1. Moved from another topic:-

I don't think it's possible for anything to reach an event horizon because it would require an infinite amount of proper time and acceleration because time dilation and length contraction approach infinity as the event horizon is approached. This is what the Schwarzschild coordinates clearly show any coordinates that allow an object to reach an event horizon in a finite amount of proper time are in direct contradiction to this coordinate system. Changing between two valid coordinate systems doesn't allow an event that can never happen in one to happen in another.

2.

3. Post by Markus Hanke:

This is a very common misconception, but it is wrong. The proper time it takes for something to fall to the event horizon (and onwards into the black hole) is finite and well defined. The crucial point here is to understand that a coordinate system is basically the “point of view” of a specific observer - in the case of Schwarzschild coordinates, that would be a stationary observer very far away from the black hole. The coordinate in-fall time diverging to infinity for Schwarzschild coordinates physically means that such observers never see/measure anything reach the horizon.

However, Schwarzschild coordinate time does not tell us anything about the geometric length of a world line extending down to the horizon. If we examine the point of view of an observer in free fall towards the black hole, then we will find that the horizon is reached - and crossed - in a finite amount of time, as measured by a co-moving clock. This measurement is equivalent to the geometric length of a radial geodesic, so it being finite means that things do indeed reach and cross the horizon, even though a Schwarzschild observer never sees that happening. So yes, an event that does not happen for one observer may indeed happen for another observer - because we are in a curved spacetime.

There is no contradiction here either, because the in-fall time as measured by a co-moving clock is a proper quantity, i.e. something that all observers agree on, whereas the infinity of the Schwarzschild observer is merely an artefact of the coordinate system, and hence observer-specific.

To put it succinctly, in order to check whether the horizon can be reached in a finite amount of time, you need to look at a watch in free fall, not at a watch stationary at infinity. In curved space-times, “time” is a purely local concept, so a far-away clock cannot tell us anything about what happens locally at the horizon, or at any point in between.

Understanding the difference between coordinate quantities and proper quantities is absolutely crucial in General Relativity.

4. Yes I know the difference between coordinate dependent time and proper time, I'm also well aware of what a change in coordinate systems can do in regards to how time is measured and what it most definitely can't do.

The time interval between events and the order in which they occur is an arbitrary choice of how you choose to measure the system but if an event does occur in one coordinate system then it's a physically real event that happens in every coordinate system (although the event may be outside of the coordinates used it will be in that coordinate system). If an event never occurs in one coordinate system than it never happens in any coordinate system.

It takes takes an infinite amount of proper time in Schwarzchild coordinates for an object to reach an event horizon (proper time as in the time measured on the distant observer's watch) so it never ever happens from the perspective of a distant observer. But a distant observer is any observer outside of the event horizon. How close to the black hole does an observer have be in order to 'see' an object crossing the horizon? It doesn't make any difference how close it is.

No matter how long a black hole lasts for, it will never be enough time for an object to reach the horizon from a distant observer's perspective. This isn't simply a case of what the distant observer sees, from their perspective it's always possible to for an object to accelerate away. That means however long the black hole hole exists for, it can never be too late for an object to escape and therefore no amount of proper time on any watch allows an object reach the horizon.

If it were possible to reach an event horizon from the perspective of a falling observer then at what point does the Schwarzchild coordinate system become invalid, how close to the event horizon? Schwarzchild coordinates must always be invalid if this is the case and it's just a matter of degree, they can't suddenly stop working at a certain distance. The Schwarzchild coordinates are valid though and show that time dilation and length contraction approach infinity at the horizon.

A coordinate system that allows an object to cross the event horizon is in direct contradiction to the Schwarzchild coordinates. It can't be anything to do with the distant observer accelerating away from the horizon in order to maintain a constant distance as opposed to the falling object only being under gravitational influence because if a distant observer were to fall towards the black hole they still could never observe the closer object reaching the horizon.

It can't be possible for a falling observer to reach the event horizon from the perspective of a more distant falling observer either because the more distant observer could accelerate away and the closer object would have to reemerge from inside the horizon. So an object can never reach an event horizon from any distance under any amount of acceleration, gravitational or otherwise. In other words an object can never reach the event horizon until you do, and you can't until another object does, and so on.

The Schwarzchild coordinates do tell us something about the geometric length of a world-line extending towards an event horizon. It tells us that due to time dilation and length contraction, the world-line is infinite in length before ever reaching the horizon.

5. Posts by Markus Hanke:

It takes takes an infinite amount of proper time in Schwarzchild coordinates for an object to reach an event horizon
No, it takes an infinite amount of coordinate time for the Schwarzschild observer, meaning that he never sees or measures anything reaching the horizon. The proper time however, i.e. what a clock physically records as it falls, is quite finite, even in Schwarzschild coordinates, as shown below.

The physical event takes place regardless of which coordinate system you choose, it’s not a frame-dependent object. It’s just that the Schwarzschild observer at infinity cannot see, measure, or calculate it using his own clocks and rulers, since his notions of time and space or not the same as the ones locally at the event horizon.

Let’s look at this in more detail. The proper distance from some shell r1 down to the event horizon, in Schwarzschild coordinates, is

The proper time this in-fall takes, still in Schwarzschild coordinates (geometrised units), is

Both of these - proper distance as well as proper time - are finite. What is not finite is the coordinate in-fall time for a Schwarzschild observer, which corresponds to what he measures on his own clock that remains stationary infinitely far away from the event horizon. Since Schwarzschild observers are by definition infinitely far away from the horizon, it is not really a surprise that their coordinate in-fall time diverges. The proper measurements on the other hand correspond to what a clock and ruler physically measure when they actually travel along the free-fall geodesic, which is of finite length - all observers agree on this, including the Schwarzschild observer.

Since both the above expressions are covariant quantities, if they are finite in one coordinate system, they are finite in all coordinate systems.

So, while the Schwarzschild observer argues that - according to his own clock - the in-falling test particle never reaches the horizon, a clock co-moving with the particle disagrees, and reaches the horizon in a finite time as given above. This is not a contradiction - they are both right, since measurements of time and space are purely local quantities. Simply put, the Schwarzschild observer never sees anything reach the horizon (he just sees it slowing down, dimming, and being red-shifted to invisibility), but the test particle itself still reaches it in finite time, as demonstrated above. Strictly speaking, a Schwarzschild observer does not even exist in the real world, the coordinate time is simply a bookkeeping device, and does not correspond to anyone’s physical clock reading.

See Taylor/Wheeler “Exploring Black Holes” chapter 3, as well as Misner/Thorne/Wheeler “Gravitation” chapter 31, for the precise derivation and physical justification of the above expressions. Also see Baez (2nd paragraph) for a plain-text explanation : What happens to you if you fall into a black holes

P.S. Apologies for the bad formatting of the maths above, the LaTeX rendering isn’t brilliant on this old forum software.

if an event does occur in one coordinate system then it's a physically real event that happens in every coordinate system
An event is physically real regardless of any coordinate system. The coordinate labels are an arbitrary choice. But that does not necessarily mean that everyone can observe, measure, or calculate that event.

it never ever happens from the perspective of a distant observer
Correct. But it does happen from the perspective of the in-falling observer himself, and it does so in finite time. There is no contradiction.

therefore no amount of proper time on any watch allows an object reach the horizon.
That is incorrect, see the maths in the last post. All observers agree that the in-fall happens in a finite amount of proper time; what differs are only the coordinate times.

The Schwarzchild coordinates do indeed tell us something about the geometric length of a world-line extending towards an event horizon. It tells us that due to time dilation and length contraction, the world-line is infinite in length before ever reaching the horizon
No. The geometric length of the world lines is finite. See the expressions in the last post, and refer to the sources I quoted for their detailed derivation and explanation. Time dilation and length contraction on the other hand are observer-dependent quantities, and both diverge at the horizon in the case of Schwarzschild observers. Do note though that Schwarzschild coordinate measurements do not correspond to anyone’s physical clock or ruler in the real world.

P.S. If you want to be absolutely sure, you can also explicitly calculate the in-fall geodesics from the geodesic equation, and then integrate up to obtain their geometric length. This is done in (e.g.) T. Fliessbach, “General Relativity”. As expected, the result - i.e. proper time - matches the expressions given earlier, and is quite finite.

6. It takes takes an infinite amount of proper time in Schwarzchild coordinates for an object to reach an event horizon
No, it takes an infinite amount of coordinate time for the Schwarzschild observer, meaning that he never sees or measures anything reaching the horizon. The proper time however, i.e. what a clock physically records as it falls, is quite finite, even in Schwarzschild coordinates, as shown below.

The physical event takes place regardless of which coordinate system you choose, it’s not a frame-dependent object. It’s just that the Schwarzschild observer at infinity cannot see, measure, or calculate it using his own clocks and rulers, since his notions of time and space or not the same as the ones locally at the event horizon.

Let’s look at this in more detail. The proper distance from some shell r1 down to the event horizon, in Schwarzschild coordinates, is

The proper time this in-fall takes, still in Schwarzschild coordinates (geometrised units), is

I don't doubt the result of the equations you're using, I doubt their applicability to this situation. Using equations derrived from the perspective of a distant observer, the proper time required on their clock for any object to reach the horizon is infinite (so is the distance in space because of length contraction but sticking with time is fine to keep it simple). I'm not talking about a hypothetical observer at infinite an distance, I'm talking about a real observer at any distance from the horizon.

Both of these - proper distance as well as proper time - are finite. What is not finite is the coordinate in-fall time for a Schwarzschild observer, which corresponds to what he measures on his own clock that remains stationary infinitely far away from the event horizon. Since Schwarzschild observers are by definition infinitely far away from the horizon, it is not really a surprise that their coordinate in-fall time diverges. The proper measurements on the other hand correspond to what a clock and ruler physically measure when they actually travel along the free-fall geodesic, which is of finite length - all observers agree on this, including the Schwarzschild observer.

Since both the above expressions are covariant quantities, if they are finite in one coordinate system, they are finite in all coordinate systems.

So, while the Schwarzschild observer argues that - according to his own clock - the in-falling test particle never reaches the horizon, a clock co-moving with the particle disagrees, and reaches the horizon in a finite time as given above. This is not a contradiction - they are both right, since measurements of time and space are purely local quantities. Simply put, the Schwarzschild observer never sees anything reach the horizon (he just sees it slowing down, dimming, and being red-shifted to invisibility), but the test particle itself still reaches it in finite time, as demonstrated above. Strictly speaking, a Schwarzschild observer does not even exist in the real world, the coordinate time is simply a bookkeeping device, and does not correspond to anyone’s physical clock reading.

See Taylor/Wheeler “Exploring Black Holes” chapter 3, as well as Misner/Thorne/Wheeler “Gravitation” chapter 31, for the precise derivation and physical justification of the above expressions. Also see Baez (2nd paragraph) for a plain-text explanation : What happens to you if you fall into a black holes
The proper time that the falling clock records before reaching the event horizon is certainly not finite in Schwarzschild coordinates. The distant (any distance) Schwarzschild observer never sees the falling observer reach the horizon, never see's them stop moving a a gradually slower relative velocity towards it and never sees the ever slowing clock stop ticking. We can therefore confidently say that the amount of proper time required to reach the horizon from any distance on both the distant observer's and the faller's clock is infinte.

if an event does occur in one coordinate system then it's a physically real event that happens in every coordinate system
An event is physically real regardless of any coordinate system. The coordinate labels are an arbitrary choice. But that does not necessarily mean that everyone can observe, measure, or calculate that event.
This (the third sentence) I disagree with. If an observer measures that an event doesn't happen in any finite amount of their own proper time then in never happens on any clock in any valid coordinate system. What we're discussing here are the effects of time dilation and length contraction on clocks and rulers. Yes the coordinate labels we assign to events are entirely arbitary but if an event happens in one coordinate system then it happens on every clock in every coordinate system after a finite amount of proper time because time dilation slows time but can never prevent the event from ever happening. Switching from a coordinate system where it takes an infinte amount of proper time to reach an event horizon on a distant observer's watch to one where it takes a finite amount of proper time on any watch is a definite contradiction.

it never ever happens from the perspective of a distant observer
Correct. But it does happen from the perspective of the in-falling observer himself, and it does so in finite time. There is no contradiction.

therefore no amount of proper time on any watch allows an object reach the horizon.
That is incorrect, see the maths in the last post. All observers agree that the in-fall happens in a finite amount of proper time; what differs are only the coordinate times.

The Schwarzchild coordinates do indeed tell us something about the geometric length of a world-line extending towards an event horizon. It tells us that due to time dilation and length contraction, the world-line is infinite in length before ever reaching the horizon
No. The geometric length of the world lines is finite. See the expressions in the last post, and refer to the sources I quoted for their detailed derivation and explanation. Time dilation and length contraction on the other hand are observer-dependent quantities, and both diverge at the horizon in the case of Schwarzschild observers. Do note though that Schwarzschild coordinate measurements do not correspond to anyone’s physical clock or ruler in the real world.

P.S. If you want to be absolutely sure, you can also explicitly calculate the in-fall geodesics from the geodesic equation, and then integrate up to obtain their geometric length. This is done in (e.g.) T. Fliessbach, “General Relativity”. As expected, the result - i.e. proper time - matches the expressions given earlier, and is quite finite.
Let's go on a journey. I'll describe what I think we'd observe, then you can tell me at what point you think I go wrong.

We accelerate towards an event horizon that has objects falling towards it. We know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. Now, we go up to the horizon so we're right next to it. Next to us are falling objects, what happens?

We stop accelerating (under our own power so we catch up to a falling object) towards the black hole and are now accelerating against gravity to hover just outside of the event horizon, do the objects fall away from us towards the event horizon? Yes! Let's follow them. Don't worry, length contraction approaches infinity as the horizon is approached so there's an infinite amount of room to play with. We now accelerate towards the falling object again and again, we know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. We do this over and over and at no point can the falling object we're chasing ever reach the horizon. We can do this for a potentially infinite amount of time and during that time the clock on the falling observer's watch keeps on ticking.

Now, how the bloody hell can that falling object ever reach the event horizon in a finite amount of its own proper time, or on any observer's watch in any coordinate system?

7. You are wrong, as Markus has explained to you.

8. Originally Posted by PhDemon
You are wrong, as Markus has explained to you.
That's really not very helpful, you can come to if you like, please be nice though.

Let's go on a journey. I'll describe what I think we'd observe, then you can tell me at what point you think I go wrong.

We accelerate towards an event horizon that has objects falling towards it. We know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. Now, we go up to the horizon so we're right next to it. Next to us are falling objects, what happens?

We stop accelerating (under our own power so we catch up to a falling object) towards the black hole and are now accelerating against gravity to hover just outside of the event horizon, do the objects fall away from us towards the event horizon? Yes! Let's follow them.

Don't worry, length contraction approaches infinity as the horizon is approached so there's an infinite amount of room to play with. We now accelerate towards the falling object again and again, we know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. We do this over and over and at no point can the falling object we're chasing ever reach the horizon. We can do this for a potentially infinite amount of time and during that time the clock on the falling observer's watch keeps on ticking.

Now, how the bloody hell can that falling object ever reach the event horizon in a finite amount of its own proper time, or on any observer's watch in any coordinate system?

9. Looks like Markus wasted his time. I'm not wasting mine...

10. Looks like you're wasting everyone's. Do you have anything meaning to contribute to the discussion or is this all you are capable of? Please don't poison the topic with negativity, it's rude and distasteful, yuk!

11. I've just read your last thread here. You are just the latest relativity crank :shrug: I won't be reading any more of your posts. Life's to short to argue with people like you.

If anyone wants a laugh it's here http://www.thescienceforum.com/trash...tml#post427338

12. That's a relief, bye then.

13. Originally Posted by A-wal
Looks like you're wasting everyone's. Do you have anything meaning to contribute to the discussion or is this all you are capable of? Please don't poison the topic with negativity, it's rude and distasteful, yuk!
What's rude and distasteful is barging into a home and passing gas loudly and unapologetically. Markus was kind and patient to show you how very wrong you are. Please learn some science and manners. Flatulence is unwelcome, aggressive flatulence even less so.

14. I'm not being aggressive. All I've done is started a new topic so I don't hijack someone else's and replied to Markus Hanke's last post with a thought experiment, I'm allowed to do that. I really don't want these kinds of posts, they're nasty for the sake of nasty and they're a distraction from the topic I'm trying to discuss in a friendly and non-confrontational way.

Consider the thought experiment a challenge (a friendly one), if you think I'm wrong then try to point out where. If I've made a mistake then it should be easy to do. If you can't then by all means post another nasty comment in the hope that it will distract people from what's a bullet proof refutation of the idea that all coordinate systems derived from GR are valid ones.

We accelerate towards an event horizon that has objects falling towards it. We know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. Now, we go up to the horizon so we're right next to it. Next to us are falling objects, what happens?

We stop accelerating (under our own power so we catch up to a falling object) towards the black hole and are now accelerating against gravity to hover just outside of the event horizon, do the objects fall away from us towards the event horizon? Yes! Let's follow them.

Don't worry, length contraction approaches infinity as the horizon is approached so there's an infinite amount of room to play with. We now accelerate towards the falling object again and again, we know that no matter how close we get to the horizon those objects will not reach the horizon before we do, if they could then they'd have to reemerge from inside the horizon if we moved away. We do this over and over and at no point can the falling object we're chasing ever reach the horizon. We can do this for a potentially infinite amount of time and during that time the clock on the falling observer's watch keeps on ticking.

Now, how can that falling object ever reach the event horizon in a finite amount of its own proper time, or on any observer's watch in any coordinate system?

15. refutation of the idea that all coordinate systems derived from GR are valid ones
Coordinate systems are not “derived from GR” - they are just arbitrary ways to label events in spacetime. For any given problem, you are welcome to use any labelling system you think is suitable for that particular problem. What is physically real is only the underlying events, and the geometry which represents how they are related; the coordinate systems are arbitrary, so long as they fulfil some basic mathematical requirements such as continuity, differentiability etc.

It is very important to understand this - the Einstein field equation is a local constraint on the metric, and hence a constraint on geometry. Coordinates are like street names - you can give streets any names you choose, it’s completely arbitrary, but it’s still going to be the same physical street. So regardless of what labels you print on a map, the geometry and outline of the street network is always the same.

Anyway, let’s go back to basics and focus on what GR is actually about, and that is the geometry of spacetime. I want to cut out personal opinions, and stick to simple maths, so I am not going to go down the road of “thought experiments”, which are largely subjective things and just lead to endless debates, which I have - quite frankly - no interest in. So, the question of whether the event horizon can be reached or not, is geometrically equivalent to asking whether there are geodesics of spacetime that connect a shell at some distance to the event horizon, while remaining of finite geometric length. This question is a geometric one, and independent of what coordinate system is used.

Without reference to a specific coordinate system, the proper distance between shells in Schwarzschild spacetime (note that this isn’t a reference to any coordinate system, just the name of this solution to the field equations) is, in terms of the metric, given by

This is not yet the length of any worldline, but simply geometric distance on the spacetime manifold. Your claim now was that this distance is infinite, so I would like you to show us how you mathematically arrived at this conclusion. I am not interested in why you think it should be infinite, I am interested only in a formal derivation, based on this particular solution to the field equations. So basically, if could evaluate the integral for us, and show how it diverges, that would be great. Once I see your mathematical derivation, I can then address the specific places where the actual problems occur, and we’ll take it from there. As you can (hopefully) see, the evaluated expression I gave earlier for this integral is quite finite, so you must have come up with a different expression, while mine is somehow wrong. I’d just like you to explicitly show this.

Can I please politely request that you restrict your response to addressing this one specific point, i.e. the mathematical evaluation of the above expression. The reason is simply that I want to put this on an objective basis, and take any subjectivity out of the discussion.

As a next step, if necessary, we can then go on and calculate the geometric length of a real-world free fall geodesic in this spacetime, step by step.

Very clear and easy to understand.

17. I agree with exchemist, Markus. Only someone stubbornly refusing to learn would quarrel with what you took pains to wriite so clearly.

18. Originally Posted by Markus Hanke
refutation of the idea that all coordinate systems derived from GR are valid ones
Coordinate systems are not “derived from GR” - they are just arbitrary ways to label events in spacetime. For any given problem, you are welcome to use any labelling system you think is suitable for that particular problem. What is physically real is only the underlying events, and the geometry which represents how they are related; the coordinate systems are arbitrary, so long as they fulfil some basic mathematical requirements such as continuity, differentiability etc.

It is very important to understand this - the Einstein field equation is a local constraint on the metric, and hence a constraint on geometry. Coordinates are like street names - you can give streets any names you choose, it’s completely arbitrary, but it’s still going to be the same physical street. So regardless of what labels you print on a map, the geometry and outline of the street network is always the same.
I don't believe I've in any way misunderstood what coordinate systems are or how they're used. GR must be used to derive coordinate systems, I don't understand what you mean when you say it isn't, unless I'm misusing the term 'derived' which is possible but I just meant that coordinates such as Schwarzschild coordinates use GR to plot events.

Originally Posted by Markus Hanke
Anyway, let’s go back to basics and focus on what GR is actually about, and that is the geometry of spacetime. I want to cut out personal opinions, and stick to simple maths, so I am not going to go down the road of “thought experiments”, which are largely subjective things and just lead to endless debates, which I have - quite frankly - no interest in.
You're happy to say that an object can reach an event horizon in a finite amount of proper time, which is in itself a very basic thought experiment, you're saying that an object falling towards a black hole would reach and cross the event horizon. All I'm asking is if objects can already have crossed the horizon from the perspective of that falling observer, and if they can then what happens if that observer accelerates away from the black hole before reaching the horizon?

Originally Posted by Markus Hanke
So, the question of whether the event horizon can be reached or not, is geometrically equivalent to asking whether there are geodesics of spacetime that connect a shell at some distance to the event horizon, while remaining of finite geometric length. This question is a geometric one, and independent of what coordinate system is used.

Without reference to a specific coordinate system, the proper distance between shells in Schwarzschild spacetime (note that this isn’t a reference to any coordinate system, just the name of this solution to the field equations) is, in terms of the metric, given by

This is not yet the length of any worldline, but simply geometric distance on the spacetime manifold. Your claim now was that this distance is infinite, so I would like you to show us how you mathematically arrived at this conclusion. I am not interested in why you think it should be infinite, I am interested only in a formal derivation, based on this particular solution to the field equations. So basically, if could evaluate the integral for us, and show how it diverges, that would be great. Once I see your mathematical derivation, I can then address the specific places where the actual problems occur, and we’ll take it from there. As you can (hopefully) see, the evaluated expression I gave earlier for this integral is quite finite, so you must have come up with a different expression, while mine is somehow wrong. I’d just like you to explicitly show this.

Can I please politely request that you restrict your response to addressing this one specific point, i.e. the mathematical evaluation of the above expression.
How can I possibly use an equation that gives a finite length to show you that it's infinite? I'm not disputing that using that solution, the geometric distance on the spacetime manifold between a distant observer and the event horizon comes out as a finite value. I'm questioning how that model describes what a falling object observes other falling objects doing when approaching an event horizon. I'd like to politely request that you restrict your response to addressing that question. It's a very simple question and I did ask first.

You're avoiding explaining how that model describes what that observer would see in regards to other objects crossing the horizon. You've agreed that objects can't reach the horizon from the perspective of a distant observer.
it never ever happens from the perspective of a distant observer
Originally Posted by Markus Hanke
Correct. But it does happen from the perspective of the in-falling observer himself, and it does so in finite time. There is no contradiction.
So the only way that can be true if objects can reach the horizon from their own perspective is if all objects that do cross the horizon do at the same time as each other from their perspective. But even then what would a distant observer see? It would mean that black holes have to last for ever if they never see any object reach the horizon and even then, at what point during the black holes life would the event of objects crossing the horizon occur?

Originally Posted by Markus Hanke
The reason is simply that I want to put this on an objective basis, and take any subjectivity out of the discussion.
Fair enough, I'll ask in an entirely unambiguous way about what is objectively observed from the perspective of specific observers in a well defined situation.

Can a falling object reach an event horizon from the perspective of a distant observer who maintains a constant distance from the event horizon? No, that's not in dispute.

Can a falling object reach an event horizon from the perspective of a more distant falling observer? If yes then what happens to that object from the perspective of the more distant falling observer if that more distant falling observer accelerates away?

So assuming that a falling object can't reach an event horizon from the perspective of a more distant falling observer, do the other falling objects reach the event horizon at the same time as the previously more distant falling observer?

19. GR must be used to derive coordinate systems, I don't understand what you mean when you say it isn't
I think this is one of the core problems in your approach to this. GR does not derive coordinate systems, it derives the metric, which represents the geometry of spacetime - not how events are labelled.
When you go about solving the field equations for a specific energy-momentum distribution (or vacuum), you start by making an “ansatz” - this means you decide how you want to label events in your spacetime, and then write down a “skeleton metric”, which is one in which the coefficients are left as unknown functions. You then insert this into the field equations, and solve for the unknown functions.
In other words, your chosen coordinate system is something you impose during the process of solving the equations, and GR then allows you to derive the metric. No matter which coordinate system you choose, you always get the same geometry as solution.

GR is a mathematical model that describes relationships between events, i.e. between points on a four-dimensional manifold. It is a model that allows you to determine the geometry of spacetime. Which labels (i.e. coordinates) you give to the individual events is completely irrelevant - because physics arise only from the relationships between events. Events are physical, coordinates are not.

All I'm asking is if objects can already have crossed the horizon from the perspective of that falling observer
This is not what the original question was. The question was whether it is possible for anything to physically cross the event horizon of a black hole, irrespective of what individual observers may or may not see or measure.
In order to answer this, we take the observer out of the question, and simply look at the geometry of spacetime. We have an event located at the horizon, and we have an event somewhere outside the black hole. We are now asking - can we link these two events via a world line of finite geometric length? So we are examining the geodesic structure of spacetime itself, which is a matter of geometry, and not subject to any “perspectives”.

You're avoiding explaining how that model describes what that observer would see in regards to other objects crossing the horizon.
Since no light escapes outwards from the horizon, no external observer - no matter his state of motion - will ever optically see anything cross it. I would have assumed that this is common sense. If you want to optically see the crossing of the horizon, you either need to be the one who is actually falling through it, or you need to be located below the horizon. Likewise, if you want to know how long it takes to fall to the horizon, you need to look at what a watch that is actually doing the falling will physically record - and not anyone else’s watch.

So the only way that can be true if objects can reach the horizon from their own perspective is if all objects that do cross the horizon do at the same time as each other from their perspective.
I honestly don’t understand what point you are trying to make here. It seems to me more and more that you are equating the occurrence of an event - such as the crossing of an horizon - to its visibility by external observers. That is no so. The event is physically real regardless of who can or cannot visually see it. If you really want to “see” the crossing, then simply fall along with the test particle, or position yourself below the horizon, and look back at the crossing happening (note that you won’t be able to remain stationary in that region).

When we say that a test particle “can reach the event horizon”, then that means the particle will reach the horizon within a finite time as measured by its own clock, not anyone else’s. Geometrically, the total time accumulated on a clock in free fall is equal to the geometric length of its world line. We don’t compare clocks here, nor is there any notion of simultaneity involved. We simply measure what a clock that travels between two events physically records.

Can a falling object reach an event horizon from the perspective of a distant observer who maintains a constant distance from the event horizon? No, that's not in dispute.

Can a falling object reach an event horizon from the perspective of a more distant falling observer? If yes then what happens to that object from the perspective of the more distant falling observer if that more distant falling observer accelerates away?

So assuming that a falling object can't reach an event horizon from the perspective of a more distant falling observer, do the other falling objects reach the event horizon at the same time as the previously more distant falling observer?
As I have said already, no external observer will “see” the crossing, regardless of their state of motion.
None of this is in any way relevant to the question, because the crossing of the horizon does not depend on what anyone else measures, it does not depend on “perspectives”, and it does not depend on notions of simultaneity. It depends only on what the clock falling with the test particle physically records - i.e. the length of its world line. And that’s finite, as I’m glad you are able to see.

20. I'm not disputing that using that solution, the geometric distance on the spacetime manifold between a distant observer and the event horizon comes out as a finite value.
Good, then the question of whether or not the horizon can be crossed is settled. Because that is exactly what it means - whether or not in-going world lines connecting external events to events at the horizon are of finite length. And they are, so clocks travelling between these events record a finite amount of time between them. Even though we haven’t explicitly shown this just yet, but of course we could, though the maths are a bit more involved than mere geometric distance. This is all I wanted to establish, and if you agree with it, then I have little else to add.

Asking what other external observers would see from their perspectives means asking about the geometry of outgoing null geodesics, which is a completely different (but valid and interesting) issue, because those outgoing geodesics aren’t the same ones as the in-fall geodesic. Let’s not conflate these questions here, but you can discuss this as a separate topic. It’s a pretty complex topic too, since photons originating at or near the horizon tend to trace out very complicated geodesics in spacetime - unlike the in-fall world line, which is pretty much simply radial.

Nonetheless, a photon being emitted outwards at exactly the point of horizon crossing cannot reach any external observer, since it is confined to the horizon surface and the region enclosed within it. Hence, no external observer (regardless of his state of motion) will ever see anything cross the horizon - but that is because there is no outgoing null geodesic that connects the event at the horizon with the external observer, and not because the event doesn’t exist. The observer who falls together with the watch, as well as observers in the interior region, can indeed see the crossing, so in that sense it certainly “happens”.

Two more notes on all of this:

1. Technically speaking, a spacetime that contains any in-falling observers at all, is no longer a Schwarzschild spacetime. So the entire question is merely academic anyway. Black holes with non-constant mass (i.e. stuff falling into them) are described by a different class of solutions called Vaidya space-times, which are geometrically much more complex than Schwarzschild ones.

2. When anything falls into a black hole, then its energy-momentum becomes part of that black hole’s mass. Hence, if you let two test particles fall in one after the other, than the second one will encounter the horizon earlier than the first one did, since the horizon will be farther out.

3. When you do the same calculation in Vaidya spacetime, you also arrive at a finite result - this is important, since it is the more correct solution to the field equations to use in this case.

21. Originally Posted by Markus Hanke
Two more notes on all of this:

1. Technically speaking, a spacetime that contains any in-falling observers at all, is no longer a Schwarzschild spacetime. So the entire question is merely academic anyway. Black holes with non-constant mass (i.e. stuff falling into them) are described by a different class of solutions called Vaidya space-times, which are geometrically much more complex than Schwarzschild ones.

2. When anything falls into a black hole, then its energy-momentum becomes part of that black hole’s mass. Hence, if you let two test particles fall in one after the other, than the second one will encounter the horizon earlier than the first one did, since the horizon will be farther out.

3. When you do the same calculation in Vaidya spacetime, you also arrive at a finite result - this is important, since it is the more correct solution to the field equations to use in this case.
If you have a Black Hole with (temporarily) no infalling objects at all, is the Black Hole entirely symmetrical? It is not "ringing "all the time with its "rings" directed inwards is it?(and its rings attenuating with time)

22. Originally Posted by geordief
If you have a Black Hole with (temporarily) no infalling objects at all, is the Black Hole entirely symmetrical? It is not "ringing "all the time with its "rings" directed inwards is it?(and its rings attenuating with time)
It will be approximately symmetrical, provided it has no electric charge or angular momentum.

23. Originally Posted by Markus Hanke
Originally Posted by geordief
If you have a Black Hole with (temporarily) no infalling objects at all, is the Black Hole entirely symmetrical? It is not "ringing "all the time with its "rings" directed inwards is it?(and its rings attenuating with time)
It will be approximately symmetrical, provided it has no electric charge or angular momentum.
What would be the effect of of any asymmetry?

24. Originally Posted by geordief
If you have a Black Hole with (temporarily) no infalling objects at all, is the Black Hole entirely symmetrical? It is not "ringing "all the time with its "rings" directed inwards is it?(and its rings attenuating with time)
I thought the 'ringdown' part of a black hole merger was both black holes smoothing out the unevenness of their merger to become one symmetrical body, so to speak. In other words, the asymmetrical shape of the merger is producing the '''rings'' of the ringdown. The rings stopping once everything is symmetrical. Maybe you knew this and I've misunderstood you.

25. Originally Posted by Andyglove
I thought the 'ringdown' part of a black hole merger was both black holes smoothing out the unevenness of their merger to become one symmetrical body, so to speak. In other words, the asymmetrical shape of the merger is producing the '''rings'' of the ringdown. The rings stopping once everything is symmetrical. Maybe you knew this and I've misunderstood you.
No,I wasn't talking about that(nor am I up to speed with it although I am familiar with the term "ringdown")

My "ringing" was just an analogy with a bell.

The Binary BH merger does seem relevant,though.

Presumably it would make a sound if there was a medium to carry it.

26. Originally Posted by geordief
No,I wasn't talking about that(nor am I up to speed with it although I am familiar with the term "ringdown")
Sorry

Originally Posted by geordief
Presumably it would make a sound if there was a medium to carry it.
LIGO detected the gravitational waves (rings) caused by the unevenness just after the merger until everything was symmetrical. I suppose I've misunderstood again

27. Originally Posted by Andyglove
Originally Posted by geordief
No,I wasn't talking about that(nor am I up to speed with it although I am familiar with the term "ringdown")
Sorry

Originally Posted by geordief
Presumably it would make a sound if there was a medium to carry it.
LIGO detected the gravitational waves (rings) caused by the unevenness just after the merger until everything was symmetrical. I suppose I've misunderstood again
Not at all Any excuse to recall that fantastic BBH merger.

I remember at the time they reconstructed the "sound" it made (a bit unspectacular)

28. Cheers for the reply Markus. I'm not really in the right frame of mind to reply at the moment but thanks for taking the time and having the patience to politely and thoroughly explain, I need to be in the right mood or it's too much effort to express myself clearly.

29. Originally Posted by A-wal
I need to be in the right mood or it's too much effort to express myself clearly.
I will try to explain for A-wal.

Reading the literature and forums I see a lot of confusion regarding all those GR coordinates. Unlike the most of the claims in my opinion Schwarzschild coordinates capture the very essence of GR. I will use one quote from the wiki article about Schwarzschild metric which states:

"The Schwarzschild coordinates therefore give no physical connection between the two patches, which may be viewed as separate solutions."

No physical connection between two patches in that solution is there to preserve the validity of law of physics for outside and then all observers. And that is the essence of GR. Gravitational time dilation of GR is there for the law of physics to hold. There were attempts to connect these two patches for in-falling observers even in Schwarzschild coordinates but what they have shown is that for in-falling finite proper time to reach at Schwarzschild radius you need infinite outside coordinate time, which is impossible. You cannot reach infinite time. Read the same quoted source! All those other coordinates (Lemaître to Kruskal–Szekeres coordinates) and their coordinate transformations are there to make the finite smooth proper time for in-falling observer to reach at Schwarzschild radius. But what all authors who work with those coordinates should explain how for static outside observer something can reach at Schwarzschild radius despite the time dilation blow up, because reaching at would be a break of physical laws for outside us. All other solutions which I have read about thick shell etc. should address the same. From the outside, the event horizon does not exist because saying that it does exist means a breakdown of law of physics, so from outside we can only operate with frozen stars. What is important is that Schwarzschild coordinates show that there are solutions in GR which fight against its break down.

The defenders of event horizon operate and compare, in for instance Kruskal–Szekeres coordinates, the two in-falling proper times of the travel all the way down to the center, but if you work in those coordinates and compare proper time of stationary outside observer with the time of in-falling observer you will also calculate asymptotic approach to Schwarzschild radius. The defenders of event horizon also say that the space is smooth there, but Schwarzschild coordinates also say that. The space outside and inside Schwarzschild radius is also smooth, but the Schwarzschild coordinates capture the behavior of the matter and energy!

I wrote this because I think that those are interesting thoughts and are worth to be addressed. I think that this is the best way to express these issues by words. So, A-wal, cheers!

Best wishes and regards,

Zlatan

30. Can a falling object reach an event horizon from the perspective of a distant observer who maintains a constant distance from the event horizon? No, that's not in dispute.

Can a falling object reach an event horizon from the perspective of a more distant falling observer? If yes then what happens to that object from the perspective of the more distant falling observer if that more distant falling observer accelerates away?

So assuming that a falling object can't reach an event horizon from the perspective of a more distant falling observer, do the other falling objects reach the event horizon at the same time as the previously more distant falling observer?
Originally Posted by Markus Hanke
As I have said already, no external observer will “see” the crossing, regardless of their state of motion.
Yes exactly. Even if the closer object set off billions of years before and has been constantly accelerating (in addition to gravitational acceleration) towards the black hole during this time, it still won't be able to reach the event horizon from the perspective of the more distant observer accelerating towards the horizon.

If there were a line of observers the length of a galaxy cluster free-falling towards an event horizon, the closest object will never reach the horizon from the perspective of any of the other falling observers. When the last one is 1mm away from the horizon the front one still hasn't reached it. Because of length contraction, that 1mm at a distance can be light years for the free-faller. There's no upper limit to time dilation and length contraction which is why the horizon is never reached.

Originally Posted by Markus Hanke
None of this is in any way relevant to the question, because the crossing of the horizon does not depend on what anyone else measures, it does not depend on “perspectives”, and it does not depend on notions of simultaneity.
Here's where we disagree. It's entirely relevant because this isn't just a case of the more distant object not being able to observer the closer object reaching the horizon. The closer object can never actually reach the horizon from the perspective of the more distant observer, it's always possible the the closer object will accelerate away from the black hole, the more distant object could always pull the closer object away with a rope.

Originally Posted by Markus Hanke
It depends only on what the clock falling with the test particle physically records - i.e. the length of its world line. And that’s finite, as I’m glad you are able to see.
Yes it's always finite but that's not the issue. The length of the worldline is defined by the amount of time dilation and length contraction the free-falling object has undergone along that wordline, not the other way around. The worldline is always of finite length but to reach the event horizon would require an infinite amount of proper time for the object moving along that worldline.

The fact that no object can ever reach the event horizon from the perspective of a more distant object means that an observer falling towards an event horizon would experience an infinite amount of time passing in the rest of the universe before reaching the horizon, so an infinite amount of time has to pass on the free-faller's watch as well before reaching the horizon.

Time dilation and length contraction are always finite, they become infinite at the event horizon but an infinite amount of proper time passes before the event horizon is reached, otherwise a closer object could reach the horizon from the perspective of a more distant falling observer. It's exactly the same as an object accelerating towards the speed of light in flat spacetime.

31. You need to go back to school...

32. Originally Posted by PhDemon
You need to go back to school...
You need to ask yourself what motivates you to come to the one type of place where people like you are allowed to get away with being that rude without the mods doing anything about it.

Do you actually understanding any of this? I'm guessing no.

Make another empty generic snotty remark if it makes you feel better about yourself.

33. You need to ask yourself why you spout nonsense from a position of ignorance... People like you, fools speaking ex ano with confidence, are more toxic than those who point it out...

34. Originally Posted by A-wal
Yes it's always finite but that's not the issue. The length of the worldline is defined by the amount of time dilation and length contraction the free-falling object has undergone along that wordline, not the other way around. The worldline is always of finite length but to reach the event horizon would require an infinite amount of proper time for the object moving along that worldline.

The fact that no object can ever reach the event horizon from the perspective of a more distant object means that an observer falling towards an event horizon would experience an infinite amount of time passing in the rest of the universe before reaching the horizon, so an infinite amount of time has to pass on the free-faller's watch as well before reaching the horizon.

Time dilation and length contraction are always finite, they become infinite at the event horizon but an infinite amount of proper time passes before the event horizon is reached, otherwise a closer object could reach the horizon from the perspective of a more distant falling observer. It's exactly the same as an object accelerating towards the speed of light in flat spacetime.
Just to be clear Markus (because that came out more statement of fact than I intended), this is my reasoning, why I think it's equivalent to accelerating to the speed of light.

If I've made a mistake somewhere I'd like to know where exactly I've gone wrong and what it is I've misunderstood.

35. Originally Posted by PhDemon
You need to ask yourself why you spout nonsense from a position of ignorance... People like you, fools speaking ex ano with confidence, are more toxic than those who point it out...
No people like you are far more toxic. You're in no position to judge the level of ignorance of anyone else on a topic that you yourself have no understanding of.

Your confidence is based not on understanding or even on a lack of it, it's entirely on the assumption that anyone who questions the validity of something is automatically wrong. You're vicious, nasty and spiteful, probably a symptom of being made to feel small and insignificant throughout your life so you come here to make yourself feel better by lashing out out people in the one place you know you can get away with it.

I didn't come here for an argument and I certainly didn't come here to be spoken to like that by somebody like you. I've reported your post, not something I would normally do. I don't expect anything to happen but people as toxic as you shouldn't be tolerated.

36. I've passed my exams, I understand the subject quite well. I am in a very good position to judge the ignorance of others (as a teacher it's crucial as I have to fix it). You are an obvious idiot. Stay stupid :shrug:

37. Originally Posted by PhDemon
I've passed my exams, I understand the subject quite well. I am in a very good position to judge the ignorance of others (as a teacher it's crucial as I have to fix it).
Oh good, this is your chance to show that you're not just a clueless, pointless troll then.

If no object can reach the reach horizon from the perspective of any external observer then we can say that it takes an infinite amount of time anywhere in the universe outside the black hole for an object to reach the event horizon. In a convoy of falling ships a trillion light years long in which the first ship set off a trillion years before the next and so on, the front ship never reaches the event horizon in front of any of the more distant ships.

So how can a falling observer reach the event horizon in a finite amount of their own proper time given that the time dilation and length contraction that the falling observer is undergoing is always a finite amount? How can infinite proper time on the watch of a more distant observer dilate to a finite amount of proper time on the faller's watch?

It's never too late in the life span of the black hole for a falling object to accelerate away, so at no point in the black hole's life does any object reach the event horizon.

This isn't a strawman, this is what the model describes. If I've made a mistake I want to know what it is.

Originally Posted by PhDemon
You are an obvious idiot. Stay stupid :shrug:
Is childish name calling acceptable in this forum?

38. Markus already explained it to you, you ignored it like a typical crank :shrug:

39. Originally Posted by PhDemon
Markus already explained it to you, you ignored it like a typical crank :shrug:
No he certainly did not.

As soon as you concede that no object can reach the event horizon from the perspective of a more distant object it's ver easy to show that it causes a paradox if an object can reach the horizon fro their own perspective. That's why I was trying to get Markus to admit this:
Originally Posted by Markus Hanke
As I have said already, no external observer will “see” the crossing, regardless of their state of motion.
If two ships free-fall towards a black hole with a rope connecting them so the slack on the rope runs out when the front ship has crossed the event horizon from the perspective of the front ship then from the front ship's perspective the ship behind won't be able to tow them away because they've crossed the horizon but from the perspective of the ship behind they can tow the front ship away because the front ship hasn't reached the horizon.

From the perspective of the ship behind, the rope would have to slacken again if it were possible for them to reach the event horizon because the front ship can't reach the horizon from their perspective before they reach it themselves. You don't need two objects to show this, with just one ship the front of the ship can't reach the event horizon from the perspective of the back of the ship before the back of the ship reaches the horizon.

Objects approaching the horizon get stretched out as the closer ones are under stronger gravity but they would have to get squashed together to zero length if they were able to reach the horizon because the front one can't reach the horizon first.

If we take a black hole's life span to be a thousand years then at 999 years all the objects that ever headed towards the black hole are still outside of the horizon, that's a fact. They could still accelerate away from the black hole from the perspective of distant objects. When there's one millisecond of the black hole's life remaining from the perspective of a distant observer, still no object has reached the horizon. The black hole dies with no object ever having reached its event horizon.

Even with an ever lasting black hole, an infinite amount of time has to pass on a distant observer's watch from the perspective of a falling object before they can reach the horizon. Time dilation and length contraction approach infinity at the event horizon, just as they do for an object approaching the speed of light but objects can't accelerate to the speed of light, they just get more and more length contracted and time dilated.

The same is obviously true of objects being accelerated by gravity towards an event horizon because there's no limit to how much potential distance there is in time or space between the falling object and the horizon, they just get more and more length contracted and time dilated.

40. Crank, :shrug:

41. Originally Posted by A-wal
Originally Posted by PhDemon
I've passed my exams, I understand the subject quite well. I am in a very good position to judge the ignorance of others (as a teacher it's crucial as I have to fix it).
Oh good, this is your chance to show that you're not just a clueless, pointless troll then.
Originally Posted by PhDemon
Crank, :shrug:
Just a clueless, pointless troll with no understanding of the topic at all then. Yea, that's what I thought.

From the moment a black hole is born to the moment it dies, no object can ever reach the horizon from the perspective of a more distant object. It's not as if the objects have crossed the horizon but the distant observers can't see them crossing, those closer objects haven't crossed the horizon, they can still accelerate away.

How can anyone accept this on the one hand and then on the other claim that an observer can reach the horizon from their own perspective?

42. Originally Posted by A-wal
From the moment a black hole is born to the moment it dies, no object can ever reach the horizon from the perspective of a more distant object. It's not as if the objects have crossed the horizon but the distant observers can't see them crossing, those closer objects haven't crossed the horizon, they can still accelerate away.

How can anyone accept this on the one hand and then on the other claim that an observer can reach the horizon from their own perspective?
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon. The distant observer never sees a time beyond the stopped clock time because they never see the infalling ship actually cross the horizon, but the infalling observer doesn't see his clock stop or even slow down and therefore sees his clock tick beyond the stopped clock time which corresponds to crossing the horizon to the inside of the blackhole.

43. Originally Posted by KJW
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon. The distant observer never sees a time beyond the stopped clock time because they never see the infalling ship actually cross the horizon, but the infalling observer doesn't see his clock stop or even slow down and therefore sees his clock tick beyond the stopped clock time which corresponds to crossing the horizon to the inside of the blackhole.

Originally Posted by KJW
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon.
No I don't believe this is true, I mean I don't believe it's officially true, it's not the accepted position.

The officially accepted position is that from the perspective of the distant observer the falling object's clock never stops, it just keeps slowing down. This is important because it means that they never reach the horizon from the perspective of the distant observer.

What you're describing is a point when objects have objectively reached the horizon from the perspective of a distant observer and if that were the case they'd be no contradiction but that's not the accepted view. It's never too late for objects to accelerate away from the black hole in the accepted model and that does contradict objects reaching the horizon from their own perspective.

44. Originally Posted by A-wal
Originally Posted by KJW
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon.
No I don't believe this is true, I mean I don't believe it's officially true, it's not the accepted position.

The officially accepted position is that from the perspective of the distant observer the falling object's clock never stops, it just keeps slowing down. This is important because it means that they never reach the horizon from the perspective of the distant observer.

What you're describing is a point when objects have objectively reached the horizon from the perspective of a distant observer and if that were the case they'd be no contradiction but that's not the accepted view. It's never too late for objects to accelerate away from the black hole in the accepted model and that does contradict objects reaching the horizon from their own perspective.
You are correct in saying that the distant observer never sees the infalling clock actually stop, that the clock just keeps slowing down. But there is a time beyond which the clock never exceeds from the perspective of the distant observer, whereas the infalling observer sees the clock tick beyond this time. In other words, the infalling observer can see times displayed on his clock that is after the lastest possible time that the distant observer can see.

45. Originally Posted by KJW
Originally Posted by A-wal
From the moment a black hole is born to the moment it dies, no object can ever reach the horizon from the perspective of a more distant object. It's not as if the objects have crossed the horizon but the distant observers can't see them crossing, those closer objects haven't crossed the horizon, they can still accelerate away.

How can anyone accept this on the one hand and then on the other claim that an observer can reach the horizon from their own perspective?
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon. The distant observer never sees a time beyond the stopped clock time because they never see the infalling ship actually cross the horizon, but the infalling observer doesn't see his clock stop or even slow down and therefore sees his clock tick beyond the stopped clock time which corresponds to crossing the horizon to the inside of the blackhole.
So the object is vanishing so how is time observed in relation to a Black hole? How is it scientists have observed this time difference I'm curious?
My own opinion a Black hole is a worm hole that leads to the former stars fuel dimension and that dimension is where planets and stars are being pulled into by the Black holes gravity field.

46. Originally Posted by A-wal
Originally Posted by KJW
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon. The distant observer never sees a time beyond the stopped clock time because they never see the infalling ship actually cross the horizon, but the infalling observer doesn't see his clock stop or even slow down and therefore sees his clock tick beyond the stopped clock time which corresponds to crossing the horizon to the inside of the blackhole.

Originally Posted by KJW
If a distant observer observes a clock onboard the infalling ship, he will see the clock slow down and eventually stop at the time that the infalling ship crosses the horizon.
No I don't believe this is true, I mean I don't believe it's officially true, it's not the accepted position.

The officially accepted position is that from the perspective of the distant observer the falling object's clock never stops, it just keeps slowing down. This is important because it means that they never reach the horizon from the perspective of the distant observer.

What you're describing is a point when objects have objectively reached the horizon from the perspective of a distant observer and if that were the case they'd be no contradiction but that's not the accepted view. It's never too late for objects to accelerate away from the black hole in the accepted model and that does contradict objects reaching the horizon from their own perspective.
Now how do they know that the ship would exist at a point where time was slower than this universe? Who came up with this theory it is interesting it could lead to time travel.

47. Originally Posted by KJW
You are correct in saying that the distant observer never sees the infalling clock actually stop, that the clock just keeps slowing down. But there is a time beyond which the clock never exceeds from the perspective of the distant observer, whereas the infalling observer sees the clock tick beyond this time. In other words, the infalling observer can see times displayed on his clock that is after the lastest possible time that the distant observer can see.
So it's never too later for the faller to accelerate away from the horizon from the perspective of the distant observer.

Doesn't that imply that (from the faller's perspective) an infinite amount of time has to pass on the watch of the distant observer before reaching the horizon?