's presentation misled the OP into an incorrect understanding of how gravitational collapse works. Which is why and others are criticizing Susskind's presentation from the standpoint of pedagogy.

 

[Yukterez]

You are trying to claim a global extension of a local observable.

Why this, I didn't claim "the black hole doesn't form in any reference frame", I only said "the black hole doesn't form in the reference frame of the stationary observer, but it does so in the frame of the infalling observer".

 

[Orodruin]

Why this, I didn't claim "the black hole doesn't form in any reference frame", I only said "the black hole doesn't form in the reference frame of the stationary observer, but it does so in the frame of the infalling observer".

Because, if you had read what you have already been told, in particular post #25, then you should understand that ”the stationary observer’s frame” is ambiguous and not uniquely defined away from that observer’s world line. Depending on the type of simultaneity convention that you choose, you will end up with different conclusions, which makes your claim coordinate dependent even if you try to refer to the ”frame of a stationary observer”.

In general you should feel very uneasy in GR whenever you find yourself referring to ”in the frame of X” as such statements will typically be coordinate dependent.

 

[Yukterez]

Depending on the type of simultaneity convention that you choose, you will end up with different conclusions

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity in which nothing falls through. Whenever you ask the external observer "where is the test-particle now?" (and this question has nothing to do with "where do you see the test-particle right now") he will always answear "just above the horizon".

For example see the trajectory of a test particle, which in it's own frame of reference falls through the horizons in a finite proper time (on the display τ propr):

while in the frame of an external observer, it freezes at the horizon and corotates there forever with the black hole, the clocks frozen short before the proper time when it penetrates the horizon:

 

[PAllen]

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity in which nothing falls through. Whenever you ask the external observer "where is the test-particle now?" (and this question has nothing to do with "where do you see the test-particle right now") he will always answear "just above the horizon".

Now is a convention for any observer, even in SR. The only psysical distinction is between causal future, causal past, on possible present i.e. spacelike separation. For any distant observer, there arises a time when part of the history of the horizon is possibly now.

Further, even insisting on the convention of 4 orthogonal geodesic definition of now, consider a distant observer oscillating radially. Then, when heading toward the BH, the horizon exists now.

 

[PAllen]

And why shouldn’t a distant observer simply use Gaussian normal coordinates, which for Schwarzschild metric are Lemaitre coordinates? Then, any external observer considers various horizon crossing events to be now.

 

[Yukterez]

And why shouldn’t a distant observer simply use Gaussian normal coordinates, which for Schwarzschild metric are Lemaitre coordinates? Then, any external observer considers various horizon crossing events to be now.

Because Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ), see Wikipedia. We are still talking about the external observer, not the infalling one.

 

[PAllen]

Because Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ), see Wikipedia. We are still talking about the external observer, not the infalling one.

We can be talking about free fall observer a billion light years from the BH, colocated with a stationary one, at relative speed 1 millimeter per century relative to the stationary observer. Then, you claim it’s objectively true that the horizon exists for one and not the other. Sorry, this is completely absurd.