How does the String Theory Propeller behave as you approach the event horizon?

[weaselman]

That is not how the OP described it. In any case, the key distinction is inertial (falling) vs. non-inertial (stationary).

Fine, he described it.
Let's not argue about this, but just agree to consider the case when Bob is inertial and distant (not falling into the black hole). In that case, will he see Alice's clock slow down or not?

Regardless of how far away Bob is there is no place where Bob would not experience some gravitational pull. Therefore if Bob is inertial then he eventually crosses the event horizon

Let's just say, he is free falling towards Earth. Or is suspended deep in the intergalatic space. He has no plans on crossing the event horizon, at least, not of that particular black hole :)

 

[JesseM]

To see this more clearly, imagine that Alice's ship is really really long. Let it be 10 light years long. Will then Alice see her propeller stop? I am sure, you'll agree, that she will.

She would see it slow down, but not stop. And if the whole ship falls through the horizon eventually, she eventually sees it cross the horizon--she'll see this at the moment she herself crosses the horizon.

 

[dubsed]

Let me verbalize and see if I am starting to get this. Bob sees Alice frozen at the horizon not because she is there forever, but because the light she emitted/reflected eons ago has undergone so much time dilation that it takes light a very long time to reach Bob because it is fighting against the flow of space.

Even though time appears to stop at the horizon as observed from a stationary observer Alice never sees this because she is moving with the flow of space eg down through the upward moving light.

Please let me know how close I am to starting to understand. And where I am understanding things and where I am not.

 

[Dale]

The classic problem of this kind involves an inertial frame falling into the black whole, and an inertial distant observer.

I have always heard it the other way, one inertial observer and the other stationary.

Let's not argue about this, but just agree to consider the case when Bob is inertial and distant (not falling into the black hole).

This is a contradiction. If he is inertial then he is falling by definition.

I hope you are not going to insist that in that case he will not see Alice's clock slow down.

He will see it slow down, but not stop.

 

[Dale]

Let me verbalize and see if I am starting to get this. Bob sees Alice frozen at the horizon not because she is there forever, but because the light she emitted/reflected eons ago has undergone so much time dilation that it takes light a very long time to reach Bob because it is fighting against the flow of space.

Even though time appears to stop at the horizon as observed from a stationary observer Alice never sees this because she is moving with the flow of space eg down through the upward moving light.

Please let me know how close I am to starting to understand. And where I am understanding things and where I am not.

I would say "curvature of spacetime" rather than "flow of space". It is more accurate and general. Other than that you are essentially correct.

 

[weaselman]

This is a contradiction. If he is inertial then he is falling by definition.

I don't see a contradiction. Like I said, he can be in the intergalactic space, or he can be falling somewhere, just not into that particular black hole.

He will see it slow down, but not stop.

Doesn't matter. I don't think anyone would see it really stop. It will become slower and slower, assympthotically approaching zero speed, but never really stop.
So, you are agreeing that the distance matters? An inertial, but distant observer will disagree with Alice's clock, right?

 

[Dale]

I don't see a contradiction.

In the Schwarzschild metric a stationary worldline is not a geodesic and therefore cannot be inertial. I can derive the proper acceleration for a stationary worldline if it is not already clear enough. Bob may either be stationary or inertial, not both, it is in fact a contradiction.

I don't think anyone would see it really stop. It will become slower and slower, assympthotically approaching zero speed, but never really stop.

The difference is that a stationary Bob, regardless of distance, will never receive a signal from Alice from at or within the event horizon even after an infinite amount of proper time, but an inertially falling Bob, regardless of distance, will receive a signal from Alice from at and within the event horizon after a finite amount of proper time. Yes, both an inertial and a stationary Bob will measure some time dilation, but the two situations are qualitatively different as described.

 

[weaselman]

In the Schwarzschild metric a stationary worldline is not a geodesic and therefore cannot be inertial. I can derive the proper acceleration for a stationary worldline if it is not already clear enough. Bob may either be stationary or inertial, not both, it is in fact a contradiction.

You are not hearing me for some reason. Schwarzschild metric is flat at a large distance from the massive body.
You would be correct if the universe was small, and consisted of a proximity of a single black hole.
In the real world though there are plenty of possibilities for an inertial observer to not be falling into a black hole (or, if he prefers to be falling, there are plenty of black holes to choose, it doesn't have to be the same Alice is falling into). Look at the Milky Way for example. It, as a whole, is very inertial, and (we all should hope) isn't falling anywhere.

The difference is that a stationary Bob, regardless of distance, will never receive a signal from Alice from at or within the event horizon even after an infinite amount of proper time, but an inertially falling Bob, regardless of distance, will receive a signal from Alice from at and within the event horizon after a finite amount of proper time. Yes, both an inertial and a stationary Bob will measure some time dilation, but the two situations are qualitatively different as described.

What about the inertially non-falling Bob, positioned somewhere on the opposite side of the universe, far from any black wholes? What will he see?

Here is a citation I found, that, perhaps, will help me get my point across :)
http://www.nicadd.niu.edu/~bterzic/PHYS652/Lecture_21.pdf
(on page 116):
"... the inertial observer (at infinity) can never witness the infalling observer reach the event horizon."

Now, if this is settled, we can move on to discussing the difference between an inertial observer, falling down (far) behind Alice, and the one, not falling at all. You are right, that the former will eventually see Alice crossing the event horizon, however, that will not happen until he himself crosses it. That is not really a qualitative difference - if the one far away just happened to be moving in the direction of the black hole, he would eventually see Alice falling in too.

Strictly speaking, the above statement is not absolutely accurate. The completely pedantic version would say "the inertial observer can bever witness the infalling observer reach the event horizon unless and until he reaches it himself."
Now, there is no difference at all between the two Bobs. Or between one of the Bobs, and Alice watching her propeller.

 

[nismaratwork]

I'm pretty sure that's the same metaphor Susskind used in his book "The Black Hole War" (I'll check when I get home), and he's one of the big theorists in this area.

I think you're misunderstanding, the proposed effect has nothing to do with frames of reference or tidal forces, it can only be understood in the context of string theory and involves observers outside the horizon seeing something different than observers who cross it (with 'see' referring to any type of coordinate-independent measurement you can think of).

In that context it's not even at the level of conjecture! Meh, I guess I'm cranky (as in grumpy, not a crank).

 

[dubsed]

If space is not actually flowing into the black hole, then is it possible to give me another metaphor to help me better understand what is actually going on? I understand that no metaphor is perfect, the one that was given in the reading James gave me was as photons as canoes paddling against a river.

I suppose it is really a bit of a miss statement to say that Bob sees Alice forever. He may be able to see small bits of light. After all you are talking about a finite amount of photons that she released on her way down gradually escaping over an infinite amount of time. becoming fewer and farther between as time progresses. I suppose this might have been where I was originally getting hung up, because to actually see Alice and recognize it as Alice you need to be getting enough photons to distinguish it as her forever, this is not possible if she is not physically there forever.

 

[weaselman]

I suppose this might have been where I was originally getting hung up, because to actually see Alice and recognize it as Alice you need to be getting enough photons to distinguish it as her forever, this is not possible if she is not physically there forever.

It is actually the other way around. She is physically there forever, but you are right that Bob will not actually really see her there. When they say "see", what they really mean is that, based on the signals, received from Alice, Bob will conclude, that she is still there.

 

[JesseM]

It is actually the other way around. She is physically there forever

In Schwarzschild coordinates yes, but not in any coordinate-independent sense.

but you are right that Bob will not actually really see her there. When they say "see", what they really mean is that, based on the signals, received from Alice, Bob will conclude, that she is still there.

I don't understand what you mean--if Bob continues to receive signals (including visual light signals) from Alice near the horizon forever, isn't it true that Bob will "actually really see her there" forever? Of course, here I'm waving aside issues like the fact that the radiation she emits is actually quantized (so they'll be some finite time when Bob receives the last photon she emitted before crossing the horizon) and that the wavelength of the light she emits would eventually become so redshifted that it'd be impossible to detect in practice even if the light were emitted in a non-quantized way.

 

[weaselman]

In Schwarzschild coordinates yes, but not in any coordinate-independent sense.

Yes, in Schwarzschild coordinates. I am not sure how to even define "forever" in coordinate-independent sense.

I don't understand what you mean--if Bob continues to receive signals (including visual light signals) from Alice near the horizon forever, isn't it true that Bob will "actually really see her there" forever? Of course, here I'm waving aside issues like the fact that the radiation she emits is actually quantized (so they'll be some finite time when Bob receives the last photon she emitted before crossing the horizon)

This is exactly what I meant. The last "signal" (photon) Bob receives will arrive at time X, and after that time he will no longer be "seeing" Alice. He will only have to conclude, that she is still on his side of the event horizon. We customarily say that he "sees" Alice stop, but what that really means is that he does not see her move.

 

[Dale]

You are not hearing me for some reason. Schwarzschild metric is flat at a large distance from the massive body.

Only at infinity. It is not flat at any finite distance. Also, as you can see from the Rindler metric, the further distance you go the more sensitive you are to small accelerations wrt the formation of event horizons even in flat spacetime. This means that although the acceleration can be made arbitrarily small as the distance increases, as the distance increases you become arbitrarily sensitive to acceleration. So I stick with my assertion that the primary difference is inertial vs. stationary, but I am glad to add the caveat about finite distances if you think it necessary.

In the real world though there are plenty of possibilities for an inertial observer to not be falling into a black hole (or, if he prefers to be falling, there are plenty of black holes to choose, it doesn't have to be the same Alice is falling into). Look at the Milky Way for example. It, as a whole, is very inertial, and (we all should hope) isn't falling anywhere.

True, but not relevant to the present discussion about the Schwarzschild metric. While other metrics could be discussed there are really not very many things that can be said in general about event horizons in arbitrary metrics. It is certainly beyond the scope of my knowledge.

You are right, that the former will eventually see Alice crossing the event horizon, however, that will not happen until he himself crosses it.

Yes, that is correct.

 

[Dale]

Yes, in Schwarzschild coordinates. I am not sure how to even define "forever" in coordinate-independent sense.

How about "'forever' is an infinite amount of proper time"? That is coordinate-independent and along the lines of what I said in #37.

 

[weaselman]

This means that although the acceleration can be made arbitrarily small as the distance increases, as the distance increases you become arbitrarily sensitive to acceleration.

So, are you saying, that an observer, suspended in intergalactic space, or one in the center of Milky Way (if that space was not already occupied) is non-inertial in any significant/measurable/noticeable way?

How about "'forever' is an infinite amount of proper time"? That is coordinate-independent and along the lines of what I said in #37.

Well, strictly speaking, the proper time is only defined locally, so I am not sure how you associate the t in the Schwarzschild metric with anyone's proper time. I am guessing, that's what JesseM had in mind with his objection.

 

[Dale]

So, are you saying, that an observer, suspended in intergalactic space, or one in the center of Milky Way (if that space was not already occupied) is non-inertial in any significant/measurable/noticeable way?

Is intergalactic space or the center of the Milky Way well described by the Schwarzschild metric and is the observer stationary in those coordinates? If so, then yes.

 

[dubsed]

I think Dale's point is that it doesn't matter how far away you are, ignoring the possibility of
falling into a closer massive body, if you have 0 kinetic energy and you do not fire any rockets you will eventually fall into the black hole.

Though this does make me wonder, at a certain distance would the expansion of the universe prevent you from getting falling closer?

 

[Dale]

Though this does make me wonder, at a certain distance would the expansion of the universe prevent you from getting falling closer?

Yes, but then you are talking about the FLRW metric instead of the Schwarzschild metric.

 

[weaselman]

I think Dale's point is that it doesn't matter how far away you are, ignoring the possibility of
falling into a closer massive body, if you have 0 kinetic energy and you do not fire any rockets you will eventually fall into the black hole.

Yeah ... It looks like he thinks that the metric just "jumps" to flatness when it hits infinity being curved all the way before that :) Oh, well ...

 

[dubsed]

Yeah ... It looks like he thinks that the metric just "jumps" to flatness when it hits infinity being curved all the way before that :) Oh, well ...

Do we really need to argue semantics? You don't like that he says its flat at infinity. I don't like that you say it actually hits infinity. Saying that it is flat at infinity is the same as saying that (1/9) + (8/9) = 1

 

[Dale]

Yeah ... It looks like he thinks that the metric just "jumps" to flatness when it hits infinity being curved all the way before that :) Oh, well ...

Do you somehow think it is flat at some finite distance? If so, where?

See here https://www.physicsforums.com/showpost.php?p=2712746&postcount=38 where I calculated the proper acceleration of a stationary observer in the Schwarzschild metric. It is non-zero for any finite r meaning that such an observer is non-inertial. And here http://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html is a page I like on Rindler coordinates and event horizons in flat spacetime.

 

[weaselman]

Do we really need to argue semantics? You don't like that he says its flat at infinity.
I don't like that you say it actually hits infinity.

I was being sarcastic :)
If it is flat at infinity (and it, of course, is), and you don't like saying that it hits infinity (and you should not), then you realize that you can make the curvature as small as you want by moving farther away. In other words, you can make a reference frame as close to inertial as you need by increasing the distance from the black hole. Insisting that such observer is not inertial is even less reasonable than saying that Alice's propeller is non-inertial, because it rotates, and the Alice herself is non-inertial, because her propeller somehow affects her acceleration.
All three statements are actually true, but the effects are so totally and completely insignificant, that it is obvious that arguing them cannot possible have any constructive goal.


And this is not semantics. Dale thinks that the difference between Bob and Alice is inertiality, which is incorrect. The real difference is the curvature of space-time that decreases with distance to the black whole. Bob spacetime is (ok, almost) flat, Alice's isn't.
At the event-horizon it becomes so curved, that the horizon itself is time-like. That is the real reason Bob cannot see anyone cross it. It does not matter if he is inertial or not (he can be stationary at some distance from black hole, orbiting around it, at infinity, or just far away), and it does not matter if Alice is (if Alice was falling non-inertialy - suppose, she was unsuccessfully trying to escape - the effects, seen by Bob would be quilitatively the same), it only matters how far from the horizon each of them is.

Same with the propeller, just on a smaller scale. Alice cannot see it cross the horizon, because the horizon is time-like. In this sense, the propeller will stop for her as well as it does for Bob, just not for as long, because as soon as she reaches horizon, it'll catch up.

Interestingly enough, they both (Alice, and propeller) will cross the horizon at the same time. And, once they do, time and space will switch places for them - the propeller will be later than Alice, but at the same spatial location

 

[weaselman]

Do we really need to argue semantics? You don't like that he says its flat at infinity.
I don't like that you say it actually hits infinity.

I was being sarcastic :)
If it is flat at infinity (and it, of course, is), and you don't like saying that it hits infinity (and you should not), then you realize that you can make the curvature as small as you want by moving farther away. In other words, you can make a reference frame as close to inertial as you need by increasing the distance from the black hole. Insisting that such observer is not inertial is even less reasonable than saying that Alice's propeller is non-inertial, because it rotates, and the Alice herself is non-inertial, because her propeller somehow affects her acceleration.
All three statements are actually true, but the effects are so totally and completely insignificant, that it is obvious that arguing them cannot possible have any constructive goal.And this is not semantics. Dale thinks that the difference between Bob and Alice is inertiality, which is incorrect. The real difference is the distance to the black whole.

The closer you are to the horizon, the slower your clock is ticking, compared to a more distant observer. At the horizon, the clock "stops", because the horizon itself is time-like. That is the real reason Bob cannot see anyone cross it. It does not matter if he is inertial or not (he can be stationary at some distance from black hole, orbiting around it, at infinity, or just far away), and it does not matter if Alice is (if Alice was falling non-inertialy - suppose, she was unsuccessfully trying to escape - the effects, seen by Bob would be quilitatively the same), it only matters how far from the horizon each of them is.

Same with the propeller, just on a smaller scale. Alice cannot see it cross the horizon, because the horizon is time-like. In this sense, the propeller will stop for her as well as it does for Bob, just not for as long, because as soon as she reaches horizon, it'll catch up.

Interestingly enough, they both (Alice, and propeller) will cross the horizon at the same time. And, once they do, time and space will switch places for them - the propeller will be later than Alice, but at the same spatial location

 

[JesseM]

The closer you are to the horizon, the slower your clock is ticking, compared to a more distant observer. At the horizon, the clock "stops", because the horizon itself is time-like.

Actually the horizon is light-like--entering the horizon is a bit like entering the future light cone of some event (and of course, once you enter an event's future light cone, you can never escape it!) As with a lot of aspects of nonrotating black holes, this is probably more intuitive if you use Kruskal-Szekeres coordinates.

Interestingly enough, they both (Alice, and propeller) will cross the horizon at the same time. And, once they do, time and space will switch places for them - the propeller will be later than Alice, but at the same spatial location

In Schwarzschild coordinates, which is what you're talking about, there is technically no time-coordinate where they cross the horizon (infinity is not a time coordinate!) And in a more physical sense, the event of Alice crossing the horizon lies in the future light cone of the event of the propeller crossing it.

 

[Dale]

I was being sarcastic :)
If it is flat at infinity (and it, of course, is), and you don't like saying that it hits infinity (and you should not), then you realize that you can make the curvature as small as you want by moving farther away. In other words, you can make a reference frame as close to inertial as you need by increasing the distance from the black hole. Insisting that such observer is not inertial is even less reasonable than saying that Alice's propeller is non-inertial, because it rotates, and the Alice herself is non-inertial, because her propeller somehow affects her acceleration.
All three statements are actually true, but the effects are so totally and completely insignificant, that it is obvious that arguing them cannot possible have any constructive goal.

I already addressed this in post 44. Yes, you can make the proper acceleration arbitrarily small, but you are incorrect that the effects are insignificant. Remember, we are looking at effects on Bob's observations of Alice, not on effects that are local to Bob. As you arbitrarily reduce the acceleration you must also arbitrarily increase the distance to Alice, which makes Bob's observations arbitrarily sensitive to his acceleration (see the Rindler metric). The net effect is that the distance cancels out and only the acceleration matters. I will work it out completely, but probably not today.

Your basic error is the assumption that because an effect is locally insignificant it must also be globally insignificant.

And this is not semantics. Dale thinks that the difference between Bob and Alice is inertiality, which is incorrect. The real difference is the distance to the black whole.

The closer you are to the horizon, the slower your clock is ticking, compared to a more distant observer. At the horizon, the clock "stops", because the horizon itself is time-like. That is the real reason Bob cannot see anyone cross it. It does not matter if he is inertial or not (he can be stationary at some distance from black hole, orbiting around it, at infinity, or just far away), and it does not matter if Alice is (if Alice was falling non-inertialy - suppose, she was unsuccessfully trying to escape - the effects, seen by Bob would be quilitatively the same), it only matters how far from the horizon each of them is.

OK, then at what initial distance does an inertial Bob have to be in order for him to never see Alice cross? And what is the minimum distance for a stationary Bob below which he will eventually see Alice cross?

Alice cannot see it cross the horizon, because the horizon is time-like. In this sense, the propeller will stop for her as well as it does for Bob

This is factually incorrect, particularly for a super-massive black hole as we have been discussing here. Alice will not notice anything unusual about the propeller crossing the event horizon provided that the tidal forces are negligible.

 

[weaselman]

Your basic error is the assumption that because an effect is locally insignificant it must also be globally insignificant.

You already agreed, that they are globally insignificant at infinity. Unless, you really believe, that they all just "jump" to insignificance when you "hit infinity", you have to admit, that they are becoming less and less significant as you approach it. There is no third possibility (no first possibility either realy, as dubsed pointed out, which leaves us only one choice). If the distance "cancels out", it can't possibly matter, even if it is infinite.

 

[Dale]

You already agreed, that they are globally insignificant at infinity. Unless, you really believe, that they all just "jump" to insignificance when you "hit infinity", you have to admit, that they are becoming less and less significant as you approach it. There is no third possibility (no first possibility either realy, as dubsed pointed out, which leaves us only one choice). If the distance "cancels out", it can't possibly matter, even if it is infinite.

Look at what I actually said in #44, you are misreading what I wrote. I agreed that the metric is locally flat "at infinity", not that the global effects wrt observations of Alice ever became insignificant. In fact, I explained exactly the fact that the global effects were significant at any finite distance.

 

[weaselman]

Look at what I actually said in #44, you are misreading what I wrote. I agreed that the metric is locally flat "at infinity", not that the global effects wrt observations of Alice ever became insignificant. In fact, I explained exactly the fact that the global effects were significant at any finite distance.

Take a look at the link I quoted in the post you were responding to. It explains the specifics of Alice's free fall from the stand point of an inertial observer at infinity. Those are exactly the observations we are discussing here (clocks/propellers slowing down and "stopping" eventually). You replied to that post and conceded that an observer at infinity would indeed be inertial and observe these effects.

 

[kj30]

Interesting thread. I'm still confused on several points:

1. If a black hole is sufficiently large to minimize the effect of tidal forces, how is it possible for someone to not notice any change in local physics after passing through the event horizon? If you were falling in feet first, how could you see your feet with your eyes? Light can't go in that direction inside the event horizon.

2. Is there a simple way to understand why someone falling through the event horizon does not experience the entire future of the universe "falling in" and burning them up in a brilliant super-high blueshift flash? If someone were suspended right above the event horizon (by massive rocket thrusters, etc.) it seems clear that this would happen. What is it about traveling into the event horizon that allows the traveler to escape this flash? I don't find the explanation in the FAQ convincing, namely that there wouldn't be enough time for light in distant parts of the universe to reach you as you pass through the event horizon. If the whole future passes before your eyes, there WILL be enough time for light anywhere to reach you. Or does that not apply if you are in an inertial frame falling freely into the black hole?