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

[PeterDonis]

I KNOW you don't feel like you're accelerating when in free-fall! That's at least the third time I've had to say that. But in free-fall you do feel the difference of gravity between the closer parts and the further parts of your body to the centre of gravity. It's normally marginal but in the case of a black hole the gravity increases so sharply that it becomes noticeable, then painful, then very painful, then death. This seems like a pretty good description of tidal force. Are you saying it's something else entirely?

It's a workable description of tidal force, with some caveats about what "the difference of gravity" actually means--I would prefer the term "tidal gravity" or even "spacetime curvature", but that's a minor point. The key is that what you've described above does not prevent objects from reaching or crossing the horizon. Tidal force is finite at the horizon (r = 2M), and below it all the way to the central singularity at r = 0; it only becomes infinite *at* r = 0.

How about a game of let's pretend? Let's pretend that general relativity has just been released, by me, and it's my version where nothing can reach the event horizon. Presumably you'd know straight away that there's an error. Now the onus is on every one else to explain to me how an object can reach the event horizon. See how you like it.

This seems to be the game we're already playing.

 

[Dale]

An area that light can never reach from the outside is an area that nothing can reach.

True, but light can reach the horizon and the interior from the exterior.

No, I said the choice of coordinate systems can't change reality. So by my logic they can't both be right. You can either reach the horizon or you can't, and I don't see how anything possibly could.

Your logic is wrong. There are two kinds of statements that you can make:
(1) statements which depend on the choice of coordinates
(2) statements which do not depend on the choice of coordinates

Both coordinate systems can be right about statements of type (1) wrt themselves, and both coordinate systems will agree about statements of type (2).

For example, suppose I have two Newtonian coordinate systems where x=X+5. If the position of an object is X=3 then the statement that the position of the object is x=8 is also right. The change in coordinates changes the statement, but BOTH statements are right. The position of the object is a statement of type (1), and so by your premise statements of type (1) are not statements about "reality".

If the position of a second object is X=5 then the distance between the two objects is X2-X1 = 5-3 = 2. In the other coordinate system the position of the second object is x=10 and the distance between the two objects is x2-x1 = 10-8 = 2. The distance between the two objects is a statement of type (2), and so by your premise statements of type (2) are statements about "reality".

Your confusion is that you believe that "it takes forever for an object to reach the event horizon" is a statement of type (2) when in fact it is a statement of type (1).

 

[A-wal]

It's a workable description of tidal force, with some caveats about what "the difference of gravity" actually means--I would prefer the term "tidal gravity" or even "spacetime curvature", but that's a minor point. The key is that what you've described above does not prevent objects from reaching or crossing the horizon. Tidal force is finite at the horizon (r = 2M), and below it all the way to the central singularity at r = 0; it only becomes infinite *at* r = 0.

So you could look at someone in a stronger gravitational field as constantly accelerating (because the time dilation and length contraction associated with it would be the same as if they were constantly accelerating in flat space-time) and although they don't feel it, they feel the increase in the rate of acceleration and so in difference in gravity is felt as proper acceleration and therefore tidal force is just g-force. That's what I was getting at.

This seems to be the game we're already playing.

If this were the game we were already playing then I'd be officially right until proven wrong and you'd have to actually show how the contradictions resolve themselves if you wanted to be taken seriously. That would be lush.

True, but light can reach the horizon and the interior from the exterior.

How when no external object can reach the horizon from the exterior? External objects can only reach the horizon from the interior. That doesn't make sense to me.

Your logic is wrong. There are two kinds of statements that you can make:
(1) statements which depend on the choice of coordinates
(2) statements which do not depend on the choice of coordinates

Both coordinate systems can be right about statements of type (1) wrt themselves, and both coordinate systems will agree about statements of type (2).

For example, suppose I have two Newtonian coordinate systems where x=X+5. If the position of an object is X=3 then the statement that the position of the object is x=8 is also right. The change in coordinates changes the statement, but BOTH statements are right. The position of the object is a statement of type (1), and so by your premise statements of type (1) are not statements about "reality".

If the position of a second object is X=5 then the distance between the two objects is X2-X1 = 5-3 = 2. In the other coordinate system the position of the second object is x=10 and the distance between the two objects is x2-x1 = 10-8 = 2. The distance between the two objects is a statement of type (2), and so by your premise statements of type (2) are statements about "reality".

Your confusion is that you believe that "it takes forever for an object to reach the event horizon" is a statement of type (2) when in fact it is a statement of type (1).

I appreciate the replies and I know exactly what you're saying but it doesn't help me understand why "it takes forever for an object to reach the event horizon" is a statement of type (1) rather than type (2). It's the same with your answers that say to just use a coordinate system that shows matter crossing the horizon. It's just saying it is that way because it is. My point is that I don't understand how any such coordinate system can reflect reality. I'm fine with two different coordinate systems giving two completely different answers as long as they don't contradict each other.

 

[Dale]

It's just saying it is that way because it is.

I don't believe that this is a fair characterization. I do not generally simply assert things and tell you to accept it because that is the way it is. I have tried over and over to give easier to understand analogies and examples. If you have not understood the analogies or examples that is one thing, but to claim that I am not attempting to teach and am simply asserting the result is misrepresenting me completely.

I'm fine with two different coordinate systems giving two completely different answers as long as they don't contradict each other.

Do you understand how one coordinate system may contradict another as to the location of an event and yet both are correct? Refering to my earlier example, do you understand that x=8 and X=3 can both be correct although they disagree. If so then what is preventing you from understanding that T=infinity and t=t0 may both be correct?

 

[PeterDonis]

So you could look at someone in a stronger gravitational field as constantly accelerating (because the time dilation and length contraction associated with it would be the same as if they were constantly accelerating in flat space-time) and although they don't feel it, they feel the increase in the rate of acceleration and so in difference in gravity is felt as proper acceleration and therefore tidal force is just g-force. That's what I was getting at.

And this is *wrong*. Relative tidal acceleration (meaning, the relative acceleration of two objects, one of which is in a stronger gravitational field than the other--for example, closer to a black hole's horizon) is *not* the same as constant acceleration in flat spacetime; it's not even analogous to it; the two have no useful features in common. The difference in gravity (i.e., the relative tidal acceleration) is *not* felt as "proper acceleration" (more precisely, it isn't if the two objects are both free-falling; yes, if you take a single rigid object with enough radial extension to see the tidal force, the ends of the object will feel some force because the internal binding forces of the object are pulling them out of their natural free-fall paths, but that still is a very different phenomenon than the "constant acceleration" required to hover at a constant distance above the horizon). Tidal force is *not* just "g-force".

I appreciate the replies and I know exactly what you're saying but it doesn't help me understand why "it takes forever for an object to reach the event horizon" is a statement of type (1) rather than type (2). It's the same with your answers that say to just use a coordinate system that shows matter crossing the horizon. It's just saying it is that way because it is. My point is that I don't understand how any such coordinate system can reflect reality. I'm fine with two different coordinate systems giving two completely different answers as long as they don't contradict each other.

The quote you were responding to here was actually from DaleSpam, not me, but I'll respond anyway because I think this is an important point. A similar point came up in another thread, to which I responded in this post:

https://www.physicsforums.com/showpost.php?p=2957772&postcount=106

In the other thread we were discussing the point with reference to Rindler observers and the Rindler horizon, but the same point I made there applies here. Even though an observer far away from the black hole can't actually *see* a free-falling observer reach or cross the horizon, he can tell that there must be events on the free-falling observer's worldline that occur after the portion of the worldline that he can observe. How? By taking the formula for the free-falling observer's proper time, as a function of Schwarzschild coordinate time t (which is the time experienced by the observer very far away from the hole), integrating it out to t= infinity, and finding that the integral converges to a finite value--in other words, the free-falling observer experiences only a finite amount of proper time even when coordinate time t goes to infinity. Another way of expressing this is to say that the portion of the free-falling observer's worldline that the faraway observer can see has only a finite length, even when extended out to infinite coordinate time t.

But worldlines don't just stop at a finite length. *Something* has to happen to the free-falling observer after he has traversed the finite length of worldline that the faraway observer can see. He can't just cease to exist (for one thing, that would violate conservation of energy). The only possible explanation is that there is a further portion of the free-falling observer's worldline after the finite portion that the faraway observer can see. But the faraway observer has no way of assigning coordinates to that further portion, because he's already gone out to infinity in his coordinate time just to cover the finite portion he can see. This is what we mean when we say that the faraway observer's coordinates (the Schwarzschild exterior coordinates) "don't cover the entire manifold"--we can show that there *must* be another part of spacetime (the part where the free-falling observer's worldline goes) that simply can't be assigned coordinates in that system.

The above is why statements like "it takes forever for an object to reach the event horizon" are statements of type (1): because the answer is only "infinite" (or "forever") with respect to Schwarzschild coordinate time t--it's *finite* with respect to the proper time of the falling object.

 

[A-wal]

I don't believe that this is a fair characterization. I do not generally simply assert things and tell you to accept it because that is the way it is. I have tried over and over to give easier to understand analogies and examples. If you have not understood the analogies or examples that is one thing, but to claim that I am not attempting to teach and am simply asserting the result is misrepresenting me completely.

Don't take it so personally. It wasn't meant that way. I was referring just to the specific examples of saying that I need to use a coordinate system which includes the interior of the horizon and that I was making a coordinate dependant statement. Also, I really wish that it wasn't instantly assumed that because I question something it means I don't understand what I've just read.

Do you understand how one coordinate system may contradict another as to the location of an event and yet both are correct? Refering to my earlier example, do you understand that x=8 and X=3 can both be correct although they disagree. If so then what is preventing you from understanding that T=infinity and t=t0 may both be correct?

Of course I understand that one coordinate system can give different results to another. That's basically what time dilation length contraction is. That doesn't mean that they contradict each other. If something happens then of course when and where it happens are relative. T=infinity and t=t0 on the other hand is a contradiction because one say it happens and the other says it doesn't.

And this is *wrong*. Relative tidal acceleration (meaning, the relative acceleration of two objects, one of which is in a stronger gravitational field than the other--for example, closer to a black hole's horizon) is *not* the same as constant acceleration in flat spacetime; it's not even analogous to it; the two have no useful features in common. The difference in gravity (i.e., the relative tidal acceleration) is *not* felt as "proper acceleration" (more precisely, it isn't if the two objects are both free-falling; yes, if you take a single rigid object with enough radial extension to see the tidal force, the ends of the object will feel some force because the internal binding forces of the object are pulling them out of their natural free-fall paths, but that still is a very different phenomenon than the "constant acceleration" required to hover at a constant distance above the horizon). Tidal force is *not* just "g-force".

If you replace constant velocity with constant acceleration (so that an object undergoing constant acceleration feels no g-force) and just use an increase in acceleration rather than velocity to produce g-force then this isn't analogous to a distant observers perspective of a free-faller even though the free-faller doesn't feel as though they're accelerating but does feel the increase in acceleration. It seems very analogous.

The quote you were responding to here was actually from DaleSpam, not me, but I'll respond anyway because I think this is an important point. A similar point came up in another thread, to which I responded in this post:

https://www.physicsforums.com/showpost.php?p=2957772&postcount=106

In the other thread we were discussing the point with reference to Rindler observers and the Rindler horizon, but the same point I made there applies here. Even though an observer far away from the black hole can't actually *see* a free-falling observer reach or cross the horizon, he can tell that there must be events on the free-falling observer's worldline that occur after the portion of the worldline that he can observe. How? By taking the formula for the free-falling observer's proper time, as a function of Schwarzschild coordinate time t (which is the time experienced by the observer very far away from the hole), integrating it out to t= infinity, and finding that the integral converges to a finite value--in other words, the free-falling observer experiences only a finite amount of proper time even when coordinate time t goes to infinity. Another way of expressing this is to say that the portion of the free-falling observer's worldline that the faraway observer can see has only a finite length, even when extended out to infinite coordinate time t.

Yes, that's exactly *in* line with what I've *been* saying.

But worldlines don't just stop at a finite length. *Something* has to happen to the free-falling observer after he has traversed the finite length of worldline that the faraway observer can see. He can't just cease to exist (for one thing, that would violate conservation of energy). The only possible explanation is that there is a further portion of the free-falling observer's worldline after the finite portion that the faraway observer can see. But the faraway observer has no way of assigning coordinates to that further portion, because he's already gone out to infinity in his coordinate time just to cover the finite portion he can see. This is what we mean when we say that the faraway observer's coordinates (the Schwarzschild exterior coordinates) "don't cover the entire manifold"--we can show that there *must* be another part of spacetime (the part where the free-falling observer's worldline goes) that simply can't be assigned coordinates in that system.

Show *me*. How can we can show that there *must* be another part of spacetime (the part where the free-falling observer's worldline goes) *that* simply can't be assigned coordinates in that system? The world line of an object heading towards a black hole is heading into an area of time-dilated and length contracted space relative to wherever they start from. So you can work *out* how long it looks like it's going to take you to reach the horizon but it will always take longer and longer the closer you get as time dilation extends the time it would take from the perspective *of* your starting position and length contraction increases the distance. As your falling in the clock at your previous position speeds *up* shortening the life span of the black hole. I can't see there ever being enough time*.*

The above is why statements like "it takes forever for an object to reach the event horizon" are statements of type (1): because the answer is only "infinite" (or "forever") with respect to Schwarzschild coordinate time t--it's *finite* with respect to the proper time of the falling object.

Yes I know. How can the proper time of the falling object possibly be infinite anyway? That would need a black hole with infinite mass. What I don't understand is how any valid coordinate system can take you past the event horizon. You can plot a world line that goes into and back out of the black hole if you want to. It would take an infinite amount of energy to escape and an infinite amount of time to get to in the first place. That doesn't mean that I think there's an infinite amount of space-time around a black hole. That would be silly (but not as silly as assuming that objects can reach an area that's obviously not reachable:). I suppose you could define the space-time contraction/dilation as the speed of light from the edge of the horizon to you're current position*time. I don't see how there can be any space-time beyond the horizon. That's what makes it an event horizon.

 

[Dale]

I really wish that it wasn't instantly assumed that because I question something it means I don't understand what I've just read.

After more than 200 posts nobody is instantly assuming anything.

T=infinity and t=t0 on the other hand is a contradiction because one say it happens and the other says it doesn't.

Do you understand the idea that a coordinate chart need not cover the entire manifold? If so, then I don't understand why you would consider this to be a contradiction.

Remember that a coordinate chart is a diffeomorphic mapping from an open subset U on a manifold to an open subset V in R(n). The most a single coordinate chart can say about the existence of a particular point on the manifold is whether or not it exists in U. In other words, while the existence of some point on the manifold is a coordinate-independent statement, whether or not that point has coordinates is a coordinate-dependent statement because it depends on whether or not the point is an element of U. So one chart says "it happens in my region of support, U" and another chart says "it does not happen in mine". There is no coordinate independent contradiction between these two statements.

 

[PeterDonis]

If you replace constant velocity with constant acceleration (so that an object undergoing constant acceleration feels no g-force) and just use an increase in acceleration rather than velocity to produce g-force then this isn't analogous to a distant observers perspective of a free-faller even though the free-faller doesn't feel as though they're accelerating but does feel the increase in acceleration. It seems very analogous.

If by "constant acceleration" you mean "constant acceleration with respect to a particular coordinate system", then I can make a kind of sense of this statement, but even then, "the free-faller doesn't feel as though they're accelerating but does feel the increase in acceleration" is self-contradictory as it stands. Either an observer feels a real acceleration or they don't; if they do, they're not free-falling. If by "feel the increase in acceleration" you mean "can observe an increasing separation from a neighboring free-faller, even though they feel no actual acceleration", that would be consistent, but I would not use the words "feel acceleration" in any form to describe a free-falling observer.

In any case, if you're just talking about "acceleration" relative to a particular coordinate system, that would be a statement of type 1 (by DaleSpam's categorization), and would not be relevant at all to the question of whether objects can reach or cross the event horizon, which involves statements of type 2.

Show *me*. How can we can show that there *must* be another part of spacetime (the part where the free-falling observer's worldline goes) *that* simply can't be assigned coordinates in that system?

Take another look at this particular statement I made:

But worldlines don't just stop at a finite length. *Something* has to happen to the free-falling observer after he has traversed the finite length of worldline that the faraway observer can see.

If the distant observer computes that the portion of the free-falling observer's worldline that he can see has a finite length, then he *has* to conclude that there is a further portion he can't see. The free-faller's worldline can't just stop at the end of the finite length. Do you agree with that statement? If you don't, then I think we need to dig further into the physics behind it, because this seems to me like a "fulcrum point" of the main question at issue.

 

[PeterDonis]

The world line of an object heading towards a black hole is heading into an area of time-dilated and length contracted space relative to wherever they start from. So you can work *out* how long it looks like it's going to take you to reach the horizon but it will always take longer and longer the closer you get as time dilation extends the time it would take from the perspective *of* your starting position and length contraction increases the distance. As your falling in the clock at your previous position speeds *up* shortening the life span of the black hole. I can't see there ever being enough time*.*

To expand a little bit on my last post, I think you're missing the point of the "finite length" I'm referring to. The finite length of the free-falling observer's worldline prior to reaching the event horizon is an *invariant*--it's the same regardless of what coordinate system we use to compute it. It doesn't *change* as the free-falling observer gets closer to the horizon, because it's just an invariant geometric quantity, and those don't change: they're global properties of the spacetime (in this case, of a particular curve in the spacetime). Another way of putting this is that the computation of the finite length already accounts for all the length contraction and time dilation that will occur as the free-falling observer traverses that entire region. The computation of the finite length means that, even *after* all those effects are taken into account, the free-falling observer will experience only a finite amount of proper time before reaching the horizon.

 

[A-wal]

Do you understand the idea that a coordinate chart need not cover the entire manifold?.

Yep.

If so, then I don't understand why you would consider this to be a contradiction..

Because if you extend the coordinates then it still won't include beyond the horizon, no matter how much you extend it by. In other words there is no space-time beyond the horizon.

Remember that a coordinate chart is a diffeomorphic mapping from an open subset U on a manifold to an open subset V in R(n). The most a single coordinate chart can say about the existence of a particular point on the manifold is whether or not it exists in U. In other words, while the existence of some point on the manifold is a coordinate-independent statement, whether or not that point has coordinates is a coordinate-dependent statement because it depends on whether or not the point is an element of U. So one chart says "it happens in my region of support, U" and another chart says "it does not happen in mine". There is no coordinate independent contradiction between these two statements.

The coordinate chart doesn’t include an area beyond the horizon, but you can still move into that area? How?

If by "constant acceleration" you mean "constant acceleration with respect to a particular coordinate system", then I can make a kind of sense of this statement, but even then, "the free-faller doesn't feel as though they're accelerating but does feel the increase in acceleration" is self-contradictory as it stands. Either an observer feels a real acceleration or they don't; if they do, they're not free-falling. If by "feel the increase in acceleration" you mean "can observe an increasing separation from a neighboring free-faller, even though they feel no actual acceleration", that would be consistent, but I would not use the words "feel acceleration" in any form to describe a free-falling observer.

In any case, if you're just talking about "acceleration" relative to a particular coordinate system, that would be a statement of type 1 (by DaleSpam's categorization), and would not be relevant at all to the question of whether objects can reach or cross the event horizon, which involves statements of type 2.

They do feel it though. It's called tidal force. Take your example of the two free-fallers - "can observe an increasing separation from a neighboring free-faller, even though they feel no actual acceleration". The force that's increasing the space between them would break their atoms apart if it was strong enough. The same thing happens with normal acceleration. If the engine's too powerful it will break away from whatever it's attached to. Replace constant velocity with constant acceleration so that now constant acceleration isn't felt. An increase in the rate of acceleration would be felt as proper acceleration.

Take another look at this particular statement I made:

If the distant observer computes that the portion of the free-falling observer's worldline that he can see has a finite length, then he *has* to conclude that there is a further portion he can't see. The free-faller's worldline can't just stop at the end of the finite length. Do you agree with that statement? If you don't, then I think we need to dig further into the physics behind it, because this seems to me like a "fulcrum point" of the main question at issue.

I do agree. The free-faller's worldline would of course be continuous but there would be no way of reaching a point on that world line that takes you past the horizon before the black hole's gone because the free-faller is slowing down through time and moving through ever more compressed space the closer they get. The further you go the harder it is to go the same distance. We've already covered that an object in free-fall will reach a point when a more distant object or a signal traveling at c wouldn’t be able to catch the closer object. If light can't even catch the closer object then no distant observer will ever reach a point when they cross the horizon because the further one can't reach the closer one and the closer one can't reach the horizon from the further ones perspective.

To expand a little bit on my last post, I think you're missing the point of the "finite length" I'm referring to. The finite length of the free-falling observer's worldline prior to reaching the event horizon is an *invariant*--it's the same regardless of what coordinate system we use to compute it. It doesn't *change* as the free-falling observer gets closer to the horizon, because it's just an invariant geometric quantity, and those don't change: they're global properties of the spacetime (in this case, of a particular curve in the spacetime). Another way of putting this is that the computation of the finite length already accounts for all the length contraction and time dilation that will occur as the free-falling observer traverses that entire region. The computation of the finite length means that, even *after* all those effects are taken into account, the free-falling observer will experience only a finite amount of proper time before reaching the horizon.

And that finite amount of time should always be greater than the life span of the black hole because the black holes dimensions aren't an invariant geometric quantity.

 

[Dale]

Because if you extend the coordinates then it still won't include beyond the horizon, no matter how much you extend it by.

What do you mean by "extend the coordinates"? I am not certain, but I suspect that what you mean will result in a mapping which is not diffeomorphic and therefore not a valid coordinate system.

In other words there is no space-time beyond the horizon.

You are incorrectly associating spacetime with the coordinate system instead of the manifold. See:
http://en.wikipedia.org/wiki/Spacetime
http://arxiv.org/abs/gr-qc/9712019

If the manifold extends beyond the horizon then the spacetime does also (by definition) regardless of whether or not a specific coordinate chart does.

The coordinate chart doesn’t include an area beyond the horizon, but you can still move into that area? How?

By using a different coordinate chart, or even by using a coordinate-free (aka component-free) treatment.

 

[PeterDonis]

They do feel it though. It's called tidal force. Take your example of the two free-fallers - "can observe an increasing separation from a neighboring free-faller, even though they feel no actual acceleration". The force that's increasing the space between them would break their atoms apart if it was strong enough. The same thing happens with normal acceleration. If the engine's too powerful it will break away from whatever it's attached to.

If the objects are freely falling, *there is no force on them*. If there is a force on them, *they are not freely falling*. The two nearby freely falling objects that observe increasing separation between them *are not connected*; that's how they can both be freely falling. If they were connected--if there were something holding them together that could be broken apart--then at least one of them could *not* be freely falling. (The simplest case would be if the common center of mass of both objects was freely falling, but that would still mean that most of the atoms in each object would *not* be freely falling--they *would* feel acceleration, even at the initial instant before any increase in separation had occurred.)

Replace constant velocity with constant acceleration so that now constant acceleration isn't felt. An increase in the rate of acceleration would be felt as proper acceleration.

Constant acceleration (meaning, constant *proper* acceleration) *is* felt as acceleration, period. And that's the only kind of acceleration that would make your second statement true (as the rate of proper acceleration increases, the increase is *felt* as increasing acceleration).

If by "constant acceleration" you mean instead "constant acceleration relative to some particular coordinate system", then you can say it "isn't felt" if you adopt coordinates such that a freely falling object "appears" to be accelerating. But then your second statement in the quote above is false; an object can free-fall in a gravity field, with its coordinate acceleration continually increasing relative to, say, the Schwarzschild coordinates of a distant observer, and never feel any acceleration at all. For the relative "acceleration" of two nearby freely falling objects in this case, see my response above.

And that finite amount of time should always be greater than the life span of the black hole because the black holes dimensions aren't an invariant geometric quantity.

Hmm...so what about the simpler case where the black hole lasts forever? (Which is what the Schwarzschild solution actually describes--the Hawking radiation that causes the black hole to eventually evaporate is a quantum effect, and we don't have a complete theory of quantum gravity that describes it; all we have are approximate models that *seem* to indicate that the black hole will evaporate in finite time. In General Relativity, leaving quantum effects out, the black hole is eternal.) In that case where the black hole lasts forever--call it an idealized case if you like--do you agree that the free-falling observer would reach the horizon, r = 2M, in a finite proper time? And that he would then *continue*, in a finite proper time by his clock, to the region of spacetime *below* the horizon, which the distant observer can't see?

 

[PeterDonis]

And that finite amount of time should always be greater than the life span of the black hole because the black holes dimensions aren't an invariant geometric quantity.

Looking back through this thread, I see you've made statements along these lines before. Defining what is meant by "the black hole's dimensions" is problematic, because the black hole isn't really an "object"--it's part of the spacetime geometry. In so far as we can define physical "dimensions" for it, for example the circumference or area of the event horizon, those dimensions *are* invariant--the event horizon is an invariant, global feature of the spacetime geometry. You'll note that the formulas for these features of the event horizon don't include any coordinate-dependent variables--they all depend only on the mass M of the hole, which is an invariant.

Strictly speaking, what I just said applies to the "eternal" black hole of classical General Relativity, which never evaporates. As I said in my last post, we don't have a complete theory of quantum gravity, but the approximate models we do have indicate that the effect of evaporation, at least until almost all of the black hole's original mass has been radiated away (i.e., when the remaining mass of the hole approaches the Planck mass), is simply to make the mass M weakly time-dependent--slowly decreasing with time, as the Hawking radiation slowly carries away energy. Then, on time scales that are short compared to the time scale over which M changes (which is many orders of magnitude longer than the lifetime of the universe for a typical stellar-mass black hole), what I said above still applies--we can treat M as effectively constant. And we can also show that there are many possible worldlines for observers free-falling into the black hole that reach the horizon in a proper time that is also short compared to the time scale over which M changes--meaning that they reach the horizon long, long before the black hole evaporates. (And again, as I said before, all this is true *after* all the effects of time dilation and length contraction as you approach r = 2M have been taken into account in the computation of the finite proper time.)

 

[A-wal]

What do you mean by "extend the coordinates"? I am not certain, but I suspect that what you mean will result in a mapping which is not diffeomorphic and therefore not a valid coordinate system.

Length contraction would allow you to lengthen the space between any hovering observer and the event horizon (from your perspective) by as much as you like if the black hole lasted forever, in exactly the same way as length contraction works with acceleration in flat space-time. Just keep getting closer. I'm not sure if that makes it diffeomorphic.

You are incorrectly associating spacetime with the coordinate system instead of the manifold. See:
http://en.wikipedia.org/wiki/Spacetime
http://arxiv.org/abs/gr-qc/9712019

If the manifold extends beyond the horizon then the spacetime does also (by definition) regardless of whether or not a specific coordinate chart does.

I get it. The manifold includes all of space-time but a coordinate chart doesn't necessarily, but the coordinate charts we're using cover all the space-time between a falling observer and the horizon and that should be all that's needed to describe what happens when approaching a black hole.

By using a different coordinate chart, or even by using a coordinate-free (aka component-free) treatment.

Wouldn't you need at least a basic coordinate chart to see the distance between the faller and the horizon? Besides, the coordinate chart would include an ever increasing amount of space-time between the observer and the horizon due to length contraction and time dilation.

If the objects are freely falling, *there is no force on them*. If there is a force on them, *they are not freely falling*. The two nearby freely falling objects that observe increasing separation between them *are not connected*; that's how they can both be freely falling. If they were connected--if there were something holding them together that could be broken apart--then at least one of them could *not* be freely falling. (The simplest case would be if the common center of mass of both objects was freely falling, but that would still mean that most of the atoms in each object would *not* be freely falling--they *would* feel acceleration, even at the initial instant before any increase in separation had occurred.)

Yes, I know and understand nearly all of that. There can be no true free-falling if you look at it like that because there'll always be more force on the side closest to the gravitational source pulling at the rest, even with a single atom. What I'm saying is that the difference in gravity can always be felt to some extent. That's tidal force and it's also proper acceleration, so tidal force is a form of proper acceleration. The bit I don't get from that paragraph is how the atoms within a falling object would feel acceleration even in the instant before they start to separate? Surely it's the separation that they feel?

Constant acceleration (meaning, constant *proper* acceleration) *is* felt as acceleration, period. And that's the only kind of acceleration that would make your second statement true (as the rate of proper acceleration increases, the increase is *felt* as increasing acceleration).

If by "constant acceleration" you mean instead "constant acceleration relative to some particular coordinate system", then you can say it "isn't felt" if you adopt coordinates such that a freely falling object "appears" to be accelerating. But then your second statement in the quote above is false; an object can free-fall in a gravity field, with its coordinate acceleration continually increasing relative to, say, the Schwarzschild coordinates of a distant observer, and never feel any acceleration at all. For the relative "acceleration" of two nearby freely falling objects in this case, see my response above.

I meant that IF you couldn't feel constant acceleration then you would still feel the increase in acceleration AS proper acceleration. That what happens in free-fall, and an observer in free-fall accelerates relative to a more distant hovering observer even without taking into account the increase in acceleration as they fall. It's the equivalent to what I was saying about an object using constant acceleration to move at a constant velocity relative to a non-constant c. That's what made me think of it. I don't see how the Rindler horizon is equivalent to the event horizon though. The Rindler horizon would be the equivalent to a falling object reaching a point when a more distant observer can't catch the free-falling regardless of how fast the chasing object moves. That way it's an individual horizon depending on the two objects involved rather then "global" in the same way as the Rindler horizon. My version of the event horizon of a black hole isn't globally constant but it would depend entirely on distance rather than distance and relative velocity. In other words it wouldn't make any difference to the relative position of the event horizon if the observer accelerates, other than the fact that acceleration would influence the observers distance.

Hmm...so what about the simpler case where the black hole lasts forever? (Which is what the Schwarzschild solution actually describes--the Hawking radiation that causes the black hole to eventually evaporate is a quantum effect, and we don't have a complete theory of quantum gravity that describes it; all we have are approximate models that *seem* to indicate that the black hole will evaporate in finite time. In General Relativity, leaving quantum effects out, the black hole is eternal.) In that case where the black hole lasts forever--call it an idealized case if you like--do you agree that the free-falling observer would reach the horizon, r = 2M, in a finite proper time? And that he would then *continue*, in a finite proper time by his clock, to the region of spacetime *below* the horizon, which the distant observer can't see?

Would a constantly accelerating observer ever reach c?

Looking back through this thread, I see you've made statements along these lines before. Defining what is meant by "the black hole's dimensions" is problematic, because the black hole isn't really an "object"--it's part of the spacetime geometry. In so far as we can define physical "dimensions" for it, for example the circumference or area of the event horizon, those dimensions *are* invariant--the event horizon is an invariant, global feature of the spacetime geometry.

How can they possibly be invariant? Time dilation would have to shorten the length of time it exists for and length contraction would have to do the same thing to its size.

You'll note that the formulas for these features of the event horizon don't include any coordinate-dependent variables--they all depend only on the mass M of the hole, which is an invariant.

Maybe that's why they're wrong then. :)

Strictly speaking, what I just said applies to the "eternal" black hole of classical General Relativity, which never evaporates. As I said in my last post, we don't have a complete theory of quantum gravity, but the approximate models we do have indicate that the effect of evaporation, at least until almost all of the black hole's original mass has been radiated away (i.e., when the remaining mass of the hole approaches the Planck mass), is simply to make the mass M weakly time-dependent--slowly decreasing with time, as the Hawking radiation slowly carries away energy. Then, on time scales that are short compared to the time scale over which M changes (which is many orders of magnitude longer than the lifetime of the universe for a typical stellar-mass black hole), what I said above still applies--we can treat M as effectively constant. And we can also show that there are many possible worldlines for observers free-falling into the black hole that reach the horizon in a proper time that is also short compared to the time scale over which M changes--meaning that they reach the horizon long, long before the black hole evaporates. (And again, as I said before, all this is true *after* all the effects of time dilation and length contraction as you approach r = 2M have been taken into account in the computation of the finite proper time.)

After relative acceleration and the increase in the rate of acceleration as the observer falls have been taken into account? Are you sure?

 

[PeterDonis]

Length contraction would allow you to lengthen the space between any hovering observer and the event horizon (from your perspective) by as much as you like if the black hole lasted forever, in exactly the same way as length contraction works with acceleration in flat space-time.

I don't know what you mean by this. The proper length from the horizon at r = 2M to a hovering observer at any r > 2M is finite, and gets shorter the closer r gets to 2M. The computation is similar to the computation that shows that the proper time for an observer to fall from any r > 2M to r = 2M is finite. This is also true for any accelerating observer in flat spacetime: the proper distance from the observer to the Rindler horizon is finite, and gets shorter as the acceleration gets larger.

Yes, I know and understand nearly all of that. There can be no true free-falling if you look at it like that because there'll always be more force on the side closest to the gravitational source pulling at the rest, even with a single atom.

This gets into a separate issue, which is whether any real object can qualify as a "test body" in the idealized sense that is used in GR, meaning an object that is too small to feel any tidal forces, and also too small to have any effect on the spacetime geometry. I've been assuming that, for purposes of discussion, we were talking about "test bodies" in this sense, because it keeps things simple. A real object that does feel tidal forces can still free-fall through the horizon, but it's more complicated to describe it, and it doesn't add anything to the root question, which is how anything can free-fall through the horizon. I'm pretty sure you would say that even a "test body", an idealized point particle that feels no tidal force, could not reach the horizon, correct? If so, why bring all the complications of tidal force into it?

The bit I don't get from that paragraph is how the atoms within a falling object would feel acceleration even in the instant before they start to separate? Surely it's the separation that they feel?

Any extended object that can be usefully treated as a single object (as opposed to just a cloud of separate particles that happen to be moving near each other) is held together by internal forces of some sort, such as the interatomic forces that hold atoms in their places in a solid body. So the parts of the object are always feeling some force; but in many cases we can separate out the motion of the object's center of mass and treat that as free-falling, geodesic motion. The parts of the object that are not at the center of mass would *not* be moving on geodesics in such a case; they would be on accelerated worldlines and would feel acceleration, which we would interpret as being caused by the internal forces on each part by the other parts. If such an object were also affected by tidal gravity, the parts away from the center of mass would feel some additional force due to the tidal gravity.

That's the kind of case I was describing, but as I noted just now, it's *not* the kind of case I think we should be discussing, because it brings in complications that don't affect the main question.

How can they possibly be invariant? Time dilation would have to shorten the length of time it exists for and length contraction would have to do the same thing to its size.

Nope. Read carefully what I said: I said the *circumference* and *area* of the horizon are invariant. Those quantities are purely in the "angular" direction; they are not affected by the "length contraction" and "time dilation" that occur in the radial direction. In any case, you're mistaken about how the "length contraction" and "time dilation" work. I commented about the length contraction above. With time dilation, it's not true that it shortens the length of time that something exists. An accelerated observer hovering just above the horizon of a black hole still experiences an infinite amount of future proper time; there is a sense in which he "experiences it slower" than an observer far away from the hole, but that doesn't stop the total amount of his future proper time from being infinite. Infinite quantities don't behave like finite ones: you can divide an infinite number by a very large finite number and it will still be infinite.

After relative acceleration and the increase in the rate of acceleration as the observer falls have been taken into account? Are you sure?

Yes, I'm sure. Do you want to go through the computation?

 

[Dale]

Length contraction would allow you to lengthen the space between any hovering observer and the event horizon (from your perspective) by as much as you like if the black hole lasted forever, in exactly the same way as length contraction works with acceleration in flat space-time. Just keep getting closer. I'm not sure if that makes it diffeomorphic.

Certainly, you could do that. It would amount to a remapping of your radial coordinate such that as the Schwarzschild radial coordinate approaches the Schwarzschild radius your remapped radial coordinate approaches -infinity. Such a coordinate transform would not change a coordinate dependent statement into a coordinate independent statement. Also, although you could write that transform for the Schwarzschild radius you can also write such a transform for any other arbitrary radius. JesseM did an example of exactly that in flat spacetime earlier during this thread.

I get it. The manifold includes all of space-time but a coordinate chart doesn't necessarily, but the coordinate charts we're using cover all the space-time between a falling observer and the horizon and that should be all that's needed to describe what happens when approaching a black hole.

Yes, the Schwarzschild coordinates are all that's needed to describe what happens when approaching the event horizon. But not all that's needed to describe what happens when reaching or crossing the event horizon. And again, the presence of these coordinates do not change coordinate dependent statements into coordinate independent statements.

Wouldn't you need at least a basic coordinate chart to see the distance between the faller and the horizon? Besides, the coordinate chart would include an ever increasing amount of space-time between the observer and the horizon due to length contraction and time dilation.

You do not need a coordinate chart to determine the proper distance between the faller and the horizon. That is a coordinate-independent geometric quantity, and it is finite.

 

[A-wal]

I don't know what you mean by this. The proper length from the horizon at r = 2M to a hovering observer at any r > 2M is finite, and gets shorter the closer r gets to 2M. The computation is similar to the computation that shows that the proper time for an observer to fall from any r > 2M to r = 2M is finite. This is also true for any accelerating observer in flat spacetime: the proper distance from the observer to the Rindler horizon is finite, and gets shorter as the acceleration gets larger.

What I mean is that you can make a hovering observer change distance from the black hole from your perspective even if they're hovering from their own perspective, but you can never make them cross the horizon.

This gets into a separate issue, which is whether any real object can qualify as a "test body" in the idealized sense that is used in GR, meaning an object that is too small to feel any tidal forces, and also too small to have any effect on the spacetime geometry. I've been assuming that, for purposes of discussion, we were talking about "test bodies" in this sense, because it keeps things simple. A real object that does feel tidal forces can still free-fall through the horizon, but it's more complicated to describe it, and it doesn't add anything to the root question, which is how anything can free-fall through the horizon. I'm pretty sure you would say that even a "test body", an idealized point particle that feels no tidal force, could not reach the horizon, correct? If so, why bring all the complications of tidal force into it?

Because it describes something important. Free-falling without tidal force is the equivalent to moving freely, while tidal force describes acceleration. You can plot a course and say there's enough proper time to reach the horizon before it dies, but I think it would change as you get closer. You'll have less and less time until it always reaches zero at the horizon, no matter how long the black hole lasts.

Any extended object that can be usefully treated as a single object (as opposed to just a cloud of separate particles that happen to be moving near each other) is held together by internal forces of some sort, such as the interatomic forces that hold atoms in their places in a solid body. So the parts of the object are always feeling some force; but in many cases we can separate out the motion of the object's center of mass and treat that as free-falling, geodesic motion. The parts of the object that are not at the center of mass would *not* be moving on geodesics in such a case; they would be on accelerated worldlines and would feel acceleration, which we would interpret as being caused by the internal forces on each part by the other parts. If such an object were also affected by tidal gravity, the parts away from the center of mass would feel some additional force due to the tidal gravity.

That's the kind of case I was describing, but as I noted just now, it's *not* the kind of case I think we should be discussing, because it brings in complications that don't affect the main question.

You've lost me. "... would feel acceleration, which we would interpret as being caused by the internal forces on each part by the other parts. If such an object were also affected by tidal gravity..." Also affected by tidal gravity? Same thing isn't it?

Nope. Read carefully what I said: I said the *circumference* and *area* of the horizon are invariant. Those quantities are purely in the "angular" direction; they are not affected by the "length contraction" and "time dilation" that occur in the radial direction. In any case, you're mistaken about how the "length contraction" and "time dilation" work. I commented about the length contraction above. With time dilation, it's not true that it shortens the length of time that something exists. An accelerated observer hovering just above the horizon of a black hole still experiences an infinite amount of future proper time; there is a sense in which he "experiences it slower" than an observer far away from the hole, but that doesn't stop the total amount of his future proper time from being infinite. Infinite quantities don't behave like finite ones: you can divide an infinite number by a very large finite number and it will still be infinite.

I know what you said. How can the *circumference* and *area* be invariant if they're not in the radial direction? How is it not true that time dilation shortens the length of time something exists? Length contraction would extend the length between the observer and the singularity, and therefore the distance between the observer and the rainbow horizon. The "experiences it slower" means that time dilation makes everything else speed up. It doesn't matter if an observers proper future time is infinite. I would dispute that, but it's besides the point. Even an observer with infinite potential future time will be unable to reach an event horizon in any given amount of finite time.

Yes, I'm sure. Do you want to go through the computation?

Go on then, let's see it.

Certainly, you could do that. It would amount to a remapping of your radial coordinate such that as the Schwarzschild radial coordinate approaches the Schwarzschild radius your remapped radial coordinate approaches -infinity. Such a coordinate transform would not change a coordinate dependent statement into a coordinate independent statement. Also, although you could write that transform for the Schwarzschild radius you can also write such a transform for any other arbitrary radius. JesseM did an example of exactly that in flat spacetime earlier during this thread.

If you say so.

Yes, the Schwarzschild coordinates are all that's needed to describe what happens when approaching the event horizon. But not all that's needed to describe what happens when reaching or crossing the event horizon. And again, the presence of these coordinates do not change coordinate dependent statements into coordinate independent statements.

If everything is described correctly in the approuch then there shouldn't be a need to describe any crossing. It's impossible to reach the horizon in Schwarzschild coordinates.

You do not need a coordinate chart to determine the proper distance between the faller and the horizon. That is a coordinate-independent geometric quantity, and it is finite.

Proper distance? Proper by definition is coordinate-independent isn't it? It's finite but not constant, and it will never be short enough. The distance would increase as you got closer. You know what I mean.

 

[Dale]

If everything is described correctly in the approuch then there shouldn't be a need to describe any crossing.

That is a weird comment. So if you were working a problem for a car crashing into a brick wall you would consider that as long as you described the approach correctly there wouldn't be a need to describe any collision?

Proper distance? Proper by definition is coordinate-independent isn't it? It's finite but not constant, and it will never be short enough. The distance would increase as you got closer. You know what I mean.

No, it will decrease. You should work this problem, it is easy enough. Set theta to 0 and phi to 90 deg (or whatever is needed to make those terms drop out. You will then have an expression in t and r which you can integrate for constant t from r to r0 (the Schwarzschild radius).

 

[A-wal]

That is a weird comment. So if you were working a problem for a car crashing into a brick wall you would consider that as long as you described the approach correctly there wouldn't be a need to describe any collision?

No. That's silly. But if the approach to the brick wall describes getting closer to the wall without ever reaching it then it obviously can't be reached.

No, it will decrease. You should work this problem, it is easy enough. Set theta to 0 and phi to 90 deg (or whatever is needed to make those terms drop out. You will then have an expression in t and r which you can integrate for constant t from r to r0 (the Schwarzschild radius).

Decrease? Maybe you didn't know what I meant. The distance would be greater than if there were no length contraction, and the increase gets sharper the closer you get. Obviously it would never reach the point where you're going backwards. That would be extremely silly. It would just take more and more energy to cover the same distance. Of course that's from a hovering observers perspective of something else approaching the black hole. There would be more and more space to cover from the perspective of the approaching observer. Combine this with time dilation and there should always be too much space to cover in the time you have because you have less and less time to reach the horizon as you get closer.

 

[PeterDonis]

Because it describes something important. Free-falling without tidal force is the equivalent to moving freely, while tidal force describes acceleration.

I *think* you're saying the same thing that I was saying in what you referred to here:

You've lost me. "... would feel acceleration, which we would interpret as being caused by the internal forces on each part by the other parts. If such an object were also affected by tidal gravity..." Also affected by tidal gravity? Same thing isn't it?

However, I'm not sure, and it's irrelevant to the main point (whether or not a free-falling observer can cross the horizon). So for the time being, anyway, I'd like to table the whole issue of tidal "force". Let's assume for the time being that we are dealing with a black hole that is so large that the tidal gravity at the horizon is negligible. (We can do this because the tidal gravity at the horizon goes like $1 / M^{2}$, so we can make the tidal gravity as small as we like by making M large enough.)

You can plot a course and say there's enough proper time to reach the horizon before it dies, but I think it would change as you get closer.

The proper time along a particular curve in spacetime *doesn't change*--it's a geometric invariant. Saying it can change as you get closer to the horizon is like saying the distance from New York to Los Angeles can change while you're passing through Las Vegas. But instead of just repeating this, since you asked me to go through the computation, here it is. Once we've gone through it, I think a lot of your other questions will be easily answered.

(BTW, I'm basing this on Misner, Thorne, and Wheeler, Chapter 25, but you'll find similar calculations in just about any textbook on General Relativity.)

The general method we will use is as follows: we define a curve in spacetime along which we want to integrate the metric to obtain a "length"--either proper time elapsed for an observer who has that curve as his worldline (if the curve is timelike), or proper distance seen by an observer moving on a worldline orthogonal to the curve (if the curve is spacelike). We then write the metric in appropriate form, using the equation for the curve to reduce the number of integration variables to one. This will (hopefully) allow us to evaluate the integral.

For the case of proper time elapsed for a freely falling observer to reach the horizon from some radius R > 2M, we first define the curve along which we will do the integral as a purely radial curve, so theta = pi / 2 (this makes the math easiest) and phi = 0 all along the curve. We start at r = R (some value >> 2M) and t = 0. We end at t = infinity. (Minor technical point: I have idealized the curve we're using so that, strictly speaking, it is the worldline of an observer "falling in from rest at infinity"--that is, at any finite value of r, this worldline has a non-zero inward velocity. This doesn't change the essential conclusion--that an observer falling from any finite radius will reach r = 2M in a finite proper time--but it makes the calculation simpler.)

The key thing we need to complete the definition of the curve is a function r(t)--or t(r)--that will let us write everything in terms of a single coordinate. For a freely falling, purely radial curve, this function is implicitly given by the equation for the coordinate velocity:

$$\frac{dr}{dt} = - \sqrt{\frac{2M}{r}} \left( 1 - \frac{2M}{r} \right)$$

This let's us write the metric as follows:

$$d\tau^{2} = dt^{2} \left[ \left( 1 - \frac{2M}{r} \right) - \left( \frac{dr}{dt} \right)^{2} \frac{1}{1 - \frac{2M}{r}} \right] = dt^{2} \left( 1 - \frac{2M}{r} \right)^{2}$$

which gives, on taking the square root and integrating,

$$\tau = \int_{0}^{\infty} \left( 1 - \frac{2M}{r} \right) dt$$

As it stands, this integral can't be evaluated, because we still have r in the integrand. To do the integral directly we would need to first solve the differential equation above in dr/dt (which is not at all straightforward) in order to get an explicit expression for r(t) to substitute into the integrand (which would still leave us with a very messy integral to do).

The usual way of dealing with this is to rewrite the integral in terms of dr, which makes it easy to evaluate. However, before doing this, we need to satisfy ourselves that as t -> infinity, r -> 2M. We can show this by noting two things about the equation above for dr / dt: first, dr / dt is negative for any r > 2M, so r will decrease with t until r = 2M is reached (i.e., for any finite value of t, if r > 2M, r will still be decreasing, so it will be closer to 2M at some larger finite value of t); second, as r -> 2M, dr / dt -> 0, so we don't expect r to actually reach 2M at any finite value of t (because the closer it gets, the slower it approaches), but only asymptotically as t -> infinity. (There's probably a slick mathematically way of proving this formally.)

Having shown that, we can invert the equation for dr / dt to obtain:

$$\frac{dt}{dr} = - \sqrt{\frac{r}{2M}} \frac{1}{1 - \frac{2M}{r}}$$

We can then rewrite the metric as:

$$d\tau^{2} = dr^{2} \left[ \frac{r}{2M} \frac{1}{1 - \frac{2M}{r}} - \frac{1}{1 - \frac{2M}{r}} \right] = dr^{2} \frac{1}{1 - \frac{2M}{r}} \left( \frac{r}{2M} - 1 \right) = dr^{2} \frac{r}{2M}$$

Taking the square root and integrating gives:

$$\tau = \int_{2M}^{R} \sqrt{\frac{r}{2M}} dr = \frac{2}{3 \sqrt{2M}} \left[ \left( R \right)^{\frac{3}{2}} - \left( 2M \right)^{\frac{3}{2}} \right]$$

This is often re-written to put all the variables in dimensionless form, which makes things look more elegant:

$$\frac{\tau}{2M} = \frac{2}{3} \left[ \left( \frac{R}{2M} \right)^{\frac{3}{2}} - 1 \right]$$

I know what you said. How can the *circumference* and *area* be invariant if they're not in the radial direction? How is it not true that time dilation shortens the length of time something exists? Length contraction would extend the length between the observer and the singularity, and therefore the distance between the observer and the rainbow horizon. The "experiences it slower" means that time dilation makes everything else speed up. It doesn't matter if an observers proper future time is infinite. I would dispute that, but it's besides the point. Even an observer with infinite potential future time will be unable to reach an event horizon in any given amount of finite time.

Using the general method outlined above, here are the answers to the substantive questions you've raised. I'll take them in a somewhat different order from the order in which you raise them.

First, time dilation. Consider the following two curves: (A) r = some value close to 2M, theta = pi / 2, phi = 0, t = 0 to infinity; (B) r = some value much much larger than 2M, theta = pi / 2, phi = 0, t = 0 to infinity. Both of these curves have infinite "potential future time". However, if I pick any positive finite value of t (say t = 100), the proper time elapsed along curve A up to that value of t will be much less than the proper time elapsed along curve B up to that value of t. This is what we mean when we say that there is "time dilation" close to the horizon, and it's *all* we mean.

In particular, note that, for an infalling observer, the r coordinate is *not* constant, and (as we'll see below) the "lines of simultaneity" for the infalling observer are *not* the same as those for the hovering observer, so there's no straightforward way to say what event on the hovering observer's worldline "corresponds" to a given event on the infalling observer's worldline. This means that there's no straightforward way to assess the "time dilation" of the infalling observer relative to the hovering observer (or to one far away). In particular, as we've seen, the fact that there is a sense in which "time dilation becomes infinite" at the horizon does *not* imply that an infalling observer cannot reach the horizon in a finite proper time--he does.

Now consider a third curve: (C) t = 0, theta = pi / 2, phi = 0, r = R (some value > 2M) to 2M. This is a spacelike curve whose length is the "proper distance" from curve (A) to the horizon (as seen by a "hovering" observer--see below). The integral is similar to the one we did above: the only term in the metric that matters is the $dr^{2}$ term, so we have:

$$s = \int_{2M}^{R} \frac{1}{\sqrt{1 - \frac{2M}{r}}} dr = \int_{2M}^{R} \sqrt{\frac{r}{r - 2M}} dr$$

This is not the sort of integral that one can do "by inspection" (at least, I can't), but that's what tables of integrals are for. Consulting one, we find that the indefinite integral is:

$$\left[ \sqrt{r \left( r - 2M \right)} + 2M ln \left( \sqrt{r} + \sqrt{r - 2M} \right) \right]$$

which we then have to evaluate from r = 2M to r = R. The 2M endpoint is simple since everything but one term is zero, and we end up with:

$$s = \sqrt{R \left( R - 2M \right)} + 2M ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)$$

where we have used the fact that subtracting logarithms is the same as dividing the arguments. In dimensionless form, this becomes:

$$\frac{s}{2M} = \sqrt{\frac{R}{2M} \left( \frac{R}{2M} - 1 \right)} + ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)$$

Armed with this, we can now investigate "length contraction". The answer we just derived was for a "line of simultaneity" in the frame of a hovering observer (which will also be a line of simultaneity for an observer very far away--in fact, for any observer who remains forever at a constant r coordinate; the only difference will be the specific value of R that we plug into the above formula). The line of simultaneity of an infalling observer will be different because that observer is changing his radial coordinate with time, and so the "proper distance" to the horizon seen by such an observer might also be different.

To calculate this, we need an equation for the line of simultaneity of the infalling observer. This will be a spacelike geodesic that is orthogonal to the infalling observer's worldline. We could laboriously write out an equation for that geodesic in Schwarzschild coordinates, but it's easier to transform to Painleve coordinates, in which the metric is:

$$ds^{2} = - \left( 1 - \frac{2M}{r} \right) dT^{2} + 2 \sqrt{\frac{2M}{r}} dT dr + dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\varphi^{2} \right)$$

where T is the "Painleve time" defined by

$$dT = dt - \sqrt{\frac{2M}{r}} \frac{1}{1 - \frac{2M}{r}} dr$$

In these coordinates, lines of constant T are the "lines of simultaneity" for the infalling observer, who moves on a worldline defined by

$$\frac{dr}{dT} = - \sqrt{\frac{2M}{r}}$$

This makes it easy to calculate the proper distance to the horizon from a radius R for the infalling observer; if dT = 0 (and also theta and phi are constant, as above), the metric is simply [itex]ds = dr[/tex], and we have

$$s = R - 2M$$

or, in dimensionless form,

$$\frac{s}{2M} = \frac{R}{2M} - 1$$

for the proper distance from radius R to the horizon at r = 2M. Note that this number is *smaller* than the proper distance for a hovering observer that we derived above. So the proper distance to the horizon *is* "length contracted" for an infalling observer. But, as we've seen, the "length contraction" and "time dilation" do *not* imply that the infalling observer can't reach the horizon. (By the way, you can also use the Painleve metric to check that I did the proper time integral correctly above; just integrate the metric from r = R to r = 2M, using the equation for $dr / dT$, in inverted form, to write the integral in terms of dr. You should get the same answer I got above.)

Finally, let's look at the circumference and area of the horizon. The circumference of the horizon, as seen by a "hovering" observer (any observer with a constant r coordinate), is simply the invariant length of the curve: r = 2M, t = constant (any value will do), theta = pi / 2 (as above, this value makes the math easiest since it makes a "great circle" around the horizon a curve where only phi varies), phi = 0 to 2 pi. Since only phi changes, the integral is easy: we can ignore every term in the metric except the d phi term, so we get:

$$s = \int_{0}^{2 \pi} r sin \theta d\varphi = 4 \pi M.$$

The area calculation is similar; we just need to vary theta from 0 to pi as well as phi over the range just given.

What will this calculation look like for the infalling observer? It will be the same except that we look at a constant "Painleve time" T instead of a constant "Schwarzschild time" t. But, as we saw from the Painleve metric above, the r-coordinate is still a constant, 2M, in both cases. So the integral for the infalling observer *looks exactly the same* as the one for the hovering observer. The circumference of the horizon does *not* "length contract". (Neither does the area, for the same reason.) Fundamentally, the circumference and area are the same for all observers because the r-coordinate of the horizon is the same for all observers.

 

[Dale]

No. That's silly. But if the approach to the brick wall describes getting closer to the wall without ever reaching it then it obviously can't be reached.

Xeno's paradox is exactly such a description.

In fact, you can make many coordinates which cannot approach a brick wall. If you have a standard inertial system t and x with a wall at x=0 then you could define a new coordinate system X=1/x or X=ln(x) both of which would never reach the brick wall.

Decrease? Maybe you didn't know what I meant. The distance would be greater than if there were no length contraction, and the increase gets sharper the closer you get. Obviously it would never reach the point where you're going backwards. That would be extremely silly. It would just take more and more energy to cover the same distance. Of course that's from a hovering observers perspective of something else approaching the black hole. There would be more and more space to cover from the perspective of the approaching observer. Combine this with time dilation and there should always be too much space to cover in the time you have because you have less and less time to reach the horizon as you get closer.

If you understand that the proper distance is finite and the proper time is finite then what is the problem?

 

[A-wal]

However, I'm not sure, and it's irrelevant to the main point (whether or not a free-falling observer can cross the horizon). So for the time being, anyway, I'd like to table the whole issue of tidal "force". Let's assume for the time being that we are dealing with a black hole that is so large that the tidal gravity at the horizon is negligible. (We can do this because the tidal gravity at the horizon goes like $1 / M^{2}$, so we can make the tidal gravity as small as we like by making M large enough.)

Okay, bigger area means a gentler curve. It's a bigger curve though so the tidal force is just more spread out. It's still there and there's still just as much of it. You'll be undergoing less acceleration but it will start further out, so the total acceleration from flatish space-time to an event horizon will always be the same.

The proper time along a particular curve in spacetime *doesn't change*--it's a geometric invariant.

What about an ever-sharpening curve though? The sharpness of the curve would be the equivalent to velocity (free-fall). The rate of change would be equivalent to acceleration (tidal force).

Saying it can change as you get closer to the horizon is like saying the distance from New York to Los Angeles can change while you're passing through Las Vegas.

The distance from New York to Los Angeles would change while your passing though Las Vegas. It would depend on your velocity.

But instead of just repeating this, since you asked me to go through the computation, here it is. Once we've gone through it, I think a lot of your other questions will be easily answered.

(BTW, I'm basing this on Misner, Thorne, and Wheeler, Chapter 25, but you'll find similar calculations in just about any textbook on General Relativity.)

The general method we will use is as follows: we define a curve in spacetime along which we want to integrate the metric to obtain a "length"--either proper time elapsed for an observer who has that curve as his worldline (if the curve is timelike), or proper distance seen by an observer moving on a worldline orthogonal to the curve (if the curve is spacelike). We then write the metric in appropriate form, using the equation for the curve to reduce the number of integration variables to one. This will (hopefully) allow us to evaluate the integral.

For the case of proper time elapsed for a freely falling observer to reach the horizon from some radius R > 2M, we first define the curve along which we will do the integral as a purely radial curve, so theta = pi / 2 (this makes the math easiest) and phi = 0 all along the curve. We start at r = R (some value >> 2M) and t = 0. We end at t = infinity. (Minor technical point: I have idealized the curve we're using so that, strictly speaking, it is the worldline of an observer "falling in from rest at infinity"--that is, at any finite value of r, this worldline has a non-zero inward velocity. This doesn't change the essential conclusion--that an observer falling from any finite radius will reach r = 2M in a finite proper time--but it makes the calculation simpler.)

The key thing we need to complete the definition of the curve is a function r(t)--or t(r)--that will let us write everything in terms of a single coordinate. For a freely falling, purely radial curve, this function is implicitly given by the equation for the coordinate velocity:

And so on.

Just as I suspected, pure gibberish. No wonder you lot are confused. Seriously though, I appreciate the effort but even if I were to spend the time necessary to be able to express it in that way, it would take me further away from understanding it. I would be able to define individual parts in detail but I'm not bothered about doing that. It's not worth it.

What will this calculation look like for the infalling observer? It will be the same except that we look at a constant "Painleve time" T instead of a constant "Schwarzschild time" t. But, as we saw from the Painleve metric above, the r-coordinate is still a constant, 2M, in both cases. So the integral for the infalling observer *looks exactly the same* as the one for the hovering observer. The circumference of the horizon does *not* "length contract". (Neither does the area, for the same reason.) Fundamentally, the circumference and area are the same for all observers because the r-coordinate of the horizon is the same for all observers.

I don't see how it can be the same for all observers, sorry.

Xeno's paradox is exactly such a description.

In fact, you can make many coordinates which cannot approach a brick wall. If you have a standard inertial system t and x with a wall at x=0 then you could define a new coordinate system X=1/x or X=ln(x) both of which would never reach the brick wall.

You need to pick one. Either the wall can be reached or it can't. I refuse to believe that reality can change depending on how you choose to measure it (quantum mechanics aside). That's basically the whole point of this thread.

If you understand that the proper distance is finite and the proper time is finite then what is the problem?

Finite, yes. Not constant. If you were to measure the distance between yourself and a gravitational source to be one light year and there's a checkpoint half way between it and you then I would expect the distance to be more than half a light year if I traveled to the checkpoint and measured it again because it would now be in an area that was length contracted from my previous position. It's they only way it can make sense.


I think that you're describing free-fall when it can't actually exist. Normally it doesn't really matter because it's negligible. But in the case of approaching an event horizon I think it matters a lot. It's the same as the fact that the idealised "flat" space-time used in special relativity can't actually exist anywhere, other than maybe single points in just the right points of space-time.

 

[PeterDonis]

No wonder you lot are confused.

I think it's a bit presumptuous of you to assume that we, not you, are the ones who are confused when you make statements like these:

You'll be undergoing less acceleration but it will start further out, so the total acceleration from flatish space-time to an event horizon will always be the same.

What does "total acceleration...to an event horizon" even mean?

What about an ever-sharpening curve though? The sharpness of the curve would be the equivalent to velocity (free-fall). The rate of change would be equivalent to acceleration (tidal force).

Tidal acceleration is *not* the same as the Newtonian acceleration due to gravity.

The distance from New York to Los Angeles would change while your passing though Las Vegas. It would depend on your velocity.

But you claim later on that even when you're at rest the distance can change; see my comment further below.

I don't see how it can be the same for all observers, sorry.

Then you don't have any basis for saying that we, not you, are confused. What you don't see is perfectly obvious, not just to the rest of us here, but to generations of researchers and students who have worked these problems. What I'm saying here is at about the same level of contentiousness, to the relativity community, as saying that water is H2O.

Finite, yes. Not constant. If you were to measure the distance between yourself and a gravitational source to be one light year and there's a checkpoint half way between it and you then I would expect the distance to be more than half a light year if I traveled to the checkpoint and measured it again because it would now be in an area that was length contracted from my previous position. It's they only way it can make sense.

No, it doesn't make sense. You talked earlier on (in reference to the distance from NY to LA while passing through Las Vegas) as though the length contraction was due to velocity; but now you're saying that even if you're at rest relative to the checkpoint, the distance can change. *That* doesn't make sense.

I think that you're describing free-fall when it can't actually exist. Normally it doesn't really matter because it's negligible. But in the case of approaching an event horizon I think it matters a lot. It's the same as the fact that the idealised "flat" space-time used in special relativity can't actually exist anywhere, other than maybe single points in just the right points of space-time.

Saying that spacetime isn't flat is *not* the same as saying that free-fall can't exist. Free fall is the natural condition of motion for most objects in the Universe. It's only strange folk like us who spend our existence on the surfaces of planets that think being under acceleration is "natural". The Moon is in free fall around the Earth. The Earth itself is in free fall around the Sun. The Sun is in free fall around the center of the galaxy. The galaxy as a whole is in free fall through the Universe. Asteroids, comets, orbiting satellites, etc., etc.--all in free fall. And all of these objects would still be in free fall, and would behave, physically, just as they do now, if they were freely falling through the horizon of a large black hole (large enough that tidal effects were negligible, as I said).

Seriously though, I appreciate the effort but even if I were to spend the time necessary to be able to express it in that way, it would take me further away from understanding it. I would be able to define individual parts in detail but I'm not bothered about doing that. It's not worth it.

Well, you did ask me to post the computation. What I posted is, as I noted, entirely commonplace in relativity physics; many thousands of students have turned in homework problems containing computations similar to what I posted. Feynman once said that "to understand Nature, you must learn the language she speaks in," meaning mathematics.

That said, if you want a quick non-mathematical sketch of what the computation is telling us, here it is: yes, radial distances "stretch" and times "slow down" as you free-fall towards the horizon. But physically, they do *not* stretch infinitely or slow down infinitely for observers that are freely falling towards the hole. It *seems* like they do in Schwarzschild coordinates, but that is an illusion; and you can tell it's an illusion because even though Schwarzschild coordinate time goes to infinity as you approach the horizon, the path length of each little segment of your worldline (if you're freely falling towards the horizon) that is traversed in a constant interval of coordinate time dt goes to zero *faster* than the time t goes to infinity. (This is the same sort of thing that happens in the "Achilles and the tortoise" problem, which I think has been mentioned in this thread.) The result is a finite sum, meaning a finite path length from a finite radius r > 2M to the horizon at r = 2M.

The fact that Schwarzschild coordinate time goes to infinity over this finite path length tells us something else: Schwarzschild coordinate time is simply *not* a good time coordinate for observers falling into the hole. That's why other coordinate systems were invented to study the experience of such observers; for example, the Painleve coordinates, which I mentioned and gave a transformation for. Basically, Schwarzschild coordinate time as you approach the horizon at r = 2M behaves something like the tangent function, which goes to infinity at a finite value of its argument (pi/2). That doesn't mean the universe just stops at that finite value; it means we need to pick a better function to study that portion of spacetime.

But at least we all appear to agree on one thing:

You need to pick one. Either the wall can be reached or it can't. I refuse to believe that reality can change depending on how you choose to measure it (quantum mechanics aside). That's basically the whole point of this thread.

I agree; it is. And what we're claiming the "reality" is ("we" who are having this discussion with you) has always been the same. The brick wall in DaleSpam's example *can* be reached, but, as he showed, by adopting an unsuitable coordinate system you can make it *appear* that it can't--but that's an illusion. Similarly, for a black hole, the horizon *can* be reached, and crossed. That's the physical reality. The observer far away can't *see* it being reached, but that is an illusion caused by adopting an unsuitable coordinate system.

 

[PeterDonis]

The observer far away can't *see* it being reached, but that is an illusion caused by adopting an unsuitable coordinate system.

On re-reading, this was not a good choice of words. Let me try a better way of stating this point: the observer far away can't see the horizon being reached, but that is an illusion caused by the curvature of the spacetime and its effect on the paths of light rays.

To expand on this a bit (this may already have been discussed earlier in the thread, if so sorry for the repetition, but it bears repeating): suppose an observer freely falling towards the black hole emits light pulses radially outward at fixed intervals of his proper time (say once every second by his clock). An observer hovering at fixed radial coordinate r far away from the hole will see these light pulses arrive at gradually increasing time intervals by *his* clock (which, if he's far enough away, will basically tick at the same rate as Schwarzschild coordinate time). As the freely falling observer gets close to the horizon, his light pulses arrive at the hovering observer farther and farther apart (meaning, later and later in time--it takes longer and longer for them to get out to the hovering observer's location).

Finally, when the freely falling observer emits a light pulse exactly *on* the horizon as he falls through it, the pulse *stays* at the horizon forever--it *never* gets out to the hovering observer (or to any observer at r > 2M). So as far as the hovering observer is concerned, that light pulse "arrives" at infinite Schwarzschild coordinate time. But that isn't because the horizon isn't there or can't be reached; it's because spacetime is curved enough at the horizon to bend radially outgoing light paths into "vertical" lines--lines that stay at the same radius forever. Close to the horizon, but just outside it, the light paths are bent *almost* vertical, so they only get out to larger radii very slowly; as they get farther and farther away, the bending decreases, so their paths look more and more like the "usual" 45 degree lines that light rays travel on in flat spacetime.

 

[A-wal]

I think it's a bit presumptuous of you to assume that we, not you, are the ones who are confused when you make statements like these:

Read the very next word.

What does "total acceleration...to an event horizon" even mean?

Curvature increase=acceleration. Curvature increases slower with a higher mass black hole because it's more spread out, but that means you start acceleration sooner (from a greater distance). Same.

Tidal acceleration is *not* the same as the Newtonian acceleration due to gravity.

Yes, I know, *again*. That's not what I meant. A curve that's constant would be free-fall. But that never happens. If your moving towards the gravitational source then the curvature is constantly increasing and it's that increase that causes the acceleration that's felt as tidal force.

But you claim later on that even when you're at rest the distance can change; see my comment further below.

I didn't mean it would change while you're at rest.

Then you don't have any basis for saying that we, not you, are confused. What you don't see is perfectly obvious, not just to the rest of us here, but to generations of researchers and students who have worked these problems. What I'm saying here is at about the same level of contentiousness, to the relativity community, as saying that water is H2O.

I'm an idiot, humour me. And I'm not a member of the "relativity community" so what's the problem? I didn't even know there was a secret club.

No, it doesn't make sense. You talked earlier on (in reference to the distance from NY to LA while passing through Las Vegas) as though the length contraction was due to velocity; but now you're saying that even if you're at rest relative to the checkpoint, the distance can change. *That* doesn't make sense.

I was referring to special relativity when I said that the distance from New York to Los Angeles would change while you're passing though Las Vegas, depending on your velocity. Change compared to the first measurement, from a different location.

Saying that spacetime isn't flat is *not* the same as saying that free-fall can't exist. Free fall is the natural condition of motion for most objects in the Universe. It's only strange folk like us who spend our existence on the surfaces of planets that think being under acceleration is "natural". The Moon is in free fall around the Earth. The Earth itself is in free fall around the Sun. The Sun is in free fall around the center of the galaxy. The galaxy as a whole is in free fall through the Universe. Asteroids, comets, orbiting satellites, etc., etc.--all in free fall. And all of these objects would still be in free fall, and would behave, physically, just as they do now, if they were freely falling through the horizon of a large black hole (large enough that tidal effects were negligible, as I said).

I meant that not only is space-time not flat, but the curvature is never constant, so every object will always be under some acceleration and so not really falling freely. Normally negligible, yes, but not in the case of an event horizon.

Well, you did ask me to post the computation. What I posted is, as I noted, entirely commonplace in relativity physics; many thousands of students have turned in homework problems containing computations similar to what I posted.

Yea because they think they already know the final answer and then work backwards from there.

Feynman once said that "to understand Nature, you must learn the language she speaks in," meaning mathematics.

Maths is just short hand, and it can't be misinterpreted like language. But it's meaningless unless you know what the symbols and values represent. Maths itself can't describe anything.

That said, if you want a quick non-mathematical sketch of what the computation is telling us, here it is: yes, radial distances "stretch" and times "slow down" as you free-fall towards the horizon. But physically, they do *not* stretch infinitely or slow down infinitely for observers that are freely falling towards the hole. It *seems* like they do in Schwarzschild coordinates, but that is an illusion; and you can tell it's an illusion because even though Schwarzschild coordinate time goes to infinity as you approach the horizon, the path length of each little segment of your worldline (if you're freely falling towards the horizon) that is traversed in a constant interval of coordinate time dt goes to zero *faster* than the time t goes to infinity. (This is the same sort of thing that happens in the "Achilles and the tortoise" problem, which I think has been mentioned in this thread.) The result is a finite sum, meaning a finite path length from a finite radius r > 2M to the horizon at r = 2M.

I never said anything about stretching or slowing down infinitely. You would need a black hole with infinite mass for that. They shape of the world line would change. If you see ten units of space between you and the destination and the strength of gravity increases as you approach then if there was a checkpoint half way to the destination (from your original position) then there would be more than five lights between you and your final destination when you reach the checkpoint.

The fact that Schwarzschild coordinate time goes to infinity over this finite path length tells us something else: Schwarzschild coordinate time is simply *not* a good time coordinate for observers falling into the hole. That's why other coordinate systems were invented to study the experience of such observers; for example, the Painleve coordinates, which I mentioned and gave a transformation for. Basically, Schwarzschild coordinate time as you approach the horizon at r = 2M behaves something like the tangent function, which goes to infinity at a finite value of its argument (pi/2). That doesn't mean the universe just stops at that finite value; it means we need to pick a better function to study that portion of spacetime.

I don't think the universe just stops. Is all the warping of space-time using Schwarzschild coordinates an illusion? A what point does it stop describing reality? You don't see the changing coordinate systems as moving the goal posts then?

But at least we all appear to agree on one thing:

You need to pick one. Either the wall can be reached or it can't. I refuse to believe that reality can change depending on how you choose to measure it (quantum mechanics aside). That's basically the whole point of this thread.

I agree; it is. And what we're claiming the "reality" is ("we" who are having this discussion with you) has always been the same. The brick wall in DaleSpam's example *can* be reached, but, as he showed, by adopting an unsuitable coordinate system you can make it *appear* that it can't--but that's an illusion. Similarly, for a black hole, the horizon *can* be reached, and crossed. That's the physical reality. The observer far away can't *see* it being reached, but that is an illusion caused by adopting an unsuitable coordinate system.

Without using length contraction/time dilation explain why the far away observer can't *see* the horizon being reached.

 

[A-wal]

On re-reading, this was not a good choice of words. Let me try a better way of stating this point: the observer far away can't see the horizon being reached, but that is an illusion caused by the curvature of the spacetime and its effect on the paths of light rays.

To expand on this a bit (this may already have been discussed earlier in the thread, if so sorry for the repetition, but it bears repeating): suppose an observer freely falling towards the black hole emits light pulses radially outward at fixed intervals of his proper time (say once every second by his clock). An observer hovering at fixed radial coordinate r far away from the hole will see these light pulses arrive at gradually increasing time intervals by *his* clock (which, if he's far enough away, will basically tick at the same rate as Schwarzschild coordinate time). As the freely falling observer gets close to the horizon, his light pulses arrive at the hovering observer farther and farther apart (meaning, later and later in time--it takes longer and longer for them to get out to the hovering observer's location).

Finally, when the freely falling observer emits a light pulse exactly *on* the horizon as he falls through it, the pulse *stays* at the horizon forever--it *never* gets out to the hovering observer (or to any observer at r > 2M). So as far as the hovering observer is concerned, that light pulse "arrives" at infinite Schwarzschild coordinate time. But that isn't because the horizon isn't there or can't be reached; it's because spacetime is curved enough at the horizon to bend radially outgoing light paths into "vertical" lines--lines that stay at the same radius forever. Close to the horizon, but just outside it, the light paths are bent *almost* vertical, so they only get out to larger radii very slowly; as they get farther and farther away, the bending decreases, so their paths look more and more like the "usual" 45 degree lines that light rays travel on in flat spacetime.

I hadn't read that post when I wrote the posted my reply. Nice explanation. That's exactly how I see it so it still doesn't help me. If "spacetime is curved enough at the horizon to bend radially outgoing light paths into "vertical" lines" then it's curved enough to stop anything reaching the horizon in any given amount of time. The matter and light share the same space-time so you can't have one rule for one and another for the other (where time dilation and length contraction are concerned).

 

[PeterDonis]

I hadn't read that post when I wrote the posted my reply. Nice explanation. That's exactly how I see it so it still doesn't help me. If "spacetime is curved enough at the horizon to bend radially outgoing light paths into "vertical" lines" then it's curved enough to stop anything reaching the horizon in any given amount of time. The matter and light share the same space-time so you can't have one rule for one and another for the other (where time dilation and length contraction are concerned).

Read what I said again, carefully. I said radially *outgoing* light paths are bent to vertical at the horizon. I didn't say anything about *ingoing* light paths being vertical. They are still ingoing, and there is still "room" between the ingoing and outgoing radial light paths for ingoing timelike worldlines.

Since you posted your previous post before reading what you quoted above, I won't comment on most of that previous post, but there are a few things in there that I still want to address:

Yes, I know, *again*. That's not what I meant. A curve that's constant would be free-fall. But that never happens. If your moving towards the gravitational source then the curvature is constantly increasing and it's that increase that causes the acceleration that's felt as tidal force.

I see a couple of misconceptions here. The first is this: you say "A curve that's constant would be free-fall." I'm not sure what you mean by "constant", but I *think* you mean it as "straight" in the sense of what would be a "straight line" in flat, Minkowski spacetime. If so, your statement is false for curved spacetime; in fact that's the whole point of curved spacetime, that worldlines that we would call "curved" in an ordinary Newtonian sense (such as the worldline of an object falling towards the Earth under the influence of Earth's gravity alone, no other "forces" acting) are actually "straight" in the sense of (a) being freely falling--objects following these worldlines are weightless, they feel no acceleration--and (b) being geodesics of the curved spacetime, i.e., the closest thing to a "straight line" that you can have in a curved manifold.

The second misconception is one I've talked about before: tidal gravity does *not* necessarily have to be "felt" by objects as "tidal force". Two objects which are both in free fall can still be experiencing "tidal acceleration" relative to one another. Again, this happens because of the curvature of spacetime. Take two objects which, at some instant of time, are both at rest above the Earth, one slightly higher than the other. As both objects fall, their spatial separation in the radial direction will increase; this is due to the relative "tidal acceleration". However, both objects are freely falling--they are both weightless, and neither one feels any acceleration. (If you attached accelerometers to both objects, they would both read zero.)

I go into all this because later on, you say:

I meant that not only is space-time not flat, but the curvature is never constant, so every object will always be under some acceleration and so not really falling freely. Normally negligible, yes, but not in the case of an event horizon.

This displays the same misconceptions I just described, and which may be getting in the way of your having a correct understanding of what happens at the event horizon. Once again, just to be clear: objects can *freely fall* through the event horizon, just as they can freely fall towards the Earth. If I start two objects at rest above the horizon, one slightly higher than the other, and then release them both, they will both freely fall through the horizon, and as they fall, their radial separation will increase, just as the two objects falling towards the Earth--but they will both be in free fall the whole time (attach accelerometers to both objects and they will both read zero).

I also wanted to comment on this:

You don't see the changing coordinate systems as moving the goal posts then?

No, of course not. Is it "moving the goal posts" to use a map of the Earth drawn using a Mercator projection for one purpose, and then use a map drawn using stereographic projection for another purpose? Does changing the map change the actual, physical geometry of the Earth's surface? Of course not. Neither does changing the "map" one draws of events in spacetime--which is all a "coordinate system" really is--change the actual, physical geometry of spacetime. But different coordinate systems can be more or less suitable for different purposes; one coordinate system may "distort" a particular aspect of the geometry in a way that hinders understanding of some particular phenomenon.

For example, take the statement I made above, that while outgoing light rays are "bent to vertical" at the horizon, ingoing light rays are still ingoing. One of the reasons Schwarzschild coordinates are not suitable at or near the horizon is that they don't show this clearly; they distort the light cone structure near the horizon so much (actually, "infinitely much" at the horizon itself) that you can't clearly see the structure of the spacetime geometry. Switching to a more suitable coordinate system that doesn't have this distortion, such as Painleve coordinates (or ingoing Eddington-Finkelstein coordinates), can greatly clarify what's going on. But the actual, physical spacetime geometry is the same either way: outgoing light rays stay at r = 2M forever, ingoing light rays are still ingoing. It's just that Schwarzschild coordinates don't show it properly, just as a Mercator projection of the Earth doesn't show the North and South Poles properly.

 

[Buckleymanor]

Film someone throwing a ball high in the air and then catching it when it falls. Then run the film backwards. Ignoring the throw and the catch, while the ball is in the air, can you tell if the film is going forwards or backwards?

Not sure if I miss the point but I reckon you can tell if you use two balls of different masses and apply the same force when you throw them.
If you let them fall from a height and released them at the same time they would hit the ground together and you would not be able to tell.
However if you threw them up in the air together, with the same force, the more massive one would hit the ground first.

 

[Dale]

You need to pick one. Either the wall can be reached or it can't. I refuse to believe that reality can change depending on how you choose to measure it (quantum mechanics aside). That's basically the whole point of this thread.

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.

 

[A-wal]

I hadn't read that post when I wrote the posted my reply.

What the hell does that mean? I think I'm slightly dyslexic.

Read what I said again, carefully. I said radially *outgoing* light paths are bent to vertical at the horizon. I didn't say anything about *ingoing* light paths being vertical. They are still ingoing, and there is still "room" between the ingoing and outgoing radial light paths for ingoing timelike worldlines.

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. 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. It seems as if the black hole's life span has been shortend/accelerated for the free-faller. I don't know why the Schwarzschild coordinates are considered an illusion. Ingoing light is not slowed down. It's constant after all. It's time that's slowed down, by the acceleration.

I see a couple of misconceptions here. The first is this: you say "A curve that's constant would be free-fall." I'm not sure what you mean by "constant", but I *think* you mean it as "straight" in the sense of what would be a "straight line" in flat, Minkowski spacetime. If so, your statement is false for curved spacetime; in fact that's the whole point of curved spacetime, that worldlines that we would call "curved" in an ordinary Newtonian sense (such as the worldline of an object falling towards the Earth under the influence of Earth's gravity alone, no other "forces" acting) are actually "straight" in the sense of (a) being freely falling--objects following these worldlines are weightless, they feel no acceleration--and (b) being geodesics of the curved spacetime, i.e., the closest thing to a "straight line" that you can have in a curved manifold.

No that's not what I meant. 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.

The second misconception is one I've talked about before: tidal gravity does *not* necessarily have to be "felt" by objects as "tidal force". Two objects which are both in free fall can still be experiencing "tidal acceleration" relative to one another. Again, this happens because of the curvature of spacetime. Take two objects which, at some instant of time, are both at rest above the Earth, one slightly higher than the other. As both objects fall, their spatial separation in the radial direction will increase; this is due to the relative "tidal acceleration". However, both objects are freely falling--they are both weightless, and neither one feels any acceleration. (If you attached accelerometers to both objects, they would both read zero.)

Not with an sensitive enough accelerometer they wouldn't! How is tidal acceleration any different to tidal force? 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.

This displays the same misconceptions I just described, and which may be getting in the way of your having a correct understanding of what happens at the event horizon. Once again, just to be clear: objects can *freely fall* through the event horizon, just as they can freely fall towards the Earth. If I start two objects at rest above the horizon, one slightly higher than the other, and then release them both, they will both freely fall through the horizon, and as they fall, their radial separation will increase, just as the two objects falling towards the Earth--but they will both be in free fall the whole time (attach accelerometers to both objects and they will both read zero).

I just meant that every object in free-fall has to be undergoing some tidal force. In fact, that covers every object.

No, of course not. Is it "moving the goal posts" to use a map of the Earth drawn using a Mercator projection for one purpose, and then use a map drawn using stereographic projection for another purpose? Does changing the map change the actual, physical geometry of the Earth's surface? Of course not. Neither does changing the "map" one draws of events in spacetime--which is all a "coordinate system" really is--change the actual, physical geometry of spacetime. But different coordinate systems can be more or less suitable for different purposes; one coordinate system may "distort" a particular aspect of the geometry in a way that hinders understanding of some particular phenomenon.

For example, take the statement I made above, that while outgoing light rays are "bent to vertical" at the horizon, ingoing light rays are still ingoing. One of the reasons Schwarzschild coordinates are not suitable at or near the horizon is that they don't show this clearly; they distort the light cone structure near the horizon so much (actually, "infinitely much" at the horizon itself) that you can't clearly see the structure of the spacetime geometry. Switching to a more suitable coordinate system that doesn't have this distortion, such as Painleve coordinates (or ingoing Eddington-Finkelstein coordinates), can greatly clarify what's going on. But the actual, physical spacetime geometry is the same either way: outgoing light rays stay at r = 2M forever, ingoing light rays are still ingoing. It's just that Schwarzschild coordinates don't show it properly, just as a Mercator projection of the Earth doesn't show the North and South Poles properly.

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?

Not sure if I miss the point but I reckon you can tell if you use two balls of different masses and apply the same force when you throw them.
If you let them fall from a height and released them at the same time they would hit the ground together and you would not be able to tell.
However if you threw them up in the air together, with the same force, the more massive one would hit the ground first.

Film two people, each throwing a ball high into the air and then catching them when they fall. Then run the film backwards. Ignoring the throws and the catches, while the balls are in the air, can you tell if the film is going forwards or backwards?

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.

Define realize. Accepting what I'm told despite that it's still not how I see it? It's frustrating for me too. What exactly did I misunderstand?

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.

LOL I hadn't even heard of this Schwarzschild bloke before he was mentioned here, and I'd never used a coordinite system before (you know what I mean). It's not a reason for me thinking anything. It describes how I thought of it before I even came here. 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.

 

[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.