The Arrow of Time: The Laws of Physics and the Concept of Time Reversal

[Passionflower]

It's worth noting that the GPS system is basically a re-run of the H-K experiment with more clocks and a wider range of relative velocities, and it confirms the GR predictions with a significantly higher accuracy than the H-K experiment did. This paper on the living reviews site gives a good overview:

http://relativity.livingreviews.org/Articles/lrr-2003-1/

I am in no way questioning the validity of GR, but in my opinion the H-K experiment could not have concluded it with the available test results, I have a similar opinion on the experiment by Eddington on Principe. In both cases the margins of error where too large in my opinion to make a conclusion.

 

[PeterDonis]

I am in no way questioning the validity of GR, but in my opinion the H-K experiment could not have concluded it with the available test results, I have a similar opinion on the experiment by Eddington on Principe. In both cases the margins of error where too large in my opinion to make a conclusion.

I definitely agree in the case of Eddington's eclipse observations in 1919. In that case, IIRC, the error bars were large enough that the results could not rule out the possibility that there was *zero* light bending by the Sun.

I'm not sure I agree in the case of the original H-K experiment. It was consistent with the GR predictions to within the accuracy of both the theoretical calculations and the experimental measurements, and it was definitely *not* consistent with no effect being present. (That is, the differences in clock readings were *not* zero to a very high degree of statistical significance.) It was also definitely *not* consistent with the known SR time dilation due to relative velocity being present (I say "known" because it had been verified by many other experiments by that time), but the predicted GR time dilation due to altitude difference not being present; only both effects together gave a prediction consistent with the experimental results. The Wikipedia page gives the values for the SR and GR predictions separately:

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

I believe similar numbers are given in MTW (don't have my copy handy right now to check exactly where). In view of all this, I would say the H-K experiment gave significantly better evidence for GR than Eddington's eclipse observations did.

 

[A-wal]

Huh? You are saying that if I pass a light beam that is moving towards me at c, that requires me to accelerate to c myself? How do you figure that?

What? That’s not what I’m saying. I meant that it would be the equivalent of c because it’s the point when c wouldn’t be fast enough to move away. It’s the point when gravity would have accelerated you all the way to c. Mass can’t do that any more than energy can. The speed that the event horizon moves inwards is c at the horizon and slows down as an inverse square of the objects distance, so you would have to accelerate past c to catch it. The reason it moves inwards is because there’s a limit to how fast objects can move relative to other objects but there’s no limit to mass or energy, so it can keep on accelerating you towards it but you’ll never reach a relative velocity of c, so you get an event horizon.

Really? So if I take two identical rocket engines with identical thrust, and attach one to a hundred ton object (by "ton" I mean a ton of mass, a thousand kilograms) and one to a two hundred ton object, and turn both engines on, both objects will accelerate identically? How do you figure that?

That’s not what I said. In that situation it’s two objects pushing against each other and the difference in mass determines how the energy is spread out between them and which one moves fastest relative to a third object. What I meant was simply that if you could turn off gravity it would take no more effort to lift Mount Everest than it would to lift your arms, as long as you had the Earth to push against. That’s why objects with different masses fall at the same rate. Objects with very different masses will move at the same speed if pushed in 0g and objects with very different masses will move (fall) at the same speed if there’s no energy.

Not necessarily. You'll note that I nowhere specified the actual proper distances between A and C and B and C. I only specified that C was at the center of mass of the object as a whole. If the object extends far enough radially that the difference in the strength of gravity is noticeable, then the object's structure will also change in response to that difference, and therefore so will the location of its center of mass.

It wouldn’t fall faster if it was longer? Do you mean that it gets ‘weighted’ so that B moves closer to a keeping the difference in gravitational strength between A and C the same as the difference in gravitational strength between C and B?

No, it isn't. It's the internal forces between the different parts of the object, that are causing those parts to move on non-geodesic worldlines. The only reason tidal gravity comes into it at all is that tidal gravity makes the geodesic worldlines diverge, which is equivalent to making worldlines that remain at a constant proper distance from one another non-geodesic worldlines. But if there were no internal forces between the parts of objects then they would not feel any force at all, even though tidal gravity was present; the parts would simply diverge as they all traveled along their separate geodesics.

Yes I know. Tidal gravity would separate the parts and if they had something that was holding them together then they’d feel tidal force. Tidal gravity causes any extended object (all of them) to feel tidal force. What’s the problem?

What is your basis for this? How can the rocket push on the object if the object feels no force?

I’m trying to show that I think the proper acceleration could be smoothed out to match gravity, making proper acceleration and tidal force equivalent.

I'm sure that standard GR predicts that an idealized point-like object would feel proper acceleration if a rocket engine were strapped to it and turned on. Obviously there are no idealized point-like objects, but subatomic particles like electrons, which as far as we know have no internal structure, have been accelerated in particle accelerators for decades and they show all the effects of feeling acceleration. I guess to know for sure we'd have to figure out how to attach accelerometers to them, but no physicists that I'm aware of doubt what the result would be if we could run that experiment.

I still don’t really see how a point-like object fells anything. It’s not just a technicality.

Once again you've got this backwards. An object that is behind the Rindler horizon can't send a signal that will catch up with the *accelerator*. The Rindler horizon has nothing to do with where signals sent *from* the accelerator can or can't go. Why do you keep mixing this up?

I wasn’t talking about a Rindler horizon. I was talking about an event horizon and its flat space-time equivalent, c!

No, there is no such point in either case, flat or curved spacetime. The statement I made is correct exactly as I made it.

What? Now I’m really confused. There is a point in flat space-time when no signal sent from an object will reach an accelerator, when they pass the accelerators Riindler horizon. There’s also an equivalent in curved space-time when an object has passed the free-fallers equivalent of the Rindler horizon.

Since I disagree with you about this equivalence, obviously I'm going to disagree about any conclusions you draw from it. In standard GR the free-faller in curved spacetime is equivalent to the free-faller in flat spacetime, and the accelerator (hoverer) in curved spacetime is equivalent to the accelerator in flat spacetime. (D'oh.)

In curved space-time the hoverer is equivalent to the free-faller/inertial observer because their acceleration from energy and their acceleration from mass are balanced. That’s why they hover. An accelerator free-falling towards the black hole is equivalent to an accelerator in flat space-time.

The nature of the singularity is a separate question from whether or not the horizon can be reached and crossed, so I'd prefer to start a separate thread if you really want to debate it. But the statement I made is correct if you take standard GR at face value. Most physicists believe that the presence of the singularity is telling us that GR is no longer valid when you get too close to the singularity, and some new physics (like quantum gravity) comes into play. But we don't know for sure at this stage of our knowledge.

I think the presence of the singularity is telling us that matter has collapsed enough to prevent any other matter reaching a certain distance of that point in space-time in a finite amount of proper time because they’d have to move faster than c, creating an event horizon.

This looks fine to me except for the part about "curvature goes past 90 degrees". I don't think that's a good description of what happens at the horizon. 90 degrees relative to what? (If you had said "the outgoing side of the light cone goes past 90 degrees", that would be a little better, but I would still want you to clarify 90 degrees relative to what?)

The dimensions in flat space-time are at 90 degree angles to each other. The curvature of gravity/acceleration decreases the angle causing distance shortening. At c or an event horizon the angle is 0, so it’s changed by 90 degrees.

I don't disagree with this as it's stated but I don't think it means what you think it means. Obviously flat spacetime is not identical in every respect to curved spacetime; the fact that a curved spacetime can have an invariant event horizon that is there regardless of any observer's state of motion, while a flat spacetime can't, is one of the respects in which the two are not identical. But that doesn't prevent us from defining a family of "accelerators" in flat spacetime in such a way that they can be usefully viewed as equivalent to "hoverers" above a black hole in curved spacetime, with the Rindler horizon of the flat spacetime accelerators being equivalent to the event horizon of the black hole. It just means the analogy between the two is not complete in every respect. But so what? It's still a useful analogy. It's not meant to be anything more than that.

It is absolutely identical. I can’t stress that enough. When I use the analogy between the two pairs of horizons it’s meant to mean a lot more than that.

No, this is *not* what the equivalent would be. The equivalent in flat spacetime is the free-faller crossing the Rindler horizon of an appropriately defined accelerating observer. The free-faller does not catch up with his own light, in either the flat or the curved spacetime case, from any observer's perspective. The correct equivalent statement in the flat spacetime case is that from the perspective of the accelerating observer, in flat spacetime, the free-faller never crosses the Rindler horizon (again, with the appropriate definition of "the perspective of the accelerating observer"), but the free-faller does cross the Rindler horizon from his own perspective.

The four horizons:

The Rindler horizon in flat space-time is the point when no signal sent will be able to catch an accelerator. It’s starts far away and follows the accelerator at a constant distance if they keep their acceleration constant. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. The Rindler horizon can never catch up to them. It gets harder to close the gap the harder they accelerate.

It’s the equivalent of c. C is a horizon in front of the accelerator that starts far away, which the accelerator follows at a constant distance if they keep their acceleration constant. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. They can never catch up it. It gets harder to close the gap the harder they accelerate. The rate they gain on the light is the same in every frame, even accelerated ones. The two horizons would meet at the same place as the accelerator if they were able to reach c.

It’s the exact equivalent of curved space-time. The Rindler horizon in curved space-time is the point when no signal sent will be able to catch a free-faller. It’s starts far away and follows the free-faller, getting closer to them as the strength of gravity increases. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. The Rindler horizon equivalent can never catch up to them. It gets harder to close the gap the harder they free-fall.

It’s the equivalent of the event horizon. The event horizon starts far away in front of the free-faller, which the free-faller approaches at an increasing rate to start with as the strength of gravity increases, then at a decreasing rate as it gets harder to close the gap the harder the free-fall. They can never catch up it. It gets harder to close the gap the harder they free-fall. The rate they gain on the light is the same in every frame, even accelerated ones. The two horizons would meet at the same place as the free-faller if they were able to reach c.

This is all true, but as you show with your final phrase, you recognize that it applies to an observer that is being accelerated by energy. Such an observer *feels* the acceleration, and has a Rindler horizon and so forth only as long as he *feels* acceleration. When the energy that is accelerating him stops, he stops feeling acceleration.

None of that applies to an observer that is in free fall the whole time. Yes, in a coordinate sense such an observer can be "accelerated" by gravity, but he never *feels* any acceleration. This is why the physical distinction between observers in free fall and observers that feel acceleration is so crucial in standard GR. The rest of your discussion here denies this crucial distinction, and that is a fatal flaw in my view; it is simply not valid physics to claim that an observer that feels acceleration (which has lots of observable effects that aren't observed in free fall) is equivalent to an observer in free fall.

All of it applies to an observer that’s in free-fall the whole time. Just replace energy with gravity:

An object can keep on free-falling towards an event horizon for as long as the black hole lasts, making their space-time more and more ‘distance shortened’. They can use this to travel huge distances in space-time from the perspective of their starting frame it a short amount of proper time for them, but when they’ve stopped accelerating they never reached c from their original frame, they just kept on accelerating and covering more and more space-time over less and less proper time until the mass that was accelerating them stopped.

Tidal gravity can be present even though it is not felt by any observer. I have repeatedly described such scenarios.

How can tidal gravity be present but not felt? There’s no such thing as an actual point-like object except a singularity, and even they don’t exist in for any amount of time so they can’t feel anything.

This makes it fundamentally different from proper acceleration. The standard GR definition of "curvature" recognizes two distinct types of curvature, as I've said before. Tidal gravity is curvature of spacetime itself; proper acceleration is curvature of a particular worldline. They are different things.

How are they different things? Describing them differently doesn’t make them different.

Once they're below the horizon, they can accelerate all they want to but it won't bring them back above the horizon. It *will* cause them to fall more slowly (assuming they accelerate radially outward) than objects free-falling inward, but they will still be falling.

That’s not what I meant. They accelerate away before they reach the event horizon. If they can accelerate “all they want” then they can escape. That’s what accelerate all they want means! I don’t know why you are okay with this mystical attitude towards gravity? It accelerates objects in exactly the same way that energy does. Neither can accelerate you to c!

How can an object that is accelerating match velocities with an object that is free-falling, at least for more than a single instant?

Easy. They know the mass of the object pulling in the free-faller, so they know exactly how fast they accelerate needs to move relative to the hoverer to keep it symmetric.

I really wish you would draw a spacetime diagram showing how you think this works. It is true that, after a certain point on the hoverer's (O's) worldline, he will not be able to send any signal to Z that will reach Z before Z crosses the black hole horizon (call this point A for further use below). There is also a further point on O's worldline after which he will not be able to send a signal to Z that will reach Z before Z hits the singularity (call this point B for further use below). I suppose either of those points could be used to define a kind of "horizon" for Z, but I don't know if there would be any useful equivalence between such a horizon and a Rindler horizon. I would have to see more details (like a diagram) to better understand what you're imagining.

That IS the Rindler horizon! The diagram would look identical to a Rindler in flat space-time because a Rindler horizon is equivalent to THIS horizon and NOT an event horizon.

This is not correct. If O emits the light beam before point A on his worldline (as defined above), the light beam will pass Z before Z crosses the black hole horizon. If O emits the light beam before point B on his worldline, the beam will pass Z after Z crosses the black hole horizon, but before Z hits the singularity. Only if O emits the light beam after point B will it never reach Z (because Z will hit the singularity first). So in so far as there is a "horizon" associated with Z, it does pass the light beam if the light beam is emitted soon enough.

Z never crosses the event horizon. Z is the hoverer. I’ll call them F and H.

Hovering observer H emits a light beam *towards* the black hole. The light beam moves away from the free-falling observer in exactly the same way it would in flat space-time, then slows as it approaches the event horizon and from the perspective of H the light beam has a slower velocity relative to the light beam. The Rindler horizon passes the hoverer but not the light beam because you’re between it and the horizon and the horizon never reaches you, just like the event horizon and Rindler horizon and c in flat space-time.

Sorry, but even if all that was hard work, I can't say I can see it now, because none of it gets any additional useful information across to me. Why can't you draw a diagram?

A diagram of what?

On re-reading I realized I should comment on this in more detail. I see at least one glaring problem with this line of reasoning:

If the "strength of acceleration" gets reduced as the inverse square of the distance, then the *difference* in strength between points at two different distances, which is what you say is actually felt, goes as the inverse *cube* of the distance, times the difference in distances. The equivalent for gravity would be that, for example, the weight you read on your bathroom scale would have to be, not the "acceleration due to gravity" at Earth's surface, 9.8 meters per second squared, times your mass, but that force times the *ratio* of your height to the radius of the Earth. This ratio is about one to three million (assuming you are about 2 meters tall; the Earth's radius is about 6 million meters), so your reasoning gives an answer for an easily observable everyday fact, the weight people observe on their bathroom scales, that is too small by a factor of three million. Too bad you're allergic to math.

You’ve completely lost me. Why cube? All I meant was if the acceleration was distributed through the object to match exactly with the way gravity is distributed through a free-faller then proper acceleration and tidal force would be exactly equivalent.

 

[A-wal]

I said that this whole part of your post didn't convey any additional useful information to me, but on re-reading I do want to comment on this. If you mean this to describe what standard SR says (as opposed to something in your personal model that may or may not match what standard SR says), then it's wrong.

I wish you’d stop calling it my model. It’s worked the way it has forever whether I’m right or wrong. I didn’t make it. I found it, but apparently I’m the only sod who can see it so I’ve got to try to explain exactly what it looks like.

The accelerating ship *never* "catches up" to its own light from *either* perspective, its own or that of the free-faller that sees the accelerator's velocity approaching c but never quite reaching it. From *both* perspectives, a light ray emitted by the accelerating ship will continually move *away* from the ship; its distance from the ship will continually *increase*, from *both* perspectives, and the velocity of the light beam will always be faster than the velocity of the ship, from both perspectives. There is *never* a time when the distance from the accelerating ship to the light beam *decreases*, or when the accelerating ship appears to move faster than the light (which is what "catching up" would mean), from either perspective.

No, catchING. They can never actually match its speed. It will always pull away. As I’ve said repeatedly, it gets harder to close the gap the harder you accelerate. From the free-faller/accelerators perspective it’s because the ‘distance shortening’ would have to be infinite to reach c. From a hoverer/inertial observers perspective light moves at a constant speed and it’s because the free-faller/accelerators mass increases.

I wish you wouldn’t call an inertial observer a free-faller. It’s very misleading. It implies an object at rest is equivalent to an object being accelerated by gravity. Giving them the same name doesn’t make them equivalent.

Why can’t you apply that logic to an event horizon? Look:

The free-falling ship *never* "catches up" to its own light from *either* perspective, its own or that of the hoverer that sees the free-fallers velocity approaching c but never quite reaching it. From *both* perspectives, a light ray emitted by the free-falling ship will continually move *away* from the ship; its distance from the ship will continually *increase*, from *both* perspectives, and the velocity of the light beam will always be faster than the velocity of the ship, from both perspectives. There is *never* a time when the distance from the free-falling ship to the light beam *decreases*, or when the free-falling ship appears to move faster than the light (which is what reaching the event horizon would mean), from either perspective.


The accelerator almost reaches the speed of their own light from the perspective of the inertial observer, and from their own perspective. As they accelerate they send a series of signals, one every second of the accelerators proper time. The signals received by the inertial observer get progressively less frequent, but they never stop coming because the accelerator can never reach the speed of their own light.

The free-faller almost reaches the speed of their own light from the perspective of the hoverer, and from their own perspective. As they free-fall they send a series of signals, one every second of the free-fallers proper time. The signals received by the hoverer get progressively less frequent, but they never stop coming because the free-faller can never reach the event horizon.

I wasn't trying to claim that either proper time or coordinate time can "run backwards" from the viewpoint of an actual observer. The original question was about whether the laws of physics are time symmetric. If they are, that means, roughly speaking, that if I have a certain sequence of events S that conforms to the laws of physics, if I consider another sequence of events S', which is the exact reverse of S, S' must also conform to the laws of physics.

Doesn’t work you run an object crossing an event horizon backwards though does it?

This is incorrect. We discussed this already and I thought you had agreed on this point (see your post 471). There is no local inertial frame where the velocity of an object free-falling across the event horizon becomes c. I.e. a free-falling object's worldline is timelike at all points including the event horizon, it never becomes lightlike.

The event horizon is the point when the even the speed of light isn’t enough to move you towards a hoverer. It is exactly equivalent. Working out acceleration relative to a hoverer in free-fall is no different from acceleration relative to an inertial observer in flat space-time. You know how acceleration works, the fact that it gets harder to accelerate the faster your relative velocity. Why do you think gravity can accelerate an object to a relative velocity of c? That would take a huge jump, an infinite jump in fact, of energy. It couldn’t happen smoothly because up until the event horizon acceleration and gravity are equivalent because you can always move away using energy.

This doesn't make sense. c is a speed, it doesn't have a direction. What direction is 100 kph?

A lightlike object can move towards, away, or tangentially to an observer, and all of those motions are at c.

The event horizon moves inwards because gravity pulls inwards, and c moves outwards because energy pushes outwards. They both move away.

Because the equations match the results of many experiments in related situations. That is the only reason we should believe any equations of physics.

In particular, the Schwarzschild spacetime and it's GR predictions have been demonstrated to be a good model for the Pound-Rebka experiment, GPS, the Shapiro delay, the deflection of light, the anomalous precession of Mercury, and the Hafele-Keating experiment. GR is the simplest theory of gravity existing which quatitatively agrees with experiment.

What I’ve been saying is consistent with Schwarzschild space-time and it's GR predictions.


There’s still some important things I just don’t get. How does your model explain these:

(1) What happens if you free-fall towards the black hole with your light in front of you crossing the event horizon, then pull away before you reach the horizon? Your light would have to turn round and come back across the event horizon! WFT!

(2) How can two coordinate systems that say completely different things (one saying an object crosses a black hole and one saying it doesn’t) possibly be consistent with each other? I still haven’t had an explain of how they could both be right. This is the most glaring problem.

(3) What is the difference between flat and curved space-time? The reason you’re getting these ridiculous predictions about gravity is because GR distinguishes between the movement of an object and a change in the amount of space-time between them. What’s the difference? The difference between flat and curved space-time is in flat space-time it's the falling object that moves and gravity is a force, whereas in curved space-time the faller is at rest. It's relative. There is no distinction! Gravity works just like acceleration does!

(4) You know that an object in free-fall outside the event horizon can’t reach a velocity of c relative to an object accelerating away from the black hole at almost c. Gravity can keep on accelerating them but it gets harder as their relative velocity increases. The way ‘distance shortening’ works is to create less space-time for the accelerator to travel through from their perspective and more from the perspective of a hoverer/inertial observer. So the free-faller is constantly traveling through a smaller and smaller amount of space-time from their perspective compared to the perspective of a distant observer, just like accelerating in flat space-time. Do you really believe it makes sense if the tiniest amount more gravity could accelerate a free-faller to c relative to a hoverer when the energy of a hydrillion cyanoid big bangs couldn’t?

 

[PeterDonis]

What? That’s not what I’m saying. I meant that it would be the equivalent of c because it’s the point when c wouldn’t be fast enough to move away.

So far, so good.

It’s the point when gravity would have accelerated you all the way to c.

Nope. As DaleSpam and I have both said repeatedly, an object free-falling through the horizon is timelike, not lightlike. It never moves faster than c (it always moves within the light cones).

The speed that the event horizon moves inwards is c at the horizon

The event horizon moves outwards, not inwards.

The reason it moves inwards is because there’s a limit to how fast objects can move relative to other objects but there’s no limit to mass or energy, so it can keep on accelerating you towards it but you’ll never reach a relative velocity of c, so you get an event horizon.

I have no idea what this means. Why can't you draw a diagram?

What I meant was simply that if you could turn off gravity it would take no more effort to lift Mount Everest than it would to lift your arms, as long as you had the Earth to push against.

Really? How do you figure that? Suppose I'm in zero g (way out in deep space somewhere), and I run the following experiment: I pick a certain force, F, and I exert that force both on Mount Everest and on my arms. (How I exert the force is immaterial; I can exert it by pushing against the Earth, or by using a small rocket engine, or whatever.) Since Mount Everest's mass is much greater than the mass of my arms, the acceleration induced on Mount Everest by force F will be much smaller than that induced on my arms. So I don't see how "it would take no more effort to lift Mount Everest than it would to lift your arms" in zero g, unless by "no more effort" you simply mean I can exert the same force on both objects without caring that the accelerations induced are vastly different. But I don't think that's what you meant, because you say:

Objects with very different masses will move at the same speed if pushed in 0g

If by "pushed" you mean "pushed with the same force", this is false.

Do you mean that it gets ‘weighted’ so that B moves closer to a keeping the difference in gravitational strength between A and C the same as the difference in gravitational strength between C and B?

More or less, yes. But I am not trying to say that this will continue to hold regardless of how long the object is; I'm limiting discussion to objects which are very short compared to the size of the gravitational source.

Tidal gravity would separate the parts and if they had something that was holding them together then they’d feel tidal force. Tidal gravity causes any extended object (all of them) to feel tidal force. What’s the problem?

You just stated that the objects only feel tidal force when something is holding them together. So clearly the tidal *force* is not caused by tidal *gravity* (which would separate the parts), but by the something that's holding them together.

I’m trying to show that I think the proper acceleration could be smoothed out to match gravity, making proper acceleration and tidal force equivalent.

And I'm saying you haven't succeeded.

I still don’t really see how a point-like object fells anything. It’s not just a technicality.

I'll comment on this near the end when I discuss the definition of tidal gravity.

I wasn’t talking about a Rindler horizon. I was talking about an event horizon and its flat space-time equivalent, c!

No, the flat spacetime equivalent of the event horizon *is* the Rindler horizon. It's not "c"--saying the event horizon is "equivalent to c" doesn't even make sense.

What? Now I’m really confused. There is a point in flat space-time when no signal sent from an object will reach an accelerator, when they pass the accelerators Riindler horizon. There’s also an equivalent in curved space-time when an object has passed the free-fallers equivalent of the Rindler horizon.

You keep on mixing up the case of *outgoing* motion with the case of *ingoing* motion. The original statement of yours that I responded to was about *ingoing* motion--you were saying that there was a point where the hoverer couldn't send signals to the free-faller. Now you're talking about *outgoing* motion, about the free-faller sending signals to the hoverer. Those are two different cases, and if you would stop mixing them up it would help you to avoid confusion.

In curved space-time the hoverer is equivalent to the free-faller/inertial observer because their acceleration from energy and their acceleration from mass are balanced. That’s why they hover. An accelerator free-falling towards the black hole is equivalent to an accelerator in flat space-time.

Continuing to repeat this doesn't make it right. You still have not explained how an observer that feels acceleration (the hoverer) can possibly be physically equivalent to an observer that doesn't (the free-faller). Unless you can address that issue, what you are saying here is simply not valid physics, and no amount of repetition will make it so.

The dimensions in flat space-time are at 90 degree angles to each other. The curvature of gravity/acceleration decreases the angle causing distance shortening.

How do you figure this? I think you're confusing the "angles of the dimensions" (not very good terminology, but I think I understand what you mean by it), which do *not* change with curvature, with the angle between the outgoing and ingoing sides of the light cones, which *does* change with curvature. But even the latter does not collapse to zero at the horizon.

It is absolutely identical. I can’t stress that enough. When I use the analogy between the two pairs of horizons it’s meant to mean a lot more than that.

So you are claming that flat spacetime is absolutely identical to curved spacetime? Then how do you explain the fact that tidal gravity, an observable physical phenomenon, is present in curved spacetime but not in flat spacetime?

The Rindler horizon in flat space-time is the point when no signal sent will be able to catch an accelerator.

This is ok, although I'm not sure I understand the rest of what you say about it, with the acceleration changing. I think you're putting in too many complications at once, which makes it difficult to focus on particular issues.

C is a horizon in front of the accelerator that starts far away, which the accelerator follows at a constant distance if they keep their acceleration constant.

I have no idea what this means; I am not aware of any "horizon" in *front* of an accelerator in flat spacetime, much less one that somehow moves at c and yet remains at a constant distance in front of the accelerator. Why can't you draw a diagram?

The Rindler horizon in curved space-time is the point when no signal sent will be able to catch a free-faller.

And as I've repeatedly said, there is no such point, at least not in standard GR. Why can't you draw a diagram that shows how you think this works?

The event horizon starts far away in front of the free-faller, which the free-faller approaches at an increasing rate to start with as the strength of gravity increases,

Ok, more or less; it would be nice if you could define what "approaches at an increasing rate" actually means, because it might help you to see why this...

then at a decreasing rate as it gets harder to close the gap the harder the free-fall.

...is false.

All of it applies to an observer that’s in free-fall the whole time.

But an observer in flat spacetime who is accelerating due to "energy" is *not* in free fall! So how can he possibly be equivalent to a free-faller in curved spacetime (or anywhere else)?

How can tidal gravity be present but not felt?

I have repeatedly described how tidal gravity works. It causes freely falling trajectories to converge or diverge. That's the definition of tidal gravity. So by definition, it is not felt; it applies to objects in free fall. See next comment.

There’s no such thing as an actual point-like object

No, but there are certainly objects that are in free fall. Yes, "point-like object" is an idealization; it assumes that any internal structure of the object can be ignored, so it can be treated as though it were a single point located at its center of mass. The idealization works because the actual internal forces in objects in free fall are so small that they can be ignored for practical purposes in many problems; the motion of the object *is*, to within the precision of our measurements, the same as what we predict using the idealization of a "point-like object". If you weren't allergic to math you could actually work some problems and see this.

I suppose you could claim that, as long as there are *any* internal forces, then there is no such thing as true "free fall". However, even that would not get you off the hook; you would still have to make a quantitative prediction about what "tidal forces" your model says should be felt, and then actually measure the tiny internal forces that standard GR idealizes away in many cases, and show that those tiny internal forces are what your model predicts they should be. If you're able to do this, by all means go ahead.

How are they different things? Describing them differently doesn’t make them different.

Showing a physical observable that differs (whether or not a force is felt) does.

If they can accelerate “all they want” then they can escape. That’s what accelerate all they want means!

No it doesn't. "Accelerate all they want" means that the measured proper acceleration (measured by an accelerometer) is unbounded; it can be as large as you like. It does *not* mean that that unbounded proper acceleration can put you anywhere in spacetime that you want.

I don’t know why you are okay with this mystical attitude towards gravity? It accelerates objects in exactly the same way that energy does.

No, it doesn't. Acceleration due to energy causes you to feel acceleration (and you can measure it with an accelerometer). "Acceleration" due to gravity does not; an accelerometer reads zero. I don't know why you are okay with claiming that these two cases, with an obvious difference in an obvious physical observable, are somehow magically the same.

Easy. They know the mass of the object pulling in the free-faller, so they know exactly how fast they accelerate needs to move relative to the hoverer to keep it symmetric.

I don't see how this answers my question. We have one object that is feeling acceleration and one that is not. How can the two possibly match velocities for more than an instant?

That IS the Rindler horizon! The diagram would look identical to a Rindler in flat space-time because a Rindler horizon is equivalent to THIS horizon and NOT an event horizon.

No it isn't. Draw the diagram and you will see why.

Hovering observer H emits a light beam *towards* the black hole. The light beam moves away from the free-falling observer in exactly the same way it would in flat space-time, then slows as it approaches the event horizon and from the perspective of H the light beam has a slower velocity relative to the light beam. The Rindler horizon passes the hoverer but not the light beam because you’re between it and the horizon and the horizon never reaches you, just like the event horizon and Rindler horizon and c in flat space-time.

Again, draw the diagram.

A diagram of what?

Did you see the diagram I posted a while back? It was a spacetime diagram; time on the vertical axis and space (one dimension of it, but that's enough for what we're discussing) on the horizontal axis. Then a bunch of worldlines, plots of time vs. space for various objects and for any horizons that you claim are present. That's the kind of diagram I'm talking about. If you really have a clear mental picture of your model, you should be able to draw such diagrams for every case you've described.

You’ve completely lost me. Why cube? All I meant was if the acceleration was distributed through the object to match exactly with the way gravity is distributed through a free-faller then proper acceleration and tidal force would be exactly equivalent.

Once again, too bad you're allergic to math. If you weren't you would see that it's obvious that the derivative of 1/r^2 is proportional to 1/r^3. You are saying that "acceleration is distributed" as 1/r^2; that means the *change* in acceleration (its spatial derivative) is proportional to 1/r^3. So over a given spatial separation dr, the difference in acceleration, which is what you claim is actually felt, goes like 1/r^3 * dr.

If that's still too abstruse for you, think of it this way. I'm standing on the surface of the Earth, so the acceleration at my feet goes like 1/R^2, where R is the radius of the Earth. My height is h, which is much less than R. So at my head, the acceleration is 1/(R + h)^2. You are saying that it is the *difference* between these two that is what I actually feel as weight. Calculate the numbers and you'll see that the difference between those two numbers is a factor of about three million too small. All the bit about the cube was just a quick way of calculating that difference, roughly, without going through the full computation.

 

[PeterDonis]

I wish you’d stop calling it my model. It’s worked the way it has forever whether I’m right or wrong. I didn’t make it. I found it, but apparently I’m the only sod who can see it so I’ve got to try to explain exactly what it looks like.

Can you give a reference for where you found it? It doesn't resemble any model I've seen anywhere else, as far as I can tell.

No, catchING. They can never actually match its speed. It will always pull away.

If by "pull away" you mean "distance constantly increasing", then I have no issue, but that doesn't seem consistent with previous things you've said, which have strongly implied, at least to me, that you believe the accelerating ship can somehow *decrease* the distance between itself and the light beam if it accelerates long enough.

I wish you wouldn’t call an inertial observer a free-faller. It’s very misleading. It implies an object at rest is equivalent to an object being accelerated by gravity. Giving them the same name doesn’t make them equivalent.

I use the term free-faller as a simple physical description: a free-faller is an observer who doesn't feel any acceleration (accelerometer reads zero). Yes, that does imply that all such observers are equivalent in an important physical sense, even if one is at rest in flat spacetime and the other is moving solely under the influence of gravity in curved spacetime. That physical equivalence is one of the cornerstones of GR, so I don't have any intention of changing terminology. To me, it's you who are using misleading terminology, by saying that an object in free fall is being "accelerated by gravity" when it clearly feels zero acceleration.

The accelerator almost reaches the speed of their own light from the perspective of the inertial observer,

True.

and from their own perspective.

How so? From the accelerator's perspective, the light beam he emits always moves away from him at the same speed; his speed relative to it never changes. It's only from the inertial observer's perspective that the accelerator's speed changes relative to the light.

As they accelerate they send a series of signals, one every second of the accelerators proper time. The signals received by the inertial observer get progressively less frequent, but they never stop coming because the accelerator can never reach the speed of their own light.

You are mixing things up again. You're describing the accelerator sending signals to the free-faller, but as we'll see in a moment, that is *not* the analogous situation to the free-faller sending signals to the hoverer in curved spacetime.

If we turn this situation around and consider a free-faller sending signals to the accelerator, once every second of his proper time, we will find that the signals from the free-faller *do* stop coming; there is a "last signal" that the free-faller sends, the last second of his proper time before he crosses the accelerator's Rindler horizon and no signal he sends can reach the accelerator any more.

The free-faller almost reaches the speed of their own light from the perspective of the hoverer,

Again, true.

and from their own perspective.

Again, how so? From the free-faller's perspective, the light always moves away from him at c. It's only from the hoverer's perspective that the free-faller's speed appears to approach c. Furthermore, this similarity between the free-faller and the accelerator in flat spacetime is misleading, because this...

As they free-fall they send a series of signals, one every second of the free-fallers proper time. The signals received by the hoverer get progressively less frequent, but they never stop coming because the free-faller can never reach the event horizon.

...is *not* true! As with the free-faller crossing the Rindler horizon of an accelerator in flat spacetime, if a free-faller falling into a black hole sends signals outward to a hoverer, once every second of his proper time (remember, you specified *proper* time), then those signals *will* stop coming after a finite time by the hoverer's clock! There will be a "last signal" that the free-faller sends, the last second of his proper time before he crosses the horizon and no signal he sends can reach the hoverer any more.

Doesn’t work you run an object crossing an event horizon backwards though does it?

Sure it does. The time reverse of a black hole horizon is a "white hole" horizon, where objects can pass outward but not inward.

 

[Dale]

The event horizon is the point when the even the speed of light isn’t enough to move you towards a hoverer.

While this is true it does not imply that a free-faller becomes lightlike.

It is exactly equivalent. Working out acceleration relative to a hoverer in free-fall is no different from acceleration relative to an inertial observer in flat space-time.

You seem to have the analogy backwards. The free falling Schwarzschild observer is analogous to the inertial observer in flat spacetime, and the hovering Schwarzschild observer is analogous to the accelerating observer in flat spacetime.

The event horizon moves inwards because gravity pulls inwards

I don't understand your reasoning here, this is in fact why the event horizon moves outward locally.

What I’ve been saying is consistent with Schwarzschild space-time and it's GR predictions.

Then derive what you have been saying using GR.

(1) What happens if you free-fall towards the black hole with your light in front of you crossing the event horizon, then pull away before you reach the horizon? Your light would have to turn round and come back across the event horizon! WFT!

I don't know why you think this. Light isn't attached to you by some sort of rubber band that would make it turn around simply because you did. This is a silly idea and certainly is not a prediction of GR.

(2) How can two coordinate systems that say completely different things (one saying an object crosses a black hole and one saying it doesn’t) possibly be consistent with each other? I still haven’t had an explain of how they could both be right. This is the most glaring problem.

You have had many explanations of this, but just didn't understand any of them. Your lack of comprehension is not a failure of GR, it is a failure of you to even want to learn the required math. If you choose not to make the effort to learn the math that is fine, but then you have chosen not have the qualifications to assess the self-consistency of GR. You cannot have it both ways. The mathematical framework is precisely what ensures the self-consistency of GR, so if you avoid the math then you cannot claim that inconsistencies exist.

(3) What is the difference between flat and curved space-time?

The difference between flat and curved space-time is the presence of tidal forces.

(4) You know that an object in free-fall outside the event horizon can’t reach a velocity of c relative to an object accelerating away from the black hole at almost c. Gravity can keep on accelerating them but it gets harder as their relative velocity increases. The way ‘distance shortening’ works is to create less space-time for the accelerator to travel through from their perspective and more from the perspective of a hoverer/inertial observer. So the free-faller is constantly traveling through a smaller and smaller amount of space-time from their perspective compared to the perspective of a distant observer, just like accelerating in flat space-time. Do you really believe it makes sense if the tiniest amount more gravity could accelerate a free-faller to c relative to a hoverer when the energy of a hydrillion cyanoid big bangs couldn’t?

No, I don't believe it makes sense.

 

[PeterDonis]

You seem to have the analogy backwards.

He certainly does. I've been going back and forth with him about it for many posts now.

 

[A-wal]

I have got a half written reply but I’m not going to be finishing it any time soon so I’ll get straight to the point.

You claim that objects can reach even horizons in a finite amount of their own proper time. If a line of objects were continuously falling into a black hole like a conveyer belt then this would cause a few problems. None of the objects in front can reach the horizon before they themselves do. So they all have to cross the horizon at the same time. If everything that will have to reach a horizon has to do it at a specific time then at what point in the black holes life will that be? Just after it's gone!

If we do the same thing with a row of accelerating objects that accelerate harder the closer they are to a distant object and started them off equally spaced then they would start to separate, and at a quicker rate the closer they are to the object that they're heading towards. An objects velocity is always c / the distance squared to the destination object. None of these objects would be able to reach the stationary object that they're heading towards. They'd just get more and more time dilated and length contracted as they get closer at a slower and slower rate from any distance.

Using the river model; there will never be a point when the river is moving at or faster than c relative to the singularity. It's no different than adding relative velocities in flat space-time.

15-20 minutes that took. That's better.

 

[Dale]

Hi A-wal, welcome back!

You claim that objects can reach even horizons in a finite amount of their own proper time. If a line of objects were continuously falling into a black hole like a conveyer belt then this would cause a few problems. None of the objects in front can reach the horizon before they themselves do. So they all have to cross the horizon at the same time.

You are mixing up proper time and Schwarzschild coordinate time. Proper time alone cannot be used to establish an ordering of events on different worldlines. You need some synchronization convention for that.

Using the Schwarzschild synchronization convention they all reach the horizon in the limit as t->∞. If you want to think of that as "at the same time" then go ahead.

 

[A-wal]

Still, no object in front of you can reach the horizon before you do, meaning all objects reach the horizon at the exact moment that you do. This clearly doesn't work. It's like trying to reach c in the other example. If they were able to reach the object they're accelerating towards then they'd all do it at the same time.

Hi A-wal, welcome back!

Thanks. I seem to be able to think clearer when I'm in pain.

 

[PAllen]

Still, no object in front of you can reach the horizon before you do

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

 

[Passionflower]

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

You mean red shifted.

 

[PAllen]

You mean red shifted.

Yes, I was thinking of static observers near the horizon, though discussing in-falling observers. For in-falling observers, the reality is redshift/blue shift of distant stars is direction dependent. If falling straight in, maximum red shift for those behind you, maximum blue shift for those in front of you, others in between.

 

[Passionflower]

Yes, I was thinking of static observers near the horizon, though discussing in-falling observers. For in-falling observers, the reality is redshift/blue shift of distant stars is direction dependent. If falling straight in, maximum red shift for those behind you, maximum blue shift for those in front of you, others in between.

By the way radially free falling observers can observe far away stars blue shifted, but the closer they get to the EH the less of a possibility that is as their proper velocity based Doppler factor overruns the gravitationally based Doppler factor.

One could plot this out taking several radially free falling observers with different energies. For all the proper velocity at the EH will be c but their proper velocities on their way to the EH will differ and thus the red/blue shift factor wrt to far away stars will also.

 

[PAllen]

By the way radially free falling observers can observe far away stars blue shifted, but the closer they get to the EH the less of a possibility that is as their proper velocity based Doppler factor overruns the gravitationally based Doppler factor.

One could plot this out taking several radially free falling observers with different energies. For all the proper velocity at the EH will be c but their proper velocities on their way to the EH will differ and thus the red/blue shift factor wrt to far away stars will also.

Right. If instead of in-falling from far, one imagined the radial in-fall of static, near horizon, observer whose rocket ran out of fuel, they would initially see all stars blue-shifted.

 

[Passionflower]

Right. If instead of in-falling from far, one imagined the radial in-fall of static, near horizon, observer whose rocket ran out of fuel, they would initially see all stars blue-shifted.

Exactly.

 

[PeterDonis]

If we do the same thing with a row of accelerating objects that accelerate harder the closer they are to a distant object and started them off equally spaced then they would start to separate, and at a quicker rate the closer they are to the object that they're heading towards. An objects velocity is always c / the distance squared to the destination object. None of these objects would be able to reach the stationary object that they're heading towards. They'd just get more and more time dilated and length contracted as they get closer at a slower and slower rate from any distance.

I don't understand what you're describing here. You are saying that the accelerating objects are supposed to be accelerating *towards* the distant object? If so, how are you coming up with the conclusion that they can never reach it?

Also, what accelerations are the objects feeling? Are they all feeling the same acceleration? Or does the acceleration they feel vary with their starting distance from the distant object? If so, how?

 

[Dale]

Still, no object in front of you can reach the horizon before you do, meaning all objects reach the horizon at the exact moment that you do.

You have to define "in front" and "before". If you use coordinate independent definitions (e.g. Based on light cones) then you will get coordinate independent results. If you simply use the Schwarzschild coordinates then obviously you get coordinate-dependent results. There is nothing particularly meaningful about that.

 

[Dale]

So they all have to cross the horizon at the same time.

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think that you would have to find the difference in coordinate time between when the two observers cross a given radius and then take the limit of that as the radius approaches the Schwarzschild radius. That limit might be zero, but it isn't obvious. In any case it is coordinate dependent

 

[A-wal]

Wrong. For a chain of infalling observers, each would see the lower object crossing the horizon 'a little' before they cross, all in very finite time on their own 'watch'. Assuming an ancient, stable, super-massive black hole, nothing would be distinguishable about the horizon, locally. Starting from well before the horizon, distant stars would be blue shifted and optically distorted, but there would be no further sudden change reflecting passage of the horizon.

That can’t be right. If they see them fall past the horizon before they reach it themselves then the object that fell in would have to come back out if the other one pulled away.

I don't understand what you're describing here. You are saying that the accelerating objects are supposed to be accelerating *towards* the distant object? If so, how are you coming up with the conclusion that they can never reach it?

Because if they reached it then they would be moving at c relative to it. They become more and more ‘distance shortened’ as they get closer.

Also, what accelerations are the objects feeling? Are they all feeling the same acceleration? Or does the acceleration they feel vary with their starting distance from the distant object? If so, how?

They’re accelerating completely smoothly so that each individual atom is accelerating independently based on its distance from the object. The only acceleration they feel is the difference between the different parts of those objects, which doesn’t become noticeable until they get fairly close.

You have to define "in front" and "before". If you use coordinate independent definitions (e.g. Based on light cones) then you will get coordinate independent results. If you simply use the Schwarzschild coordinates then obviously you get coordinate-dependent results. There is nothing particularly meaningful about that.

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think that you would have to find the difference in coordinate time between when the two observers cross a given radius and then take the limit of that as the radius approaches the Schwarzschild radius. That limit might be zero, but it isn't obvious. In any case it is coordinate dependent

From the perspective of an in-falling observer; no observer in front of you can possibly reach the horizon before you do!

 

[PeterDonis]

Because if they reached it then they would be moving at c relative to it. They become more and more ‘distance shortened’ as they get closer.

I still don't understand. Are you trying to describe (1) what you think GR actually says, or (2) what your non-GR model says, or (3) what happens in a scenario in flat spacetime (no gravity) that you think is analogous to the scenario of a black hole?

If it's #1, your description is simply wrong. GR does not say what you are claiming it says.

If it's #2, how does your model make the predictions you are describing?

If it's #3, then I don't see how the following fits in:

They’re accelerating completely smoothly so that each individual atom is accelerating independently based on its distance from the object. The only acceleration they feel is the difference between the different parts of those objects, which doesn’t become noticeable until they get fairly close.

This can't happen in flat spacetime; you're basically saying the only "acceleration" felt by the objects is tidal acceleration, but there is no tidal acceleration in flat spacetime. Which brings me back to either #1 or #2 above, but see my questions about those.

 

[PeterDonis]

Actually, now that I think about it I am not sure this is true even in Schwarzschild coordinates. Infinity minus infinity isn't 0, it is undefined.

I think this is a better way of putting it, because the important point, to me, is that Schwarzschild coordinates are singular at the horizon. That means you can't use them to even describe relationships like "before", "after", "in front of", "in back of" at the horizon in the first place. A-wal is assuming that t = infinity at the horizon in Schwarzschild coordinates implies that all infalling objects reach the horizon "at the same time", when what it actually means is that Schwarzschild coordinates can't be used to describe the horizon at all.

 

[PAllen]

That can’t be right. If they see them fall past the horizon before they reach it themselves then the object that fell in would have to come back out if the other one pulled away.

No, what I said is correct. The answer to your objection is simply that if A and B are a pair of infallers, with B at a little higher r value, then B sees (absorbs) light from A crossing the horizon at the moment B crosses the horizon. So, indeed, no light leaves the horizon. This is no different from any normal situation where if I see light from you now, and you are moving in a known direction, I interpret that 'now' you are further in the direction you were moving. That is, when B crosses the horizon, it looks like A crossed before (and A is, indeed, inside already), and is now a bit inside. Note that light emitted directly away from the singularity at the horizon remains frozen at the horizon - waiting, if you will, for the next infaller.

If B pulls away any time before crossing the horizon, then, indeed, they never see light from A at or inside the horizon. This would look a lot like a Rindler horizon. B suddenly accelerates madly to escape, A gets red shifted and frozen, just like a Rindler horizon.

One final observation is that B never sees A reach the singularity. At the moment B reaches the singularity, the last light they see from A is from a little before A reached the singularity.

 

[A-wal]

I still don't understand. Are you trying to describe (1) what you think GR actually says, or (2) what your non-GR model says, or (3) what happens in a scenario in flat spacetime (no gravity) that you think is analogous to the scenario of a black hole?

If it's #1, your description is simply wrong. GR does not say what you are claiming it says.

If it's #2, how does your model make the predictions you are describing?

If it's #3, then I don't see how the following fits in:

It’s 3.

This can't happen in flat spacetime; you're basically saying the only "acceleration" felt by the objects is tidal acceleration, but there is no tidal acceleration in flat spacetime. Which brings me back to either #1 or #2 above, but see my questions about those.

I’m saying that they have energy applied to them in such a way that any object or individual part of any of the objects has a velocity equal to c / the distance between that part of the object and the object they’re all heading towards, squared. In terms of actual acceleration it’s slightly different because the individual parts of an object are held together by an electro-magnetic field which creates resistance, which is felt as tidal force in the first scenario and proper acceleration in the second one.

I think this is a better way of putting it, because the important point, to me, is that Schwarzschild coordinates are singular at the horizon. That means you can't use them to even describe relationships like "before", "after", "in front of", "in back of" at the horizon in the first place. A-wal is assuming that t = infinity at the horizon in Schwarzschild coordinates implies that all infalling objects reach the horizon "at the same time", when what it actually means is that Schwarzschild coordinates can't be used to describe the horizon at all.

I’m assuming that if no object in front of you can reach the horizon before you do from the perspective of an in-falling observer, then obviously if you can’t reach the horizon before they do neither then they all reach at the same time. To say that they reach the horizon separately in a finite time own their own clocks doesn’t make sense. Either make it make sense or admit that you’ve all made a terrible mistake.

No, what I said is correct. The answer to your objection is simply that if A and B are a pair of infallers, with B at a little higher r value, then B sees (absorbs) light from A crossing the horizon at the moment B crosses the horizon. So, indeed, no light leaves the horizon. This is no different from any normal situation where if I see light from you now, and you are moving in a known direction, I interpret that 'now' you are further in the direction you were moving. That is, when B crosses the horizon, it looks like A crossed before (and A is, indeed, inside already), and is now a bit inside. Note that light emitted directly away from the singularity at the horizon remains frozen at the horizon - waiting, if you will, for the next infaller.

Right. So you’re saying when B crosses the horizon B sees the light from A (and so presumably all the other objects that went in before them) cross at the same time that they do, and then they all jump to whatever distance resembles the last frame that made sense, when they were all in a line?

If B pulls away any time before crossing the horizon, then, indeed, they never see light from A at or inside the horizon. This would look a lot like a Rindler horizon. B suddenly accelerates madly to escape, A gets red shifted and frozen, just like a Rindler horizon.

Then you agree that A cannot cross the horizon until B does?

One final observation is that B never sees A reach the singularity. At the moment B reaches the singularity, the last light they see from A is from a little before A reached the singularity.

If you’d replace the word singularity with the word horizon in those sentences then I’d agree. The singularity is the horizon at zero distance.

 

[PAllen]

Right. So you’re saying when B crosses the horizon B sees the light from A (and so presumably all the other objects that went in before them) cross at the same time that they do, and then they all jump to whatever distance resembles the last frame that made sense, when they were all in a line?

If I'm on a train with a rows of seats ahead of me, at any moment I see light from all rows ahead of me. I image it so it looks likes the rows are spaced out in front of me (which is the correct reality). The moment of B crossing the horizon is no different. They see light from A (and prior infallers) and it images just like rows of seats on a train.

Then you agree that A cannot cross the horizon until B does?

No, I completely disagree. Light and the emitter of light (A) are not the same thing. The event of B crossing the horizon and receiving light from when A crossed, is not the same event as the event of A crossing the horizon. This is no different from any normal situation - my receiving your signal is different from you sending it. From A's point of view, A has moved on before B crosses the horizon. A can see when B crosses the horizon after (behind) him (if they are close enough). B also sees (visually interprets, correctly) that A crossed before.

If you’d replace the word singularity with the word horizon in those sentences then I’d agree. The singularity is the horizon at zero distance.

I was speaking of the actual singularity, not the horizon. See above for the horizon behavior. The actual singularity is really a time, not a place. B is seeing A some finite distance away the moment B ceases to exist 'when the singularity occurs' for B.

 

[A-wal]

If I'm on a train with a rows of seats ahead of me, at any moment I see light from all rows ahead of me. I image it so it looks likes the rows are spaced out in front of me (which is the correct reality). The moment of B crossing the horizon is no different. They see light from A (and prior infallers) and it images just like rows of seats on a train.

That doesn’t work! They can’t see the light from any prior in-fallers crossing the horizon before they themselves reach it because if they were then to pull away, they would observer any objects that they saw cross coming back out from inside the horizon.

No, I completely disagree. Light and the emitter of light (A) are not the same thing. The event of B crossing the horizon and receiving light from when A crossed, is not the same event as the event of A crossing the horizon. This is no different from any normal situation - my receiving your signal is different from you sending it. From A's point of view, A has moved on before B crosses the horizon. A can see when B crosses the horizon after (behind) him (if they are close enough). B also sees (visually interprets, correctly) that A crossed before.

How? A would have to jump forward after B reaches the horizon. ?

I was speaking of the actual singularity, not the horizon. See above for the horizon behavior. The actual singularity is really a time, not a place. B is seeing A some finite distance away the moment B ceases to exist 'when the singularity occurs' for B.

I know what you meant. It’s an infinitesimally small portion of space-time. The closest anything can ever get to it is arbitrarily close to the event horizon.

 

[Passionflower]

A-Wal you are completely mistaken.
Perhaps it will help if you study Gullstrand–Painlevé coordinates.

It seems you completely mix up that some observers cannot see an event with an event actually happening, which is the whole point of an event horizon.

Suit yourself, but you will not learn if you do not accept what people are trying to tell you here.

 

[PAllen]

That doesn’t work! They can’t see the light from any prior in-fallers crossing the horizon before they themselves reach it because if they were then to pull away, they would observer any objects that they saw cross coming back out from inside the horizon.

How? A would have to jump forward after B reaches the horizon. ?

I know what you meant. It’s an infinitesimally small portion of space-time. The closest anything can ever get to it is arbitrarily close to the event horizon.

Let me try one last(?) to explain. Let's introduce C as well as A and B. A, B and C are free falling directly towards a supermassive black hole, A starts out closer than B, who is closer than C. Let's also be clear they are quite close together (it makes things simpler). From A,B,C perspective, the black hole horizon is rushing ever faster toward them. When it reaches A, it is moving at the same speed as the light A emits towards B. When it (the horizon) reaches B, it is moving at the same speed as the light from B towards C. To clarify further what C sees, let's introduce a fanciful element - that anything crossing the event horizon turns pink. So what C sees is A and B some distance away, pink horizon rushing towards them. At some moment, though A and B continue to appear some distance away, perfectly normal, they both turn pink at the same time (and so does C), from C's point of view. The visual 'turning pink all at once' occurs simply because the horizon is moving at the speed of light past A, then B, then C.

[EDIT: Note that if C interprets what he sees in a normal way, then even though he 'sees' A and B turn pink simultaneously, he reasons that since A is further away than B, A must have turned pink slightly before B. If it were really simultaneous from C's point of view, C would expect to see B turn pink first.]

[EDIT2: And to show the similarity to a Rindler horizon, let's say, instead of a horizon, we simply have A, B, and C sitting in a row. A light source beyond A (in the same line) suddenly starts emitting pink light. C visually sees A, B and himself turn pink at the same time, but interprets that A turned pink, then B, then C. Up until the moment the pink light reaches C, C has the option of starting frantic acceleration away from B. Given sufficient acceleration, the pink light never reaches C, and C never sees A and B turn pink (though they do). This is a Rindler horizon and it is quite analogous to the black hole horizon.]

 

[PeterDonis]

It’s 3.

I’m saying that they have energy applied to them in such a way that any object or individual part of any of the objects has a velocity equal to c / the distance between that part of the object and the object they’re all heading towards, squared.

As you've stated it, this makes no sense. The units don't even work; c is a velocity, so you can't divide it by a distance to get another velocity.

Anyway, in flat spacetime it's impossible for an object to be accelerating towards another object which is not accelerating itself, and not reach the other object eventually. So whatever scenario you think you're constructing as an analogy, it won't work.

 

[Dale]

I’m assuming that if no object in front of you can reach the horizon before you do from the perspective of an in-falling observer, then obviously if you can’t reach the horizon before they do neither then they all reach at the same time. To say that they reach the horizon separately in a finite time own their own clocks doesn’t make sense. Either make it make sense or admit that you’ve all made a terrible mistake.

I thought you were basing your conclusions on Schwarzschild coordinates, which definitely do not represent the perspective of an in-falling observer. If you want to do that then you would be much better off using GP coordinates as Passionflower suggested. However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the persepective of an in-falling observer.

 

[Passionflower]

However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the perspective of an in-falling observer.

Indeed, radially in-falling at escape velocity to be exact.

 

[Dale]

Good point, you could certainly consider radial free-falling observers that were falling at velocities other than escape velocity. The GP coordinates would not reflect their perspective.

 

[A-wal]

A-Wal you are completely mistaken.
Perhaps it will help if you study Gullstrand–Painlevé coordinates.

It seems you completely mix up that some observers cannot see an event with an event actually happening, which is the whole point of an event horizon.

Suit yourself, but you will not learn if you do not accept what people are trying to tell you here.

What people are trying to tell me here is that I need to use coordinate systems to put it into terms they understand. I’m not talking about what they see. I’m talking about crossing an event horizon. You can account for the delay in what they see.

Let me try one last(?) to explain. Let's introduce C as well as A and B. A, B and C are free falling directly towards a supermassive black hole, A starts out closer than B, who is closer than C. Let's also be clear they are quite close together (it makes things simpler). From A,B,C perspective, the black hole horizon is rushing ever faster toward them. When it reaches A, it is moving at the same speed as the light A emits towards B. When it (the horizon) reaches B, it is moving at the same speed as the light from B towards C. To clarify further what C sees, let's introduce a fanciful element - that anything crossing the event horizon turns pink. So what C sees is A and B some distance away, pink horizon rushing towards them. At some moment, though A and B continue to appear some distance away, perfectly normal, they both turn pink at the same time (and so does C), from C's point of view. The visual 'turning pink all at once' occurs simply because the horizon is moving at the speed of light past A, then B, then C.

[EDIT: Note that if C interprets what he sees in a normal way, then even though he 'sees' A and B turn pink simultaneously, he reasons that since A is further away than B, A must have turned pink slightly before B. If it were really simultaneous from C's point of view, C would expect to see B turn pink first.]

C sees that A is further away than B? For what you’re saying to make any sense you must mean that C is able to see A and B cross the event horizon? If the answer is yes then that’s not what I’ve been told before, and if it’s no then the horizon can’t be reached by A or B before it’s reached by C, so everything reaches it at the same time, never.

[EDIT2: And to show the similarity to a Rindler horizon, let's say, instead of a horizon, we simply have A, B, and C sitting in a row. A light source beyond A (in the same line) suddenly starts emitting pink light. C visually sees A, B and himself turn pink at the same time, but interprets that A turned pink, then B, then C. Up until the moment the pink light reaches C, C has the option of starting frantic acceleration away from B. Given sufficient acceleration, the pink light never reaches C, and C never sees A and B turn pink (though they do). This is a Rindler horizon and it is quite analogous to the black hole horizon.]

That’s the equivalent to the point when no signal sent will be able to catch a free-faller.

As you've stated it, this makes no sense. The units don't even work; c is a velocity, so you can't divide it by a distance to get another velocity.

Of course you can.

Anyway, in flat spacetime it's impossible for an object to be accelerating towards another object which is not accelerating itself, and not reach the other object eventually. So whatever scenario you think you're constructing as an analogy, it won't work.

Okay, so you tell me how much proper time it would take me to reach an object 100 light years away if my velocity were equal to c / the distance between me and it, squared?

I thought you were basing your conclusions on Schwarzschild coordinates, which definitely do not represent the perspective of an in-falling observer. If you want to do that then you would be much better off using GP coordinates as Passionflower suggested. However, those results would still be coordinate-dependent, but at least they would be coordinates that reflect the persepective of an in-falling observer.

I'm simply talking from the perspective of the free-faller. I don't know enough about coordinate systems to tell you which ones you should be using.

 

[PAllen]

C sees that A is further away than B? For what you’re saying to make any sense you must mean that C is able to see A and B cross the event horizon? If the answer is yes then that’s not what I’ve been told before, and if it’s no then the horizon can’t be reached by A or B before it’s reached by C, so everything reaches it at the same time, never.

You have a chain of infallers. At all times C sees A further away than B. What I described is correct. If you want to simply deny the factual predictions of GR, there is obviously nothing to discuss. Please read and think about what I wrote. The moment C crosses, they see that A and B crossed before. They can't see that A and B crossed until they crossed for the simple reason that the horizon is keeping pace with the light from A at the point where A crossed, and also in step with the light from B when B crossed.

That’s the equivalent to the point when no signal sent will be able to catch a free-faller.

No, it's equivalent to the fact that until the infaller crosses the horizon (from the free faller's point of view: until the horizon passes them at the speed of light), they have the choice to start accelerating frantically to stay 'ahead' of the event horizon that would pass them at the speed of light if they did not accelerate away from it. If they do so, they will never see light from beyond the horizon. The every day scenario I described of an expanding light front captures all essential features of a chain of free fallers approaching (being approached by) the event horizon.

 

[PeterDonis]

Of course you can.

You have got to be either kidding or extremely confused. See next comment.

Okay, so you tell me how much proper time it would take me to reach an object 100 light years away if my velocity were equal to c / the distance between me and it, squared?

I can't because, as I said, if I divide c by a distance I don't get a velocity. c is in meters per second, or m s^-1; distance is in meters. If I divide c by a distance I get s^-1, which is a frequency, not a velocity. If I square it, I get a frequency squared, not a velocity. Or if I interpret the "squared" as applying to the "distance" part only, I divide m s^-1 by m^2, which gives me m^-1 s^-1, which is something I don't even think there's a common word for. It's certainly not a velocity. So what you're asking makes no sense.

 

[A-wal]

You have a chain of infallers. At all times C sees A further away than B. What I described is correct. If you want to simply deny the factual predictions of GR, there is obviously nothing to discuss. Please read and think about what I wrote. The moment C crosses, they see that A and B crossed before. They can't see that A and B crossed until they crossed for the simple reason that the horizon is keeping pace with the light from A at the point where A crossed, and also in step with the light from B when B crossed.

Your explanation doesn’t work. You say that at the moment an in-falling observer reaches the horizon they see all the objects in front of them at the exact instant that they reached the horizon? The horizon’s moving outwards, so you’re seeing them as they were crossing the horizon when the horizon was where they are now? So the horizon is everywhere inside the black hole all at once? Right, so the horizon moves outwards (wrong btw, it’s moving inwards by the time you ‘see’ it) at c locally, and obviously slower the further you are away from it? Two problems.

First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong.

And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself.

No, it's equivalent to the fact that until the infaller crosses the horizon (from the free faller's point of view: until the horizon passes them at the speed of light), they have the choice to start accelerating frantically to stay 'ahead' of the event horizon that would pass them at the speed of light if they did not accelerate away from it. If they do so, they will never see light from beyond the horizon. The every day scenario I described of an expanding light front captures all essential features of a chain of free fallers approaching (being approached by) the event horizon.

No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them.

You have got to be either kidding or extremely confused. See next comment.

I don't think I'm the one getting confused.

I can't because, as I said, if I divide c by a distance I don't get a velocity. c is in meters per second, or m s^-1; distance is in meters. If I divide c by a distance I get s^-1, which is a frequency, not a velocity. If I square it, I get a frequency squared, not a velocity. Or if I interpret the "squared" as applying to the "distance" part only, I divide m s^-1 by m^2, which gives me m^-1 s^-1, which is something I don't even think there's a common word for. It's certainly not a velocity. So what you're asking makes no sense.

Just give them a starting velocity of zero and make them accelerate smoothly so that their velocity stays equal to c / the distance between them and the object they’re heading towards, squared. If you halve the distance you multiply it by four, because it’s squared. WTF wouldn’t you be able to do that? Start them at c and work backwards to zero where they start. Now work out the proper time it would take to reach the object. It's infinite! I thought I was the one who was crap at maths.

 

[PAllen]

Your explanation doesn’t work. You say that at the moment an in-falling observer reaches the horizon they see all the objects in front of them at the exact instant that they reached the horizon? The horizon’s moving outwards, so you’re seeing them as they were crossing the horizon when the horizon was where they are now? So the horizon is everywhere inside the black hole all at once? Right, so the horizon moves outwards (wrong btw, it’s moving inwards by the time you ‘see’ it) at c locally, and obviously slower the further you are away from it? Two problems.

First the event horizon can only move slower than the speed of the light from the objects in front of you until it reaches you, so you would still only be able to see them cross the horizon before the horizon reaches you, which is wrong.

And also if as you cross you’re seeing everything as it crosses the horizon then the horizon moves depending on what you’re looking at. It can only be in one place at a time. If it’s right in front of you for example and the in-falling objects are all still neatly lined up at varying distances then you’ve seen the ones in front of you cross the horizon! Stop trying to have it both ways. You’re contradicting yourself.

No it isn’t. That happens with free-fallers too. It’s exactly the same. There’s a point when no signal sent will reach a free-faller. That horizon is equivalent the Rindler horizon and it’s further away the further away the free-faller is. It gets closer to them as they fall in, but it never overtakes or reaches them.

We have reached the point where you are simply making false statements with no basis. You are doing the equivalent of saying arithmetic does not state that 1+1=2 because you disagree. I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

 

[PeterDonis]

Just give them a starting velocity of zero and make them accelerate smoothly so that their velocity stays equal to c / the distance between them and the object they’re heading towards, squared.

Now you're saying something *different*, which still doesn't make sense. Call the moving object O and the target object T. Suppose O starts out 1000 meters from T. You say the starting velocity is zero; but c divided by 1000 meters, or c divided by 1000 meters squared, is not zero (even if I ignore the fact that the units don't work). So I can't start O out consistently with what you're saying, even if I ignore the units (which I can't because the units are part of the physics, and if they aren't balancing, there is something wrong).

If you halve the distance you multiply it by four, because it’s squared.

In which case the velocity would go to infinity as O approached T (see below). This is inconsistent with SR, which you have said you accept.

WTF wouldn’t you be able to do that? Start them at c and work backwards to zero where they start.

*Now* you're saying something different from the above *again*. You're saying that the velocity of O should be c when it is co-located with T. But when it is co-located with T, its distance from T is zero, so going by your previous prescription, its velocity would be c / 0 = infinite.

It's infinite! I thought I was the one who was crap at maths.

You are certainly not winning any prizes for logic or clarity.

 

[PeterDonis]

I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

Welcome to the reason why this thread is 540 posts long. :sigh:

 

[PeterDonis]

Call the moving object O and the target object T. Suppose O starts out 1000 meters from T.

I suppose I ought to try and expedite this discussion by laying out what SR actually says about this type of scenario, for what it's worth. A-wal appears to be saying that O should start out with velocity zero, and should accelerate towards T. That, at least, is a scenario capable of consistent analysis. However, he also appears to want to stipulate that O should accelerate in such a way that its velocity is c when its distance from T is zero, which of course is impossible. I'm pretty sure A-wal knows *that*, anyway.

A-wal appears to want to draw the conclusion, however, that all this proves somehow that O can't reach T. In SR, of course, it proves no such thing. O can accelerate towards T with any acceleration profile he likes, and two things still will be true: (1) O will reach T in some finite proper time T by O's clock, and also in some finite time t by T's clock (which is the time in the inertial frame in which T is at rest, and in which O is initially at rest--call this frame T's frame); (2) O's velocity, as seen in T's frame, will always be *less* than c (if he maintains a non-zero acceleration, regardless of the specific acceleration profile, his velocity in T's frame will continually increase and approach c more and more closely, but will never reach it).

 

[Passionflower]

Notice that when O is accelerating away from T with an increasing proper acceleration the distance between O and T only approach zero in the limit but never become zero.

 

[PeterDonis]

Notice that when O is accelerating away from T with an increasing proper acceleration the distance between O and T only approach zero in the limit but never become zero.

I assume you mean "distance between O and T, as seen in O's comoving frame"? If O's proper acceleration increases without bound, yes, this distance will approach zero asymptotically if O is accelerating away from T.

However, A-wal has explicitly used the word "towards" several times (see, for example, post #525). So I really think he has in mind a scenario where O's proper acceleration is *towards* T.

 

[Passionflower]

However, A-wal has explicitly used the word "towards" several times (see, for example, post #525). So I really think he has in mind a scenario where O's proper acceleration is *towards* T.

Of course, I just thought it'd be interesting to highlight.

 

[Dale]

I'm simply talking from the perspective of the free-faller. I don't know enough about coordinate systems to tell you which ones you should be using.

GP would definitely be better than Schwarzschild. In GP coordinates the observers fall through the horizon in a finite coordinate time as well as a finite proper time.

 

[Dale]

What people are trying to tell me here is that I need to use coordinate systems to put it into terms they understand. I’m not talking about what they see. I’m talking about crossing an event horizon.

In relativity, the words "X's perspective" is shorthand for "the coordinate system where X is at rest", usually X is an inertial observer and it is implied that they are using an inertial coordinate system. It isn't what they visually see, but what they determine happens after accounting for visual effects like the finite speed of light.

So what I am trying to tell you is that since you are interested in a free fallers perspective you are interested in a coordinate system like GP where free fallers are at rest. They mean the same thing.

 

[Lost in Space]

What puzzles me is the time dilation thing. I mean, from an outsider's perspective (us in other words) no object can actually pass over the event horizon but just appears to hover on the edge forever. But black holes feed, and their event horizons grow as they swallow more mass, don't they? We can observe and measure their gravitational influence caused by their mass even if we can't see them directly so does this mean that effectively all of the mass they have previously absorbed is (as far as we're concerned) piled up on the event horizon, when in reality it has already crossed over? And also, if this is correct, shouldn't black holes then be extremely bright objects?

 

[A-wal]

We have reached the point where you are simply making false statements with no basis. You are doing the equivalent of saying arithmetic does not state that 1+1=2 because you disagree. I don't know how to proceed from here because lack of knowledge or understanding is no longer the issue - denial of facts is impossible to discuss.

No we’ve reached the point were you don’t have an answer so you’re just claiming I don’t understand. Either answer the question or admit that you can’t. A lack of understanding on my part has never been the issue, it’s why I first came here. I think you should try to answer my question. Are you saying that before an object crosses an event horizon that object observes any of the objects in front of them that they can see as being on the other side of the event horizon but they’re seeing them as they were before they reached the horizon, and as the object crosses the horizon it sees them as they were the moment they crossed the horizon? Just trying to be clear because I don’t think any of you even know what you think would happen. So how far in can they observe the objects in front of them? All the way to the singularity? How close do they have to be to observe objects on the other side of the horizon?

Now you're saying something *different*, which still doesn't make sense. Call the moving object O and the target object T. Suppose O starts out 1000 meters from T. You say the starting velocity is zero; but c divided by 1000 meters, or c divided by 1000 meters squared, is not zero (even if I ignore the fact that the units don't work). So I can't start O out consistently with what you're saying, even if I ignore the units (which I can't because the units are part of the physics, and if they aren't balancing, there is something wrong).
...
In which case the velocity would go to infinity as O approached T (see below). This is inconsistent with SR, which you have said you accept.
...
*Now* you're saying something different from the above *again*. You're saying that the velocity of O should be c when it is co-located with T. But when it is co-located with T, its distance from T is zero, so going by your previous prescription, its velocity would be c / 0 = infinite.
...
You are certainly not winning any prizes for logic or clarity.

I’ve been crystal clear about the scenario. Your lack of understanding at this point is a reflection on you, not me. It’s really not that complicated, you’re just making it seem that way. To work out the velocity of any of the objects in the second scenario simply divide c by the distance squared, so that for example if they’re half way in from their starting point then they would be moving at .25c relative to the object that they’re heading towards. Clear? Nothing ever reaches c and I never suggested it did. That’s the whole point.

Welcome to the reason why this thread is 540 posts long. :sigh:

That’s not the reason. It’s taken me this long to get to the point where I can show any who’s actually willing to listen exactly why it works the way it does and confidently challenge you, Dalspam and anyone else who thinks they know better. I’m a slow learner.

I suppose I ought to try and expedite this discussion by laying out what SR actually says about this type of scenario, for what it's worth. A-wal appears to be saying that O should start out with velocity zero, and should accelerate towards T. That, at least, is a scenario capable of consistent analysis. However, he also appears to want to stipulate that O should accelerate in such a way that its velocity is c when its distance from T is zero, which of course is impossible. I'm pretty sure A-wal knows *that*, anyway.

A-wal appears to want to draw the conclusion, however, that all this proves somehow that O can't reach T. In SR, of course, it proves no such thing. O can accelerate towards T with any acceleration profile he likes, and two things still will be true: (1) O will reach T in some finite proper time T by O's clock, and also in some finite time t by T's clock (which is the time in the inertial frame in which T is at rest, and in which O is initially at rest--call this frame T's frame); (2) O's velocity, as seen in T's frame, will always be *less* than c (if he maintains a non-zero acceleration, regardless of the specific acceleration profile, his velocity in T's frame will continually increase and approach c more and more closely, but will never reach it).

Start with a THEORETICAL velocity of c at zero distance then use the distance squared to work out the relative velocity at that time. This isn’t difficult or complicated.

GP would definitely be better than Schwarzschild. In GP coordinates the observers fall through the horizon in a finite coordinate time as well as a finite proper time.

In relativity, the words "X's perspective" is shorthand for "the coordinate system where X is at rest", usually X is an inertial observer and it is implied that they are using an inertial coordinate system. It isn't what they visually see, but what they determine happens after accounting for visual effects like the finite speed of light.

So what I am trying to tell you is that since you are interested in a free fallers perspective you are interested in a coordinate system like GP where free fallers are at rest. They mean the same thing.

Okay and what does GP tell you about seeing objects in front of you crossing the event horizon?

What puzzles me is the time dilation thing. I mean, from an outsider's perspective (us in other words) no object can actually pass over the event horizon but just appears to hover on the edge forever. But black holes feed, and their event horizons grow as they swallow more mass, don't they? We can observe and measure their gravitational influence caused by their mass even if we can't see them directly so does this mean that effectively all of the mass they have previously absorbed is (as far as we're concerned) piled up on the event horizon, when in reality it has already crossed over? And also, if this is correct, shouldn't black holes then be extremely bright objects?

They can't answer you.

 

[Dale]

Okay and what does GP tell you about seeing objects in front of you crossing the event horizon?

That they each cross in a finite coordinate time as well as a finite proper time.

 

[Passionflower]

Are you saying that before an object crosses an event horizon that object observes any of the objects in front of them that they can see as being on the other side of the event horizon but they’re seeing them as they were before they reached the horizon, and as the object crosses the horizon it sees them as they were the moment they crossed the horizon? Just trying to be clear because I don’t think any of you even know what you think would happen.

When two objects, close enough to each other, approach the EH then they will not lose contact because when, after the leading objects passes the event horizon, it sends a light signal back to the trailing object it will actually reach the trailing object only after the trailing object passes the EH as well due to the finite speed of light. However if the trailing object is too far removed from the leading object then because the trailing object does not reach the EH in time to catch the signal from the leading object it will lose contact with the leading object. We can calculate how far the objects can be separated for this to happen or not.