Just a minor quibble here -

As the falling observer inches (millimeters) closer to the EH (as seen by us), the time delay of his light signals increases towards infinity even though he has moved a very small distance, so I would expect a large apparent time dilatation superimposed on the gravitational time dilatation.

But I haven't done the calculations.

But I found your post very clear and understandable, Especially compared to Richard and Ken's esoteric maths (my problem, not theirs!). Ken and Richard are having an interesting discussion, but I fear are leaving most of us behind.

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Mike

As I understand it, (call it an if I understand it), there shouldn't be a time delay as c is the same as viewed from all frames. There is a massive red shift due to gravity and dilation. However, since c is c for each frame of reference and the surface of the EH is at a known distance which doesn't change, there should not be shifting in the delay of the received signal because the light is travelling essentially the same distance as viewed by the distant observer and so taking the same amount of time. Light doesn't change speed (like a bullet) as it leaves a gravitational well, it shifts down in frequency (red shifts - because it does lose energy as it climbs out of the well).

This means we should be seeing real time (minus essentially a constant T for the light to traverse the distance X) as we watch our astronut (they'd have to be a nut) start to fall into the EH where we observe the clock on his wall slowing down to a stop as the colors shift further red. As the emitting object reaches the EH, the photon wavelengths approach infinity and the photon energy approaches 0 but it's speed for the whole trip should be c and the time delay from emission to reception should still be essentially the same constant T. Somewhere before the astronut or object reaches the BH, the time dilation will go so low there is no more information detectable coming from it but there will always be something coming back, dim and extremely red shifted and delayed by that time T.

Charles

Charles, I understand what you are saying, but Ken would immediately respond with "c is always c when you measure it locally in your own time frame".

How would we measure c at the falling observer? Lets say he sends a light beam across the inside of his spaceship. It travels a distance d in time d/c = t2 - t1, the times of emission and reception. But our clock measures a much greater interval between t1 and t2 due to the time dilatation, and as d is across the spaceship, there should be no lorentz contraction or anything to alter the distance. So our measure of c in his spaceship is much lower.

This would apply if he is hovering near the EH. If he is falling in, it gets a little more complicated, but the idea still holds.

But I agree that when the light reaches us, it is travelling at c, but is red-shifted.

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Mike

You came to the right place-- folks around here will bring you up to speed pronto!

I don't believe that's actually correct-- and there are three ways to see why T is actually infinite, depending on the coordinates you use (and Richard will correct me if I make a mistake). If you choose a coordinate system that is locally measurable, you can measure distance and time at each point along the path of a freefalling observer just the way that freefalling observer would. I've argued these coordinates are the most "physical" because the quantities are not just coordinates, they are actual measurables, and all for the same observer. In these coordinates, the distance to the event horizon is effectively infinite, because of the extreme length contraction. Thus although c is always c in any locally measurable coordinates, the time is infinite if the distance is infinite. What makes these hard to use in practice is the desynchronizing effects of having the observer be moving through a gravity field, so this one would take the most relativity to get right, I'd imagine.

Or, you can choose a simpler coordinatization that is purely mathematical, and is a kind of conceptual extrapolation of the measurable distance and time of we on Earth. Those are the coordinates that have the event horizon be at a fixed known distance, but the speed of light slows down to a standstill due to the extreme time dilation near the event horizon. Those are the coordinates Richard likes to use because they are mathematically the cleanest, but I've argued they are not useful for making physical statements about "how things are", they are just good for getting the answers right.

Or, you can choose a coordinatization that is kind of a hybrid of these, as it is locally measurable, but only for a chain of different observers who are "hovering" at a fixed location relative to the black hole. Those coordinates have some similarities to the free-fall coordinates, but experience both time dilation and length contraction to the point that the speed of light stays the same (as for any local measurable). Again the time is infinite, because of the length contraction, but I'm not sure what that time really means because it is cobbled together from a bunch of different observers.

Anyway, no matter how you slice it, T is infinite, and that, moreso than the infinite redshift, is why we also lose contact with someone falling into a black hole. There is also a "last word" that we can send to them-- they will not see our entire future simply because their time appears to be stopping to us (and that's one reason I've argued that their time is not "really stopping for us"-- if it were, they should be able to see our whole future play out).

If it is sent across the space craft, the gravitational pull on the light I would expect would require the light to be moving at c at an angle because it is being accelerated down. The distance across the craft would be unchanged but the fact that c is occuring along an angle means that the velocity across the craft would be based on a lower rate depending on the angle - the x component where c^2 = x^2+y^2 (where x is Vx and y is Vy). Hence the time dilation is countered by the lower Vx component of c for the local observer. This is the special relativity equivalence understanding. This would be akin to the special relativity example of a spacecraft traveling at Vx while trying to measure the round trip light time of a Y axis round trip. There is no space contraction in the Y axis like in the X axis but there is time dilation. The answer is that the light seen going from the emitter to the mirror and back to the sensor as seen by the astronaut would be the light which is going at an angle based on the x axis velocity such that the light must be traveling at an angle so as to hit the mirror as it travels along the x axis - according to the at rest observer. Since c^2 = Vy^2 + Vx^2 the actual time of flight is 'adjusted' by the fact that there is an X component in some reference frames but not all.

When you start sending light back and forth along the direction of travel, you not only have time dilation and lorentz contraction of space, you've also got time variations or differences at various places along the axis. (if I recall the spcl relativity class stuff). To try to make measurements at various levels of R near a very strong gravity well where clocks are operating at significantly different rates is starting to get rather hairy. One would think the astronut would be able to detect the clock in the rear of the ship to be speeding up compared to his clock beside him in the front. It would seem that to a finer point, one would experience pretty much the same thing on earth looking at one's local time piece versus that sitting on top of mt everest (assuming a nonrotating earth). The difference is that the astronut is not experiencing gravity since he is in freefall where as the person on earth would be experiencing gravity since he is stationary. I believe there is a relationship in GR between acceleration and gravity that makes them equivalent.

As for the astronut recognizing c being c along R even with the time dilation, I'm not sure what that mechanism is, especially at this late hour of the evening. I don't think you mentioned that case.

In any case, the distant observer knows the distance from the massive object and knows how long it takes for light to get from there to him. Also, he is far enough away such that any effects from the object are essentially zero. Consequently, the photons must be traveling at c all the way back and not changing speed as they leave. At least that is how I understand it.

Charles

Ken,

The coordinates of the radial free-faller is something I've stayed away from trying to mess with because, a) it's complicated, b) makes my head hurt, and c) I'm getting lazy. But I'm just gonna have to bite the bullet and try to learn what how things look in that frame.

I don't think the free-faller thinks there is infinite distance to the horizon. But something very weird goes on depending on some things. It may well be that he "hits the wall in his proper time" before he thinks he got to the horizon! There is something there that, unless he was "thrown down" with initial velocity within a certain window where he could catch something dropped ahead of him before it hit the singularity, he will never see the stuff ahead of him "die". It has to do with information just can't "go backwards", so you never know what happens to stuff ahead of you.

Again, it gets very weird. His local reference frame seems normal to him, as always, but in terms of our coordiantes externally, radius and time flip roles. And trying to imagine that is, well, difficult.

Forgetting about what he thinks is "ahead" of him, and only looking back at what he's leaving behind, I'm almost sure he never thinks the (his coordinate) distance back to us becomes infinite or anything. But things get very weird globally for him.

The infinite distance to the horizon comes from POV of external coordinates. We can define a "proper distance" by integrating the space-like radius of the metric:

d = Integral[ dr/(1 - R/r) ], from some r0 to R. That blows up, and one of the coordinatizations that has light speed being 'c' the whole way makes use of that and so you get it takes infinite time.

But again, that "time" there doesn't really correspond to any one observer, like you noted.

Another way to see the "infinite distance" is to consider the local radial ruler of a stationary observer that gets closer and closer to the horizon. His ruler shrinks as 1/(1 - R/r), so if he's at some r0, he thinks radial distances are longer, and the horizon gets *farther away* the closer *we say* it is.

That's the ol' fishing line "absurdity". Hover over the hole and try to drop a fishing line into the horizon. Do it slowly and you just keep feeding out line as the length "shrinks" faster than you reel it out. [You'd have to be careful, because if you let it start moving fast enough, you'd get in trouble and the line would either have to break or yank you in with it, and it might happen that you fell in before you knew it. ]

But what a free-faller thinks of his distance is something else again entirely, and I'm just not really sure.

ETA: To any lurkers reading this, I'm sure you know, but just for the record, this "hovering" near the horizon and reeling out fishing line and all that stuff is just colorful ways to describe what is basically mathematical operations. Hovering close enough to an event horizon for the fishing line thing to be stark would require you to be experiencing *billions* and even trillions of earth g's. There's nothing that could stand anywhere near that.

-RIchard

I seem to have missed this post and the 2 below where it is above..

I guess what I am really saying is that 'somehow' Time is being misplaced and that it should be 'attached' to spacetime and what is happening to it. After all it called 'spacetime' and the time dimension was embedded into the 3d by Minkowski, right? Maybe that is where the problem is? It seems that 'time' is attached to light?

But the point is that at the Event Horizon, spacetime isn't 'frozen', it is just curved inward to the point of no return, and that is what is really happening to time. Time isn't frozen, it is just curved inward and continuing on its journey down to the singularity, where time ceases to exist in our universe.

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RussT
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Everything is, as it should be, otherwise, it wouldn't be!

This is very interesting, Richard.

How many times have I said things to the effect of... everyone needs to stop Sci-Fy-ing things around black holes and only consider what can realistically be happening there from the aspect of the particles, NOT humans in impossible simulations or thought experiments at event horizons especially. At the event horizon, aren't you trying to see what the photon sees?

Okay I see now, it is not a refrence frame, it is actually dimenionless or 0 dimension for the 3d of space.

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RussT
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Everything is, as it should be, otherwise, it wouldn't be!

[quote=publius;946034 Forgetting about what he thinks is "ahead" of him, and only looking back at what he's leaving behind, I'm almost sure he never thinks the (his coordinate) distance back to us becomes infinite or anything. But things get very weird globally for him.
[/quote]What surprises me about that is, let's imagine the falling person drops from an initial point near the EH and is reeling out a rope as she falls. The rope has marks on it that were one foot apart at the original point, and has a fixed total length, which we can let get arbitrarily large if we like. Will the rope save the person from falling across the EH? Now, I hate physical objects in relativity, they are very hard to keep track of, but it seems to me the stationary hovering observers would claim the rope will save her. Then we could imagine an observer at each of these foot marks, and they all start dropping in sequence like paratroopers. Would they not agree on that the rope measures distance, and doesn't reach the EH? So it sounds like they'd think the distance was infinite. I don't know if that argument is legit, but it's what seems right to someone not accustomed to doing GR calculations.

The one thing we know is that the freefaller can apply physics to figure out that her signal "I'm now crossing the ev-e-n--t h--o---r---i----z-----o...." will never reach Earth, but the reason for that will be different in different frames. The Earth frame thinks it is because time stops for the freefaller, but that's just our conceptualization (i.e., coordinatization) of her time, it is not a uniquely physical description, even for us. What the freefaller thinks is the "reason" is what we are still grappling with.

Ken,

You know, I just realized something. This "infinite distance" notion, this fishing-line "peculiar result" , is entirely a coordinate thing, and as such, it is actually hindering our understanding, not helping! (IMHO, now, I must say). I'm glad I found this forum and starting participating in discussions and *learning myself*. Previously, I knew "GR people" would get "funny acting" about what are essentially coordinate dependent effects. I did not really appreciate what the fuss was about until now.

Anyway, the fishing line/infinite proper distance is entirely a function of *spatial curvature* in a our Schwarzschild coordinates. When we reel out the line, we're doing it slowly, and using our Schwarzschild-stationary ruler.

Consider Rindler. There's an horizon there, *but absolutely no spatial curvature at all*. There's only time coordinate curvature. The fishing line/infinite proper distance thing would not apply at all there. Hovering near the Rindler horizon, and reeling out line, all that would happen is the line would break (or pull us in) as the tension became infinite before the end reached the horizon. Same thing would happen in Schwarzschild if we could ever get a stationary reel to ever get there -- spatial curvature there just prevents that, and as such is just a interesting coordinate effect that really has nothing fundamental to do with the big picture.

Now, consider the Rindler "free faller". Globally inertial Minkowski. How does he explain how he's causally disconnected from the Rindler observer(s)? Why, his light can't catch 'em! The distance to those he left behind remains finite, although increasing ever so closer to light speed, but finite.

The Schwarzschild free-faller has that date with the wall in proper time so to speak, and he's not going to be Minkowski because there's real curvature afoot, and a curvature that increases as he gets closer to the singularity, but I'm certain past the horizon (as we would define the coordinate distance), he says "aha, light from here can't catch 'em". As far as he's concerned, *we're the ones* accelerating away from him like crazy (and for good reason).

Suppose he whizzed by while we're sitting there reeling out that fishing line. I'm certain what he would see would be close to the Minkowski observer watching some Rindler dudes lay out some length. That length would be contracting down to zero according to the Minkowski guy.

-Richard

Yeah, that makes good sense-- I think you're onto something there. So you may be right that the freefaller never thinks the distance back to Earth is infinite, but she does think the Earth is accelerating away and so will never get caught by the "I'm crossing the horizon now" signal. Indeed, I have often mentioned that the "forgotten effect" in relativity is the temporal desynchronization effect, and this messes up a lot of people, but I think what you have here is the spatial equivalent of that same effect-- as the freefaller approaches the horizon, her sense of "how far" keeps shifting, just as her sense of "now" does. The Earth reaches the point of no return before she crosses the event horizon, so regardless what is happening to her sense of "now", the signal just will never get there because it can't catch up. When she reaches the event horizon, everything behind her is pulling away so fast that her beam can't go backward at all. That may be the analog of "time and space swapping places" here-- instead of asking the question when will the light beam reach Earth, the question becomes, where will the light beam ever reach, and the answer is-- the singularity only. It's a topological issue, not an "amount of time" issue, as the latter has no physical meaning, it's just a coordinate. I think this is also what you mean about curvature-- it's the topology that curvature changes at the event horizon.

Ken,

I hope you're sastified 'cause you're making my head hurt looking at
Schwarzschild free-fall stuff. Head hurt bad. Now, drum roll please,

The radial free-faller's *spatial metric* is FLAT as a pancake -- there's only time curvature there (and big-time time funny business). So the darn radial free-faller's metric has something very much in common with Rindler, actually. But the time factor is something different and more complicated.

I'm trying to make heads or tails out of some diagrams (mixing various coordinate notions), but near as I can figure, inside the horizon, the free faller will think a beam of light he shoots "outward" will just curve around in circular (well not a true circle) path, bending right back.

Don't count on me coming up with anything concrete to say about this for a while. This is tedious stuff. Well, for me. I'm sure it is child's play for the Big Boys.

-Richard

Squashed,

Exactly at the event horizon is sort of a infinitely thin razor -- you can't be there, really. There are two sides, and things behave differently on either side sort of touching tangents at that razor edge. I can't really explain it well, but basically it would take infinite time for a photon launched outward exactly at the infinite horizon to go anywhere. Such a thing effectively doesn't exist. Or just think of "escape is possible infinitely close to the horizon, but from the horizon and on in, it is impossible".

-Richard

That's interesting, it is also reminiscent of the observed flatness of our freely falling universe (though it apparently didn't have to be). My guess is, you mean flat for a faller from rest at infinity, or in the case of the universe, flat for that faller but with time running backward. Perhaps this is why some like to view the universe as having zero total energy. Also, the role of dark energy is probably as though the metric included a kind of radial frame dragging, such that the backward-time faller had to be thrown in at a pretty high speed at infinity to make it to the EH.

If the light can travel an infinite distance from the event horizon then below the event horizon the light would still be able to achieve a "less than infinite" distance, or in other words the light can still travel some radial distance (from an outside perspective) before falling back.

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My invisible elf ??? Why he is made of dark matter and lives off of dark energy !!!
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