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

[PeterDonis]

As you travel into more and more length contracted/time dilated space you have to constantly recalculate the distance between you and the horizon (which will be non-linear and always greater than you'd expect it to be without the relative stuff) and the time the black hole lasts for in your proper time.

Nope, this is wrong. The calculation I posted already takes into account all the "length contraction" and "time dilation" that takes place as you fall to the horizon. There is no "recalculating" necessary. I think you are mistaking the frame-dependent effects of length contraction and time dilation with the frame-independent, invariant "length" of a given worldline in spacetime, which doesn't change when your frame of reference changes. The calculation I posted, of the finite proper time it takes to free-fall to the horizon from a given radius, is of the latter sort: it's a calculation of the invariant "length" of a given worldline and its result remains invariant regardless of the free-falling observer's state of motion.

This comment also applies to the next quote of yours below, but I have an additional comment about it as well:

The time it will take to reach the horizon will always go up (from a distant observers perspective) forever (or to any given time) as you get closer, until the black hole dies.

As I noted quite a few posts ago, if you don't agree that the horizon can ever be reached for an "eternal" black hole that *doesn't* evaporate, then we're not even ready to discuss the case where the black hole *does* evaporate. I don't think you mean here to concede that the horizon of an "eternal", non-evaporating black hole *can* be reached, do you?

I know that free-fall is equivalent to being at rest (or moving freely I should say) as I've said before, but I think that tidal force acts as acceleration, meaning you can't have a straight line in free-fall. I was thinking of a curved line for free-fall before though, but from the perspective of a distant observer. A line that's curve is constant from this view would be the geodesics you're talking about. That curve would get sharper and sharper from this perspective as the object accelerates (not talking about accelerating up to terminal velocity because that's not true acceleration) while in free-fall. The falling object feels this acceleration as tidal force.

Not with an sensitive enough accelerometer they wouldn't!

Yes, they would. As I believe I've already noted, I think this misunderstanding on your part is a big factor in your misunderstanding of how a black hole horizon works. This claim of yours is *wrong*, pure and simple. Two freely falling objects that separate due to tidal gravity feel *zero* acceleration--not just "too small to measure without really sensitive instruments because Earth gravity is too weak", but *zero*. See next comment.

It's like if you accelerate two identical cars and give them both exactly the same amount of throttle, but one sets off just before the other. The first one to start will constantly pull away from the second one. Your example of tidal force is exactly the same accept it's strong enough to separate "solid objects". The front end of the object will be in length contracted/time dilated space-time relative to the back end. So the front will be trying to move faster than the back even though they both measure their own speeds to be the same, because the front thinks the back is moving too slowly and the back thinks the front is moving too quickly.

Nope, this is wrong, the two cases are *not* the same; they are not even analogous. I'm not sure why you think the two cases are the same, so I'm not sure how to explain why they're not, except to say what I've been repeating for some time now, that freely falling objects separated by tidal gravity are *freely falling*. That means there is no rocket attached to them, nothing to push on them, nothing to exert force on them--whereas in your example, the two cars *are* being pushed, by the force on their wheels. Or, in the more usual example of the "Bell spaceship paradox", the two rockets are each firing their engines, which is why they feel acceleration. For the examples I gave of freely falling objects affected by tidal gravity, there is *nothing* analogous to the car engines/wheels or the rockets in the Bell spaceship paradox; these objects have no "propulsion" mechanism, hence they are freely falling. I'm not sure I can say much more until I understand how you can possibly see a physical similarity between objects in free fall and objects being accelerated by a rocket (or a car engine/wheel, or any other propulsion mechanism).

Right, so the Schwarzschild coordinates are misleading and the time dilation and length contraction that the coordinates show are an illusion? Presumably it's always an illusion then, or does it suddenly become an illusion at the horizon?

The "distortion" of Schwarzschild coordinates gets larger the closer you get to the horizon, and becomes "infinite" *at* the horizon. Far enough away from the black hole, the distortion is negligible--it goes to zero at spatial infinity. The distortion on a coordinate chart does not have to be the same everywhere--for example, a Mercator projection gives zero distortion at the Earth's equator (I'm assuming the "standard" projection which is centered on the equator), and gradually increasing distortion as you get closer to the poles, going to infinite distortion *at* the poles.

 

[Dale]

Accepting what I'm told despite that it's still not how I see it? It's frustrating for me too.

You have had 3 or 4 people spending hundreds of posts and coming up with as many different ways to explain it to you as they could think of. You seem to be completely unwilling/unable to learn. You seem to think that you will never have to make any mental effort at all, that it is the responsibility of us and the universe to conform to how you see it, which is set in stone and immovable. Over the course of the years I have explained this stuff to many people, none of whom have been as unteachable as you, even people like Anamitra who were almost openly antagonistic to GR concepts.

1). Agreed. 2). Agreed. 3). That's not very scientific. What do you actually mean by that? 4). Are those coordinate charts dynamic or fixed? They should change as you move to a gravitational field of different strength. Just like they change when you accelerate in flat space-time and move to a different frame. 5). Yea it would be.

Point 3) means that the Schwarzschild coordinates are not diffeomorphic at the event horizon so they do not include the event horizon. Given that you agree with points 1)-5) the only logical conclusion is that if you believe that you can reach the brick wall then you must concede that you can reach the event horizon also.

I would recommend that you leave off studying black holes and the Schwarzschild coordinates and instead learn about Rindler coordinates in flat spacetime. If you are interested in trying that as a new approach then start a new thread on the topic and I will try once again.

 

[PeterDonis]

I would recommend that you leave off studying black holes and the Schwarzschild coordinates and instead learn about Rindler coordinates in flat spacetime.

I would add that doing this will also remove the confusion about tidal gravity, which is completely absent in flat spacetime. I've tried to remove it by specifying that we're talking about a black hole with a large enough mass that tidal gravity at the horizon is negligible, but it doesn't seem to have taken; perhaps seeing how there can be a horizon in completely flat spacetime, with no tidal effects at all, will help to separate these two distinct issues.

 

[Passionflower]

A couple of facts about radial and stationary observers in the Schwarzschild solution that might help A-wal:

• A free falling observer will observe no proper acceleration.
• A free falling observer will observe inertial acceleration wrt the black hole (he would for instance observe that his velocity wrt stationary observers increases in time).
• A stationary observer is an observer who neutralizes the inertial acceleration by using a proper acceleration in the opposite direction and has a zero velocity wrt the black hole.
• A free falling observer will reach the event horizon and singularity in finite proper time (e.g. the time on his clock).
• Two separated free falling observers will observe an inertial acceleration wrt each other. The magnitude is directly related to the amount of proper time removed from the singularity and is independent of the size of the black hole.
• A free falling observer will observe a smaller distance to the event horizon than a co-located stationary observer.
• Co-located free falling observers can have a non zero velocity wrt each other.

 

[Dale]

[*]Two separated free falling observers will observe an inertial acceleration wrt each other. The magnitude is directly related to the amount of proper time removed from the singularity and is independent of the size of the black hole.

Interesting! I didn't know that one.

 

[A-wal]

Why is everyone so quick to assume that I've misunderstood what I've been told whenever I say something that goes against what you've been told? I've had my points misunderstood a lot more than anyone else (my fault for not being clear enough), but I am learning this as I go along. I may very well be missing something very important in my understanding, but if I am then I'm almost sure it's not something that's been pointed out to me yet. In fact I came in here with a vague idea and virtually everything I've been told has backed it up despite their intention to argue the opposite.

Nope, this is wrong. The calculation I posted already takes into account all the "length contraction" and "time dilation" that takes place as you fall to the horizon. There is no "recalculating" necessary. I think you are mistaking the frame-dependent effects of length contraction and time dilation with the frame-independent, invariant "length" of a given worldline in spacetime, which doesn't change when your frame of reference changes. The calculation I posted, of the finite proper time it takes to free-fall to the horizon from a given radius, is of the latter sort: it's a calculation of the invariant "length" of a given worldline and its result remains invariant regardless of the free-falling observer's state of motion.

How can the calculation for proper time possibly be invariant (unaffected by further time dilation and length contraction) when it's a measurement of how much time something takes to cover a certain distance? Are you saying that the calculation you posted takes into account the fact that time dilation and length contraction increase as you get closer to the gravitational source? If you want to view your personal time and length (what you would call a worldline) as invariant then you have to view the space-time you're traveling through as variable. The spacetime between the faller and the black hole would have to contract more and more as you fall, making your invariant worldline vary in relation to the space-time between you and the horizon.

As I noted quite a few posts ago, if you don't agree that the horizon can ever be reached for an "eternal" black hole that *doesn't* evaporate, then we're not even ready to discuss the case where the black hole *does* evaporate. I don't think you mean here to concede that the horizon of an "eternal", non-evaporating black hole *can* be reached, do you?

As I noted quite a few posts ago, the question of whether or not you can reach the horizon of an eternal black hole doesn't make sense. I'm saying that nothing can reach in any given amount of time. It's exactly the same as asking whether something that accelerates forever will reach c.

Yes, they would. As I believe I've already noted, I think this misunderstanding on your part is a big factor in your misunderstanding of how a black hole horizon works. This claim of yours is *wrong*, pure and simple. Two freely falling objects that separate due to tidal gravity feel *zero* acceleration--not just "too small to measure without really sensitive instruments because Earth gravity is too weak", but *zero*. See next comment.

This simply cannot be! If they are separating due to tidal gravity then the objects are separating (or trying to) from themselves as well. The front of the objects are pulling the rest forward. That's acceleration.

Nope, this is wrong, the two cases are *not* the same; they are not even analogous. I'm not sure why you think the two cases are the same, so I'm not sure how to explain why they're not, except to say what I've been repeating for some time now, that freely falling objects separated by tidal gravity are *freely falling*. That means there is no rocket attached to them, nothing to push on them, nothing to exert force on them--whereas in your example, the two cars *are* being pushed, by the force on their wheels. Or, in the more usual example of the "Bell spaceship paradox", the two rockets are each firing their engines, which is why they feel acceleration. For the examples I gave of freely falling objects affected by tidal gravity, there is *nothing* analogous to the car engines/wheels or the rockets in the Bell spaceship paradox; these objects have no "propulsion" mechanism, hence they are freely falling. I'm not sure I can say much more until I understand how you can possibly see a physical similarity between objects in free fall and objects being accelerated by a rocket (or a car engine/wheel, or any other propulsion mechanism).

They're not the same but how are they not analogous? Looking purely at the way the two objects move relative to each other, it's exactly the same.

The "distortion" of Schwarzschild coordinates gets larger the closer you get to the horizon, and becomes "infinite" *at* the horizon. Far enough away from the black hole, the distortion is negligible--it goes to zero at spatial infinity. The distortion on a coordinate chart does not have to be the same everywhere--for example, a Mercator projection gives zero distortion at the Earth's equator (I'm assuming the "standard" projection which is centered on the equator), and gradually increasing distortion as you get closer to the poles, going to infinite distortion *at* the poles.

It looks distorted from a distance. If someone were to fly through a strong gravitational field and come back then you could presumably use the distortion of Schwarzschild coordinates to work out their proper time. It looked distorted to you but it didn't to them, you did. It seems like you lot are taking the Schwarzschild coordinates at face value when it suits you and looking at it as an illusion when it doesn't.

You have had 3 or 4 people spending hundreds of posts and coming up with as many different ways to explain it to you as they could think of. You seem to be completely unwilling/unable to learn. You seem to think that you will never have to make any mental effort at all, that it is the responsibility of us and the universe to conform to how you see it, which is set in stone and immovable. Over the course of the years I have explained this stuff to many people, none of whom have been as unteachable as you, even people like Anamitra who were almost openly antagonistic to GR concepts.

Point 3) means that the Schwarzschild coordinates are not diffeomorphic at the event horizon so they do not include the event horizon. Given that you agree with points 1)-5) the only logical conclusion is that if you believe that you can reach the brick wall then you must concede that you can reach the event horizon also.

Only if the brick wall has no event horizon.

I would recommend that you leave off studying black holes and the Schwarzschild coordinates and instead learn about Rindler coordinates in flat spacetime. If you are interested in trying that as a new approach then start a new thread on the topic and I will try once again.

I would add that doing this will also remove the confusion about tidal gravity, which is completely absent in flat spacetime. I've tried to remove it by specifying that we're talking about a black hole with a large enough mass that tidal gravity at the horizon is negligible, but it doesn't seem to have taken; perhaps seeing how there can be a horizon in completely flat spacetime, with no tidal effects at all, will help to separate these two distinct issues.

I don't really see the point of starting a new thread for a single point. The Rindler horizon is the one thing I've actually looked up and as I understand it, it occurs when an object can no longer possibly catch an accelerating observer (assuming it keeps accelerating at the same rate of course). You can substitute acceleration for gravity and it seems to back up my point of view. There will be a point during an objects free-fall when an external object won't ever be able to catch the free-faller, no matter how much they accelerate. The horizon will obviously be further in than any of these objects from they perspective of an external observer. So the event horizon is uncatchable from any external perspective. It seems to me that an event horizon is a Rindler horizon that applies to everything outside it.

 

[A-wal]

A couple of facts about radial and stationary observers in the Schwarzschild solution that might help A-wal:

• A free falling observer will observe no proper acceleration.
• A free falling observer will observe inertial acceleration wrt the black hole (he would for instance observe that his velocity wrt stationary observers increases in time).
• A stationary observer is an observer who neutralizes the inertial acceleration by using a proper acceleration in the opposite direction and has a zero velocity wrt the black hole.
• A free falling observer will reach the event horizon and singularity in finite proper time (e.g. the time on his clock).
• Two separated free falling observers will observe an inertial acceleration wrt each other. The magnitude is directly related to the amount of proper time removed from the singularity and is independent of the size of the black hole.
• A free falling observer will observe a smaller distance to the event horizon than a co-located stationary observer.
• Co-located free falling observers can have a non zero velocity wrt each other.

Thanks.

• "A free falling observer will observe no proper acceleration."

They feel tidal force!

• "A free falling observer will observe inertial acceleration wrt the black hole (he would for instance observe that his velocity wrt stationary observers increases in time)."

Of course. The way I look at there's no actual movement involved because it's the space between them that's affected rather than the objects themselves.

• "A stationary observer is an observer who neutralizes the inertial acceleration by using a proper acceleration in the opposite direction and has a zero velocity wrt the black hole."

What I've been referring to as "hovering".

• "A free falling observer will reach the event horizon and singularity in finite proper time (e.g. the time on his clock)."

How? Where does this postulate actually come from? If you work out how much time it would take from one position, then move closer and work it out again then you'll get a different answer. It takes more time the closer you get.

• "Two separated free falling observers will observe an inertial acceleration wrt each other. The magnitude is directly related to the amount of proper time removed from the singularity and is independent of the size of the black hole."

Just like two accelerating cars, but apparently it's not the same, or even analogous.

• "A free falling observer will observe a smaller distance to the event horizon than a co-located stationary observer."

Okay. That's special relativity isn't it? Because of their velocity relative to the black hole? I still don't ever see it reaching zero.

• "Co-located free falling observers can have a non zero velocity wrt each other."

You mean if their initial velocities are different then they will remain different, and if they both start off hovering at the same distance then it would be zero?

 

[PeterDonis]

This responds to both of A-wal's recent posts:

Why is everyone so quick to assume that I've misunderstood what I've been told whenever I say something that goes against what you've been told?

We're not assuming you've misunderstood; we're deducing it from the fact that you make statements that are factually incorrect, yet you assert that they are a necessary part of the picture as you understand it. See next comment.

I may very well be missing something very important in my understanding, but if I am then I'm almost sure it's not something that's been pointed out to me yet.

I disagree with this as you've stated it; you've had several things pointed out to you that, IMO, include something important that you are missing. However, I would have to agree that no one has yet succeeded in stating what you are missing in the right way to let you see *how* you are missing it.

That said, there's at least one *factual* item that has come up several times now, which has been pointed out to you several times, yet which you continue to mis-state. It is this (from the 2nd of your two recent posts):

• "A free falling observer will observe no proper acceleration."

They feel tidal force!

Sorry to shout, but this has already been repeated to you several times: NO, THEY DO NOT FEEL ANY FORCE! This is a straightforward factual question to which you keep on giving the wrong answer. Once again:

• "Co-located free falling observers can have a non zero velocity wrt each other."

You mean if their initial velocities are different then they will remain different, and if they both start off hovering at the same distance then it would be zero?

Passionflower here is describing the same scenario I have described several times in this thread. Start with two objects, both at rest relative to a large massive body (e.g., the Earth or a black hole), at slightly different radial coordinates (one at r = R, the other at r = R + epsilon). Therefore both bodies also start out at rest relative to each other (zero relative velocity). Their spatial separation in the radial direction at time t = 0 is some value s (if the radial coordinate R is much larger than the Schwarzschild radius, s is very close to epsilon, but as R gets closer to 2M, ds gets larger for a given value of epsilon--but all that is irrelevant to what I'm about to say). At time t = 0, both objects are released and start freely falling towards the large massive body. As time t increases, the spatial separation between the bodies *increases* from the starting value of s (and therefore their relative velocity also becomes nonzero and increases), even though both objects, being freely falling, feel *no* acceleration (i.e., no "force", no stress, no weight, etc.).

If you don't agree or can't understand what I've just said, then we have a straightforward factual disagreement that we need to resolve before we can make progress on anything else. Part of your confusion may be because (as I've also pointed out before) you are confusing the scenario I just described with a *different* scenario, where you have a single extended body subject to tidal gravity; that is what you describe here:

This simply cannot be! If they are separating due to tidal gravity then the objects are separating (or trying to) from themselves as well. The front of the objects are pulling the rest forward. That's acceleration.

The scenario I described above, with two freely falling objects separating spatially but feeling no force, idealizes both objects as single, point-like "test objects" with no internal structure. That allows us to focus on the fact that tidal gravity causes spatial separation of freely falling geodesics, without being confused by other effects due to the internal structure of objects. It's those other effects that you're referring to when you say that objects are "separating (or trying to) from themselves"--only extended objects with internal structure can do that. An extended object with internal structure *can* feel actual forces, internally, due to tidal gravity, but in that case the individual pieces that make up the body (its atoms, or whatever) are *not* all traveling on freely falling geodesic worldlines. This makes the situation much more complicated and difficult to analyze, without really adding anything to our understanding of tidal gravity itself. That's why we've tried to idealize the situation so it focuses solely on the effects of tidal gravity, without throwing in all the complications due to the internal structure of objects.

Until we get this issue resolved, I don't want to comment on most of the other things you've said, because we'll be coming from different and incompatible understandings of this fundamental point. However, there is one other thing you said that I do want to comment on, because you mentioned the Rindler horizon again:

The Rindler horizon is the one thing I've actually looked up and as I understand it, it occurs when an object can no longer possibly catch an accelerating observer (assuming it keeps accelerating at the same rate of course). You can substitute acceleration for gravity and it seems to back up my point of view. There will be a point during an objects free-fall when an external object won't ever be able to catch the free-faller, no matter how much they accelerate.

You've mis-stated it slightly: the Rindler horizon is the path of a *light ray* that cannot catch up to an accelerating observer (provided they keep accelerating forever with the same proper acceleration). Also, you switched it around when you translated to "gravity" terms: the horizon of a black hole is the path of a light ray that can never "catch up" with any observer hovering at a radius r > 2M (the radius of the horizon). It has nothing to do with accelerating (hovering) observers being unable to catch free-falling observers.

 

[PeterDonis]

An extended object with internal structure *can* feel actual forces, internally, due to tidal gravity, but in that case the individual pieces that make up the body (its atoms, or whatever) are *not* all traveling on freely falling geodesic worldlines.

Let me elaborate on this a little, so that it's clear what I have in mind. For reasons that will hopefully become clear in a moment, I'm going to start with *three* objects, starting at radius R (object A), R - b (object B), and R + c (object C), where b, c << R. The starting spatial separation between each adjacent pair of objects is s (so the separations A to B and A to C start out as s, and the separation B to C starts out as 2s)--note that this means that b will, in general, *not* be exactly equal to c (because of the radial change in the metric coefficient g_{rr}, meaning that the amount of radial coordinate change corresponding to a given proper length s is not exactly the same for b and c). However, the actual (proper) spatial separation is all we're concerned with here. All three objects start out at rest at time t = 0 (at rest relative to each other, and relative to the large mass M that the radial coordinates are relative to); at that time they are released and fall radially towards the large mass.

First consider the case where all three objects move freely, with no connection between them. Then each object's center of mass moves on a freely falling geodesic worldline (I say "center of mass" to eliminate any consideration of the internal structure of the objects--see below), so the spatial separation between A and B, and between A and C, will increase with time as they all fall radially.

Now consider a case where objects A and B, and A and C, are connected by springs. This means we can consider all three together as one large "object" O, with internal parts that can exert forces on each other. Each spring has an equilibrium proper length s, so at time t = 0, when all three objects are at rest, both springs are in equilibrium and there is no force anywhere in the system. The (rest) masses of all three objects, A, B, and C, are equal, and the masses of the springs are equal too, so object A is at the center of mass of the whole extended object O at time t = 0; and because O is exactly symmetrical in the radial direction, A will remain at the center of mass of O for all time.

How will O move? At time t = 0, again, everything is released to fall freely in the radial direction. Because A is at the center of mass, it will follow the same geodesic worldline as it followed in the previous case, when there were no springs. However, objects B and C will *not* follow the same worldlines as before; they will not follow geodesics, because as they move, the tidal gravity will start to stretch the springs, which will pull back on B and C and cause them to stay closer to A than they would if the springs weren't there. That means objects B and C will feel a net proper acceleration; but the proper accelerations will be *towards* object A in both cases. So there will be an equilibrium configuration of this extended object O, in which the center of mass (A) falls on a radial geodesic, but the ends (B and C) are accelerated; B will be accelerated radially outward (towards A), while C will be accelerated radially *inward* (towards A). B and C will then, in this equilibrium, maintain a constant spatial separation from A, which will be somewhat larger than s (because the springs need to be stretched in order to exert the force towards A that accounts for the net proper acceleration of B and C).

What about object A? How can it fall on a geodesic worldline when the springs are exerting force on it? It is true that there will be a nonzero *stress* at A due to the springs, but since the spring forces at A are equal and opposite, there is no *net* force on A, so there is no *net* proper acceleration, and A moves on a freely falling geodesic worldline. So the whole extended object O moves such that its center of mass follows a freely falling geodesic, but its other internal parts do not because of the internal forces set up by tidal gravity. This is the sort of thing I was talking about when I referred to the complications due to the internal structure of objects. All these internal forces, though, do not add anything to our understanding of how tidal gravity works: we can understand that solely by looking at the motion of the center of mass, A, [edit] and comparing it with the motion of the center of mass of other objects that are radially separated and do not exchange any interaction forces with A [/edit].

(Incidentally, the above also shows why the "acceleration" due to tidal gravity, for an extended object with internal structure, is *not* the same as the acceleration required to hover over a black hole or other large mass. The "hovering" proper acceleration is always radially outward, and it applies at the center of mass of the object--it applies even to an idealized point-like "test object" with no internal structure, and it causes the center of mass of the object to move on a non-geodesic worldline. The radial proper acceleration of internal parts of an extended object due to tidal gravity is always towards the center of mass, which can be radially outward or inward depending on the location of the part relative to the center of mass; and *at* the center of mass, the proper acceleration due to tidal gravity is zero, so the CoM always moves on a freely falling geodesic.)

 

[A-wal]

Sorry to shout, but this has already been repeated to you several times: NO, THEY DO NOT FEEL ANY FORCE! This is a straightforward factual question to which you keep on giving the wrong answer. Once again:

Passionflower here is describing the same scenario I have described several times in this thread. Start with two objects, both at rest relative to a large massive body (e.g., the Earth or a black hole), at slightly different radial coordinates (one at r = R, the other at r = R + epsilon). Therefore both bodies also start out at rest relative to each other (zero relative velocity). Their spatial separation in the radial direction at time t = 0 is some value s (if the radial coordinate R is much larger than the Schwarzschild radius, s is very close to epsilon, but as R gets closer to 2M, ds gets larger for a given value of epsilon--but all that is irrelevant to what I'm about to say). At time t = 0, both objects are released and start freely falling towards the large massive body. As time t increases, the spatial separation between the bodies *increases* from the starting value of s (and therefore their relative velocity also becomes nonzero and increases), even though both objects, being freely falling, feel *no* acceleration (i.e., no "force", no stress, no weight, etc.).

"You mean if their initial velocities are different then they will remain different, and if they both start off hovering at the same distance then it would be zero?" was right than. What's epsilon?

If you don't agree or can't understand what I've just said, then we have a straightforward factual disagreement that we need to resolve before we can make progress on anything else. Part of your confusion may be because (as I've also pointed out before) you are confusing the scenario I just described with a *different* scenario, where you have a single extended body subject to tidal gravity; that is what you describe here:

The scenario I described above, with two freely falling objects separating spatially but feeling no force, idealizes both objects as single, point-like "test objects" with no internal structure. That allows us to focus on the fact that tidal gravity causes spatial separation of freely falling geodesics, without being confused by other effects due to the internal structure of objects. It's those other effects that you're referring to when you say that objects are "separating (or trying to) from themselves"--only extended objects with internal structure can do that. An extended object with internal structure *can* feel actual forces, internally, due to tidal gravity, but in that case the individual pieces that make up the body (its atoms, or whatever) are *not* all traveling on freely falling geodesic worldlines. This makes the situation much more complicated and difficult to analyze, without really adding anything to our understanding of tidal gravity itself. That's why we've tried to idealize the situation so it focuses solely on the effects of tidal gravity, without throwing in all the complications due to the internal structure of objects.

Until we get this issue resolved, I don't want to comment on most of the other things you've said, because we'll be coming from different and incompatible understandings of this fundamental point. However, there is one other thing you said that I do want to comment on, because you mentioned the Rindler horizon again:

I'm not sure there's anything to resolve here. I've already stated that the only time tidal force wouldn't apply is to a point-like object.

You've mis-stated it slightly: the Rindler horizon is the path of a *light ray* that cannot catch up to an accelerating observer (provided they keep accelerating forever with the same proper acceleration). Also, you switched it around when you translated to "gravity" terms: the horizon of a black hole is the path of a light ray that can never "catch up" with any observer hovering at a radius r > 2M (the radius of the horizon). It has nothing to do with accelerating (hovering) observers being unable to catch free-falling observers.

Same thing isn't it? If a light ray can reach the horizon then nothing can.

Let me elaborate on this a little, so that it's clear what I have in mind.

I appreciate the depth of the response but I already get all that and it doesn't have anything to do with what I'm trying to get across. I don't believe that an object can even accelerate into a black hole, but I'm being told that it doesn't even have to accelerate, it can free-fall in. I'm thinking tidal force could be the answer. Imagine a very long object free-falling towards a black hole. It's at an angle so that (if we ignore length contraction for a moment) if it's ten metres long then the back will be ten metres further from the black hole than the front. There has to be an easier way of explaining that. There's probably a single word that describes that angle. Anyway, the front will be pulling the back along, even though the front is just trying to free-fall. What's free-fall speed at the front isn't the same as what's free-fall speed at the back. The difference is caused by the fact that the front end is more time dilated/length contracted than the back. From the backs perspective the front is accelerating, so objects closer than whoever's observing them do accelerate from that observers perspective. In other words objects do accelerate when they free-fall. As they accelerate they become time dilated/length contracted and the rest of the universe (including the life span of the black hole) appears to speed up. Anything at the event horizon would, in effect, have accelerated to c. I don't see how a free-faller could gradually reach infinite time dilation/length contraction. It doesn't matter how long the black hole lasts because they journey inwards would always last longer. I think tidal force comes from the fact that you can't assign a set proper time or length in which anything can happen when acceleration's involved because you'd have to constantly recalculate as you go. You'd have to take into account that what's a metre or a second at the speed your traveling now won't be the same after you accelerate. I think we all know that's true in special relativity and I don't see how it works any differently with gravity.

 

[Passionflower]

I appreciate the depth of the response but I already get all that and it doesn't have anything to do with what I'm trying to get across. I don't believe that an object can even accelerate into a black hole, but I'm being told that it doesn't even have to accelerate, it can free-fall in. I'm thinking tidal force could be the answer. Imagine a very long object free-falling towards a black hole. It's at an angle so that (if we ignore length contraction for a moment) if it's ten metres long then the back will be ten metres further from the black hole than the front. There has to be an easier way of explaining that. There's probably a single word that describes that angle. Anyway, the front will be pulling the back along, even though the front is just trying to free-fall. What's free-fall speed at the front isn't the same as what's free-fall speed at the back. The difference is caused by the fact that the front end is more time dilated/length contracted than the back. From the backs perspective the front is accelerating, so objects closer than whoever's observing them do accelerate from that observers perspective. In other words objects do accelerate when they free-fall. As they accelerate they become time dilated/length contracted and the rest of the universe (including the life span of the black hole) appears to speed up. Anything at the event horizon would, in effect, have accelerated to c. I don't see how a free-faller could gradually reach infinite time dilation/length contraction. It doesn't matter how long the black hole lasts because they journey inwards would always last longer. I think tidal force comes from the fact that you can't assign a set proper time or length in which anything can happen when acceleration's involved because you'd have to constantly recalculate as you go. You'd have to take into account that what's a metre or a second at the speed your traveling now won't be the same after you accelerate. I think we all know that's true in special relativity and I don't see how it works any differently with gravity.

You may think you get it but really I would urge you to take some advice, clearly your understanding is flawed.

A free falling observer will reach the event horizon and singularity in finite proper time (e.g. the time on his clock)."

How? Where does this postulate actually come from? If you work out how much time it would take from one position, then move closer and work it out again then you'll get a different answer. It takes more time the closer you get.

Again you are wrong, a free falling observer reaches the singularity in finite proper time, and the closer he gets, the less time he has left.

Do you perhaps want to discuss the formulas that show this?

 

[PeterDonis]

"You mean if their initial velocities are different then they will remain different, and if they both start off hovering at the same distance then it would be zero?" was right than. What's epsilon?

No, I mean that their initial velocities start out both being zero, but as the two objects fall, their velocities *become* different (the lower object falls faster than the higher object), even though they are both freely falling and both feel no force.

Epsilon is just the initial difference in radial coordinates of the two objects.

I'm not sure there's anything to resolve here. I've already stated that the only time tidal force wouldn't apply is to a point-like object.

And yet you keep on trying to bring in tidal force when I and others have told you multiple times that an object can free-fall through the horizon even in the case where tidal force is negligible (the object's length is so short compared to the size of the black hole that the tidal gravity of the hole at the horizon has no appreciable effect on the object's motion). See further comment below.

Same thing isn't it? If a light ray can reach the horizon then nothing can.

I never said a light ray can't (I assume that's what you meant, not "can") reach the horizon; I said the horizon *is* the path of a light ray--an outgoing light ray at r = 2M.

I'm thinking tidal force could be the answer.

No, tidal force is a red herring that's distracting you from the real physics involved. That's why we keep recommending that you consider a Rindler horizon, which works like a black hole horizon in key respects but occurs in flat spacetime, where there is *no* tidal gravity to confuse things.

 

[clitvin]

" Imagine a very long object free-falling towards a black hole. From the backs perspective the front is accelerating, so objects closer than whoever's observing them do accelerate from that observers perspective.

If the black hole is massive enough the front and the back will accelerate at the same speed.

Very large black holes will only experience tidal forces beyond the EH closer to the singularity.

I don't see how a free-faller could gradually reach infinite time dilation/length contraction.

You do not experience any time dilation. Time is always contant for yourself. Outside observers would see an image of you "get stuck" just beyond the event horizon but only because the light coming towards them would make it appear so. However even this won't last forever as eventually the light wavelenght would become so long it is no longer inside the visible spectrum.

 

[PeterDonis]

Imagine a very long object free-falling towards a black hole...What's free-fall speed at the front isn't the same as what's free-fall speed at the back.

This is true, but irrelevant. As I showed in a recent post, even for an extended object like this one, subject to tidal gravity, the center of mass of the object moves on a freely falling worldline--an accelerometer mounted at the center of mass of the object would read zero proper acceleration. This alone is enough to prove that the object will reach the horizon in a finite proper time, starting from a finite radius R > 2M. (The proof is what I posted earlier, since that proof applies to *any* freely falling worldline, including the one followed by the center of mass of an extended object moving solely under the influence of gravity.) This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

 

[Passionflower]

This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

I completely agree with Peter.

For a black hole the tidal force between the bottom and the ceiling of a free falling object traveling is directly related to the remaining time until the singularity is hit, the location of the EH, and thus the black hole's mass, is totally irrelevant to this.

 

[yuiop]

You clearly misunderstood; I posted that example as a reductio ad absurdum counterargument to your position. I am glad that you recognized that the argument is wrong in the case of a brick wall, but it is frustrating that even after more than 200 posts you don't realize that it is just as wrong in the case of an event horizon also.

Here are some incontrovertible mathematical facts:
1) the proper time along an infalling radial geodesic is finite in all coordinate systems
2) there are coordinate systems where the coordinate time is also finite
3) the Schwarzschild chart does not smoothly cover the event horizon
4) other coordinate charts do
5) the coordinate time is infinite in the Schwarzschild coordinates

Your position has been that because of 5) no observer can cross the event horizon despite 1)-4). In other words, your position is that an object cannot cross the horizon for no other reason than the fact that there exists some coordinate system where the coordinate time goes to infinity. If this logic were correct then it would apply to other coordinate systems in other spacetimes also, and therefore you would not be able to hit a brick wall because there exists some coordinate system where the coordinate time goes to infinity. This is your argument as you have described it here.

Wikipedia http://en.wikipedia.org/wiki/Hawking_radiation gives the time t_{ev} for a black hole to evaporate as:

t_{ev} = \frac{5120 \pi G^2 M^3}{\hbar c^4}

This evaporation time is proportional to the cube of the black hole mass and if finite for any finite mass black hole. The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction). Now coordinate measurements *exactly* to or at the event horizon are contentious because of the impossibility of having a stationary observer at that location so the event horizon is considered "off the Schwarzschild chart". This is not a problem, because we can always find a local stationary observer outside but arbitrarily close to the event horizon that will say that when the free falling observer arrived the black hole and event horizon had evaporated. In other words we can always find a point *outside* the event horizon where a local observer will confirm that the black hole evaporated before the free falling object's arrival. The proper time for a free falling observer that falls from a finite height less than infinite but greater than 2m is finite. Let us say that this finite free falling proper time is 16 seconds, then the black hole evaporates in less than 16 seconds of the free faller's proper time. This means, if we believe Hawking, that in practice no free falling particle has ever fallen to or through an event horizon, so points 1 to 4 in the list become academic because they never happened in nature and in principle never will.

 

[PeterDonis]

The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction).

This is not correct. The infinite coordinate time for an observer to free-fall to the event horizon assumes an "eternal" black hole, one that never evaporates; that is what the Schwarzschild solution to the EFE describes. When you add in evaporation a la Hawking, the spacetime as a whole is no longer the exact Schwarzschild spacetime, so it no longer takes infinite coordinate time for an observer to free-fall to the horizon. Instead, the coordinate time at which the hole finally evaporates (the formula you gave) is also the coordinate time at which any free-falling observer hits the horizon. However, you have to be very careful drawing implications from that statement; it is *not* the same as saying that all free-falling observers hit the horizon at the same event.

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

So basically, at the event of final evaporation of the hole, a flash is emitted that contains images of all the free-falling observers that crossed the horizon, just as they were crossing it. (Unfortunately, the free-falling observers themselves have already fallen into the singularity, since that is still to the future of any timelike worldline that crosses the horizon.) That single flash therefore gets assigned a single coordinate time in the "approximately Schwarzschild" coordinate system used by a distant observer, which will be the coordinate time of the final evaporation *and* of all the "crossing events" whose images are contained in the flash.

 

[surajc0504]

that means will an isentropic process be always time reversible ??

 

[yuiop]

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality. For example, take a look at this adaptation of the diagram you linked to:

The diagram shows light rays coming from the surface (orange curve) of a star collapsing to form the black hole. Examination of the path of observer A (green curve) shows he sees the star continue to shine all the way until he hits the singularity. Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist. It seems that no observer sees a region of space that is gravitationally attractive that is not shining and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated, so its seems no one ever sees the black hole as black, although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash. Does that seem right?

 

[PeterDonis]

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality.

When you look at the Penrose diagrams for, e.g., the maximally extended Schwarzschild spacetime (which includes the "white hole" and the second exterior region), I agree the diagrams are showing things which, although they follow mathematically from the given solution to the EFE, are probably not physically reasonable. However, I don't think that's the case for this particular diagram; it presents a picture which seems, at least to me, to be consistent and in principle physically realizable, although perhaps counterintuitive.

Examination of the path of observer A shows he sees the star continue to shine all the way until he hits the singularity.

Agreed.

Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist.

Yes, it has just evaporated--the second observer will see the final evaporation at the same time he sees the star stop shining, where "stop shining" means "just crossing the horizon" (since light from both events is coming to him along the same null worldline), and the term "stop shining" is justified by the fact that this light ray is the *last* one that will reach the second observer from any event on the star's worldline.

It seems that no observer sees a region of space that is gravitationally attractive that is not shining...

Not quite true, because there's a long period of time before all those light rays reach the second observer, during which he can definitely see the effects of the black hole's gravity but can't yet see any of the light coming from events close to the horizon. During that time he will indeed see a region of space that is gravitationally attractive but "not shining".

...and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated...

Yes.

...so its seems no one ever sees the black hole as black...

No, see above.

...although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash.

Yes.

 

[A-wal]

You may think you get it but really I would urge you to take some advice, clearly your understanding is flawed.

LOL. My understanding is flawed is it? I don't think I'm the one under a false impression of understanding to be honest. If I understood it perfectly they'd be no point in me being here. What don't I get now? I hate it when I'm told I don't get it but can't be told what it is I don't get for some reason. What's the advice then? My understanding is flawed? That's really more of an observation than advice. I'll give you some advice instead. Don't assume that you're the one that must be right when clearly you either don't even know what you're talking about or can't just be bothered to explain yourself. Which is it?

Again you are wrong, a free falling observer reaches the singularity in finite proper time, and the closer he gets, the less time he has left.

Again you're being a lot less than helpful. I don't think you even understood what I meant. You're just saying it is because that's the way it is. How does that sentence bring anything at all to the discussion. I'm beginning to think I've accidentally stumbled into a new age religious website

Do you perhaps want to discuss the formulas that show this?

I'm not sure it will mean much to me. We can try though. I just need to see if it's frame dependant really. If the formula for it shows that an object can reach an horizon then I can't be. At least I don't see how it can be.

No, I mean that their initial velocities start out both being zero, but as the two objects fall, their velocities *become* different (the lower object falls faster than the higher object), even though they are both freely falling and both feel no force.

Okay, that's twice now that you've said "No I meant this..." and then repeated what I've said but worded it slightly differently. I'm getting used to it but you've now done it twice to the same statement.

And yet you keep on trying to bring in tidal force when I and others have told you multiple times that an object can free-fall through the horizon even in the case where tidal force is negligible (the object's length is so short compared to the size of the black hole that the tidal gravity of the hole at the horizon has no appreciable effect on the object's motion). See further comment below.

Saying it multiple times doesn't make it right, and it definitely doesn't help me to understand why or how it's possible. It doesn't make sense.

I never said a light ray can't (I assume that's what you meant, not "can") reach the horizon; I said the horizon *is* the path of a light ray--an outgoing light ray at r = 2M.

Oops. Yes that's what I meant. But light can't reach the horizon because no object can be seen reaching the horizon.

No, tidal force is a red herring that's distracting you from the real physics involved. That's why we keep recommending that you consider a Rindler horizon, which works like a black hole horizon in key respects but occurs in flat spacetime, where there is *no* tidal gravity to confuse things.

But as I've already said twice: The equivalent to the Rindler horizon using gravity just places a limit on the distance an object can be in order to catch the free-faller before the black hole's gone. This distance is individual and depends on the distance of the free-faller to the other object and to the black hole. Maybe I've misunderstood the Rindler horizon?

If the black hole is massive enough the front and the back will accelerate at the same speed.

Very large black holes will only experience tidal forces beyond the EH closer to the singularity.

The front and the back will never accelerate at the same speed because that doesn't make sense. You can reduce the difference by increasing the mass, but it's still there. Very large black holes don't experience less tidal force. It's just more spread out.

You do not experience any time dilation. Time is always contant for yourself. Outside observers would see an image of you "get stuck" just beyond the event horizon but only because the light coming towards them would make it appear so. However even this won't last forever as eventually the light wavelenght would become so long it is no longer inside the visible spectrum.

You do not experience time dilation and time is always constant for yourself you say. Thanks for that. Yes I know time is always constant for yourself. It would be ridiculous if it wasn't. If you want it from the fallers perspective then the black hole will evaporate quicker and quicker as you get closer to it and you will never have time to reach it.

This is true, but irrelevant. As I showed in a recent post, even for an extended object like this one, subject to tidal gravity, the center of mass of the object moves on a freely falling worldline--an accelerometer mounted at the center of mass of the object would read zero proper acceleration. This alone is enough to prove that the object will reach the horizon in a finite proper time, starting from a finite radius R > 2M. (The proof is what I posted earlier, since that proof applies to *any* freely falling worldline, including the one followed by the center of mass of an extended object moving solely under the influence of gravity.) This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

Maybe I wasn't clear enough. I'm going to explain this one more time as simply as I can. My thinking is that objects do accelerate in free-fall, kind of. That's what the Schwarzschild coordinates show. This acceleration has to be felt and I think it's felt as tidal force because tidal force is caused by the difference in velocity between objects large enough to notice the difference. Tidal force is just the difference in the rate of acceleration. Constant acceleration (as perceived by a distant observer) due to free-fall isn't felt as acceleration (so please no-one bother to point this out again, it's getting very frustrating). It's the equivalent to being at rest but an increase in the rate of acceleration is felt, as proper acceleration. This is all tidal force is. The rate of acceleration determines the value of length contraction and time dilation. The event horizon is the closest possible Rindler horizon to the singularity.

I completely agree with Peter.

For a black hole the tidal force between the bottom and the ceiling of a free falling object traveling is directly related to the remaining time until the singularity is hit, the location of the EH, and thus the black hole's mass, is totally irrelevant to this.

What? One of the factors is the location of the event horizon, which is determined by the black holes mass.

Wikipedia http://en.wikipedia.org/wiki/Hawking_radiation gives the time t_{ev} for a black hole to evaporate as:

t_{ev} = \frac{5120 \pi G^2 M^3}{\hbar c^4}

This evaporation time is proportional to the cube of the black hole mass and if finite for any finite mass black hole. The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction). Now coordinate measurements *exactly* to or at the event horizon are contentious because of the impossibility of having a stationary observer at that location so the event horizon is considered "off the Schwarzschild chart". This is not a problem, because we can always find a local stationary observer outside but arbitrarily close to the event horizon that will say that when the free falling observer arrived the black hole and event horizon had evaporated. In other words we can always find a point *outside* the event horizon where a local observer will confirm that the black hole evaporated before the free falling object's arrival. The proper time for a free falling observer that falls from a finite height less than infinite but greater than 2m is finite. Let us say that this finite free falling proper time is 16 seconds, then the black hole evaporates in less than 16 seconds of the free faller's proper time. This means, if we believe Hawking, that in practice no free falling particle has ever fallen to or through an event horizon, so points 1 to 4 in the list become academic because they never happened in nature and in principle never will.

I appreciate the backup but i don't think hawking radiation is needed to show that no object can possibly reach an EV, it's common sense. The fact that in-falling objects can reach a point when they can no longer be caught should be a big hint. Everything will reach the RH before reaching the EV. The EV is the closest any possible RH can be to the singularity.

This is not correct. The infinite coordinate time for an observer to free-fall to the event horizon assumes an "eternal" black hole, one that never evaporates; that is what the Schwarzschild solution to the EFE describes. When you add in evaporation a la Hawking, the spacetime as a whole is no longer the exact Schwarzschild spacetime, so it no longer takes infinite coordinate time for an observer to free-fall to the horizon. Instead, the coordinate time at which the hole finally evaporates (the formula you gave) is also the coordinate time at which any free-falling observer hits the horizon. However, you have to be very careful drawing implications from that statement; it is *not* the same as saying that all free-falling observers hit the horizon at the same event.

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

So basically, at the event of final evaporation of the hole, a flash is emitted that contains images of all the free-falling observers that crossed the horizon, just as they were crossing it. (Unfortunately, the free-falling observers themselves have already fallen into the singularity, since that is still to the future of any timelike worldline that crosses the horizon.) That single flash therefore gets assigned a single coordinate time in the "approximately Schwarzschild" coordinate system used by a distant observer, which will be the coordinate time of the final evaporation *and* of all the "crossing events" whose images are contained in the flash.

What? How can the free-falling observers have already fallen into the singularity? If you're ten light minutes away from them I they don't cross in the next ten minutes then they hadn't already crossed the horizon. The fact that you'll never see them cross means they never cross the bloody horizon! I'll say it again: The reason the light from those objects slows as they approach the horizon is because time slows as they approach the horizon. It's not an illusion!

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality. For example, take a look at this adaptation of the diagram you linked to:

The diagram shows light rays coming from the surface (orange curve) of a star collapsing to form the black hole. Examination of the path of observer A (green curve) shows he sees the star continue to shine all the way until he hits the singularity. Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist. It seems that no observer sees a region of space that is gravitationally attractive that is not shining and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated, so its seems no one ever sees the black hole as black, although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash. Does that seem right?

Time's frozen at the horizon so of course some light from the star will always take longer then the black hole's been there for as you get close to the horizon.

When you look at the Penrose diagrams for, e.g., the maximally extended Schwarzschild spacetime (which includes the "white hole" and the second exterior region), I agree the diagrams are showing things which, although they follow mathematically from the given solution to the EFE, are probably not physically reasonable. However, I don't think that's the case for this particular diagram; it presents a picture which seems, at least to me, to be consistent and in principle physically realizable, although perhaps counterintuitive.

That exact same logic is used to try to show that books like the bible make some kind of sense. "Well I like the final conclusion so I'll work backwards from there, even though this forced reverse engineering will lead to things that seem 'counterintuitive'."

 

[PeterDonis]

Okay, that's twice now that you've said "No I meant this..." and then repeated what I've said but worded it slightly differently. I'm getting used to it but you've now done it twice to the same statement.

But the "slightly different words" indicate a *very* different meaning, which you keep refusing to accept. In this case, the key item you refuse to accept is that the two objects are both freely falling and both feel no force even though their radial separation is changing with time. See further comment below.

But as I've already said twice: The equivalent to the Rindler horizon using gravity just places a limit on the distance an object can be in order to catch the free-faller before the black hole's gone. This distance is individual and depends on the distance of the free-faller to the other object and to the black hole. Maybe I've misunderstood the Rindler horizon?

Two things here: first, yes, you *are* misunderstanding the Rindler horizon (or at least you're mis-stating something), because the Rindler horizon (in the case of flat spacetime, no gravity) marks the limit of the distance an object can be in order to catch an *accelerating* object, meaning an object that is undergoing proper acceleration, due to rockets firing or some other physical mechanism that changes its 4-momentum. (More precisely, it's the limit of the distance an object can be and still emit a light ray that can catch an accelerating object).

Second, you keep on bringing in the evaporation of the black hole in the gravity case, and you still haven't answered the question I keep asking: in the case where the black hole is eternal--it never evaporates--do you agree that a freely falling observer *would* reach the horizon (and then the singularity after that)? If you do, then we can refocus this whole discussion on whether or not having the black hole evaporate eventually makes a difference. If you don't, then the evaporation of the black hole is irrelevant because you don't believe the horizon is reachable even when there is no evaporation.

Maybe I wasn't clear enough. I'm going to explain this one more time as simply as I can. My thinking is that objects do accelerate in free-fall, kind of. That's what the Schwarzschild coordinates show. This acceleration has to be felt and I think it's felt as tidal force because tidal force is caused by the difference in velocity between objects large enough to notice the difference. Tidal force is just the difference in the rate of acceleration. Constant acceleration (as perceived by a distant observer) due to free-fall isn't felt as acceleration (so please no-one bother to point this out again, it's getting very frustrating). It's the equivalent to being at rest but an increase in the rate of acceleration is felt, as proper acceleration. This is all tidal force is. The rate of acceleration determines the value of length contraction and time dilation. The event horizon is the closest possible Rindler horizon to the singularity.

You've been quite clear enough about this multiple times, and as I and others have told you multiple times, your thinking here is *wrong*. Period. (I've put the particular statement that zeroes in on the wrong idea in bold.) As long as you refuse to accept that, and keep on hanging on to this wrong idea, you will remain stuck and unable to understand what we're telling you about how the black hole horizon actually works.

 

[A-wal]

But the "slightly different words" indicate a *very* different meaning, which you keep refusing to accept. In this case, the key item you refuse to accept is that the two objects are both freely falling and both feel no force even though their radial separation is changing with time. See further comment below.

Yes, I definitely refuse to accept that (unless they're point like objects). If they're freely falling and separating as they do it then that same force that's separating them will also affect the objects themselves. It my be marginal but that's not the point. How could there possibly be a force acting on the space between them but not the space that they occupy?

Two things here: first, yes, you *are* misunderstanding the Rindler horizon (or at least you're mis-stating something), because the Rindler horizon (in the case of flat spacetime, no gravity) marks the limit of the distance an object can be in order to catch an *accelerating* object, meaning an object that is undergoing proper acceleration, due to rockets firing or some other physical mechanism that changes its 4-momentum. (More precisely, it's the limit of the distance an object can be and still emit a light ray that can catch an accelerating object).

Right! And in the case of a black hole it marks the limit of the distance an object can be in order to catch a *free-falling* object. The event horizon marks the tightest possible Rindler horizon.

Second, you keep on bringing in the evaporation of the black hole in the gravity case, and you still haven't answered the question I keep asking: in the case where the black hole is eternal--it never evaporates--do you agree that a freely falling observer *would* reach the horizon (and then the singularity after that)? If you do, then we can refocus this whole discussion on whether or not having the black hole evaporate eventually makes a difference. If you don't, then the evaporation of the black hole is irrelevant because you don't believe the horizon is reachable even when there is no evaporation.

I've answered that question more than once. If I was to accelerate away from Earth forever then I would still never reach c, just as it's impossible to reach an EV even if it lasts forever. Think about it. If it wasn't possible to reach any event horizon of a radiating BH then it can't be possible to reach the EV of an everlasting one because an object would have to cross at a given time (from any coordinate system, it doesn't matter). The length of time between the start of the black holes life and the time when the object crosses would obviously have to be finite so it's a nothing question.

You've been quite clear enough about this multiple times, and as I and others have told you multiple times, your thinking here is *wrong*. Period. (I've put the particular statement that zeroes in on the wrong idea in bold.) As long as you refuse to accept that, and keep on hanging on to this wrong idea, you will remain stuck and unable to understand what we're telling you about how the black hole horizon actually works.

I don't mind if I'm *wrong*. I'm just trying to make sure you know what I mean, and therefore know exactly what it is that you're refuting. And I don't see how hanging on to this idea will make me any less open minded to what I'm being told. I'll stop hanging on to it when it no longer makes sense to me for whatever reason.

 

[PeterDonis]

Yes, I definitely refuse to accept that (unless they're point like objects). If they're freely falling and separating as they do it then that same force that's separating them will also affect the objects themselves. It my be marginal but that's not the point. How could there possibly be a force acting on the space between them but not the space that they occupy?

You're assuming that there is a force acting on the space between them. There isn't. The *spacetime* in question does not have any "force" acting on it. It's just curved. The curvature of the spacetime is what causes the two freely falling objects to separate, even though they both feel no force. Your mental model of what's going on is giving you wrong answers because it uses the concept of "force" as a basic concept, instead of the concept of "curved spacetime". (In the curved spacetime view, the term "force", strictly speaking, should not be used for what we've been calling, informally, "tidal force"; that's why I've tried to use the term "tidal gravity" instead where possible.)

I realize that I've simply stated a contrary view, without *explaining* why you should prefer that view to yours. Nor does it help, apparently, to point out that my view gives correct answers to other questions (such as the current one, whether or not a black hole's event horizon is reachable) and yours doesn't, because you simply deny that your view's answer is incorrect--your view seems logically consistent to you, so you accept the answers it gives you even when the contradict the answers given by standard general relativity. Since we can't directly test the specific question about black holes experimentally, I'm not sure how to resolve the discrepancy. But see further comments below.

Right! And in the case of a black hole it marks the limit of the distance an object can be in order to catch a *free-falling* object. The event horizon marks the tightest possible Rindler horizon.

No, this is wrong. The analogy between a black hole horizon (which occurs in a curved spacetime, the Schwarzschild spacetime) and a Rindler horizon (which occurs in flat spacetime) is exact on this point: both represent the limit of the distance an object can be in order to catch (more precisely, send a light ray to catch) an *accelerating* object.

It *is* also true that an object inside a black hole's horizon can't send a light ray to catch outgoing free-falling objects *outside* the horizon (for example, an object moving at almost the speed of light directly radially outward, just outside the black hole horizon, could, if it were moving fast enough, escape to infinity; a light ray launched outward at or beneath the horizon can't ever get outside the horizon). But this feature of the black hole spacetime has nothing to do with the horizon being analogous to a Rindler horizon.

I've answered that question more than once. If I was to accelerate away from Earth forever then I would still never reach c, just as it's impossible to reach an EV even if it lasts forever. Think about it. If it wasn't possible to reach any event horizon of a radiating BH then it can't be possible to reach the EV of an everlasting one because an object would have to cross at a given time (from any coordinate system, it doesn't matter). The length of time between the start of the black holes life and the time when the object crosses would obviously have to be finite so it's a nothing question.

After reading back through the thread, you're correct, you have answered this before; but I think the review was helpful. Once again, your analysis is wrong, and I've highlighted the particular phrase that zeroes in on the point where you're wrong in bold. The length of time in question does *not* have to be finite in every coordinate system; in particular, in Schwarzschild coordinates, the "Schwarzschild time" t at which any object that crosses the horizon does so is plus infinity. But the region labeled "plus infinity" in a Schwarzschild coordinate diagram is *not* a single point (i.e., not a single event); it's an infinite line of events, all of which get labeled "plus infinity" by Schwarzschild coordinates. That's why Schwarzschild coordinates are not good ones to use to study what happens at the horizon. Other coordinate systems work better because they give all those events on the horizon separate labels, so that they are "visible" and can be analyzed. In Painleve coordinates, for example, the horizon corresponds to a line (r = 2M, T = minus infinity to plus infinity); each event on this line is a *different* event at which some object can cross the horizon. The region covered by the Schwarzschild coordinates (r = 2M, t = minus infinity to plus infinity) is all shoved down "below" T = minus infinity in Painleve coordinates.

Once again, I've just stated a contrary view, without really explaining why you should accept it. And again, in the absence of actual experimental data about black hole horizons, I'm not sure how to resolve the discrepancy (but see next comment). However, I do emphasize that what I've said above is not at all controversial; I've stated the position of standard general relativity, accepted and used by everyone that works in the field, as best I can (and others are welcome to correct or clarify if I've misstated something). I should also emphasize that, however unusual it sounds, what I've said above is perfectly consistent, logically and mathematically.

I don't mind if I'm *wrong*. I'm just trying to make sure you know what I mean, and therefore know exactly what it is that you're refuting. And I don't see how hanging on to this idea will make me any less open minded to what I'm being told. I'll stop hanging on to it when it no longer makes sense to me for whatever reason.

I think I have a decent understanding of two points on which you hold views that are inconsistent with the standard view of general relativity, and therefore lead you to answers to certain questions that are inconsistent with those of standard general relativity (to put it as neutrally as possible and avoid using the word "wrong"). Just to recap, the two points are:

(1) You believe that two freely falling objects separating due to tidal gravity must feel some force (because there must be a "force on the space between them"). In standard GR (and in Newtonian gravity as well), there is no such force; the two objects, separating due to tidal gravity, are both freely falling and feel no force. (In Newtonian gravity, both objects are responding to the "force" of gravity, and they separate because of the spatial variation in that force; but even in Newtonian physics, the "force" of gravity is not *felt* as a force--objects moving only under gravity are in free fall and feel no force, they are weightless. A person standing on the surface of the Earth feels weight because the Earth is pushing up on them, not because gravity is pulling them down, even in Newtonian physics.)

(2) You believe that any object that crosses a black hole horizon must do so at a finite value of the time coordinate in any coordinate system. In standard GR, this is false (I explained the standard GR view in my last comment just above).

As I said above, without experimental data on black hole horizons, it's impossible to directly test point (2). However, there is a lot of indirect evidence that GR holds just fine in strong gravity regimes fairly close to objects that we can't explain, on our current knowledge, unless they are black holes. And also, as I said, mathematically the standard GR view on point (2) is perfectly consistent, so if you're having trouble seeing how it can be so, looking further into how the various coordinate charts of Schwarzschild spacetime work might help.

As far as point (1) is concerned, I would think that there would be plenty of direct tests that freely falling objects separating due to tidal gravity are both weightless, but I haven't been able to find any decent treatment of such tests on the web. But, once again, mathematically, the curved spacetime model is perfectly consistent, so if you're having trouble understanding how it can be so, a better understanding of how it works might help.

 

[A-wal]

You're assuming that there is a force acting on the space between them. There isn't. The *spacetime* in question does not have any "force" acting on it. It's just curved. The curvature of the spacetime is what causes the two freely falling objects to separate, even though they both feel no force. Your mental model of what's going on is giving you wrong answers because it uses the concept of "force" as a basic concept, instead of the concept of "curved spacetime". (In the curved spacetime view, the term "force", strictly speaking, should not be used for what we've been calling, informally, "tidal force"; that's why I've tried to use the term "tidal gravity" instead where possible.)

Okay but if you look at it as curved space-time rather than a force then I'm not really sure how that changes anything. It's just semantics. I do normally look at it as curved space-time btw when there's more than two objects because it becomes easier. Looking at it in this way the space that the objects occupy is curved as well as the space between them so there'll still be a force acting on the objects themselves.

No, this is wrong. The analogy between a black hole horizon (which occurs in a curved spacetime, the Schwarzschild spacetime) and a Rindler horizon (which occurs in flat spacetime) is exact on this point: both represent the limit of the distance an object can be in order to catch (more precisely, send a light ray to catch) an *accelerating* object.

RIGHT! I didn't mean an actual Rindler horizon because that's already been defined. I'm saying you can use the same principle for gravity by substituting acceleration for free-fall because of tidal acceleration. An observer in flat space-time who sends a message from beyond the Rindler horizon to an accelerating observer shouldn't expect the message to ever reach the accelerating observer until their acceleration slows. The same thing happens if you send a message to a free-falling object. The free-faller will appear to slow from the distant observers perspective but so does the message.

After reading back through the thread, you're correct, you have answered this before; but I think the review was helpful. Once again, your analysis is wrong, and I've highlighted the particular phrase that zeroes in on the point where you're wrong in bold. The length of time in question does *not* have to be finite in every coordinate system; in particular, in Schwarzschild coordinates, the "Schwarzschild time" t at which any object that crosses the horizon does so is plus infinity. But the region labeled "plus infinity" in a Schwarzschild coordinate diagram is *not* a single point (i.e., not a single event); it's an infinite line of events, all of which get labeled "plus infinity" by Schwarzschild coordinates. That's why Schwarzschild coordinates are not good ones to use to study what happens at the horizon. Other coordinate systems work better because they give all those events on the horizon separate labels, so that they are "visible" and can be analyzed. In Painleve coordinates, for example, the horizon corresponds to a line (r = 2M, T = minus infinity to plus infinity); each event on this line is a *different* event at which some object can cross the horizon. The region covered by the Schwarzschild coordinates (r = 2M, t = minus infinity to plus infinity) is all shoved down "below" T = minus infinity in Painleve coordinates.

Okay, maybe not in any coordinate system then. I really just meant that if it's impossible to reach the EV of a black hole then it's impossible no matter how long it lasts because it would have to happen at a certain time and what if the black hole's life is finite but it last longer than this?

I think I have a decent understanding of two points on which you hold views that are inconsistent with the standard view of general relativity, and therefore lead you to answers to certain questions that are inconsistent with those of standard general relativity (to put it as neutrally as possible and avoid using the word "wrong"). Just to recap, the two points are:

(1) You believe that two freely falling objects separating due to tidal gravity must feel some force (because there must be a "force on the space between them"). In standard GR (and in Newtonian gravity as well), there is no such force; the two objects, separating due to tidal gravity, are both freely falling and feel no force. (In Newtonian gravity, both objects are responding to the "force" of gravity, and they separate because of the spatial variation in that force; but even in Newtonian physics, the "force" of gravity is not *felt* as a force--objects moving only under gravity are in free fall and feel no force, they are weightless. A person standing on the surface of the Earth feels weight because the Earth is pushing up on them, not because gravity is pulling them down, even in Newtonian physics.)

Yea that's how I look at it. The Earth accelerates outward pushing us downward. Length contraction means the Earth isn't expand as this happens. I got shot down for looking at it that way a few posts ago. I've already relied to the first bit.

(2) You believe that any object that crosses a black hole horizon must do so at a finite value of the time coordinate in any coordinate system. In standard GR, this is false (I explained the standard GR view in my last comment just above).

Not really, I think it's just that I worded it wrong. I shouldn't have said “in any coordinate system”. I should of said “has to happen at a given time regardless of the coordinate system you want to use”. If it is possible to pass the event horizon then you should be able to work out the time this happens according to the watch of the external observer even though they won't be able to see it yet. If that true then they'll be able to see the object reach the horizon if they wait long enough. It's not a self-consistent idea. Not to me anyway.

As I said above, without experimental data on black hole horizons, it's impossible to directly test point (2). However, there is a lot of indirect evidence that GR holds just fine in strong gravity regimes fairly close to objects that we can't explain, on our current knowledge, unless they are black holes. And also, as I said, mathematically the standard GR view on point (2) is perfectly consistent, so if you're having trouble seeing how it can be so, looking further into how the various coordinate charts of Schwarzschild spacetime work might help.

Coordinate systems aren't how I like to look at it. By definition they give an extremely limited perspective. I don't see how the standard view is self-consistent. Nobody's posted an answer to the paradox I posted ages ago yet.

As far as point (1) is concerned, I would think that there would be plenty of direct tests that freely falling objects separating due to tidal gravity are both weightless, but I haven't been able to find any decent treatment of such tests on the web. But, once again, mathematically, the curved spacetime model is perfectly consistent, so if you're having trouble understanding how it can be so, a better understanding of how it works might help.

Weightless? The very front would be but not the rest of the object. The rest is being dragged along at least slightly.

 

[PeterDonis]

Okay but if you look at it as curved space-time rather than a force then I'm not really sure how that changes anything. It's just semantics. I do normally look at it as curved space-time btw when there's more than two objects because it becomes easier. Looking at it in this way the space that the objects occupy is curved as well as the space between them so there'll still be a force acting on the objects themselves.

Nope. At least, if that's how you are using the concept "curved spacetime", then you are using it wrong. Or if you object to the word "wrong", then I'd say that the concept you are labeling with the term "curved spacetime" is *not* the same as the concept that standard GR labels with the term "curved spacetime". In standard GR, spacetime (flat *or* curved) does not exert any force on anything. All it does is define what states of motion are freely falling (geodesic) at each event and what states of motion are not. Which particular state of motion any actual object is actually *in* depends on the object and what it is doing (for example, is it firing rockets or not), not on "spacetime".

A general comment in view of what I've just said: as I see it, this discussion has gone on so long because every time someone tries to explain how standard GR treats a certain concept, you come back with your own personal version of that concept, instead of trying to understand the standard GR version.

RIGHT! I didn't mean an actual Rindler horizon because that's already been defined. I'm saying you can use the same principle for gravity by substituting acceleration for free-fall because of tidal acceleration. An observer in flat space-time who sends a message from beyond the Rindler horizon to an accelerating observer shouldn't expect the message to ever reach the accelerating observer until their acceleration slows. The same thing happens if you send a message to a free-falling object. The free-faller will appear to slow from the distant observers perspective but so does the message.

I can sort of see what you're describing here in the case of free fall towards a black hole, but it still has nothing to do with the black hole horizon being analogous to a Rindler horizon, and the idea of "substituting acceleration for free-fall because of tidal acceleration" makes no sense in standard GR.

Yea that's how I look at it. The Earth accelerates outward pushing us downward. Length contraction means the Earth isn't expand as this happens. I got shot down for looking at it that way a few posts ago.

"Length contraction" due to spacetime curvature has nothing to do with the Earth not expanding even though it's pushing objects outward. That's why you got shot down. (Also, the Earth pushes us *upward*--or outward--not *downward*.)

If it is possible to pass the event horizon then you should be able to work out the time this happens according to the watch of the external observer even though they won't be able to see it yet. If that true then they'll be able to see the object reach the horizon if they wait long enough. It's not a self-consistent idea. Not to me anyway.

Once again, your reasoning is based on a false premise (I've highlighted it in bold again in the quote above). In standard GR, the "time this happens" (i.e., the event of an observer crossing the horizon) is plus infinity according to the watch of any external observer (outside the horizon). They will *never* see the object crossing the horizon, even if they wait an infinite amount of time. This is all perfectly self-consistent; it just uses a model that does not satisfy your intuition.

I don't see how the standard view is self-consistent. Nobody's posted an answer to the paradox I posted ages ago yet.

Can you re-post it or point me to it? I don't remember what it was. (Maybe it wasn't in this thread?)

Weightless? The very front would be but not the rest of the object. The rest is being dragged along at least slightly.

I've asked several times already to restrict discussion to idealized, point-like "test objects" that have no internal parts, and therefore no internal forces between the parts. For such objects, there's no such thing as the front of the object "dragging the rest along", so what you've said above would not apply. As I've also said several times already, the reason this idealization is not unreasonable is that even for a non-idealized object, the motion of its *center of mass* will still be "weightless" motion; the object will feel no weight (no force, no acceleration) at its center of mass, and therefore will undergo no proper acceleration as a whole (though parts of it may accelerate and feel force relative to other parts). It's the center-of-mass motion that is being referred to when standard GR (and Newtonian physics too) say that bodies moving only under gravity (including tidal gravity) are weightless.

 

[PeterDonis]

I've asked several times already to restrict discussion to idealized, point-like "test objects"

Looking back at your post #263, I realized that you did include the qualifier "unless they're both point-like objects". So you *do* agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time due to tidal gravity? If so, does that mean you agree that a point-like object *could* reach a black hole horizon?

 

[A-wal]

Nope. At least, if that's how you are using the concept "curved spacetime", then you are using it wrong. Or if you object to the word "wrong", then I'd say that the concept you are labeling with the term "curved spacetime" is *not* the same as the concept that standard GR labels with the term "curved spacetime". In standard GR, spacetime (flat *or* curved) does not exert any force on anything. All it does is define what states of motion are freely falling (geodesic) at each event and what states of motion are not. Which particular state of motion any actual object is actually *in* depends on the object and what it is doing (for example, is it firing rockets or not), not on "spacetime".

What the hell were you reading? I didn't think I was being that unclear. You're so eager to keep disagreeing with everything I say that you're taking everything I write to mean something different to what I've actually said. This is semantics, as I said in the last post. You can look at it as a force acting between two objects. If it works then it's right. That's what right means. It may be more accurate to think of it as curved space-time but that's the same thing here. Instead of a force acting on the objects you could say the same-time between them feels the force rather than the objects themselves. But if you want to take that view then the space occupied by the objects is curved as well as the space between them, effectively replacing the force acting on them.

A general comment in view of what I've just said: as I see it, this discussion has gone on so long because every time someone tries to explain how standard GR treats a certain concept, you come back with your own personal version of that concept, instead of trying to understand the standard GR version.

This discussion has gone on for so long because virtually everything I post gets misunderstood or taken out of context so I spend most of the time on here rewording posts that I've already written.

I can sort of see what you're describing here in the case of free fall towards a black hole, but it still has nothing to do with the black hole horizon being analogous to a Rindler horizon, and the idea of "substituting acceleration for free-fall because of tidal acceleration" makes no sense in standard GR.

If there was an object that we could see as being one millimetre away from the horizon then would it reach the horizon if it was to move two millimetres forward from its own perspective? No, because that's not what it would look like if you were the one approaching the horizon. You'd be in length contracted space so you could move forwards without crossing the horizon. The millimetre from the previous perspective could be a mile or a light year or whatever if you were actually there, depending on the mass of the black hole. Bigger ones last longer but there's more length contracted space to pass through before you get to the horizon, and you'd be more time dilated so it would take longer to get there. I don't see how any conversion formula that's accurate for moving between the two perspectives can possibly claim that you could cross the horizon if you were there but not from a distance.

"Length contraction" due to spacetime curvature has nothing to do with the Earth not expanding even though it's pushing objects outward. That's why you got shot down. (Also, the Earth pushes us *upward*--or outward--not *downward*.)

LOL. I'm fairly positive it does and it doesn't, respectively.

Once again, your reasoning is based on a false premise (I've highlighted it in bold again in the quote above). In standard GR, the "time this happens" (i.e., the event of an observer crossing the horizon) is plus infinity according to the watch of any external observer (outside the horizon). They will *never* see the object crossing the horizon, even if they wait an infinite amount of time. This is all perfectly self-consistent; it just uses a model that does not satisfy your intuition.

Well it's not self-consistent as far as I can see. See below.

Can you re-post it or point me to it? I don't remember what it was. (Maybe it wasn't in this thread?)

A ship is deliberately pulled into a black hole. It crosses the horizon when the black hole is a certain size and there's a second observer who follows close behind but doesn't allow themselves to be pulled in. There's a very strong rope linking the two. The second observer can never witness the first one reaching the event horizon so it can never be too late for them to find the energy to pull the first observer away from the black hole even after the it's shrunk to a smaller than when the first one crossed from it's own perspective. If the closer one always has the potential to escape the black hole under its own power from the further ones perspective then it should always be possible for the closer one to escape under the further ones power. So the first ship can't escape from it's own perspective but it does from the second ships perspective. Paradox!

I've asked several times already to restrict discussion to idealized, point-like "test objects" that have no internal parts, and therefore no internal forces between the parts. For such objects, there's no such thing as the front of the object "dragging the rest along", so what you've said above would not apply. As I've also said several times already, the reason this idealization is not unreasonable is that even for a non-idealized object, the motion of its *center of mass* will still be "weightless" motion; the object will feel no weight (no force, no acceleration) at its center of mass, and therefore will undergo no proper acceleration as a whole (though parts of it may accelerate and feel force relative to other parts). It's the center-of-mass motion that is being referred to when standard GR (and Newtonian physics too) say that bodies moving only under gravity (including tidal gravity) are weightless.

Looking back at your post #263, I realized that you did include the qualifier "unless they're both point-like objects". So you *do* agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time due to tidal gravity? If so, does that mean you agree that a point-like object *could* reach a black hole horizon?

Yes, I agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time. Of course this has no effect on whether a point like object can reach the horizon. I don't see how anything can reach the horizon including light. It's because of the curvature of space-time that no object can reach the horizon. There'll be no time and an infinite amount of potential space as the object heads into ever more length contracted/time dilated space.

 

[PeterDonis]

If it works then it's right. That's what right means.

And if it doesn't work, then it's not right. Your view of tidal gravity as a force doesn't work; it leads you to incorrect conclusions about what happens at the horizon of a black hole. Or at least, it does if your view of tidal gravity as a force is contributing to your claim that nothing can reach a black hole horizon. See further comment below on post #263 and your response to my question about it.

I should clarify that the view of tidal gravity as a force can work in more restricted domains; for example, it works fine for understanding how the Moon causes ocean tides on the Earth. But the curved spacetime view also works fine in those more restricted domains, *and* it works in domains where the view of tidal gravity as a force doesn't. It's because we're talking about the latter type of domain that I've been saying the view of tidal gravity as a force is wrong.

If there was an object that we could see as being one millimetre away from the horizon then would it reach the horizon if it was to move two millimetres forward from its own perspective? No, because that's not what it would look like if you were the one approaching the horizon. You'd be in length contracted space so you could move forwards without crossing the horizon. The millimetre from the previous perspective could be a mile or a light year or whatever if you were actually there, depending on the mass of the black hole. Bigger ones last longer but there's more length contracted space to pass through before you get to the horizon, and you'd be more time dilated so it would take longer to get there. I don't see how any conversion formula that's accurate for moving between the two perspectives can possibly claim that you could cross the horizon if you were there but not from a distance.

You don't see how, but it's true. I've posted the calculation already in this thread--both the calculation that the proper time for an object to fall to the horizon from a finite radius is finite, *and* the calculation that the proper distance to the horizon from an object at a finite radius is finite. Those calculations already take into account *all* of the "length contraction" and "time dilation" caused by gravity as the horizon is approached and reached; which means that length contraction and time dilation do *not* "go to infinity" as the horizon is approached and reached, even though your intuitive picture tells you that they do. You appear to prefer your intuitive picture to the actual math, but that doesn't make your intuitive picture right. I've tried in previous posts to give an alternate intuitive picture that helps to illustrate how the math arrives at the result it does, but the intuitive picture is not why physicists accept the result. They accept it because of the math, and because wherever we've checked the math against experiment, it's been confirmed.

LOL. I'm fairly positive it does and it doesn't, respectively.

So you believe that the force exerted on your feet by the ground is pushing you downward? Remember that I'm speaking from the GR spacetime perspective, where "gravity" is not a force, so the only force the Earth is exerting on you is the push from the ground on your feet.

A ship is deliberately pulled into a black hole. It crosses the horizon when the black hole is a certain size and there's a second observer who follows close behind but doesn't allow themselves to be pulled in. There's a very strong rope linking the two. The second observer can never witness the first one reaching the event horizon so it can never be too late for them to find the energy to pull the first observer away from the black hole even after the it's shrunk to a smaller than when the first one crossed from it's own perspective. If the closer one always has the potential to escape the black hole under its own power from the further ones perspective then it should always be possible for the closer one to escape under the further ones power. So the first ship can't escape from it's own perspective but it does from the second ships perspective. Paradox!

The answer to this has already been posted in this thread, but it's been a while and it was spread over several posts, so I'll recap it here. But first, you keep bringing in evaporation even though you've said (or at least I think you've said) that you believe the horizon of an eternal black hole, one that never evaporates and has the same mass for all time, is also unreachable. If that's the case, the evaporation is irrelevant and you should rephrase the paradox to apply to the case of an eternal black hole. In what follows I'll treat the paradox in that form.

The error in your argument is in bold above. The key point you're missing is that the ability of the second observer to see the first one reaching the horizon depends on the behavior of *outgoing* light rays, whereas the ability of the second observer to affect the trajectory of the first observer depends on the behavior of *ingoing* light rays (and other ingoing objects or causal influences, like infalling observers and pulls on ropes). Those two behaviors are different in the spacetime around a black hole. Outgoing light rays that start at the horizon, r = 2M, stay at r = 2M forever; but ingoing light rays (and other ingoing things) can reach the horizon from outside it just fine. So the fact that the second observer can never receive an *outgoing* light signal from the event of the first observer crossing the horizon, does not prevent the *ingoing* first observer from falling through the horizon; nor does it prevent there from being a "last point" on the second observer's worldline where an *ingoing* causal influence can be emitted that will reach the first observer before he crosses the horizon.

You may ask what happens if the second observer pulls on the rope *after* the "last point" has passed. The answer is that the rope will already have broken, so no causal influence transmitted down the rope can reach the first observer any more. The rope will break at some point *before* the "last point" is reached, from the second observer's perspective; a rope that breaks just before the "last point" is reached is an "idealized" rope with the maximum possible tensile strength allowed by relativity, a tensile strength so high that the speed of sound in the rope equals the speed of light. A real rope, of course, will break long before that.

And what force is it that breaks the rope? Not "tidal gravity". In fact it's just the force exerted by the second observer, who has to be exerting a force on himself and anything attached to him in order to hover at a constant radius R above the horizon. This force holds one end of the rope and forces the rope to stretch as the first observer free-falls; eventually (just before the first observer crosses the horizon, if the rope is the "idealized" rope--much earlier, for any real rope), that stretching force will overcome the rope's tensile strength and the rope will break.

Yes, I agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time. Of course this has no effect on whether a point like object can reach the horizon.

Does this mean you agree that, if we limit the discussion to point-like objects, tidal gravity is irrelevant to the question of whether an object can reach the horizon? Or do you still think tidal gravity plays a role somehow? I'm not clear about the role that your view of tidal gravity as a force is playing in your logic.

I don't see how anything can reach the horizon including light. It's because of the curvature of space-time that no object can reach the horizon. There'll be no time and an infinite amount of potential space as the object heads into ever more length contracted/time dilated space.

See my comment above. The time is not zero and the space is not infinite.

 

[A-wal]

And if it doesn't work, then it's not right. Your view of tidal gravity as a force doesn't work; it leads you to incorrect conclusions about what happens at the horizon of a black hole. Or at least, it does if your view of tidal gravity as a force is contributing to your claim that nothing can reach a black hole horizon. See further comment below on post #263 and your response to my question about it.

I should clarify that the view of tidal gravity as a force can work in more restricted domains; for example, it works fine for understanding how the Moon causes ocean tides on the Earth. But the curved spacetime view also works fine in those more restricted domains, *and* it works in domains where the view of tidal gravity as a force doesn't. It's because we're talking about the latter type of domain that I've been saying the view of tidal gravity as a force is wrong.

I wasn't talking about tidal gravity, I was just talking about gravity. I don't view gravity as a force unless it's easier. You're arguing with yourself again. And the view of tidal gravity isn't contributing to my claim that nothing can reach a black hole horizon. It's the other way round. I'm trying to bridge the gap between what I'm being told and how I think of it.

You don't see how, but it's true. I've posted the calculation already in this thread--both the calculation that the proper time for an object to fall to the horizon from a finite radius is finite, *and* the calculation that the proper distance to the horizon from an object at a finite radius is finite. Those calculations already take into account *all* of the "length contraction" and "time dilation" caused by gravity as the horizon is approached and reached; which means that length contraction and time dilation do *not* "go to infinity" as the horizon is approached and reached, even though your intuitive picture tells you that they do. You appear to prefer your intuitive picture to the actual math, but that doesn't make your intuitive picture right. I've tried in previous posts to give an alternate intuitive picture that helps to illustrate how the math arrives at the result it does, but the intuitive picture is not why physicists accept the result. They accept it because of the math, and because wherever we've checked the math against experiment, it's been confirmed.

If the equations work then that doesn't necessarily tell the whole story. People thought Newtons equations worked until they were applied to the orbit of Mercury. It's what the equations represent that has to make sense. Equations can work but if what they represent doesn't than the they aren't complete. I don't know how anyone can be satisfied with a bunch of numbers and symbols if they can't understand what they're actually supposed to represent.

So you believe that the force exerted on your feet by the ground is pushing you downward? Remember that I'm speaking from the GR spacetime perspective, where "gravity" is not a force, so the only force the Earth is exerting on you is the push from the ground on your feet.

No I believe the force exerted on my feet by the ground is pushing me upward while length contraction stops me from going anywhere.

The answer to this has already been posted in this thread, but it's been a while and it was spread over several posts, so I'll recap it here. But first, you keep bringing in evaporation even though you've said (or at least I think you've said) that you believe the horizon of an eternal black hole, one that never evaporates and has the same mass for all time, is also unreachable. If that's the case, the evaporation is irrelevant and you should rephrase the paradox to apply to the case of an eternal black hole. In what follows I'll treat the paradox in that form.

The error in your argument is in bold above. The key point you're missing is that the ability of the second observer to see the first one reaching the horizon depends on the behavior of *outgoing* light rays, whereas the ability of the second observer to affect the trajectory of the first observer depends on the behavior of *ingoing* light rays (and other ingoing objects or causal influences, like infalling observers and pulls on ropes). Those two behaviors are different in the spacetime around a black hole. Outgoing light rays that start at the horizon, r = 2M, stay at r = 2M forever; but ingoing light rays (and other ingoing things) can reach the horizon from outside it just fine. So the fact that the second observer can never receive an *outgoing* light signal from the event of the first observer crossing the horizon, does not prevent the *ingoing* first observer from falling through the horizon; nor does it prevent there from being a "last point" on the second observer's worldline where an *ingoing* causal influence can be emitted that will reach the first observer before he crosses the horizon.

You may ask what happens if the second observer pulls on the rope *after* the "last point" has passed. The answer is that the rope will already have broken, so no causal influence transmitted down the rope can reach the first observer any more. The rope will break at some point *before* the "last point" is reached, from the second observer's perspective; a rope that breaks just before the "last point" is reached is an "idealized" rope with the maximum possible tensile strength allowed by relativity, a tensile strength so high that the speed of sound in the rope equals the speed of light. A real rope, of course, will break long before that.

And what force is it that breaks the rope? Not "tidal gravity". In fact it's just the force exerted by the second observer, who has to be exerting a force on himself and anything attached to him in order to hover at a constant radius R above the horizon. This force holds one end of the rope and forces the rope to stretch as the first observer free-falls; eventually (just before the first observer crosses the horizon, if the rope is the "idealized" rope--much earlier, for any real rope), that stretching force will overcome the rope's tensile strength and the rope will break.

I already answered that. From the perspective of the distant observer it will always be possible to pull the free-faller back because the free-faller will always be outside of the horizon from the perspective of the distant observer (as well as their own). So it is always possible to pull anyone away from a black hole in a finite amount of time, using a finite amount of energy and a finite strength rope.

Does this mean you agree that, if we limit the discussion to point-like objects, tidal gravity is irrelevant to the question of whether an object can reach the horizon? Or do you still think tidal gravity plays a role somehow? I'm not clear about the role that your view of tidal gravity as a force is playing in your logic.

I don't think tidal gravity is the reason why objects can't reach the horizon. I think it's a consequence of it.

See my comment above. The time is not zero and the space is not infinite.

But we know the time is zero and the space is infinite. Look at the outgoing light rays. At the horizon they're frozen. The only way light can be frozen like that is if time has come to a complete stop at the horizon. The speed of light is a measurement of the speed of time. I don't see how it could possibly work one way but not the other. The version you're pushing contradicts itself when it switches between ingoing light rays that aren't effected by time dilation as much as outgoing ones, and also when switching between the views of a distant observer who does witness infinite time dilation at the horizon and the free-faller who doesn't. Look at it in terms of light years/minutes etc. The free-faller would measure the distance between themselves and the horizon by the local speed of light. At the horizon light stops so there's always an infinite amount of light years between any object and any event horizon. So if there's an infinite amount of potential space between you and the horizon htf are you meant to reach it? I really do believe that it's the equivalent of trying to reach c. The reason you can't reach c is because c is effectively infinitely fast, so how are you supposed to go from a finite speed to an infinite one? You obviously can't, but it doesn't look infinite from any given rest frame. So even though the speed of light appears reachable it's not because it effectively changes as you change frame in order to remaining constant. The distance between a free-faller and the horizon always looks finite but it isn't.