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EDG
2010-Mar-30, 02:56 AM
Imagine that we had a spaceship that could instantly traverse the 4.3 lightyears between Sol and Alpha Centauri, and that it spends three days out there, and then jumps back instantly to Sol. Assume that we also have a very powerful detector at Sol that can see the ship when it arrives at Alpha C.

So, the timeline of events here (as I reckon) should be like this:

year 0, day 0: Ship jumps out from Sol. Instantly arrives at Alpha C.
year 0, day 3: Ship jumps out from Alpha C. Instantly arrives back at Sol.
year 4.3 + 0 days: Detector spots ship arriving at Alpha C, sees ship at Alpha C for the next 3 days.
year 4.3 + 3 days: Detector spots ship departing from Alpha C.

As far as I can see, the detector is only observing the image of the ship at Alpha C, since the real ship has obviously been back at Sol for the past four and a bit years. So it's not like the ship is really in two places at the same time.

But scenarios like this (I think) are usually said to be a violation of causality, aren't they? How is that the case, exactly? I'm clearly missing something here (I think it has something to do with 'lightcones' but I'm not sure what the problem is).

korjik
2010-Mar-30, 03:15 AM
The ship is exerting gravity from two places at once to the observer during those 3 days that it can see the ship in two places at once. Assuming that the ship is still by the observer

EDG
2010-Mar-30, 03:42 AM
While the ship's at Sol, the observer is feeling its gravity and seeing light reflected/emitted from its surface. While the ship's at AC, the observer at Sol isn't feeling anything or seeing anything of it, because that takes 4.3 years to get to travel to the observer from AC. And then when the ship jumps back to Sol three days after it gets there, the observer will see and feel the ship at Sol again. Then 4.3 years later the observer will see and feel the signal from AC from the ship for three days.

Everything's still linear, just disjointed somewhat. Granted, for three days (4.3 years later) there is a gravity and light signal from the ship originating both from AC and Sol, but why is that actually problematic? It's not like the image of the ship at AC can influence anything happening 4.3 years in its future (is it??).

pzkpfw
2010-Mar-30, 03:43 AM
It's not so much that a specific example does or doesn't (appear to) violate causality, it's that one of the components ("instant" travel) would allow causality to be violated.

If Sol and Alpha Centauri are in motion with respect to each other (which we can expect), then they do not have the same concept of time (each sees the other as having a slower clock).

"Instant" travel then allow a spaceship from one star to jump to the past. A spaceship from each, jumping to the past of each other, allows the kind of "kill your own grandfather" paradox - i.e. what if each spaceship jumps to the past of each other, and blows up the other spaceship?

pzkpfw
2010-Mar-30, 03:45 AM
See post #9 here: http://www.bautforum.com/against-mainstream/75007-instant-communications.html


Previous discussions:

http://www.bautforum.com/space-astronomy-questions-answers/94941-why-impossibility-ftl-such-certainty.html

http://www.bautforum.com/space-exploration/78066-hypothetical-ftl-method-doesnt-violate-causality.html

Jens
2010-Mar-30, 03:49 AM
Everything's still linear, just disjointed somewhat. Granted, for three days (4.3 years later) there is a gravity and light signal from the ship originating both from AC and Sol, but why is that actually problematic? It's not like the image of the ship at AC can influence anything happening 4.3 years in its future (is it??).

Well, you know that gravity exerts itself throughout the universe, so you've just magically increased the total mass of the universe. That seems problematic.

EDG
2010-Mar-30, 03:54 AM
It's not so much that a specific example does or doesn't violate causality, it's that one of the components ("instant" travel) would allow causality to be violated.

If Sol and Alpha Centauri are in motion with respect to each other (which we can expect), then they do not have the same concept of time (each sees the other as having a slower clock).

"Instant" travel then allow a spaceship from on star to jump to the past. A spaceship from each, jumping to the past of each other, allows the kind of "kill your own grandfather" paradox - i.e. what if each spaceship jumps to the past of each other, and blows up the other spaceship?

OK... so I guess what I'm talking about here only works if there's a single universal frame of reference? (or if somehow the ship can sync up to whatever is the local timeframe at its destination during the jump)?

But then what about things like quantum entanglement? I thought it's been shown that if you change the state of one thing that's entangled with another, then the other responds to the change instantaneously? Or is that only on small scales like what we observe on Earth (i.e. if we separated the two entangled things by a lightyear, would one still 'flip' instantaneously?)

EDG
2010-Mar-30, 03:54 AM
Well, you know that gravity exerts itself throughout the universe, so you've just magically increased the total mass of the universe. That seems problematic.

That does seem a bit problematic, I'll grant you that :).

Van Rijn
2010-Mar-30, 03:55 AM
Imagine that we had a spaceship that could instantly traverse the 4.3 lightyears between Sol and Alpha Centauri, and that it spends three days out there, and then jumps back instantly to Sol. Assume that we also have a very powerful detector at Sol that can see the ship when it arrives at Alpha C.

So, the timeline of events here (as I reckon) should be like this:



Who's timeline?



year 0, day 0: Ship jumps out from Sol. Instantly arrives at Alpha C.
year 0, day 3: Ship jumps out from Alpha C. Instantly arrives back at Sol.


"Year 0, day 0" in who's reference frame? "Instantly" in who's reference frame? "Year 0, day 3" in who's reference frame?



year 4.3 + 0 days: Detector spots ship arriving at Alpha C, sees ship at Alpha C for the next 3 days.


Year 4.3 + 0 days in who's reference frame? Where is this telescope (or whatever it is)?



year 4.3 + 3 days: Detector spots ship departing from Alpha C.


See above.

EDG
2010-Mar-30, 03:58 AM
I guess it'd have to be in Sol's timeframe (unless, as I mentioned earlier, I'm actually creating a universal external timeframe that isn't actually possible in relativity).

Van Rijn
2010-Mar-30, 04:43 AM
I guess it'd have to be in Sol's timeframe (unless, as I mentioned earlier, I'm actually creating a universal external timeframe that isn't actually possible in relativity).

Well, it's easy to assume an absolute reference frame, since for everyday life and language it's usually assumed.

You could have (as a thought experiment) a ship next to a giant telescope near Earth. The ship disappears and reappears in three days next to the telescope. In 4.3 years, the telescope sees a spacecraft appear near Alpha Centauri and it disappears after three days . . . but this doesn't talk about the situation from the point of view of the ship's crew, or what other observers could have seen.

EDG
2010-Mar-30, 05:42 AM
but this doesn't talk about the situation from the point of view of the ship's crew, or what other observers could have seen.

What would the crew and those observers see though?

Wouldn't the crew just be at Sol, then blip out and see AC for three days, and then blip back and be at Sol? What would they see that's different?

And what about an observer at say, Barnard's Star (5.963 ly from Sol, 6.475 ly from AC)?

(BTW I'm genuinely curious here, this isn't leading up to some ATM thing ;) )

Schneibster
2010-Mar-30, 06:44 AM
pzkpfw's got it right.

Another way to look at this is, if you can "jump" to a remote location faster than light can get there, then there exists a valid frame of reference in which you arrived before you started, and that is an explicit causality violation. Furthermore, if you can "move" faster than light relative to nearby objects, then there exists a valid frame of reference who will observe your time going backwards, which is also an explicit causality violation. These are simple geometric consequences of the finite speed of light.

The postulates of relativity are few and simple. It supports them admirably. Most folks who are "unsure" if "relativity is right" are unclear on what either GR or SR actually say. It should be stressed that there is a point of view in which the main statements of SR and GR are really obvious things like "you can't see something going faster than light." Which is about as obvious as it gets if you think about it for three seconds.

It really is a very simple matter of whether you believe what you see or not.

EDG
2010-Mar-30, 08:27 AM
Another way to look at this is, if you can "jump" to a remote location faster than light can get there, then there exists a valid frame of reference in which you arrived before you started, and that is an explicit causality violation. Furthermore, if you can "move" faster than light relative to nearby objects, then there exists a valid frame of reference who will observe your time going backwards, which is also an explicit causality violation. These are simple geometric consequences of the finite speed of light.

I guess the part I'm fuzzy on is the "time goes backward" part. In what frame of reference could you arrive before you started?



The postulates of relativity are few and simple. It supports them admirably. Most folks who are "unsure" if "relativity is right" are unclear on what either GR or SR actually say. It should be stressed that there is a point of view in which the main statements of SR and GR are really obvious things like "you can't see something going faster than light." Which is about as obvious as it gets if you think about it for three seconds.

Well, I'm not doubting GR or SR here - I'm just trying to get my head around them :). In my example I don't think it's about seeing something going faster than light though, because I'm talking about the light from the ship at AC travelling at lightspeed back to the detector at Sol. I don't see a situation in my scenario where anything except the ship itself is travelling faster than light (and tecnnically it isn't - it's just going from A to B without travelling through the intervening space. It's actual velocity at either end could be a few m/s for all I know).


Though one question springs to mind - when people say "nothing travels faster than the speed of light", they mean "nothing travels faster than about 300,000 km/s", right? Could there be any situation where light itself can travel faster than 300,000 km/s (most likely due to weird things happening to space-time, I guess?)?

grant hutchison
2010-Mar-30, 09:43 AM
I guess the part I'm fuzzy on is the "time goes backward" part. In what frame of reference could you arrive before you started? In the frame of anyone who is in motion from Earth towards Alf Cen when you make the first jump, and anyone who is moving from Alf Cen towards Earth when you make the return jump. Such observers have a different measure of simultaneity from the one you're using in the (approximately) shared rest frame of your origin and destination.

Grant Hutchison

pzkpfw
2010-Mar-30, 07:27 PM
I think this is pretty much how it would work...




1. Spaceships A and B at rest, crew members c and d are aboard.

Ac


Bd


2. The ships start moving at some speed relative to each other.


Ac -->


<-- Bd

3. Both c and d now see/think/know/understand/calculate that
the clock of the other is lagging behind their own.

Ac -->


<-- Bd

c: "My clock says 10 pm but d's clock must be at 8 pm"
d: "My clock says 10 pm but c's clock must be at 8 pm"

4. Both c and d "instantly teleport" to the other ship.

Ad -->


<-- Bc

c: "The clock on this ship says 8 pm, I sort of went back in time!"
d: "The clock on this ship says 8 pm, I sort of went back in time!"

5. Both c and d see/think/know/understand/calculate that
the clock of the other ship (the one they came from)
is lagging behind the ship they are - now - aboard...

c: "The clock on this ship says 8 pm, the other must be at 6 pm."
d: "The clock on this ship says 8 pm, the other must be at 6 pm."

6. So when c and d "instantly teleport" back to their original ship....

Ac -->


<-- Bd

c: "Wow. I left my ship when its clock said 10 pm, and now
I'm back and it says 6 pm. I really did go back in time."
d: "Wow. I left my ship when its clock said 10 pm, and now
I'm back and it says 6 pm. I really did go back in time."



The conclusion is that "instant" teleportation (or any travel faster than light) can't be possible; even though a different scenario could be made that didn't raise these causality issues.

EDG
2010-Mar-31, 01:07 AM
OK. so instant doesn't work. What if the jump wasn't instant, but rather took as long as required to get the ship's timeframe in sync with the local timeframe of the destination (assuming that's even possible)?

Schneibster
2010-Mar-31, 01:32 AM
I guess the part I'm fuzzy on is the "time goes backward" part. In what frame of reference could you arrive before you started? The one in which you're going faster than the speed of light, of course.


Well, I'm not doubting GR or SR here - I'm just trying to get my head around them . Sure- and being able to imagine this is the first step.

It might make it easier to talk about rapidity instead of velocity. You can actually represent velocity as a four-vector in Minkowski spacetime, and if you do, then you can also represent it as a three-vector plus a rotation about one, two, or three invisible axes of rotation; the Lorentz transform allows you to make this conversion in the Minkowski case, and the geometric equivalent uses hyperbolic trigonometric functions and shows how to make the same conversion in that representation. The functions are the direct analogs of those used in calculating ordinary foreshortening caused by normal rotations. The three invisible axes are the axes of the xt, yt, and zt planes of rotation; the axes of the xy, yz, and xz planes are all explicitly visible, but the xt, yt, and zt axes are not. They can only be seen by their foreshortening effects, as specified by the Lorentz transform. Go google up "rapidity relativity" and see what you get; if it's not helpful, come back and I'll explain in more detail.


In my example I don't think it's about seeing something going faster than light though, because I'm talking about the light from the ship at AC travelling at lightspeed back to the detector at Sol. I don't see a situation in my scenario where anything except the ship itself is travelling faster than light (and tecnnically it isn't - it's just going from A to B without travelling through the intervening space. It's actual velocity at either end could be a few m/s for all I know). It doesn't matter what you see. What matters is that there's an allowed frame that sees what you think is perfectly OK as against causality. Nature doesn't permit that, as far as we can tell, and so far we haven't found an exception.


Though one question springs to mind - when people say "nothing travels faster than the speed of light", they mean "nothing travels faster than about 300,000 km/s", right? Could there be any situation where light itself can travel faster than 300,000 km/s (most likely due to weird things happening to space-time, I guess?)?No. It's simply not possible; why is a major line of research in current physics, but the first and most obvious answer appears to be "because nature doesn't permit causality violations." Keep in mind that the conservation laws-- mass/energy, momentum, angular momentum, for good examples-- are the mathematical result of the constancy of physical law across, respectively, spacetime location, spatial position, and angular position, by Noether's Theorem. This implies that the very dimensionality of the universe directly requires causal consistency, since the major physical definition of causality involves obedience to the conservation laws.

grav
2010-Mar-31, 01:58 AM
Actually, even if it were possible to travel faster than the speed of light, it still wouldn't make one go backward in time, so that is mostly just sci-fi folklore. An observer sees the time that passes on a ship drop toward zero with speeds approaching the speed of light with a time dilation for the ship of sqrt[1 - (v/c)^2]. At c, events that would occur on the ship stop completely and time appears frozen for the ship to all observers in the universe, since all observers would measure the same speed of c for the ship. Since time has stopped upon the ship, there is no mechanical way for the ship observers to make the ship accelerate to a greater speed since the ship observers and all controls are frozen in time at this point. However, even if the ship were somehow made to go faster, then the time dilation would become the square root of a negative number, so complex or imaginary. In order to go backwards in time, the observed time dilation of the ship would have to be negative, not complex. A complex result means that it is not possible to travel faster than c to begin with, so such a result generally just signifies an impossibility.

Schneibster
2010-Mar-31, 02:03 AM
Actually, even if it were possible to travel faster than the speed of light, it still wouldn't make one go backward in time, so that is mostly just sci-fi folklore. Actually, what it would do is make there exist a frame that is not an FTL frame that would perceive one going backward in time, and it's neither sci-fi nor folklore, but simple geometric fact if you know anything about relativity at all.

EDG
2010-Mar-31, 03:12 AM
It might make it easier to talk about rapidity instead of velocity. You can actually represent velocity as a four-vector in Minkowski spacetime, and if you do, then you can also represent it as a three-vector plus a rotation about one, two, or three invisible axes of rotation; the Lorentz transform allows you to make this conversion in the Minkowski case, and the geometric equivalent uses hyperbolic trigonometric functions and shows how to make the same conversion in that representation. The functions are the direct analogs of those used in calculating ordinary foreshortening caused by normal rotations. The three invisible axes are the axes of the xt, yt, and zt planes of rotation; the axes of the xy, yz, and xz planes are all explicitly visible, but the xt, yt, and zt axes are not. They can only be seen by their foreshortening effects, as specified by the Lorentz transform. Go google up "rapidity relativity" and see what you get; if it's not helpful, come back and I'll explain in more detail.

I appreciate the attempt to be helpful, but this was pretty much incomprehensible to me :). I did actually try to look up "rapidity" but even the wiki page on that was pretty impenetrable to me. The language people use when discussing relativity is all too often too dense for me to follow well.


No. It's simply not possible; why is a major line of research in current physics, but the first and most obvious answer appears to be "because nature doesn't permit causality violations." Keep in mind that the conservation laws-- mass/energy, momentum, angular momentum, for good examples-- are the mathematical result of the constancy of physical law across, respectively, spacetime location, spatial position, and angular position, by Noether's Theorem. This implies that the very dimensionality of the universe directly requires causal consistency, since the major physical definition of causality involves obedience to the conservation laws.

So what about quantum entanglement? It seems there's a contradiction between the quantum scale and relativity - if one entangled particle can instantaneously 'flip' to another state because its partner thousands of miles away has changed, and if this can happen over the scale of lightyears and still be instantanous (I don't know if it can - can it?) then it seems that causality is being violated right there.

Jens
2010-Mar-31, 05:25 AM
So what about quantum entanglement? It seems there's a contradiction between the quantum scale and relativity - if one entangled particle can instantaneously 'flip' to another state because its partner thousands of miles away has changed, and if this can happen over the scale of light years and still be instantaneous (I don't know if it can - can it?) then it seems that causality is being violated right there.

I don't know the official position, but I can tell you for sure that this is not necessarily true. I can give you another possibility -- though I think it is also wrong -- where it could happen without causality being violated. The two particles could have been destined to flip at that time. But again, I don't understand entanglement.

grav
2010-Mar-31, 06:01 AM
Actually, what it would do is make there exist a frame that is not an FTL frame that would perceive one going backward in time, and it's neither sci-fi nor folklore, but simple geometric fact if you know anything about relativity at all.Okay, I'll bite. Let's say that to an observer on the ground, a ship can travel faster than c, say 1.1 c. What relative speed will the observer on the ship say the ground observer is moving? Will each say the other's clock and all physical processes are going backwards in time? Also, what do you mean by "there exist a frame that is not a FTL frame that would perceive one going back in time"?

Schneibster
2010-Mar-31, 12:47 PM
I appreciate the attempt to be helpful, but this was pretty much incomprehensible to me :). Well, thanks for being nice about it. :D

OK, let's try this a different way. You know what two dimensions is like, right? Flatland. Now, if you make a circle, where does its axis go? Out of flatland, right? This is a general principle. The axes of geometric figures need not exist within the space that contains the figures. Now, if we go to three dimensions, well, we have axes that appear within those three dimensions for any geometric figure, right? So we have a chauvanism, or bias, or whatever you want to call it, that such axes just simply have to exist within the dimensions we're already talking about. But we forget something important if we do this.

Time is a dimension, too, right? We can speak of a length of time, and that makes sense, right? So here's a question: what do you think the speed of time is?

OK, now let's talk about rotation. Most people think about these axes. We've seen how that winds up. Physicists, however, think about rotations in terms of the plane of rotation; for example, the xy plane. That's our flatland example from above. Who cares what the axis is? We're interested in the plane. Now, how many planes of rotation are there in a manifold? (A manifold is the thing that has dimensions; it's generalized, it can have any number, like two, or five, or twenty-three. That's not an exact definition, but it's good enough for this). The answer is, the number of combinations of two dimensions. So for a two dimensional manifold, there's only one plane of rotation; but for a three-dimensional one, there's three. And most importantly, in our four-dimensional universe, there's six. As I said in my previous post, for three dimensions, there's xy, yz, and xz; and for four, you add to that xt, yt, and zt.

We'll take this a couple steps at a time. Once you have the above down, we'll move on. Ask me any questions you have about the speed of time, and about this concept of planes of rotation.


I did actually try to look up "rapidity" but even the wiki page on that was pretty impenetrable to me. The language people use when discussing relativity is all too often too dense for me to follow well.I've learned to start with the more advanced stuff, it cuts to the chase if you're talking to someone knowledgeable and it helps place someone's level if not so you can communicate effectively with them too. Don't worry about it, you'll understand it all in just a little bit. I know of a way to think about it that's a lot easier than the math.

Rapidity is just a word. It's terminology. If you don't know it, then you will when we're done. Start with the above on the speed of time and on planes of rotation.


So what about quantum entanglement? Quantum entanglement is irrelevant at this point. We're talking about relativity. If you change the subject in the middle of the conversation, you're sure to get confused.


It seems there's a contradiction between the quantum scale and relativity - if one entangled particle can instantaneously 'flip' to another state because its partner thousands of miles away has changed, and if this can happen over the scale of lightyears and still be instantanous (I don't know if it can - can it?) then it seems that causality is being violated right there.Whether or not that's actually happening is an open question; it's one I have an opinion on, one shared by the majority of physicists, but that's neither here nor there in a conversation about relativity.

Schneibster
2010-Mar-31, 12:53 PM
Okay, I'll bite. No, you won't. It is generally considered impolite to contradict someone directly, and in addition, it's usually wisest to determine the strength of the opposition prior to engaging in battle. Sometimes it's a good idea to try to cooperate rather than to fight, and once you've been impolite it's much more difficult. I suggest you give this serious thought.


Let's say that to an observer on the ground, a ship can travel faster than c, say 1.1 c. Let's not. Nothing can go faster than light, and that's a postulate of relativity. Period. Your example doesn't work. It's unphysical. We've never seen anything behave as you postulate here, and we've seen lots of things behave the way we'd expect if, in fact, the speed of light is finite and maximal. All the evidence we have (and it's a considerable amount) says it's impossible, period, end of conversation.

grant hutchison
2010-Mar-31, 01:18 PM
No, you won't. It is generally considered impolite to contradict someone directly, and in addition, it's usually wisest to determine the strength of the opposition prior to engaging in battle. Sometimes it's a good idea to try to cooperate rather than to fight, and once you've been impolite it's much more difficult. I suggest you give this serious thought.I think grav was simply trying to enter into a dialogue about something that wasn't intuitively obvious to him. He likes worked examples. From long acquaintance, I very much doubt if he was trying to contradict you. :)

Grant Hutchison

grav
2010-Mar-31, 03:11 PM
I think grav was simply trying to enter into a dialogue about something that wasn't intuitively obvious to him. He likes worked examples. From long acquaintance, I very much doubt if he was trying to contradict you. :)

Grant HutchisonRight, thanks Grant. Although I will sometimes debate certain stances and otherwise just ask questions in case I am missing something about the perspective of what the other poster is saying, I was trying not to put put anyone on the spot with my original post in this case by not quoting anyone directly, so just making a general statement about the "time goes backward" part that was mentioned, although which posts it stems from could still be inferred I suppose. In any case, time going backward is a mathematical impossibility, so that's what I am looking at, and I have yet to see any actual mathematics worked out that says otherwise, so it would be interesting to see someone attempt it, even allowing as a given that a ship can travel faster than c.

It's good to see you around again. It's been a while since I've seen you on the board. :)

grant hutchison
2010-Mar-31, 03:23 PM
In any case, time going backward is a mathematical impossibility ...What a bizarre thing to say. Time reverses very easily, mathematically.
And you should recall that in this case you're plugging in a physical impossibility under SR, in the form of faster than light travel, so you shouldn't be surprised to get strange stuff out again.

If you draw yourself a Minkowski diagram, it's pretty easy to show yourself that the order of events along the worldline of an FTL traveller can be reversed by a suitable choice of frame for an external observer. It's all part of the pattern of causality violation: if I observe you to arrive at your destination before you depart, then I also must observe your time running backwards during your journey.

Grant Hutchison

EDG
2010-Mar-31, 03:45 PM
We'll take this a couple steps at a time. Once you have the above down, we'll move on. Ask me any questions you have about the speed of time, and about this concept of planes of rotation.

Thanks, all that was much easier to understand - I'm following you so far :).



Quantum entanglement is irrelevant at this point. We're talking about relativity. If you change the subject in the middle of the conversation, you're sure to get confused.

Well, I think it's relevant in that it seems to directly contradict what relativity predicts. But let's get to that eventually.[/QUOTE]

grav
2010-Mar-31, 04:57 PM
What a bizarre thing to say. Time reverses very easily, mathematically.Well, you have taken a stance there that I don't intuitively see, so of course you know I am going to debate it in order to draw out the conclusions on a level in which it can be reckoned with. :)


And you should recall that in this case you're plugging in a physical impossibility under SR, in the form of faster than light travel, so you shouldn't be surprised to get strange stuff out again.Right.


If you draw yourself a Minkowski diagram, it's pretty easy to show yourself that the order of events along the worldline of an FTL traveller can be reversed by a suitable choice of frame for an external observer. It's all part of the pattern of causality violation: if I observe you to arrive at your destination before you depart, then I also must observe your time running backwards during your journey.Well, that is a different given than a ship simply travelling at a speed greater than c. If a ship were to arrive at a destination before it departs, which again is a different given altogether, then the ship would have to be at travelling faster than even an infinite speed to the observer if the observer's clock runs forward, but which would have the effect of rolling things back around to a negative speed for the ship, since travelling in a positive direction over a negative time on the observer's clock will give a negative speed. That will result in a positive time dilation for the ship if the ship travels at a speed between -c and 0, but still gives a complex time dilation if it travels at a speed beyond -c, which indicates its impossibility. So taking a speed for the ship to be between -c and 0, the time dilation will be real and positive, and multiplying that by the negative time that the observer measures between the events, the time that has passed for the ship in this case will be real and negative, so the ship will have arrived at the destination before it left from the perspective of both the observer and the ship, yes.

Okay, so that did give exactly what you claimed in that case with the particular given supplied, so is the closest I've seen so far for that, but it also requires a negative speed for the ship over a positive direction travelled. The problem I see with that is that we can have a positive speed for a positive direction travelled or a negative speed for a negative direction travelled when using positive time passing for the observer, but a negative speed for a positive direction travelled has no physical meaning, even less so than considering a positive speed greater than c when otherwise disregarding what SR would say about that. If we were to reverse the time that passes on the observer's clock also, however, then we could have a positive speed in a positive direction or negative speed in a negative direction, either way, so everything proceeds normally when looking at it from the ship's and observer's points of view, but we, taking a general overview of the situation while watching the ship's clock run backward in time as it did before but with the observer's clock running backward also, would say that it is just the same as rewinding a videotape, where all clocks will be seen to run backwards and the events occur in reverse.

Grey
2010-Mar-31, 04:59 PM
So what about quantum entanglement? It seems there's a contradiction between the quantum scale and relativity - if one entangled particle can instantaneously 'flip' to another state because its partner thousands of miles away has changed, and if this can happen over the scale of lightyears and still be instantanous (I don't know if it can - can it?) then it seems that causality is being violated right there.The answer appears to be that, if one makes a few basic assumptions that seem reasonable, this does violate causality, but since it's always concealed within quantum randomness, there is no way to use this to send or receive a message faster than light. Some other possibilities are that maybe the reasonable assumptions are not so reasonable. One of those is similar what Jens said: if you assume that all events are preordained, then there's no need for any kind of superluminal effects.

grant hutchison
2010-Mar-31, 05:16 PM
Well, you have taken a stance there that I don't intuitively see, so of course you know I am going to debate it in order to draw out the conclusions on a level in which it can be reckoned with. :)It doesn't excuse you for making bizarre claims about time being mathematically irreversible.


Well, that is a different given than a ship simply travelling at a speed greater than c. If a ship were to arrive at a destination before it departs, which again is a different given altogether ...No, it all ties together very simply.
You observe a ship moving at 1.1c. It will take 1 year to travel from the start to the finish of a 1.1 light-year journey, in your coordinates. It will therefore arrive after it departs (in the conventional manner), in your coordinates.
But (using your knowledge of the relativity of simultaneity) you are immediately able to identify an observer in whose coordinates the ship will arrive before it departs. That's anyone whose line of simultaneity slopes more steeply than 1 year per 1.1 light-years, in your coordinates. That's anyone who is moving in the same direction as the FTL ship, at more than 1 light-year per 1.1 years, or ~0.91c, in your coordinates.
And that's exactly what Schneibster was telling you with "what it would do is make there exist a frame that is not an FTL frame that would perceive one going backward in time". As soon as you see FTL travel, you can immediately find a conventional (ie, non-FTL) frame in which the traveller arrives before he departs.

Grant Hutchison

grav
2010-Mar-31, 05:36 PM
No, it all ties together very simply.
You observe a ship moving at 1.1c. It will take 1 year to travel from the start to the finish of a 1.1 light-year journey, in your coordinates. It will therefore arrive after it departs (in the conventional manner), in your coordinates.
But (using your knowledge of the relativity of simultaneity) you are immediately able to identify an observer in whose coordinates the ship will arrive before it departs. That's anyone whose line of simultaneity slopes more steeply than 1 year per 1.1 light-years, in your coordinates. That's anyone who is moving in the same direction as the FTL ship, at more than 1 light-year per 1.1 years, or ~0.91c, in your coordinates.
And that's exactly what Schneibster was telling you with "what it would do is make there exist a frame that is not an FTL frame that would perceive one going backward in time". As soon as you see FTL travel, you can immediately find a conventional (ie, non-FTL) frame in which the traveller arrives before he departs.

Grant HutchisonOkay right, that's true, and that is certainly a unique perspective for looking at it that I hadn't considered, so thanks for explaining what was meant by that. Nevertheless, it still just results in the same thing that was found in the last post. If the ship is travelling at v1 = 1.1 c to the first observer and a second observer is travelling at a speed greater than v = c / 1.1 to the first observer in the same direction as the ship, then the second observer will measure the relative speed of the ship to be v2 = (v1 - v) / (1 - v1 v / c^2) = (1.1 c - v) / (1 - (1.1 c) v / c^2), which for any speed that is greater than v = c / 1.1 (and less than c) will give a negative speed for the ship in the positive direction, so reduces to the same scenario as in the last post as the second observer measures the events taking place.

Looking at it from the perspective of the first observer, though, we can now see that it still represents a mathematical impossibility because if one year passes for the first observer when the ship is travelling at 1.1 c, then sqrt[1 - (v1/c)^2] (one year) gives a complex result for the time that has passed on the ship's clock when it reaches the destination and that is what an observer that coincides in the same place with its arrival must read upon the clock of the ship when it gets there, so the second observer must also agree upon this.

EDG
2010-Mar-31, 05:38 PM
One of those is similar what Jens said: if you assume that all events are preordained, then there's no need for any kind of superluminal effects.

Why would I want to assume that all events are preordained? Isn't that getting more into religious discussion?

Grey
2010-Mar-31, 06:08 PM
Why would I want to assume that all events are preordained? Isn't that getting more into religious discussion?I'm just saying that is one way out of needing to assume nonlocality. Bell's theorem (which is the proof that no purely local model can account for the quantum behavior we observer) assumes something called "contrafactual definiteness" or CFD. CFD is the assumption that it's meaningful to talk about things that did not happen, but that might have happened. For example, although we performed measurement A, we imagine that we could just as well have performed measurement B, and that had we done so we would have gotten results consistent with all the other measurements we've made (although we generally don't know exactly what those results would have been). If we decide to deny CFD (for example, by assuming that we never really had any choice in what measurement to perform), we can avoid the conclusion that any mechanism underlying the quantum facts must be nonlocal.

grant hutchison
2010-Mar-31, 06:15 PM
Okay right, that's true, and that is certainly a unique perspective for looking at it that I hadn't considered, so thanks for explaining what was meant by that. Nevertheless, it still just results in the same thing that was found in the last post. If the ship is travelling at v1 = 1.1 c to the first observer and a second observer is travelling at a speed greater than v = c / 1.1 to the first observer in the same direction as the ship, then the second observer will measure the relative speed of the ship to be v2 = (v1 - v) / (1 - v v1 / c^2) = (1.1 c - v) / (1 - (1.1 c) v / c^2), which for any speed that is greater than v = c / 1.1 (and less than c) will give a negative speed for the ship, so reduces to the same scenario as in the last post as the second observer measures the events taking place.The "negative speed" is simply a sign that the FTL traveller arrives before he departs: he seems to travel in the opposite direction. But during that journey, events along his timeline will be time-reversed, according to our second observer.
The first observer sees our FTL traveller depart from nearby (call that spacetime event "cause"), and arrive far away (call this spacetime event "effect"). The second observer sees the FTL traveller apparently depart from far away (but this is the "effect" event) and arrive nearby (but this is the "cause" event).


Looking at it from the perspective of the first observer, though, we can now see that it still represents a mathematical impossibility because if one year passes for the first observer when the ship is travelling at 1.1 c, then sqrt[1 - (v1/c)^2] (one year) gives a complex result for the time that has passed on the ship's clock when it reaches the destination and that is what an observer that coincides in the same place with its arrival must read upon the clock of the ship when it gets there, so the second observer must also agree upon this.Which is what you get if you use the equations of special relativity on a situation which is specifically excluded by special relativity. Garbage in, garbage out.

grant hutchison
2010-Mar-31, 06:22 PM
Why would I want to assume that all events are preordained? Isn't that getting more into religious discussion?
According to Victor Stenger, you could instead assume that the arrow of time just doesn't operate at the quantum level, but is emergent as part of decoherence. If distant detectors are able to "talk back" to the source at the quantum level, thereby subverting the macroscopic arrow of time, then the spooky entanglement stuff becomes less spooky (although at the expense of having a spooky timeless domain underlying "our" reality). Or so says Stenger. :)

Grant Hutchison

grav
2010-Mar-31, 06:31 PM
The "negative speed" is simply a sign that the FTL traveller arrives before he departs: he seems to travel in the opposite direction. But during that journey, events along his timeline will be time-reversed, according to our second observer.
The first observer sees our FTL traveller depart from nearby (call that spacetime event "cause"), and arrive far away (call this spacetime event "effect"). The second observer sees the FTL traveller apparently depart from far away (but this is the "effect" event) and arrive nearby (but this is the "cause" event).Right.


Which is what you get if you use the equations of special relativity on a situation which is specifically excluded by special relativity. Garbage in, garbage out.Yes, I agree. I have been trying to look into precisely where the contradiction took place, since all observers should agree upon the clock reading of the ship when it reaches the destination, but the only conclusion I have gained so far is that it is a clear indication that the equations of SR are simply not meant to be used in this way.

Thanks Grant. This has been a very interesting discussion indeed. :)

grav
2010-Mar-31, 09:05 PM
Alright, I finally figured out where the contradiction came from. There isn't one. We found before that if the speed of the ship as measured by the second observer is between 0 and -c, then a normal time dilation will be observed upon on the clock of the ship by the second observer, so if the second observer says a negative time has passed between the two events of the ship leaving and the ship arriving at its destination, then a negative time has passed upon the ship's clock also. However, if the speed of the ship as measured by the second observer is beyond -c, then the clock reading of the ship will be complex. According to the first observer, the clock reading of the ship will always be complex when it reaches its destination, and both the first and second observers will agree that the reading is complex when the speed of the ship as measured by the second observer is beyond -c.

Okay, so the question becomes what relative speed v would the second observer have to have to the first observer in order for the second observer to measure a speed for the ship between 0 and -c? Well, the addition of speeds gives v = (v1 - v2) / (1 - v1 v2 / c^2), so using v1 = 1.1 c and v2 = 0 to -c, let's plug in some actual numbers and find out what v must be.


v2 v

0 1.1 c
-.2 c (1.3 c) / (1 - (-.22)) = 1.3 c / 1.22
-.4 c (1.5 c) / (1 - (-.44)) = 1.5 c / 1.44
-.6 c (1.7 c) / (1 - (-.66)) = 1.7 c / 1.66
-.8 c (1.9 c) / (1 - (-.88)) = 1.9 c / 1.88
-.9999 c (2.0999 c) / (1 - (-1.09989)) = 2.0999 c / 2.09989

So as we can see here, the second observer will never measure a negative speed for the ship between 0 and -c unless the second observer also travels faster than c to the first observer. In this case, according to the first observer, a complex time will pass on the clock of the second observer also, so the second observer must agree that a complex time has passed upon the second observer's own clock. If we were to plug in a relative speed between the first and second observer that is less than c, then for v = .95 c, for example, we get v2 = -3.333 c, and for v = .91 c, we get v2 = -190 c, both of which give complex times for the clock reading of the ship according to the second observer also. So even though the time dilation that the second observer sees of the ship is real and not complex in itself when measuring the ship to have a relative speed to the second observer between 0 and -c, since the second observer's own clock reads a complex time that has passed when travelling faster than c to the first observer, that will still make the time that has passed upon the ship's clock complex as well. As a result, there is no set of SR observers in any frame that will ever read a negative time that has passed upon a clock, and travelling faster than light won't cause this to happen.

grant hutchison
2010-Mar-31, 10:00 PM
So as we can see here, the second observer will never measure a negative speed for the ship between 0 and -c unless the second observer also travels faster than c to the first observer.You're misapplying addition of velocities. All you're working out there is that the first observer can never see the second observer catch up with the ship, unless the second observer goes faster than the ship. Hardly a surprise.
This doesn't affect the reversal of cause and effect in the frame of the second observer. Draw a diagram.

Grant Hutchison

EDG
2010-Mar-31, 10:01 PM
According to Victor Stenger, you could instead assume that the arrow of time just doesn't operate at the quantum level, but is emergent as part of decoherence. If distant detectors are able to "talk back" to the source at the quantum level, thereby subverting the macroscopic arrow of time, then the spooky entanglement stuff becomes less spooky (although at the expense of having a spooky timeless domain underlying "our" reality). Or so says Stenger. :)
Grant Hutchison

Sounds interesting... do you have any links to any more detailed discussion of that idea?

grant hutchison
2010-Mar-31, 10:08 PM
Sounds interesting... do you have any links to any more detailed discussion of that idea?Here (http://www.colorado.edu/philosophy/vstenger/) is Stenger's web-page, which consists mainly of links. I don't know if there are any web discussions of this idea, but he talks about it in several of his books, probably most extensively in Timeless Reality.

Grant Hutchison

grav
2010-Mar-31, 10:15 PM
You're misapplying addition of velocities. All you're working out there is that the first observer can never see the second observer catch up with the ship, unless the second observer goes faster than the ship.The second observer doesn't catch up to the ship. In order to gain a real and positive time dilation that the second observer will measure for the ship, then according to the perspective of the first observer, if the ship is travelling away at 1.1 c, then the second observer must be travelling away at a relative speed between c and 1.1 c, but then the time that passes for the second observer will be complex, so when multiplied by the real and positive time dilation, gives a complex time that has passed upon the clock of the ship as well. If the second observer travels away from the first observer at less than c, then the second observer will still measure a complex time passing for the ship's clock as the first observer does. Either way, a complex time will pass upon any clock that travels faster than c according to all observers.

grant hutchison
2010-Mar-31, 10:21 PM
... if the ship is travelling away at 1.1 c, then the second observer must be travelling away at a relative speed between c and 1.1 c.I don't really know what you're attempting, here, but this is certainly wrong if you're seeking the conditions for time reversal seen by the second observer. You're getting the wrong answer because the velocity addition formula doesn't apply correctly to superluminal velocities: it isn't intended for that purpose. The velocity addition formula just doesn't handle the time reversal the second observer experiences. The correct answer is c/1.1 to c. Draw a diagram.

Grant Hutchison

EDG
2010-Mar-31, 11:01 PM
Also, keep in mind that my question wasn't about anyone travelling (i.e. actually moving through spacetime) at FTL speeds. It was about 'jumping' from one place and arriving instantaneously at another several lightyears away. Though I guess you could say that the ship is moving from A to B in zero time.

grant hutchison
2010-Mar-31, 11:15 PM
Also, keep in mind that my question wasn't about anyone travelling (i.e. actually moving through spacetime) at FTL speeds. It was about 'jumping' from one place and arriving instantaneously at another several lightyears away. Though I guess you could say that the ship is moving from A to B in zero time.As a corollary to this business of always finding an observer who sees cause and effect reversed for an FTL signal, there's always an observer who sees instantaneous signal transmission. And if signals are instantaneous in one frame, there are always observers who find that emission happens after reception, and observers who find that emission happens before reception. So it's all one big causality-violating deal. As Feinberg (http://adsabs.harvard.edu/abs/1967PhRv..159.1089F) pointed out long ago, this means that you can't really tell the emission of a tachyon from the reception of a tachyon: each are both, as my friend Dave is wont to say.

Grant Hutchison

Van Rijn
2010-Mar-31, 11:28 PM
Also, keep in mind that my question wasn't about anyone travelling (i.e. actually moving through spacetime) at FTL speeds.


If nobody is going at FTL speeds, there is no problem per relativity. If that's the case, though, then if a spacecraft leaves the vicinity of Earth and goes to Alpha Centauri, at, say, .9999999999999999999 C using some super space drive, the earliest anyone near Earth (with, say, a giant telescope) could see the spacecraft near Alpha Centauri would be about 8.6 years after it left. That is, enough time for it to get there, not going faster than light, and enough time for the light to get back to the telescope.

According to the people on the ship, the flight would be instant (or so close as not to matter). To other observers, though, it wouldn't be instant.




It was about 'jumping' from one place and arriving instantaneously at another several lightyears away. Though I guess you could say that the ship is moving from A to B in zero time.

But, again, this is "instantly" according to . . . ?

grav
2010-Apr-01, 12:19 AM
I don't really know what you're attempting, here, but this is certainly wrong if you're seeking the conditions for time reversal seen by the second observer. You're getting the wrong answer because the velocity addition formula doesn't apply correctly to superluminal velocities: it isn't intended for that purpose. The velocity addition formula just doesn't handle the time reversal the second observer experiences. The correct answer is c/1.1 to c. Draw a diagram.

Grant HutchisonOkay, here is a diagram of sorts. To better demonstrate, I am changing the scenario somewhat. The ship is now just a clock that travels away from the first observer A at 1.1 c. The second observer is now two observers, B and D, at each end of a ship that travels away from the first observer A at v = .91 c. A, B at the back of the ship, and the moving clock all coincide in the same place where they start out at the top of the diagram, and their times are all synchronized to T = 0. The rest frame of A and E measures the length of the ship to be d_BD, so B and D measure it as d_BD' = d_BD / sqrt[1 - (v/c)^2]. From the perspective of this rest frame of A and E, D's clock is simultaneity shifted to T_D = - d_BD' v / c^2. At the bottom of the diagram, the moving clock and D arrive at the destination at E at the same time so that they coincide in the same place at E.

So first let's find the length of the ship. According to the rest frame, the moving clock travels the distance d_AE = v1 t_AE in the same time that the ship travels the distance v t_AE when the front end of the ship coincides with E, so v1 t_AE = v t_AE + d_BD, giving d_BD = (v1 - v) t_AB. The length of the ship as the ship observers measure it, then, is d_BD' = (v1 - v) t_AB / sqrt[1 - (v/c)^2]. According to the rest frame, d_AE = 1.1 light years, so the time that the moving clock has travelled is t_AB = 1 year. The length of the ship as the ship observers measure it, then, is d_BD' = (1.1 c - .91 c) (1 year) / sqrt[1 - (.91)^2] = .45826 lt-yr.

Alright, the simultaneity shift found along the length of the ship according to the rest frame is tl = d_BD' v / c^2 = .41702 yr. That means that when the moving clock, A, and B start off with clocks synchronized to T=0, D's clock reads T_D = - .41702 yr. The time that passes in the rest frame for the moving clock and D to reach E is 1 year, so sqrt[1 - (v/c)^2] (1 yr) passes upon D's clock, so that it reads T_D' = - .41702 yr + sqrt[1 - (.91)^2] (1 yr) = -2.4119 * 10^(-3) yr when it reaches D, so that is the time that the ship observers say has passed. According to the clocks of the ship observers, then, the moving clock has reached its destination before it left. So that much is true, and is an astonishing result in itself.

Now let's look at the time that A and the ship observers will say has passed for the moving clock. A says the time that passes upon the moving clock will just be sqrt[1 - (v1/c)^2] t_AE = sqrt[1 - (1.1)^2] (1 yr) = +/- .4583 i yr. The ship observers say that the clock has just travelled from one side of the ship to the other in a time of t_BD = -2.4119 * 10^(-3) yr, so with a relative speed of v2 = d_BD' / t_BD = (.45826 lt-yr) / (-2.4119 * 10^(-3) yr) = -190 c. This can also be found with the addition of relative speeds with v2 = (v1 - v) / (1 - v1 v / c^2) = (1.1 c - .91 c) / (1 - (1.1) (.91)) = -190 c. The time dilation for the moving clock according to the ship, then, is sqrt[1 - (-190)^2] = +/- 189.997 i, so the time that the ship says passes for the clock is (+/- 189.997 i) (-2.4119 * 10^(-3) yr) = +/- .45826 i yr, the same as the rest frame measures. So any clock that travels faster than c will have a complex reading for the time when it reaches its destination, so no SR observer in any frame will ever measure a negative time to have passed upon a clock.

However, I must correct myself somewhat here. In my original post, I said "no SR observers" will read a negative time that passed upon a clock because I couldn't tell at the time what the observer that travels faster than c with the moving clock would themselves measure, having complex time upon their own clock and all, and could only see that the time dilation will be complex for that of the SR observers as the observer that travels faster than c would measure it, and of course any other non-SR observer that travels faster than c would still measure a complex time of another clock that travels faster than c also, and also that the non-SR observer cannot travel away and back to the same clock and measure a negative time that has passed for that clock. But now I can see that the observer that travels with the moving clock starts at B when B's clock reads T=0 and arrives at D's clock when D's clock reads T=-2.4119 * 10^(-3) yr, so due to the simultaneity shift along the length of the ship, that could be considered a negative time passing upon the clocks of the ship as the non-SR observer would measure them. I'm still not sure about that, though, because since a complex time passes for the non-SR observer, it is hard to tell whether that could be considered time going forward or backwards or neither for that observer, so it difficult to tell whether a non-SR observer would really consider that the ship's time even goes backwards in relation to his own time.

grav
2010-Apr-01, 12:44 AM
Okay, so let's see. According to a rest frame, an observer cannot travel away and back to the same point and arrive before they left, but they can travel to some distant destination faster than c as another frame would measure the speed and arrive at the destination before they left according to what the clocks of the rest frame will read due to the relativity of simultaneity, although the clock of the observer that travelled faster than c will read a complex time when that observer arrives, which has no meaning, but the scenario still grasps the general concept about arriving before one left as viewed from the rest frame. Does that pretty much sum it up?

Schneibster
2010-Apr-01, 03:24 AM
I think grav was simply trying to enter into a dialogue about something that wasn't intuitively obvious to him. He likes worked examples. From long acquaintance, I very much doubt if he was trying to contradict you. :)I won't contradict you. ;)

Schneibster
2010-Apr-01, 03:45 AM
Right, thanks Grant. Although I will sometimes debate certain stances and otherwise just ask questions in case I am missing something about the perspective of what the other poster is saying, I was trying not to put put anyone on the spot with my original post in this case by not quoting anyone directly, so just making a general statement about the "time goes backward" part that was mentioned, although which posts it stems from could still be inferred I suppose. Well, the statement was not "time goes backward" but "time appears to go backward." Second, it's probably not a good idea to compare something someone said with speculative fiction on a forum that's about errors and misunderstandings of known fact and mainstream theory. It has a tendency to appear aggressive. But in any case, I accept your statement that it was not your intent to be so pushy.


In any case, time going backward is a mathematical impossibility, so that's what I am looking at, and I have yet to see any actual mathematics worked out that says otherwise, so it would be interesting to see someone attempt it, even allowing as a given that a ship can travel faster than c.In fact, the time reversibility of physics is a postulate from which the Fluctuation Theorem is derived, and the Second Law of Thermodynamics is proven by the Fluctuation Theorem. That's pretty much iron-clad evidence of time-reversal symmetry in quantum mechanics, as well as thermodynamics. You will find extensive articles on time-reversal symmetry in Wikipedia, along with proof of time-reversal asymmetry in certain interactions among high-energy particles in accelerators, for example kaons and B-mesons; these are also referred to as CP-asymmetric interactions, since CP-symmetry is dual to T-symmetry under the overlying Poincaire symmetry of relativity.

In addition, if you study the "shut up and calculate" interpretation of quantum mechanics, that is, Feynman's sum over paths approach, you will find that Feynman states that a positron is simply an electron with a reversed time direction. This is fairly well known and is mentioned both in Q.E.D.: The Strange Theory of Light and Matter, Princeton, 1988, and one or more of the Five Easy Pieces books. Some later physicists tried to gloss over it, but a couple decades went by and there it is again, to bedevil a new generation. You'll also find that it's pretty important to the Cramer "Transactional interpretation" of QM, because of Feynman and Wheeler's earlier work, but the above is more than sufficient to illustrate that time reversal is common in high energy particle physics, quantum mechanics, and relativity.

Schneibster
2010-Apr-01, 03:52 AM
That's anyone who is moving in the same direction as the FTL ship, at more than 1 light-year per 1.1 years, or ~0.91c, in your coordinates.
And that's exactly what Schneibster was telling you with "what it would do is make there exist a frame that is not an FTL frame that would perceive one going backward in time". As soon as you see FTL travel, you can immediately find a conventional (ie, non-FTL) frame in which the traveller arrives before he departs.Precisely. Thank you.

Schneibster
2010-Apr-01, 04:07 AM
Thanks, all that was much easier to understand - I'm following you so far :). Excellent. OK, so here's the next little bit.

If we take it as a postulate that time is a dimension, just like height, width, and depth are, then we have four dimensions, and six planes of rotation; as I said earlier, the three from three dimension, xy, yz, xz, plus the other three introduced by the extra time dimension, xt, yt, and zt. Now, because of the character of our universe, we don't actually see time as a dimension like width. So we can't actually directly observe xt, yt, and zt as we can xy, yz, and xz. But if something rotates in those planes, we can tell it has, for two reasons. First, it foreshortens, just like if you were looking at a stick held sideways in front of you, and someone rotated it around an axis that went straight up and down- that is, rotated it in a plane that's flat and parallel to the line between your eyes and to the line of your vision. When the stick is parallel to the line between your eyes, it is at maximum length. When it's parallel to your line of vision, it is at minimum length. In exactly the same way, when something rotates in xt, yt, zt, or some combination of them, it appears to have a different length. But that length is a matter of actual fact in our coordinates, because the rotation is about invisible axes. We can't see them. But we can see that something's rotated in them, because it's foreshortened. That's the Lorentz-FitzGerald contraction. That's geometrically what causes it. This is not a theory, it is not open to question. As a matter of actual experimentally verifiable fact that is precisely and exactly what we really see.

Now, I asked earlier what the speed of time is, and now we have an answer. The speed of time is the speed of light.

OK, let's make sure you're comfortable with that part before we move on.


Well, I think it's relevant in that it seems to directly contradict what relativity predicts. But let's get to that eventually.That's the right way, I think.

EDG
2010-Apr-01, 04:50 AM
In exactly the same way, when something rotates in xt, yt, zt, or some combination of them, it appears to have a different length. But that length is a matter of actual fact in our coordinates, because the rotation is about invisible axes. We can't see them. But we can see that something's rotated in them, because it's foreshortened. That's the Lorentz-FitzGerald contraction. That's geometrically what causes it. This is not a theory, it is not open to question. As a matter of actual experimentally verifiable fact that is precisely and exactly what we really see.

So things going fast enough have their spatial dimensions compressed because they're rotating around a space-time axis that we can't see?



Now, I asked earlier what the speed of time is, and now we have an answer. The speed of time is the speed of light.

I don't think you actually mentioned the speed of light earlier, so you seem to have pulled this statement from out of nowhere. Also, I'm not sure what you mean by "the speed of time".

grant hutchison
2010-Apr-01, 03:05 PM
Okay, so let's see. According to a rest frame, an observer cannot travel away and back to the same point and arrive before they left, but they can travel to some distant destination faster than c as another frame would measure the speed and arrive at the destination before they left according to what the clocks of the rest frame will read due to the relativity of simultaneity, although the clock of the observer that travelled faster than c will read a complex time when that observer arrives, which has no meaning, but the scenario still grasps the general concept about arriving before one left as viewed from the rest frame. Does that pretty much sum it up?No.
As usual, my brain shuts down about a third of the way through your enormous run-on sentences, especially the ones with embedded plain-text equations. I honestly just can't read that stuff.
But it's quite possible for a pair of EDG_'s "instantaneous" jumps to bring a traveller back to a time before he left, if they operate according to the simultaneity conventions of SR. (And if they don't operate according to that convention, we can abandon the effort to discuss them using SR.)
Jump to a distant star. Accelerate away from home so that your acquired velocity tilts your line of simultaneity into your home's past. Jump back according to these new coordinates. Prepare to murder your own grandfather.

Grant Hutchison

grav
2010-Apr-01, 04:07 PM
No.
As usual, my brain shuts down about a third of the way through your enormous run-on sentences, especially the ones with embedded plain-text equations. I honestly just can't read that stuff.
But it's quite possible for a pair of EDG_'s "instantaneous" jumps to bring a traveller back to a time before he left, if they operate according to the simultaneity conventions of SR. (And if they don't operate according to that convention, we can abandon the effort to discuss them using SR.)
Jump to a distant star. Accelerate away from home so that your acquired velocity tilts your line of simultaneity into your home's past. Jump back according to these new coordinates. Prepare to murder your own grandfather.

Grant HutchisonOkay, sorry. Referring to the diagram again, the traveller is the black dot. A and E is the rest frame that measures the traveller moving at some speed greater than c. B and D is the frame that sees the traveller arriving before he left. Now, according to the rest frame, the time dilation for the traveller is complex, so the traveller's clock will read a complex time when he arrives at the destination. All observers in all frames must agree upon that, and they do as demonstrated in a couple of the posts earlier. So that has no meaning as far as the what the traveller's clock will read when he arrives and of course nothing can travel faster than c to any frame anyway, but ignoring those two factors along with the contraction of the traveller being complex when in motion to the rest frame also, we will just look at what is involved with the simultaneity shift.

According to the rest frame of A and E, if the traveller departs from B at some speed to the rest frame between c and infinite speed when B's clock reads T=0, then the traveller will arrive at D when D's clock reads a negative time, so the traveller has arrived before he departed according to the frame of B and D. If his grandfather were at the destination, he could kill his grandfather. Well, actually I still have issues with that because of the complex time on the traveller's clock meaning it cannot really occur, but otherwise yes. He cannot, however, travel back to the same point of departure at any positive speed to the rest frame between 0 and infinite and arrive before he left. That would require a third frame that sees the traveller at a speed between c and infinite where the frame of B and D are travelling in the opposite direction, in which case the original rest frame of A and E would measure a negative speed for the traveller. If the grandfather were standing at the point of departure in the past for that frame, he could then kill his grandfather there after making two jumps.

grav
2010-Apr-01, 04:24 PM
Okay, I'm looking at something else now. Even allowing faster than light speeds and a complex reading on the clock of the traveller, there is still something wrong about the idea of something like going back and killing one's own grandfather, because it would violate causality even from the rest frame of A and E in the diagram. In order for one to kill their own grandfather, then two related cause and effect events must be observed simultaneously even from the rest frame. For instance, let's say that the mother has just given birth to her son at the destination. So the traveller goes back and kills himself when he was born. Then from the rest frame, it must be observed that the mother has given birth and is holding her son at D, simultaneously with the older version of the same son getting ready to jump back to that point at B. That cannot be, however, because from the view of the rest frame, even with a simultaneity shift across the frame of B and D, two of the same objects cannot be observed simultaneously in two different places from the rest frame of A and E. Similarly, I am thinking that anything that would produce a cause and effect violation, like jumping back even further and killing one's own grandfather, cannot occur either, but I will have to look into it further.

grant hutchison
2010-Apr-01, 04:51 PM
Well, actually I still have issues with that because of the complex time on the traveller's clock meaning it cannot really occur, but otherwise yes.But we all know it can't occur. You chose to discuss the implications of this impossible scenario: there's no point in protesting that it's impossible.


He cannot, however, travel back to the same point of departure at any positive speed to the rest frame between 0 and infinite and arrive before he left.He certainly can't return at a positive speed, since you seem to have designated positive velocity as being in the home-to-destination direction. But he can return backwards in time in exactly the manner I described, if he obeys the simultaneity conventions of SR. A stationary observer on Earth will measure that he arrives on Earth before he departs from his distant stopping point. He follows a closed curve in spacetime and arrives on his own worldline before his departure.


Okay, I'm looking at something else now. Even allowing faster than light speeds and a complex reading on the clock of the traveller, there is still something wrong about the idea of something like going back and killing one's own grandfather, because it would violate causality even from the rest frame of A and E in the diagram.That's the point. FTL generates time-travel, which generates causality violation.

On the matter of elapsed time along the worldline of our FTL traveller, we can easily mark off a series of spacetime events along his worldline, just as we could with a conventional traveller. A series of flashes is generated as the FTL traveller passes certain points along his route, for instance.
The difference for the FTL traveller is that the sequence in which these flashes go off is not fixed for all observers: for some observers they detonate in sequence, left to right. For other observers, they detonate in sequence, right to left. The direction the traveller moves in is ambiguous; the temporal sequence of arrival and departure is ambiguous. This is what people mean when they say suitable observers can "reverse the time" of an FTL traveller. There's really no point in protesting about imaginary time on the clocks of the traveller, because that's just what happens when you apply SR calculations to non-SR situations: in this case, trying to figure the Lorentz factor for FTL.

Grant Hutchison

Grey
2010-Apr-01, 04:54 PM
According to Victor Stenger, you could instead assume that the arrow of time just doesn't operate at the quantum level, but is emergent as part of decoherence. If distant detectors are able to "talk back" to the source at the quantum level, thereby subverting the macroscopic arrow of time, then the spooky entanglement stuff becomes less spooky (although at the expense of having a spooky timeless domain underlying "our" reality). Or so says Stenger. :)Hmm, if the detector can pass information essentially backward through time, that would seem to qualify as nonlocal pretty much by definition. Still, there are other ways around Bell's conclusion. It's a proof by reductio ad absurdum, and in any such proof there's always not just the deliberately shaky assumption you're trying to disprove, but a host of other assumptions, and if you don't like the conclusion, you can always try to find a different assumption to break.

grav
2010-Apr-01, 04:57 PM
Okay, so far I can see that even if the traveller lived his entire life on the ship of B and D, he will only be seen to exist in one place at a time from any frame, so if the current place where he is observed to be is where he is about to travel back in time to any other place upon the ship, then that is the one and only place that he exists according to any frame, so he will never be able to travel back and meet up with his former self.

grant hutchison
2010-Apr-01, 05:05 PM
Okay, so far I can see that even if the traveller lived his entire life on the ship of B and D, he will only be seen to exist in one place at a time from any frame, so if the current place where he is observed to be is where he is about to travel back in time to any other place upon the ship, then that is the one and only place that he exists according to any frame, so he will never be able to travel back and meet up with his former self.He's got to turn around at some point if he wishes to intersect his own worldline, certainly. But if he does that as I described, then he will indeed be observable twice in the same place at the same time. He can do it several times over, in fact, and "fold himself" into multiple coexistent copies.
Many consider this to be a problem. :)

Grant Hutchison

grav
2010-Apr-01, 05:08 PM
But we all know it can't occur. You chose to discuss the implications of this impossible scenario: there's no point in protesting that it's impossible.But that's all I was originally saying, that it's mathematically impossible, meaning that the mathematics show it cannot ever actually occur. There is nothing that says directly that nothing can travel faster than light, but according to the mathematics of SR, a traveller that does so will not read a negative or positive time upon his own clock, but complex, which signifies its overall impossibility. I'm still enjoying the discussion, though. It's given me a lot of insight about this time travel thing.


He certainly can't return at a positive speed, since you seem to have designated positive velocity as being in the home-to-destination direction. But he can return backwards in time in exactly the manner I described, if he obeys the simultaneity conventions of SR. A stationary observer on Earth will measure that he arrives on Earth before he departs from his distant stopping point. He follows a closed curve in spacetime and arrives on his own worldline before his departure.Right.


That's the point. FTL generates time-travel, which generates causality violation.I'm not so sure about that now, but I'm still looking into it.

eburacum45
2010-Apr-01, 05:11 PM
And a later instance of the traveller can attempt to persuade an earlier instance not to travel, which would cause a paradox if successful. Or even kill them, which would cause a paradox too, of course. A tangled thread indeed.

grav
2010-Apr-01, 05:22 PM
He's got to turn around at some point if he wishes to intersect his own worldline, certainly. But if he does that as I described, then he will indeed be observable twice in the same place at the same time. He can do it several times over, in fact, and "fold himself" into multiple coexistent copies.
Many consider this to be a problem. :)Yes, like me. :) Although I haven't formulated a proof for it, or maybe I have in the past and forgot, because I have worked through numerous scenarios in the past to see if the relativity of simultaneity could ever place the same object at two different locations at the same time to some frame and it has never occured, so I convinced myself it never will. How could any rest frame ever observe the same object in two locations in another frame?

grav
2010-Apr-01, 05:26 PM
He's got to turn around at some point if he wishes to intersect his own worldline, certainly. But if he does that as I described, then he will indeed be observable twice in the same place at the same time. He can do it several times over, in fact, and "fold himself" into multiple coexistent copies.
Many consider this to be a problem. :)

Grant HutchisonOh, wait. I had always been looking at observers that remain SR observers their whole lives as in the frame of B and D. But are you saying that once the traveller jumps faster than c to his destination, although he will never intersect his own worldline from there, another frame would then see him existing in two places at once, and he could then jump again to another point and intersect his own worldline?

grav
2010-Apr-01, 05:44 PM
He's got to turn around at some point if he wishes to intersect his own worldline, certainly. But if he does that as I described, then he will indeed be observable twice in the same place at the same time. He can do it several times over, in fact, and "fold himself" into multiple coexistent copies.
Many consider this to be a problem. :)

Grant HutchisonOkay, sorry about the multiple posts for the same quote, but I'm still not seeing that either. From any frame whatsoever, SR frames anyway, the traveller can only exist in one place in a frame before he travels. Then once he travels to another location in that frame faster than light speed, he will still only continue to exist in one place in that frame to all SR frames, and this will continue no matter how many times he jumps, so he can never intersect his own worldline as viewed from any SR frame, so not to any frame in general since all frames must agree upon the traveller coinciding with his former self, which will never occur. It would still seem he should be able to do this in the frame from which he made the jump, though, since he would exist in two places at the time he jumped to within that frame, so I may be overlooking something. It has me baffled and I will not rest until I have it worked out.

grant hutchison
2010-Apr-01, 06:29 PM
But that's all I was originally saying, that it's mathematically impossible, meaning that the mathematics show it cannot ever actually occur. Well, that's not what you said, and it's not what people on this thread objected to.
You wrote:
Actually, even if it were possible to travel faster than the speed of light, it still wouldn't make one go backward in time, so that is mostly just sci-fi folklore.I've given you a solid example of how it can create backwards time travel. Imaginary time isn't really a showstopper for a tachyon, and you could use a couple of tachyon transmitters to get information into your own past in pretty much the manner I've described.

Grant Hutchison

grav
2010-Apr-01, 08:30 PM
Well, that's not what you said, and it's not what people on this thread objected to.
You wrote:

I've given you a solid example of how it can create backwards time travel. Imaginary time isn't really a showstopper for a tachyon, and you could use a couple of tachyon transmitters to get information into your own past in pretty much the manner I've described.

Grant HutchisonYes, you're right. I was more focused on the clock of the traveller themself as discussed in the rest of that post, whose own reading would be complex when he arrived, which is true, making it an overall mathematical impossibility for original SR observers anyway, although not saying much about anything involving tachyons. I was also considering that the time of a clock that a traveller moves away from and travels back to faster than light according to the frame of the clock will always still read positive as well, which is also true. But what I hadn't considered and you have since shown me is that it is possible for the traveller to jump between two points in a frame and arrive before he left according to that frame when the traveller travels at a speed greater than light relative to another frame, which will be measured as negative speed in the frame the traveller jumped from, so due to the relativity of simultaneity, will indeed allow the traveller to arrive before he left according to the perspective of that frame. It would be perceived as the traveller going backwards in time within that frame, so I was definitely wrong in that regard, which does happen sometimes just like anybody else, but thanks to your explanation and to Schneibster for mentioning it, now I know. :)

Believe it or not, that was all originally a run-on sentence, but I caught it when I considered what you said earlier and repunctuated. ;)

Schneibster
2010-Apr-01, 08:33 PM
So things going fast enough have their spatial dimensions compressed because they're rotating around a space-time axis that we can't see?Precisely. That rotation is velocity. Velocity is rotation in the fourth dimension. This is real, not somebody's guesses, not some wild theory- that's precisely what Einstein says relativity is, in his book Relativity. I can't stress that enough. If we accept time as the fourth dimension, and that's a postulate of relativity (for the pedantic, it's a subset of the continuum postulate), then this is a real description of the real situation we find ourselves in in our universe.

To distinguish that rotation from the ordinary way we refer to velocity, which is meters per second, we call it rapidity. It's measured in degrees, or grads, or radians, like all rotations. So that's what rapidity is. Remember, however, that the relation between time and the three space dimensions is not the same as the relation of those dimensions among themselves; among themselves, the geometry of the spatial dimensions is circular geometry, and uses circular trig to calculate distances from angles; however, the relation of time to each space dimension is hyperbolic, so the geometry of space with respect to time uses hyperbolic trig to calculate distances from angles, and hyperbolic trig has some very unusual characteristics. For example, there is an angle in hyperbolic trig that is infinity, and that angle is equivalent to the circular trig right angle. That characteristic is the underlying geometric reason for the speed-of-light limit; in order to reach the speed of light, your rapidity must become infinite, which is impossible.


I don't think you actually mentioned the speed of light earlier, so you seem to have pulled this statement from out of nowhere. Well, actually, it's a direct consequence of the above. The reasoning is the explanation for the term, "speed of time," and you've conveniently asked that question as well:


Also, I'm not sure what you mean by "the speed of time".Well, obviously, it's one second per second, right? However, remember that we can convert space and time into one another using hyperbolic trig; that's what Einstein said. So if we can, how long is "a second" in spatial measurement? Answer, of course, is 3.997925x10^8 m, more or less. And if we go that distance in a second, then that's the speed of light. And, of course, the speed of time.

grant hutchison
2010-Apr-01, 09:52 PM
Here's an example of how EDG_'s "instantaneous jump" allows travel to one's own past, if the jump obeys SR's simultaneity convention.
The attachment shows a spacetime diagram which follows a closed loop. Time plotted vertically, space horizontally in our initial reference frame.
Our traveller is initially at rest in this reference frame during the interval AB. At B, he jumps instantaneously to a distant destination, C. He then returns to rest in the initial reference frame for the interval CD. At D he accelerates away from home and then coasts along the worldline DE. By adopting this new moving reference frame, he shifts his measure of simultaneity pastwards along BA, and he is therefore able to perform an instantaneous jump EF which carries him into the past according to the coordinates of his original rest frame. After coasting from F, he decelerates to synchronize with his original rest frame at A, thereby rendezvousing with his previous self.

We can use the same diagram to see how a signaller following worldline FAB can send a zero-energy tachyon signal (BC) instantaneously to an assistant on worldline CD, while the two share the same rest frame. The assistant then changes velocity (DE) to enter a common reference frame with the signaller's worldline segment FA. Not only do they share the same state of motion, they are simultaneous with each other according to their shared version of simultaneity. Another pulse of zero-energy tachyons can therefore take a message back (EF) to the signaller's past.

Grant Hutchison

DrRocket
2010-Apr-01, 10:09 PM
The postulates of relativity are few and simple. It supports them admirably. Most folks who are "unsure" if "relativity is right" are unclear on what either GR or SR actually say. It should be stressed that there is a point of view in which the main statements of SR and GR are really obvious things like "you can't see something going faster than light." Which is about as obvious as it gets if you think about it for three seconds.

It really is a very simple matter of whether you believe what you see or not.

SR will prevent a massive body from traveling at c in grounds of energy conservation.

If you demand that causality be preserved, then SR also implies that information cannot be transmitted faster than c.

You have to be rather careful regarding statements about "what you see". If one takes that literally then SR is not at all that obvious. For instance, a ring that is circular in its rest frame, will be an ellipse in the frame of an observer past which it is moving at relativistic speed. But that observer will actually perceive it visually as a circle, not an ellipse, and that is due to the finite speed of light. There are several surprising things that crop up with respect to what would be recorded on a camera in relativistic situations.

grav
2010-Apr-02, 01:21 AM
Okay, yup. It's that negative speed thing that was getting me, until I realized a really simple way to look at it. From the frame of A and E, the traveller can travel from B to D at a positive speed greater than c and arrive before he left, and the speed will be negative to the frame of B and D. To the rest frame, the traveller only exists in one place, but from the frame of B and D, the traveller now exists in two places at the same time. The rest frame cannot see this because all of the locations the former self was before for the times observed of the BD frame have already passed. Then even from the frame of B and D, if we allow any speed between 0 and infinite relative to that frame, the traveller can instantly transport at infinite speed to the same place of the former self and coexist in the same place at the same time. The BD frame sees that as a positive speed now, but the rest frame will see that as a negative speed, with the traveller just popping into existence, but at an earlier time according to the rest frame, before the traveller even departed.

I'm still trying to work out the mathematics of why the traveller cannot catch up to his former self when only jumping once.

grant hutchison
2010-Apr-02, 10:18 AM
I was more focused on the clock of the traveller themself as discussed in the rest of that post, whose own reading would be complex when he arrived, which is true, making it an overall mathematical impossibility for original SR observers anyway, although not saying much about anything involving tachyons.On the topic of this imaginary time measure, I think it's just telling us that the FTL traveller's time has become spacelike. In the context of what Schneibster's been saying, there has been an extreme hyperbolic rotation, with rapidity moving "beyond" infinity.
So we can imagine that the FTL traveller's clock ticks away in units of imaginary time, the hands indicating time periods which we multiply by sqrt(-1) in order to understand them in our own reference frame. The traveller (like all travellers under SR) sees nothing unusual about his own time.
So the FTL traveller's clock marks off spacelike intervals. If I reconstruct things correctly, this produces the interesting result that a four-light-year instantaneous journey in our frame will take four years by the traveller's clock. Travellers who wish to make long journeys in short personal time periods need to stay in the vicinity of lightspeed: either just above or just below.

Grant Hutchison

grav
2010-Apr-02, 03:17 PM
In reference to the traveller living his whole life on the ship and only making one jump, there are just too many possibilities to come up with a formal proof mathematically. For instance, from the perspective of the rest frame, the traveller can jump instantly to a distance of d that would be measured in the ship's frame, which would also be simultaneity shifted to an earlier time of tl = - d v / c^2, so in order to catch up to his former self before his former self jumps, the traveller would have now have to race a distance of d within a time of d v / c^2, so at a speed of v_x = [d] / [d v / c^2] = c^2 / v, so faster than light. Since he can only make one faster than light jump, he will not catch up to his former self in time. Likewise, if the traveller lives his whole life in one spot and then moves some distance at a constant speed before jumping back to the original spot, we find that regardless of whatever the constant speed was or the distance moved before jumping, that when he jumps back to the original spot, some time will have passed at the original spot according to the rest frame after he left and before he jumped, so when the traveller arrives he will find his former self will already have left, so someone there will say "You just missed him. He went thataway.", and the traveller must still race after his former self to get to him before he jumps at a speed of v_x = (v1 - v) / (v1 v / c^2 - 1) where v1 > c, c^2 / v1 < v < c, and of course 0 < v_x < c, which produces a similar situation to that in post #39, where v_x will always be found to be greater than c.

About the only type of real proof that can be offered, then, is that from the rest frame, the former self can be seen living out his entire life upon the ship, every moment accounted for in a normal causal way, then at some point the former self becomes the traveller that will be seen to jump after the former self is gone from the ship, never to be seen again, and the movements of the traveller can then also be tracked in a causal way from the rest frame, even when travelling faster than light, and will then continue to live out his life upon the ship without ever coinciding with his former self that was gone forever once the traveller jumped. It is only from another new frame that is different from the rest frame that the traveller will be seen to be in two places at one time, since the jump will appear non-causal with a negative speed to that frame, with the traveller just popping into existence, but still never coinciding with his former self, because if he did, then all frames must agree upon that, including the original rest frame. But the traveller can then jump again in a causal way from the place that he now exists according to the new frame to the place of his former self at faster than light speed, which will now be seen as causal to the new frame, but non-causal to the original rest frame.

grav
2010-Apr-02, 03:58 PM
On the topic of this imaginary time measure, I think it's just telling us that the FTL traveller's time has become spacelike.

So the FTL traveller's clock marks off spacelike intervals.Well, the spacetime interval between the events of the departure and arrival is space-like, yes, but the clock of the traveller directly follows the events, so is actually measuring the proper time upon his own clock, although complex, just as would be with any other set of events that occur where light cannot travel fast enough between the events so gives a space-like spacetime interval. However, some rest frame will exist that doesn't measure the speed of the traveller as positive or negative, but infinitely fast, so instantaneously jumping between the points of departure and arrival, such that the events occur simultaneously, and that would be the space-like spacetime interval given as the proper distance between the events as measured by that frame.


So we can imagine that the FTL traveller's clock ticks away in units of imaginary time, the hands indicating time periods which we multiply by sqrt(-1) in order to understand them in our own reference frame. The traveller (like all travellers under SR) sees nothing unusual about his own time.Regardless of whatever the traveller thinks is happening to his clock while travelling, when the traveller arrives at the destination, all frames must agree that his clock will read now a complex time, which has no general meaning.

grant hutchison
2010-Apr-02, 04:05 PM
When the traveller arrives at the destination, all frames must agree that his clock will read now a complex time, including the traveller, which has no general meaning.Sure. And that comes about because you're putting garbage in, by switching the traveller from sub-light to FTL and back, and attempting to accumulate elapsed time.
But the fact remains that the FTL traveller comes equipped with quite conventional space and time axes in his own reference frame, and could (in principle) own a clock that ticks off proper time along his worldline. And that clock can be made to run forward or backwards by an appropriate choice of coordinates by a sublight observer.

Grant Hutchison

grav
2010-Apr-02, 04:29 PM
Sure. And that comes about because you're putting garbage in, by switching the traveller from sub-light to FTL and back, and attempting to accumulate elapsed time.
But the fact remains that the FTL traveller comes equipped with quite conventional space and time axes in his own reference frame, and could (in principle) own a clock that ticks off proper time along his worldline. And that clock can be made to run forward or backwards by an appropriate choice of coordinates by a sublight observer.

Grant HutchisonI don't see that, notatall, so you would have to show me a workable example. It's starting to sound to me like you might be pro time travel and only considered that travelling faster than light would be the one hurdle to overcome, but now complex readings resulting on the traveller's clock has thrown another hurdle in the way, and you would like it to be gone. Would that be about right, or close? :)

grant hutchison
2010-Apr-02, 05:16 PM
I don't see that, notatall, so you would have to show me a workable example. It's starting to sound to me like you might be pro time travel and only considered that travelling faster than light would be the one hurdle to overcome, but now complex readings resulting on the traveller's clock has thrown another hurdle in the way, and you would like it to be gone. Would that be about right, or close? :)No, that would be complete ********.
I'm just pointing out a fact about these space-like worldlines, which is that they can have a proper time associated with them, which can be plotted on a spacetime diagram, and which can be made to run backwards or forwards against judiciously chosen observer coordinates. Indeed it's because the worldline is spacelike that we can do this: the FTL traveller's time behaves like space in our coordinates, so we can reverse the temporal sequence of events along the worldline in a way that can't be done with conventional timelike worldlines.

If you think this is about being "pro" or "anti" time travel, then perhaps you should examine your own thinking about the problem. I'm beginning to think Schneibster may have been correct in detecting a little hostility in your approach to something that's after all no more than a maths puzzle.

Grant Hutchison

grav
2010-Apr-02, 06:17 PM
No, that would be complete ********.
I'm just pointing out a fact about these space-like worldlines, which is that they can have a proper time associated with them, which can be plotted on a spacetime diagram, and which can be made to run backwards or forwards against judiciously chosen observer coordinates. Indeed it's because the worldline is spacelike that we can do this: the FTL traveller's time behaves like space in our coordinates, so we can reverse the temporal sequence of events along the worldline in a way that can't be done with conventional timelike worldlines.

If you think this is about being "pro" or "anti" time travel, then perhaps you should examine your own thinking about the problem. I'm beginning to think Schneibster may have been correct in detecting a little hostility in your approach to something that's after all no more than a maths puzzle.

Grant HutchisonWell no, I just strongly disagree, that is all, or don't intuitively see it as you put it earlier, so would like to see the maths worked out for that. You have already shown me that my statement at the beginning of the thread was incorrect, since what would be seen as faster than light travel in a causal manner to one SR frame would be measured by another SR frame as a non-causal negative speed and going back in time, which I hadn't considered, and I appreciate you pointing that out to me, so I thought maybe you've got something else up your sleeve when it comes to how SR observers would observe the complex reading on the traveller's clock as positive or negative, but I would have to see it.

grant hutchison
2010-Apr-02, 07:16 PM
Well no, I just strongly disagree, that is all ...In which case maybe you should just say "I strongly disagree", rather than suggesting I have some idiotic agenda behind this discussion. It's easy to annoy people that way.
Take a look at what Schneibster has been writing about rapidity and hyperbolic rotation, think about the difference between a spacelike and a timelike interval, and see if some imaginary numbers don't fall out that have relevance to what I wrote.

Grant Hutchison

grav
2010-Apr-02, 07:49 PM
In which case maybe you should just say "I strongly disagree", rather than suggesting I have some idiotic agenda behind this discussion. It's easy to annoy people that way.Okay well, sorry about that. It's just that the last few posts you made were starting to read like wishful thinking to me. I can do that too sometimes and hopefully that last statement doesn't offend you or anything either, as you have always provided very good insights and I respect what you say, but I don't really know what any of this has to do with how I state things anyway, other than being incorrect with my original statement earlier in the thread, which you have promptly corrected. Maybe the forum deepens the tone a level or two, I don't know. In any case, I just want the math. "Show me da math. " :)



Take a look at what Schneibster has been writing about rapidity and hyperbolic rotation, think about the difference between a spacelike and a timelike interval, and see if some imaginary numbers don't fall out that have relevance to what I wrote.Right, the spacetime interval is space-like to all SR observers, and I can almost see how a complex proper time could be read as a space-like interval passing on the clock of the traveller also, depending upon how one looks at it, but what I want to know is how the SR observers can measure the actual time that reads on the traveller's clock as real when he reaches the destination when reading their own times as real as well. I have been following some of Schneibster's posts, but I would otherwise have no idea what I am supposed to be looking for in regards to observing a complex or real clock reading by the SR observers unless you point it out to me.

m74z00219
2010-Apr-02, 11:13 PM
Not sure if it's been explicitly pointed out already, but it seems to me that instantaneous travel would be "allowed" (mathematically and intuitively speaking, though probably not physically speaking) if there were only a single frame of reference in which this could be done. If this were the case, then there could be instantaneous travel without any paradoxes: no time traveling into the past.

M74

EDG
2010-Apr-03, 12:28 AM
What if we had the ship jump instantaneously (according to its own timeframe), but from outside the duration of the jump was however long it takes to travel between the stars at lightspeed?

i.e. A ship jumps from Sol to Alpha Centauri (4.3 ly away). For the crew, the jump is instantaneous. Our detector on Sol (which is looking at AC) would see the ship appear at AC 8.6 years later (from an outside perspective, the ship arrives at AC 4.3 years after it left Sol, and the signal of its arrival takes 4.3 years to arrive at Sol). When the ship leaves AC three days later, the jump is again instantaneous for the crew, but again takes 4.3 years from an outside perspective. So the ship returns to Sol 8.6 years (+ 3 days) after it left.

From the perspective of the crew of the ship however, they've been gone for three days.

I guess this kinda syncs up? At 8.6yrs + 3 days, the detector would see the ship vanish at AC when it left there, but then the ship arrives back at Sol at the same time (unless I've missed something here)? Would that break causality?

grav
2010-Apr-03, 12:34 AM
Not sure if it's been explicitly pointed out already, but it seems to me that instantaneous travel would be "allowed" (mathematically and intuitively speaking, though probably not physically speaking) if there were only a single frame of reference in which this could be done. If this were the case, then there could be instantaneous travel without any paradoxes: no time traveling into the past.

M74Not if things move relative to that frame. As Grant demonstrated earlier, if a ship were moving relative to the rest frame, even very slowly, there would still be some slight simultaneity shift observed between the front and the back of the ship, so if the traveller instantly jumped from the back to the front of the ship over zero time according to the rest frame, he would be jumping into the past according to the frame of the ship, so could alter any sequence of events that would otherwise take place upon the ship, even as viewed from the rest frame.

pzkpfw
2010-Apr-03, 12:38 AM
In any case, it's kind of like saying "I have this magic elixir that makes me strong enough to lift Mount Everest, but I won't be doing anything 'impossible' as long as I only lift egg-cups after I've drunk it".

"Instant" travel allows causality to be broken. Finding a scenario where it doesn't seem to break causality does not make "instant" travel possible.

grav
2010-Apr-03, 12:39 AM
What if we had the ship jump instantaneously (according to its own timeframe), but from outside the duration of the jump was however long it takes to travel between the stars at lightspeed?

i.e. A ship jumps from Sol to Alpha Centauri (4.3 ly away). For the crew, the jump is instantaneous. Our detector on Sol (which is looking at AC) would see the ship appear at AC 8.6 years later (from an outside perspective, the ship arrives at AC 4.3 years after it left Sol, and the signal of its arrival takes 4.3 years to arrive at Sol). When the ship leaves AC three days later, the jump is again instantaneous for the crew, but again takes 4.3 years from an outside perspective. So the ship returns to Sol 8.6 years (+ 3 days) after it left.

From the perspective of the crew of the ship however, they've been gone for three days.

I guess this kinda syncs up? At 8.6yrs + 3 days, the detector would see the ship vanish at AC when it left there, but then the ship arrives back at Sol at the same time (unless I've missed something here)? Would that break causality?No, that wouldn't break causality. That would just be the ship travelling at light speed and observing things as would normally be observed for that speed, although in reality, the ship could only approach light speed, so the trip wouldn't be instantaneous.

Van Rijn
2010-Apr-03, 12:41 AM
What if we had the ship jump instantaneously (according to its own timeframe), but from outside the duration of the jump was however long it takes to travel between the stars at lightspeed?


See my post above:

http://www.bautforum.com.php5-9.dfw1-2.websitetestlink.com/showthread.php/102514-FTL-and-causality?p=1709760#post1709760

There's no obvious way to make a ship go exactly at the speed of light, but that wouldn't cause time travel issues. You could imagine, I suppose, some type of teleporter that sends a transmission at the speed of light, or you could just have the ship traveling arbitrarily close to the speed of light (which is what I did). Either way, the trip would be instantaneous or so close to it as not to matter for the people on the ship, but it wouldn't cause causality issues per relativity.

grav
2010-Apr-03, 01:56 AM
Not sure if it's been explicitly pointed out already, but it seems to me that instantaneous travel would be "allowed" (mathematically and intuitively speaking, though probably not physically speaking) if there were only a single frame of reference in which this could be done. If this were the case, then there could be instantaneous travel without any paradoxes: no time traveling into the past.
Not if things move relative to that frame. As Grant demonstrated earlier, if a ship were moving relative to the rest frame, even very slowly, there would still be some slight simultaneity shift observed between the front and the back of the ship, so if the traveller instantly jumped from the back to the front of the ship over zero time according to the rest frame, he would be jumping into the past according to the frame of the ship, so could alter any sequence of events that would otherwise take place upon the ship, even as viewed from the rest frame.No wait, I'm wrong again. Oops. :sad: Well, right and wrong. The traveller would be travelling into the past according to the frame of the ship, the crew of the ship seeing him pop in and out of existence, but not from the rest frame, so no events are ever non-causal, so they aren't in the ship frame either. In order for events to become non-causal and create a paradox, a third frame would also have to see the traveller moving at a speed greater than c at some point and make the speed negative as seen from the rest frame. Otherwise, if the traveller always only travels instantaneously according to the rest frame, then all events remain causal, because the traveller can still never travel between events fast enough to change any sequence of the events that lie within a light cone in the frame of the ship in order to create a paradox, so the events remain causal to all frames.

EDG
2010-Apr-03, 03:57 AM
There's no obvious way to make a ship go exactly at the speed of light, but that wouldn't cause time travel issues. You could imagine, I suppose, some type of teleporter that sends a transmission at the speed of light, or you could just have the ship traveling arbitrarily close to the speed of light (which is what I did). Either way, the trip would be instantaneous or so close to it as not to matter for the people on the ship, but it wouldn't cause causality issues per relativity.

I don't know if grav is confusing you with his concurrent argument (which is unrelated to my question), but the ship in my example isn't doing ANYTHING at lightspeed. It's disappearing at one point and reappearing somewhere else, without travelling through the intervening space. Yes, I know it's not realistic or possible, but humour me here. But please do not assume that the ship is literally travelling through space at lightspeed or anything close to it, because it isn't. It's a straight-up instantaneous teleport from A to B. The only thing travelling at the speed of light in my example is light itself.

But anyway, it seems that as long as the teleport appears (from the outside) to take the same amount of time as light would take to travel from A to B, then that technically doesn't violate causality, right? Though it's interesting that the detector would have no warning of the ship's arrival back home - there's a lightspeed lag between the time the ship leaves Sol and the detector spots it at AC, but they see it disappear at AC and then instantly reappear next to them at Sol.

EDG
2010-Apr-03, 04:00 AM
Now, let's go back to quantum entanglement: if you separate two quantumly-entangled things by a lightyear and the state of one of them is determined, is the state of the other one defined at exactly the same time as the other, or will there a lightspeed delay? From what I'm reading about this, there won't be a delay.

Though this paper ( http://arxiv.org/abs/0808.3316 ) seems to be suggesting that the "speed of quantum information" is at least 10,000 times the speed of light (!).

grav
2010-Apr-03, 04:22 AM
I don't know if grav is confusing you with his concurrent argument (which is unrelated to my question), but the ship in my example isn't doing ANYTHING at lightspeed. It's disappearing at one point and reappearing somewhere else, without travelling through the intervening space. Yes, I know it's not realistic or possible, but humour me here. But please do not assume that the ship is literally travelling through space at lightspeed or anything close to it, because it isn't. It's a straight-up instantaneous teleport from A to B. The only thing travelling at the speed of light in my example is light itself.

But anyway, it seems that as long as the teleport appears (from the outside) to take the same amount of time as light would take to travel from A to B, then that technically doesn't violate causality, right? Though it's interesting that the detector would have no warning of the ship's arrival back home - there's a lightspeed lag between the time the ship leaves Sol and the detector spots it at AC, but they see it disappear at AC and then instantly reappear next to them at Sol.What you have described is just the same thing as travelling at light speed. The ship could never actually travel at c though, so let's take a speed near light speed so we can compare distances and times, and then that can be pushed closer and closer toward the limit of c. To Earth observers, the ship would be seen travelling at near c over some large distance, but the clocks of the ship would be time dilated to barely ticking at all, so the Earth observers will say that the ship observers will have barely noticed any time at all passing for their journey. To the ship observers, however, their clocks tick normally, and they cannot reach their destination any faster than c, so they never see the destination moving toward them any faster than c either, but the same as the Earth observers measure for the speed of the ship if AC is considered stationary to Earth, but to the ship observers, the distance between Earth and their destination has been contracted to near zero, so from their perspective also, barely any time at all passes upon their clocks during the quick journey over such a short distance travelled.

Van Rijn
2010-Apr-03, 05:22 AM
I don't know if grav is confusing you with his concurrent argument (which is unrelated to my question), but the ship in my example isn't doing ANYTHING at lightspeed.


So, from Earth, does the ship reach Alpha Centauri before light from Earth could get there, just as soon as light could get there, or some time after light could get there? (ETA: From your own post, I thought it was pretty clear you were now talking about a spacecraft that would get there no sooner than light could.)

As far as causality in relativity goes, it doesn't matter how it gets from Earth to Alpha Centauri, as long as it doesn't get there faster than light could.


But please do not assume that the ship is literally travelling through space at lightspeed or anything close to it, because it isn't. It's a straight-up instantaneous teleport from A to B.


And still again, "instantaneous" according to . . . ?

If it is just instantaneous to the people on the ship, it is the same as what I've already discussed.

m74z00219
2010-Apr-03, 07:40 AM
No wait, I'm wrong again. Oops. Well, right and wrong. The traveller would be travelling into the past according to the frame of the ship, the crew of the ship seeing him pop in and out of existence, but not from the rest frame, so no events are ever non-causal, so they aren't in the ship frame either. In order for events to become non-causal and create a paradox, a third frame would also have to see the traveller moving at a speed greater than c at some point and make the speed negative as seen from the rest frame. Otherwise, if the traveller always only travels instantaneously according to the rest frame, then all events remain causal, because the traveller can still never travel between events fast enough to change any sequence of the events that lie within a light cone in the frame of the ship in order to create a paradox, so the events remain causal to all frames.

Yeah! So there is a situation that "works". But like pzkpfw has already observed, it's like saying impossible would be possible if another impossible thing were possible as well. Well, this is a nice loop hole for creative writers at least!

Another nail in the coffin for instantaneous jumps would be that any other normal inertial frame moving relative to the special jump frame would observe this ship to be under a "crazy amount" of tension. Crazy enough that the jump would have actually caused the ship to have been atomized - or worse.

-M74

grav
2010-Apr-03, 08:31 AM
Yeah! So there is a situation that "works". But like pzkpfw has already observed, it's like saying impossible would be possible if another impossible thing were possible as well. Well, this is a nice loop hole for creative writers at least!Yes, from the rest frame's point of view, it would be just like having normal Newtonian physics where any observers can travel anywhere up to infinite speed with no limit at c. Of course, according to SR, any travellers with a speed greater than c will be contracted into complex space and their clocks will read complex times as well, but if we ignore that and just treat them as points in space, then everything's fine, just as it would normally be with any speed up to infinite in Newtonian physics. As far as the rest frame is concerned, a simultaneity shift along the length of the ship (travelling under c so it has length), is just a result of improper clock settings by the ship observers, so the rest frame says they do not represent the real times the clocks should read, so everything always proceeds causally to the rest frame, and therefore to any other frames as well. For instance, if the rest frame were to see someone assassinated at some place on the ship, they could not then send an observer instantaneously to any other point on the ship in order to try to stop it, because whatever led up to the event of the assasination has already taken place at all points on the ship, so even an infinite speed to any other point or jumping to various points cannot stop it because as far as the rest frame is concerned, all events upon the ship are taking place causally and the traveller can only affect them causally also no matter how fast he travels, just like with Newtonian physics.


Another nail in the coffin for instantaneous jumps would be that any other normal inertial frame moving relative to the special jump frame would observe this ship to be under a "crazy amount" of tension. Crazy enough that the jump would have actually caused the ship to have been atomized - or worse.If a ship accelerates until it approaches light speed, the ship observers do not observe any more tension in the ship than when they first started accelerating, because speed is only a coordinate effect between observers. If the ship were to cut off its engines at near light speed, they would observe no tension at all, since they would just be travelling inertially. The ship could accelerate very slowly over a long period of time and the tension would never change to the rest frame of the ship. The ship would observe no more tension travelling inertially away from a space station at near light speed than the space station does, considering that from the ship's point of view, the space station is also travelling inertially away from the ship at near light speed.

swampyankee
2010-Apr-03, 01:24 PM
General Relativity and Quantum Mechanics have mutual issues. Quantum entanglement may be one of them.

My gut feeling is that FTL won't cause any issues with causality; it will merely confuse a large number of physics students, and require the physicists to make some adjustments to SR and GR.

grant hutchison
2010-Apr-03, 03:53 PM
In any case, I just want the math. "Show me da math. " :)You've got the maths already. It's a matter of interpreting the maths, and choosing an appropriate point at which to throw your hands in the air and say: "Physically impossible!" It seems you're quite happy to swallow the whale (FTL flight within SR) but are choking on a fly (the time looks imaginary).
We measure proper time along an observer's worldline. The interval has to be pure time along the worldline, because the observer is always there: there's no spatial movement as far as the observer is concerned. So we can mark off events along a timelike worldline, measure the interval between them, and know that's proper time. Now all those infinities that pop up at lightspeed, including the infinity in rapidity Schneibster mentioned, are telling you that can't simply push continuously from subluminal to supraluminal flight: you can't rotate your worldline into the spacelike part of the Minkowski diagram without going through infinity. So there's a discontinuity there, which should encourage you to think differently about how time and space are represented either side of the discontinuity.
Now draw the spacelike worldline of an FTL traveller. Plot some events along it. Those events are on the worldline, and must represent events in proper time for the traveller. Measure the interval. You know it's going to be imaginary if you treat it as timelike, but the point is that it's not a timelike interval, it's spacelike, because you chose to give your traveller spacelike proper time when you pushed the button on his FTL ship. No use whimpering about it and trying to pretend it's timelike and imaginary: it's spacelike and real. The FTL traveller's clock ticks off time in a spacelike direction, and we can measure his elapsed time using a spacelike interval. Likewise, the FTL traveller has a timelike spacial dimension: his time and space axes have exchanged places.
Is this familiar from other situations? Sure. It's like a Schwarzschild observer watching someone fall into a black hole. According to the Schwarzschild observer, the faller hits a bunch of infinities at the event horizon, which prevent him crossing the event horizon in finite Schwarzschild time. Beyond the horizon, the Schwarzschild observer calculates that the faller will have imaginary proper time. But he also calculates that the space and time axes are swapped around beyond the event horizon, so that the faller has a timelike radial coordinate and spacelike time. So does the Schwarzschild observer decide it's impossible for someone to fall into a black hole, because they would experience imaginary proper time? Not if he works the problem in the faller's own coordinates, in which case he finds that the faller's clock keeps ticking off proper time quite normally until his worldline terminates at the singularity. So we see that the appearance of imaginary proper time is a flag that you're using the wrong coordinates, not that the worldline is mathematically impossible.
An analogy: if all you've ever done is measure movement along a north-south line, you'll find east-west travel difficult to deal with. "I cannot map this route into north-south coordinates," you cry. "Distance appears to have no meaning there." But you just need to rotate yourself into a new set of coordinates, and meaningful distance is recovered.
So for your problem of someone making a transition from subluminal to FTL and back again, you need to measure always along the traveller's time axis. You'll have a timelike interval before launch, a spacelike interval for the FTL flight, and a timelike interval while you wait for the engines to cool. Add them up, and you have the proper time elapsed on that traveller's clock. (But because the FTL segment is spacelike, the ordering of events along it is reversible in coordinate time, which is what Schneibster tried to tell you several pages ago.)

I expressly don't provide this as a defence of superluminal flight and time travel. I just point out that I think you're boggling in the wrong place: boggle at crossing through the lightspeed infinities; boggle at the implications for causality violation; but there's no need to boggle at the phantom of imaginary time.

Grant Hutchison

grav
2010-Apr-03, 05:18 PM
You've got the maths already. It's a matter of interpreting the maths, and choosing an appropriate point at which to throw your hands in the air and say: "Physically impossible!" It seems you're quite happy to swallow the whale (FTL flight within SR) but are choking on a fly (the time looks imaginary).Right, because there is nothing that directly disallows FTL speed in itself, until we try to determine the time dilation that would be observed upon the traveller's clock, which would be complex, so indicates the impossibility of a traveller ever attaining a speed greater than c relative to the observer from the observer's perspective in the first place, just as it is impossible for a traveller to fall into a black hole in finite time from the perspective of a distant observer. I understand what you are saying in the rest of your post about how the traveller might measure things from his own frame, but that is not what I'm asking about. I'm asking what the SR observers will measure for the reading upon the traveller's clock when he reaches his destination. For instance, let's say that the traveller takes off from the rest frame at 1.1 c at T=0, travels a distance of 1.1 light years through the rest frame, so that one year passed for the rest frame, and then comes back to rest. What time will the rest frame now directly read upon the traveller's clock?

grant hutchison
2010-Apr-03, 06:02 PM
For instance, let's say that the traveller takes off from the rest frame at 1.1 c at T=0, travels a distance of 1.1 light years through the rest frame, so that one year passed for the rest frame, and then comes back to rest. What time will the rest frame now directly read upon the traveller's clock?What's the spacelike interval? In units of years and lightyears, It's sqrt(1.12-12) = 0.458. The traveller's clock reads 0.458 years. Your calculated inverse gamma factor is 0.458i, which matches the timelike interval. So if you could watch the traveller's calendar-clock during flight, you could say that it was counting off years of its own proper time, imaginary years of the imaginary timelike interval, or lightyears of the spacelike interval.

Grant Hutchison

grav
2010-Apr-03, 06:31 PM
What's the spacelike interval? In units of years and lightyears, It's sqrt(1.12-12) = 0.458. The traveller's clock reads 0.458 years. Your calculated inverse gamma factor is 0.458i, which matches the timelike interval. So if you could watch the traveller's calendar-clock during flight, you could say that it was counting off years of its own proper time, imaginary years of the imaginary timelike interval, or lightyears of the spacelike interval.

Grant HutchisonWell, the way you stated that, I'm not sure which of those you are saying the rest frame will actually observe on the clock of the traveller when he arrives at his destination. So let me ask you this. SR says that nothing can travel faster than light speed relative to the rest frame, right? There must be a mathematical reason for this, since just saying so doesn't make it a fact. So what is it about the results of the equations of SR that demonstrates the impossibility of travelling faster than light speed to the rest frame?

grant hutchison
2010-Apr-03, 07:14 PM
Well, the way you stated that, I'm not sure which of those you are saying the rest frame will actually observe on the clock of the traveller when he arrives at his destination.A time. That's what clocks measure as you transport them along worldlines.


So what is it about the results of the equations of SR that demonstrates the impossibility of travelling faster than light speed to the rest frame?Nothing at all. Tachyons are permitted under SR. What SR tells you is that you can't take an object with real rest mass and accelerate it beyond the speed of light; it also tells you that you can't take a tachyon and slow it below the speed of light: either of those options encounters prohibitive infinities. As a separate issue, FTL travel generates ambiguous temporal order, time travel and the potential for causality violation. So proponents of tachyons have to apply considerable ingenuity to work around those problems; what they don't have is problem with SR.

Grant Hutchison

grav
2010-Apr-03, 07:38 PM
A time. That's what clocks measure as you transport them along worldlines.

Nothing at all. Tachyons are permitted under SR. What SR tells you is that you can't take an object with real rest mass and accelerate it beyond the speed of light; it also tells you that you can't take a tachyon and slow it below the speed of light: either of those options encounters prohibitive infinities. As a separate issue, FTL travel generates ambiguous temporal order, time travel and the potential for causality violation. So proponents of tachyons have to apply considerable ingenuity to work around those problems; what they don't have is problem with SR.

Grant HutchisonEven with a traveller that always travelled faster than light to the rest frame, and even if his clock reads a real time when he passes one point in the rest frame, when he passes another point some time later as the rest frame measures it, the light from his clock can still travel to an observer that resides at that point, and his clock will still be observed with a complex component added to it, which has no physical meaning at all to the rest frame, since the hands of a clock can only go backwards or forwards, not complexly. We can skate around this effect with the concept of tachyons because they do not carry any clocks we can directly observe if we only consider them as points within a frame.

I have been asking these types of questions in the usual spirit of BAUT, but I also already know how well you understand Relativity, heck you taught me some of it ;) , and we have pursued this long enough now for each of us to know what the other is saying, so I guess we'll just have to agree to disagree as the saying goes on this one particular point. Oh well, it happens. As always, I have valued your insights and I hope you have enjoyed the discussion as much as I have. :)

grant hutchison
2010-Apr-03, 08:01 PM
We can skate around this effect with tachyons because they do not carry any clocks we can directly observe if we only consider them as points within a frame, and as such points, we can also ignore the fact that they should also be contracted so that they reside in complex space, not the real space of the rest frame.Well, no, we can't skate around it with tachyons: ducking the issue by citing their inability to carry clocks would an act of quite breath-taking sophistry. We still need to think carefully about how the coordinates work when the axes are swapped. If for no other reason, then because of Gell-Mann's maxim that what is not forbidden will turn out to be required.

The interval always gives us two results, one real and one imaginary. I hereby choose to treat your proper time as being spacelike, and am disconcerted to have to tell you that you can't possibly exist under those constraints.

Grant Hutchison

grav
2010-Apr-03, 08:07 PM
Well, no, we can't skate around it with tachyons: ducking the issue by citing their inability to carry clocks would an act of quite breath-taking sophistry. We still need to think carefully about how the coordinates work when the axes are swapped. If for no other reason, then because of Gell-Mann's maxim that what is not forbidden will turn out to be required.

The interval always gives us two results, one real and one imaginary. I hereby choose to treat your proper time as being spacelike, and am disconcerted to have to tell you that you can't possibly exist under those constraints.

Grant HutchisonIf I read that correctly, then I wholeheartedly agree. :)

grant hutchison
2010-Apr-03, 09:06 PM
If I read that correctly, then I wholeheartedly agree. :)You can't agree, because I have already demonstrated that you can't exist under the coordinate choice I have adopted. <Shrug.> Sorry.

Grant Hutchison

EDG
2010-Apr-04, 12:09 AM
You can't agree, because I have already demonstrated that you can't exist under the coordinate choice I have adopted. <Shrug.> Sorry.

Grant Hutchison

In that case, who are you arguing with? ;)

This could end up like the end of John Carpenter's "Dark Star", you know ;)

eburacum45
2010-Apr-04, 08:40 AM
: And in addition to the darkness there was also me. And I moved upon the face of the darkness. And I saw that I was alone. Let there be light.
.

astromark
2010-Apr-04, 09:48 AM
You's have dug a deep hole... Which I wish to stay out of... Edg has made a unreal suggestion that can not be argued about.. Its not real.
But, If it were then what is wrong with what he said ?
The ship leaves and after just three days is back. 4.2 years later you would see it at Alpha Cent., We know it can not be... When you play lets pretend any thing you say...IS. How can you argue with this ? hes right.

grav
2010-Apr-04, 03:41 PM
You's have dug a deep hole... Which I wish to stay out of... Edg has made a unreal suggestion that can not be argued about.. Its not real.
But, If it were then what is wrong with what he said ?
The ship leaves and after just three days is back. 4.2 years later you would see it at Alpha Cent., We know it can not be... When you play lets pretend any thing you say...IS. How can you argue with this ? hes right.Right, the trip is instantaneous to the ship observers, minus the three days, but not because they are measuring themselves to be travelling at infinite speed. The Earth observers measure their speed at c, light speed, so the ship takes some time to travel the distance according to Earth's clocks, but Earth observers notice that the time dilation of the ship observers is zero, so the ship observers and their clocks appear to be frozen in place upon the ship, so zero time passes for the ship observers during the trip to AC and back. The ship observers also say that they are travelling at a relative speed of c to Earth, but the distance to AC has been contracted to zero, so zero time passes for the ship observers during the trip according to them also.

astromark
2010-Apr-04, 08:05 PM
No Grav., thats wright but wrong... Edg was clear regarding instantly being at AC. Zero travel time. Left Here arrived there. Nothing to do with observation. The light image of all this is going to arrive 4.2 years later. We can Analise this from a more practical view, and you are correct about time compression and all sorts of realities... None of it applies as 'this' is a 'what if we could'.... nothing about Edg's idea is or can be real.

grav
2010-Apr-04, 10:21 PM
No Grav., thats wright but wrong... Edg was clear regarding instantly being at AC. Zero travel time. Left Here arrived there. Nothing to do with observation. The light image of all this is going to arrive 4.2 years later. We can Analise this from a more practical view, and you are correct about time compression and all sorts of realities... None of it applies as 'this' is a 'what if we could'.... nothing about Edg's idea is or can be real.EDG said


What if we had the ship jump instantaneously (according to its own timeframe), but from outside the duration of the jump was however long it takes to travel between the stars at lightspeed?

To the ship observers, the trip would be instantaneous because they are travelling at light speed over a distance that is contracted to zero, so zero time passes for them during the trip. To the Earth observers, some time has passed on Earth, but zero time also passes upon the ship according to the Earth observers due to the time dilation involved, but the ship is still travelling at light speed relative to Earth according to both frames as EDG stipulated, not infinite speed, so the ship doesn't get there instantly, but it only seems that way to the ship observers because the distance to AC has contracted to zero as they measure it.

astromark
2010-Apr-05, 05:43 AM
Imagine that we had a spaceship that could instantly traverse the 4.3 lightyears between Sol and Alpha Centauri, and that it spends three days out there, and then jumps back instantly to Sol. Assume that we also have a very powerful detector at Sol that can see the ship when it arrives at Alpha C.

So, the timeline of events here (as I reckon) should be like this:

year 0, day 0: Ship jumps out from Sol. Instantly arrives at Alpha C.
year 0, day 3: Ship jumps out from Alpha C. Instantly arrives back at Sol.
year 4.3 + 0 days: Detector spots ship arriving at Alpha C, sees ship at Alpha C for the next 3 days.
year 4.3 + 3 days: Detector spots ship departing from Alpha C.

As far as I can see, the detector is only observing the image of the ship at Alpha C, since the real ship has obviously been back at Sol for the past four and a bit years. So it's not like the ship is really in two places at the same time.

But scenarios like this (I think) are usually said to be a violation of causality, aren't they? How is that the case, exactly? I'm clearly missing something here (I think it has something to do with 'lightcones' but I'm not sure what the problem is).

and that is what I was responding to. Post number one. It is as I said.

Van Rijn
2010-Apr-05, 07:23 AM
and that is what I was responding to. Post number one. It is as I said.

Have you read the responses to the OP and EDG's later posts? The discussion has moved on from the OP.

Van Rijn
2010-Apr-05, 07:39 AM
No Grav., thats wright but wrong... Edg was clear regarding instantly being at AC. Zero travel time. Left Here arrived there.


Unfortunately, that's not clear, hence the discussion in thread. Per relativity, there is no universal clock by which everyone can agree on "instantly" or "zero travel time." See:

http://en.wikipedia.org/wiki/Relativity_of_simultaneity

astromark
2010-Apr-05, 07:52 AM
When we deviate from the real and ever present science, and play these 'What if' or 'lets just say...' You are in trouble from the outset.
I was told that i was wrong... I was not. Yes its all moved on somewhat.. and so it should. From reading the whole four pages I do not see a resolution that would please the OP. Its a fiction and might have been better as a ATM idea and discussion. What I see is the same few facts bantered about as we like... its not a possible thing to be any place quicker than light speed. So the whole idea is faulted.. folded. kaput.

EDG
2010-Apr-05, 04:58 PM
You's have dug a deep hole... Which I wish to stay out of... Edg has made a unreal suggestion that can not be argued about.. Its not real.

Whether it's "real" or not is utterly irrelevant. Questions can still be asked that can help people understand what's going on (and despite what you claim, it has helped me) - think of it as a thought experiment if you have such a problem with 'fiction'. Bluntly, fiction and imagination are a perfectly valid tools for people to expand their knowledge of science - if you think science itself moves forward without anyone thinking "what if..." every now and then, then you'd be utterly mistaken.

As people keep telling you, if you can't make a positive, informed and useful contribution to a discussion then please stay out of it. I for one am getting rather fed up with your constantly negative posts on a lot of threads, telling people that this or that isn't possible while ignoring or being unable or unwilling to address the actual question being asked.

Webbo
2010-Apr-05, 05:25 PM
What if the ship in the OP had a drive which was able to expand the space between itself and the point of departure? Could the issues of FLT then be avoided?

grant hutchison
2010-Apr-05, 05:39 PM
What if the ship in the OP had a drive which was able to expand the space between itself and the point of departure? Could the issues of FLT then be avoided?An Alcubierre-type drive would avoid the infinities associated with passing through lightspeed. Since it's unaffected by time dilation, grav would need to deal with FTL clock readings which were undeniably real. But it would still create the problems associated with ambiguous temporal ordering, travel into the past and causality violations.

Grant Hutchison

TampaDude
2010-Apr-05, 06:55 PM
An Alcubierre-type drive would avoid the infinities associated with passing through lightspeed. Since it's unaffected by time dilation, grav would need to deal with FTL clock readings which were undeniably real. But it would still create the problems associated with ambiguous temporal ordering, travel into the past and causality violations.

Grant Hutchison

I thought a ship contained within the Alcubierre Metric would travel on a free-fall geodesic in the space surrounded by the warp bubble, and thus there would be no FTL time travel or causality issues. The prohibition against moving FTL is for objects or signals moving through space, not space itself moving. The expanding universe has regions that are accelerating away from us at speeds far greater than c, so we know this is possible, even if it never becomes technologically feasible.

grant hutchison
2010-Apr-05, 07:20 PM
I thought a ship contained within the Alcubierre Metric would travel on a free-fall geodesic in the space surrounded by the warp bubble, and thus there would be no FTL time travel or causality issues. The prohibition against moving FTL is for objects or signals moving through space, not space itself moving. The expanding universe has regions that are accelerating away from us at speeds far greater than c, so we know this is possible, even if it never becomes technologically feasible.Alcubierre sidesteps the problems of special relativity, but causality is still a problem if the ship arrives at its destination faster than a light signal. We then immediately know that there are potential observers for whom the temporal order of the ship's arrival and departure is reversed, and for whom the occupants of the ship therefore live backwards, decreasing entropy in the closed system of the Alcubierre "bubble".

Grant Hutchison

grav
2010-Apr-05, 08:22 PM
An Alcubierre-type drive would avoid the infinities associated with passing through lightspeed. Since it's unaffected by time dilation, grav would need to deal with FTL clock readings which were undeniably real.I don't have to deal with much of anything since I don't exist. :) Just kidding.

grant hutchison
2010-Apr-05, 09:06 PM
I don't have to deal with much of anything since I don't exist. :) Just kidding.:)
It's a problem for you, though. You can't insist on applying that imaginary Lorentz factor of yours to the worldline of an Alcubierre drive. Real time, ticking at a real rate, moving faster than light.

Grant Hutchison

swampyankee
2010-Apr-05, 11:34 PM
I've no idea whether either an Alcubierre drive or its various relatives, like the Krasnikov (sp?) tube are even theoretically possible. I will walk way out onto the metaphorical limb and say that the existence of an FTL drive would not imply a breakdown of causality; it would mean that something is wrong with the way we are interpreting SR and GR.

grant hutchison
2010-Apr-06, 12:14 AM
I've no idea whether either an Alcubierre drive or its various relatives, like the Krasnikov (sp?) tube are even theoretically possible. I will walk way out onto the metaphorical limb and say that the existence of an FTL drive would not imply a breakdown of causality; it would mean that something is wrong with the way we are interpreting SR and GR.Or it might mean that something is wrong with the way we are interpreting causality.
Kip Thorne's team wrote a couple of papers in the early nineties, feeling towards the idea that time-travel induced causal loops may be possible, but only if they produce consistent (ie, non-paradoxical) results. I haven't followed the development of this idea since: I've no idea if it has flown or not.

Cauchy problem in spacetimes with closed timelike curves (http://resolver.caltech.edu/CaltechAUTHORS:FRIprd90)
Billiard balls in wormhole spacetimes with closed timelike curves: Classical theory (http://resolver.caltech.edu/CaltechAUTHORS:ECHprd91)

Grant Hutchison

Sophic
2010-Apr-06, 06:55 PM
After reading through the 5 or so pages of this I'm going to try to simplify this to make sense, and anybody here can correct me where I'm wrong.

I'm going to make an example, if there is a massive power outage on the US east coast and the power comes back on at 10 PM, and you have three observers all watching Earth through a telescope, one on Mars, one on Pluto, and one at a distance of Alpha Centauri. On Mars and Pluto they have watches that are set to the exact same time as Earth, on Mars an observer would see the east coast light back up at 10:03 PM (3 light minutes away), and on Pluto he would he see it 3 AM (5 light hours away). All the way on Alpha Centauri it would take 4 years before the observer sees that event.

This is the point of my story, the event still happened at 10:00 PM regardless what anybodies clock says, it just took longer for each observer farther out to see it. It happened at 10:03 PM or 3 AM to that persons frame of reference, but in reality it happened at 10:00 PM.

The point of my post is that if a ship could be instantly transported from Earth to Pluto, lets say through a wormhole. The ship's crew member could then turn around at Pluto and watch himself board the ship and go through the wormhole. To an outside observer he would appear to be in two places at once, but this would be nothing more than an illusion caused by how fast he traveled, because he is actually on Pluto.

I guess to make an example, although not a good one, it would be like moving faster than the speed of sound, that you can see an object before you hear, moving faster than light would be like getting there before you can see it, if that makes sense. I wouldn't say this would be time travel, its simply an illusion to somebody watching because of how long it takes light to pass through space.

The ship crewman couldn't send messages to himself because it already happened, he couldn't influence events in the past because all he is looking at is light, even if he caught up to himself during the journey he would simply be an observer. Kinda like watching a 3-D video of his own journey.

pzkpfw
2010-Apr-06, 08:52 PM
This is the point of my story, the event still happened at 10:00 PM regardless what anybodies clock says, it just took longer for each observer farther out to see it. It happened at 10:03 PM or 3 AM to that persons frame of reference, but in reality it happened at 10:00 PM.

The problem isn't about when everyone sees an event (i.e. when the light from that event reaches their eyes). Time isn't the same for everyone, when they are in relative motion. As soon as you have instant transportation (as seen by someone not travelling*) you introduce the possibility for causality violations. See post #16. (Substitute Earth and Pluto for the space ships, if you wish).



(* by which I mean - a traveller going at exactly the speed of light (if that were possible) who essentially has time stop for them and therefore experiences "instant travel" - isn't an issue.)

grant hutchison
2010-Apr-06, 09:15 PM
I wouldn't say this would be time travel, its simply an illusion to somebody watching because of how long it takes light to pass through space.All you say is correct, but that isn't the problem with FTL travel. As I described earlier, it raises the possibility of someone genuinely returning to their own past; being in two places at the same time, or in the same place twice at the same time.

Grant Hutchison

astromark
2010-Apr-06, 09:18 PM
Welcome; 'Sophic' I like the way you have said that... It is as I understand it and find tolerance of instant as a fiction only. You all know that.
It would be truly great if a method could be proposed for beating the laws of reality... Understanding 'Pzkpwf's' point as true.. its about your point of reference and for that I will wait and comment no further...

Sophic
2010-Apr-06, 09:36 PM
I guess my question is, would that person actually be in the past, or would it just be an illusion of the past? Could he manipulate events of the past, or would he just be able to observe them since they already happened? Maybe, I'm just headed to far into the realm of fiction and none of this is possible anyway.

pzkpfw
2010-Apr-07, 12:16 AM
I guess my question is, would that person actually be in the past, or would it just be an illusion of the past? Could he manipulate events of the past, or would he just be able to observe them since they already happened? Maybe, I'm just headed to far into the realm of fiction and none of this is possible anyway.

In your simple example, no the person wouldn't be in the past, but the Universe isn't that simple.

Earth and Pluto are in relative motion, so they do not experience the same time as each other.

In your example you have the Pluto person seeing the Earth event at 10:03 AM (10:00 AM + 00:03 minutes flight time of the light of the event). But by whose clock is that 00:03 minutes measured?

It turns out that an Earth observer and Pluto observer won't agree that that light reached Pluto at 10:03 AM, because the Earth and Pluto observer both know that the other has a slower clock. (The difference would be too little to really notice, but that's not the point).

So what if an Earth observer "instantly" went to Pluto at 10:03 AM on the Earth clock. They'd arrive at Pluto at the "instant" that Earth thinks it is 10:03 AM - but the Pluto clock won't be 10:03 AM. (It'll be 10:02.99999999999 AM).

Vice versa, if a Pluto traveller then goes "instantly" to Earth, when the Pluto clock says 10:02.9999999 AM, what time is it on Earth? It's 10:02.8888888888 AM.

So the Pluto traveller gets to Earth at 10:02.88888888 AM - and stops the Earth travellor going to Pluto. This game of tag won't work!


(The actual numbers are made up, of course, for illustration; but I believe the effect to be accepted mainstream science.)

Sophic
2010-Apr-07, 12:32 AM
In your simple example, no the person wouldn't be in the past, but the Universe isn't that simple.

Earth and Pluto are in relative motion, so they do not experience the same time as each other.

In your example you have the Pluto person seeing the Earth event at 10:03 AM (10:00 AM + 00:03 minutes flight time of the light of the event). But by whose clock is that 00:03 minutes measured?

It turns out that an Earth observer and Pluto observer won't agree that that light reached Pluto at 10:03 AM, because the Earth and Pluto observer both know that the other has a slower clock. (The difference would be too little to really notice, but that's not the point).

So what if an Earth observer "instantly" went to Pluto at 10:03 AM on the Earth clock. They'd arrive at Pluto at the "instant" that Earth thinks it is 10:03 AM - but the Pluto clock won't be 10:03 AM. (It'll be 10:02.99999999999 AM).

Vice versa, if a Pluto traveller then goes "instantly" to Earth, when the Pluto clock says 10:02.9999999 AM, what time is it on Earth? It's 10:02.8888888888 AM.

So the Pluto traveller gets to Earth at 10:02.88888888 AM - and stops the Earth travellor going to Pluto. This game of tag won't work!


(The actual numbers are made up, of course, for illustration; but I believe the effect to be accepted mainstream science.)

Thanks for the answer, so he would essentially being traveling in time, and so there isn't any possible way to have FTL travel without time travel, correct?

Webbo
2010-Apr-07, 01:44 PM
In your simple example, no the person wouldn't be in the past, but the Universe isn't that simple.

Earth and Pluto are in relative motion, so they do not experience the same time as each other.

In your example you have the Pluto person seeing the Earth event at 10:03 AM (10:00 AM + 00:03 minutes flight time of the light of the event). But by whose clock is that 00:03 minutes measured?

It turns out that an Earth observer and Pluto observer won't agree that that light reached Pluto at 10:03 AM, because the Earth and Pluto observer both know that the other has a slower clock. (The difference would be too little to really notice, but that's not the point).

So what if an Earth observer "instantly" went to Pluto at 10:03 AM on the Earth clock. They'd arrive at Pluto at the "instant" that Earth thinks it is 10:03 AM - but the Pluto clock won't be 10:03 AM. (It'll be 10:02.99999999999 AM).

Vice versa, if a Pluto traveller then goes "instantly" to Earth, when the Pluto clock says 10:02.9999999 AM, what time is it on Earth? It's 10:02.8888888888 AM.

So the Pluto traveller gets to Earth at 10:02.88888888 AM - and stops the Earth travellor going to Pluto. This game of tag won't work!


(The actual numbers are made up, of course, for illustration; but I believe the effect to be accepted mainstream science.)

Why would it be 10:02.9999999 AM and not 10:03 when they arrive on pluto?

grant hutchison
2010-Apr-07, 01:49 PM
Why would it be 10:02.9999999 AM and not 10:03 when they arrive on pluto?As pzkpfw already said, because Earth and Pluto are in relative motion: they don't agree about clock rates, and the don't agree about simultaneity. Because the relative motion is very small, the disagreement is very small, which is what pzkpfw is indicating by that string of 9s. The actual value of the disagreement would depend on the state of relative motion of the two bodies.

Grant Hutchison

Webbo
2010-Apr-07, 02:02 PM
As pzkpfw already said, because Earth and Pluto are in relative motion: they don't agree about clock rates, and the don't agree about simultaneity. Because the relative motion is very small, the disagreement is very small, which is what pzkpfw is indicating by that string of 9s. The actual value of the disagreement would depend on the state of relative motion of the two bodies.

Grant Hutchison

Couldn't this be positive or negative? And shouldn't the observations of each other time be effected as well therefore negating the causality problem?

grant hutchison
2010-Apr-07, 02:43 PM
Couldn't this be positive or negative? And shouldn't the observations of each other time be effected as well therefore negating the causality problem?It could be positive or negative, depending on their relative state of motion. If they are approaching each other, there is no causality problem for FTL jumps between them; if they are moving apart, problems arise as pzkpfw described. What is instantaneous travel for one is travel into the past for the other.

Grant Hutchison

Webbo
2010-Apr-07, 03:16 PM
It could be positive or negative, depending on their relative state of motion. If they are approaching each other, there is no causality problem for FTL jumps between them; if they are moving apart, problems arise as pzkpfw described. What is instantaneous travel for one is travel into the past for the other.

Grant Hutchison

Surely the movement apart is only a measure of the new distance between them.

If leaving Earth at 10:03 it would arrive on Pluto at 10:03 but due to the new distance between them of 00:03.000000001 it would observe Earth time from Pluto as being 09:59:999999999. If now travelling back to Earth at Pluto time which is still 10:03, it would have a further distance to travel (assuming they are still moving apart at the same velocity). It would still arrive back at Earth at 10:03 but when observing Pluto the time would be 09:59:99999998. It's only the observations that are moving back in time because the distance between the 2 points is increasing and you are overtaking the light signal.

grant hutchison
2010-Apr-07, 03:21 PM
Surely the movement apart is only a measure of the new distance between them.No, it's an issue of the relativity of simultaneity. Because they are in relative motion, they have different measures of simultaneity. Is the concept of relativity of simultaneity something you are familiar with?

Grant Hutchison

Webbo
2010-Apr-07, 05:22 PM
No, it's an issue of the relativity of simultaneity. Because they are in relative motion, they have different measures of simultaneity. Is the concept of relativity of simultaneity something you are familiar with?

Grant Hutchison

Some, but I would still say its an effect of observation as per my example. From any particular reference point the time can be observed to be ahead or behind the other, but you can't interact with it. All you are doing is observing the relative delays to the observer of the light signal. Even if you could get there instantly, you couldn't go back in time, you just go back to time observed now (9:59:99999999 in the example) plus the currrent time light takes to travel the distance (00:03:00000001) = 10:03

grant hutchison
2010-Apr-07, 05:48 PM
Some, but I would still say its an effect of observation as per my example. From any particular reference point the time can be observed to be ahead or behind the other, but you can't interact with it. All you are doing is observing the relative delays to the observer of the light signal.No, that's a separate issue from the relativity of simultaneity. The discussion on this thread (and the causality problems arising from FTL flight) are related to simultaneity, not to time-of-flight effects in signalling.

Grant Hutchison

pzkpfw
2010-Apr-07, 08:29 PM
Some, but I would still say its an effect of observation as per my example. From any particular reference point the time can be observed to be ahead or behind the other, but you can't interact with it. ...

But that's exactly what you are doing by invoking "instant" (or FTL) travel.

Time dilation is real, it's not just a side effect of observation and the delay in seeing things due to having to wait for the sight of those things arriving. Pluto may see things that occured on Earth 3 minutes (by Earth clock) after they occured, but that's not what time dilation is, and is a different issue to Plutos clock being slower than Earths (and vice versa).

Say at 10:03 AM (Earth) the Pluto clock is at 10:02.9999 AM; if an Earth traveller "instantly" travels to Pluto at 10:03 (Earth), then that traveller will be on Pluto when the Pluto clock says 10:02.9999 AM.

The second part of this is that time dilation is reciprical.

If at 10:03 AM (Earth) the Pluto clock is at 10:02.9999 AM
Then at 10:03 AM (Pluto) the Earth clock is at 10:02.9999 AM
(And at 10:02.9999 AM (Pluto) the Earth clock is even earlier ...)

Our traveller went from 10:03 (Earth) to 10:02.9999 (Pluto).
But at 10:02.9999 (Pluto) it will be even earlier on Earth.
So if that traveller "instantly" goes back to Earth as soon as they've arrived on Pluto, they'll get back to Earth sooner than they left.


The delays in anyone seeing any of this is not relevant.

Webbo
2010-Apr-07, 08:52 PM
But that's exactly what you are doing by invoking "instant" (or FTL) travel.

Time dilation is real, it's not just a side effect of observation and the delay in seeing things due to having to wait for the sight of those things arriving. Pluto may see things that occured on Earth 3 minutes (by Earth clock) after they occured, but that's not what time dilation is, and is a different issue to Plutos clock being slower than Earths (and vice versa).

Say at 10:03 AM (Earth) the Pluto clock is at 10:02.9999 AM; if an Earth traveller "instantly" travels to Pluto at 10:03 (Earth), then that traveller will be on Pluto when the Pluto clock says 10:02.9999 AM.

The second part of this is that time dilation is reciprical.

If at 10:03 AM (Earth) the Pluto clock is at 10:02.9999 AM
Then at 10:03 AM (Pluto) the Earth clock is at 10:02.9999 AM
(And at 10:02.9999 AM (Pluto) the Earth clock is even earlier ...)

Our traveller went from 10:03 (Earth) to 10:02.9999 (Pluto).
But at 10:02.9999 (Pluto) it will be even earlier on Earth.
So if that traveller "instantly" goes back to Earth as soon as they've arrived on Pluto, they'll get back to Earth sooner than they left.


The delays in anyone seeing any of this is not relevant.

I didn't think we were discussing time dilation and don't understand why it's relevant here.

Exactly how do you establish that the time 10:03 at one point and 10:02:9999 at the other? From one point all you can establish is that the time is 10:03 here (Earth) and observed to be 10:00 at the other (Pluto) or after the trip 10:03 here (now Pluto) and observed 09:59:9999 at the other (now Earth) if you assume they are moving apart.

pzkpfw
2010-Apr-07, 09:03 PM
I didn't think we were discussing time dilation and don't understand why it's relevant here.

The thread is about FTL travel and why it breaks causality.

Causality is broken by FTL (or "instant") travel due to the effects of time dilation.


Exactly how do you establish that the time 10:03 at one point and 10:02:9999 at the other?

That's just the way the Universe works. Two things in relative motion will each know the others clock is slower than their own. This is not illusion.

While they may have somehow synchronised at 10:00 AM, by the time each gets to 10:03 AM, the others clock will be lagging behind.


From one point all you can establish is that the time is 10:03 here (Earth) and observed to be 10:00 at the other (Pluto) or after the trip 10:03 here (now Pluto) and observed 09:59:9999 at the other (now Earth) if you assume they are moving apart.

It's not about observation, it's about what's actually going on.

At 10:03 AM an Earth observer might see (with a strong telescope) a clock on Pluto that reads 10:00 AM (because it took 3 minutes (Earth time) for the image of that clock to get to Earth).

But that observer knows that time has passed on Pluto in the meantime. He or she is seeing 10:00 AM on the clock, but knows the clock "now" (of course "now" gets hard to define) the clock will have moved on.

Without time dilation, the observer might assume that on Pluto the clocks are now also reading 10:03 AM. (And for most purposes that's close enough).

But if that telescope user, at 10:03 AM on Earth "instantly" travelled to Pluto - what time is it really on Pluto? Turns out, it's not quite 10:03 AM. Plutos clock was slower than Earths (and vice versa).

Webbo
2010-Apr-07, 09:03 PM
No, that's a separate issue from the relativity of simultaneity. The discussion on this thread (and the causality problems arising from FTL flight) are related to simultaneity, not to time-of-flight effects in signalling.

Grant Hutchison

From what I can understand of the relativity of simultaneity it appears to me to be an effect of signalling and the relative distance from origin of the signals. All you can do is observe the situation when they arrive at your observation point, where contradictions & causality problems can be evident. However, there is no way of interacting with the signals prior to emission even if you could travel instantly to the source. Just because I can see an event that happened 3 mins ago from 3 light minutes away, I can still only travel 3 light minutes in an instant and can therefore only "go back" a maximum of 3 minutes so I can't arrive before the signal I have just seen.

Webbo
2010-Apr-07, 09:11 PM
That's just the way the Universe works. Two things in relative motion will each know the others clock is slower than their own. This is not illusion.

I gues I am struggling with how they know each others clock will be slower. My understanding is that they only observe each other clock to be slower. However, this appears to be different to the effect I was discussing with Grant who agreed that the effect can be positive or negative depending on their relative direction.

pzkpfw
2010-Apr-07, 09:36 PM
I gues I am struggling with how they know each others clock will be slower.

I'm not sure why this matters. It's the way the Universe works, so that's what will happen.


My understanding is that they only observe each other clock to be slower.

Nope. Time dilation is real. Different observers will experience different time.


However, this appears to be different to the effect I was discussing with Grant who agreed that the effect can be positive or negative depending on their relative direction.

I'm not sure about this myself. I'm still thinking through the +ve/-ve effect claim. (My understanding is that observers in relative motion will each see/know/understand/calculate that the others clock is slower. I didn't think the direction of that relative motion mattered, just the amount of relative motion would affect the amount of time dilation.) Still, this doesn't affect the results above.

Webbo
2010-Apr-07, 09:49 PM
I'm not sure why this matters. It's the way the Universe works, so that's what will happen.



Nope. Time dilation is real. Different observers will experience different time.



I'm not sure about this myself. I'm still thinking through the +ve/-ve effect claim. (My understanding is that observers in relative motion will each see/know/understand/calculate that the others clock is slower. I didn't think the direction of that relative motion mattered, just the amount of relative motion would affect the amount of time dilation.) Still, this doesn't affect the results above.

I still dont think this is a time dilation issue hence the confusion.

Lets try the same example but this time we start with 2 syncronised clocks on 2 adjacent ships in deep space. Both ships now travel exactly 1.5 minutes away from this start point in opposite directions at the same speed, acceleration, deceleration and stop. Now do the experiment from 1 ship to the other as I believe there shouldn't be any time dilation issues or relativity of simultaneity for that matter.

grant hutchison
2010-Apr-07, 09:50 PM
From what I can understand of the relativity of simultaneity it appears to me to be an effect of signalling and the relative distance from origin of the signals. All you can do is observe the situation when they arrive at your observation point, where contradictions & causality problems can be evident. However, there is no way of interacting with the signals prior to emission even if you could travel instantly to the source. Just because I can see an event that happened 3 mins ago from 3 light minutes away, I can still only travel 3 light minutes in an instant and can therefore only "go back" a maximum of 3 minutes so I can't arrive before the signal I have just seen.Then you're misunderstanding simultaneity under SR. A line of simultaneity can be drawn connecting all events that an observer considers simultaneous, after allowing for the time it takes signals to travel the intervening distance. Simultaneity is about spacetime coordinates, not about signal lag.
Each observer has his own standard of simultaneity, and that will not match the standard used by another observer in relative motion.
Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity4.jpg) is a spacetime diagram for two observers who are moving apart in our reference frame. Time in our reference frame is plotted vertically, space horizontally. The red arrow pointing up and left is the worldline of the "red observer", who is moving to the left with increasing time; the blue arrow is the worldline of the "blue observer", who is moving rightwards at the same speed. The slanting lines without arrowheads mark simultaneity as understood by each observer: the red lines connect events the red observer considers simultaneous after allowing for signal delays; the blue lines mark events that the blue observer considers simultaneous after allowing for signalling delays.
Someone travelling with the blue observer who makes an "instantaneous" jump will travel along a blue simultaneity line. But the red observer finds that this jump travels backwards in time as it approaches his location: it intersects progressively earlier red simultaneity lines as it approaches the red observer's worldline.
Our jumper can now match velocity with the red observer, and jump back according to the red observer's simultaneity: now he travels backwards in time according to the blue observer. An "instantaneous" jumper who follows the upper blue simultaneity line has time to match velocities with the red observer and then jump back along the lower red simultaneity line. He can therefore carry a signal into the blue observer's past.

There are no light signals in the diagram. Simultaneity is a coordinate issue, not a signalling issue.

Grant Hutchison

pzkpfw
2010-Apr-07, 09:56 PM
Lets try the same example but this time we start with 2 syncronised clocks on 2 adjacent ships in deep space. Both ships now travel exactly 1.5 minutes away from this start point in opposite directions at the same speed, acceleration, deceleration and stop. Now do the experiment from 1 ship to the other as I believe there shouldn't be any time dilation issues or relativity of simultaneity for that matter.

If they are not in relative motion, then they are in the same frame and there won't be time dilation issues.

But finding one scenario where there may be no issues does not make FTL possible.

What would happen if your two ships were in relative motion?

Webbo
2010-Apr-07, 10:08 PM
Then you're misunderstanding simultaneity under SR. A line of simultaneity can be drawn connecting all events that an observer considers simultaneous, after allowing for the time it takes signals to travel the intervening distance. Simultaneity is about spacetime coordinates, not about signal lag.
Each observer has his own standard of simultaneity, and that will not match the standard used by another observer in relative motion.
Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity4.jpg) is a spacetime diagram for two observers who are moving apart in our reference frame. Time in our reference frame is plotted vertically, space horizontally. The red arrow pointing up and left is the worldline of the "red observer", who is moving to the left with increasing time; the blue arrow is the worldline of the "blue observer", who is moving rightwards at the same speed. The slanting lines without arrowheads mark simultaneity as understood by each observer: the red lines connect events the red observer considers simultaneous after allowing for signal delays; the blue lines mark events that the blue observer considers simultaneous after allowing for signalling delays.
Someone travelling with the blue observer who makes an "instantaneous" jump will travel along a blue simultaneity line. But the red observer finds that this jump travels backwards in time as it approaches his location: it intersects progressively earlier red simultaneity lines as it approaches the red observer's worldline.
Our jumper can now match velocity with the red observer, and jump back according to the red observer's simultaneity: now he travels backwards in time according to the blue observer. An "instantaneous" jumper who follows the upper blue simultaneity line has time to match velocities with the red observer and then jump back along the lower red simultaneity line. He can therefore carry a signal into the blue observer's past.

There are no light signals in the diagram. Simultaneity is a coordinate issue, not a signalling issue.

Grant Hutchison

But its all about what the observer sees. They cannot see what is happening now at all or any coordinate apart from their own. Everything else is deduced and infered from light signals. You say there are no light signals then what exactly are they observing. It even contains the phrase "after allowing for signal delays".

Webbo
2010-Apr-07, 10:15 PM
If they are not in relative motion, then they are in the same frame and there won't be time dilation issues.

But finding one scenario where there may be no issues does not make FTL possible.

What would happen if your two ships were in relative motion?

OK so you agree there are no causality issues in this scenario. I never proposed FTL was possible. The FTL in the example was/is hypothetical I assume.

I thought we already tried it with relative motion in the planets scenario, although I still think this is just a signaling issue and you couldn't travel to the past in this way.

grant hutchison
2010-Apr-07, 10:29 PM
But its all about what the observer sees.No, it's not. That's a separate issue. You won't understand relativity of simultaneity until you stop thinking about what the observer sees, and start thinking about the coordinates.


They cannot see what is happening now at all or any coordinate apart from their own. Everything else is deduced and infered from light signals.Exactly. That's what I said. And then we discard the light signals so that we can understand what the coordinates are doing. The coordinates tell us what happens simultaneously; the light signals tell us only which events look simultaneous.

Grant Hutchison

grant hutchison
2010-Apr-07, 10:51 PM
By the way, you can see from my diagram (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity4.jpg) how reciprocal time dilation arises from relativity of simultaneity. Since the red and blue observers are moving with equal and opposite speed in our reference frame, their proper time elapses at the same rate along each worldline in the diagram. Imagine that they have synchronized their clocks as they passed each other, just off the bottom of the diagram where the red and blue worldlines would cross. Because the simultaneity lines of each observer slope into the past of the other observer, each finds that the other's clock has ticked off less time than his own, since the synchronization. (Again, this is a fact about the coordinates, which each observer can deduce once he allows for light travel time.)

But what about when they were approaching each other? Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity5.jpg) is the relevant diagram, with the red observer coming in from the right, and blue converging from the left. Simultaneity lines slope as before. But now the simultaneity lines of each observer slope into the other's future. Each judges that the other has less time to go before the synchronization event, which will occur just off the top of the diagram. So each, again, finds that the other's clock runs slower than his own. Mutual time dilation, as before. But on this occasion, with the observers approaching each other, the instantaneous jumps travel futurewards, and don't create causal paradoxes for these two observers.

Grant Hutchison

pzkpfw
2010-Apr-08, 12:15 AM
I thought we already tried it with relative motion in the planets scenario, although I still think this is just a signaling issue and you couldn't travel to the past in this way.

Try describing the scenario in detail, with clearly defined steps in the reasoning...

pzkpfw
2010-Apr-08, 12:23 AM
Thanks Grant.


... (a) So each, again, finds that the other's clock runs slower than his own. ... (b) the instantaneous jumps travel futurewards ...

I'm haing trouble reconciling these two statements. "Why" do the simultaneity lines go "forward" in the second diagram?

JMV
2010-Apr-08, 03:28 AM
"Why" do the simultaneity lines go "forward" in the second diagram?
If you look at the diagrams you'll see that the simultaneity lines slant in the same direction in both diagrams. Only difference is the worldlines have switched places.



I'm haing trouble reconciling these two statements.
That's because you still seem to think the time travel problem arises from time dilation. It's doesn't. Time dilation is just what happens to the distance between the intersection points of two parallel simultaneity lines and a worldline when the worldline is not parallel to the other worldline to whom the simultaneity lines belong to. Sounds confusing, but look at Grant's diagrams and you'll see what I mean.

How you can reconcile the two statements is by forgetting time dilation and noting which way the simultaneity line slants towards to, depending on the direction of travel. [color=blue]Towards the future to the right of a right-going worldline and towards the past to the left of a right-going worldline. Correspondingly, towards the future to the left of a left-going worldline and towards the past to the right of a left-going worldline.

When you switch places of a right-going worldline and a left-going worldline, the points at which the simultaneity lines intersect the other worldline switch from future to the past or vice versa depending on whether the worldlines are pointing towards or away from each other.

Or you can look at the equation for Lorentz transformation for time.

t' = γ ( t - vx/c2 )

See how the term vx/c2 changes sign depending on whether v and x have same signs or not. It is the sign of that term that determines whether we are talking about going into the future or into the past.

grant hutchison
2010-Apr-08, 09:44 AM
I'm haing trouble reconciling these two statements. "Why" do the simultaneity lines go "forward" in the second diagram?Sorry this isn't clear. I should have produced one big diagram showing the two observers crossing over, and then it would have been clear that the slope of the simultaneity lines doesn't change at any point. It was a bit of a rush job, though, so I ended up just doing a bit of flipping and reflecting on the first diagram in order to produce something illustrative.

The "uphill" part of the simultaneity line is always in the direction the observer is travelling. So when blue and red approach each other, they are each running into the uphill section of the other's simultaneity; each is therefore simultaneous with the other's future.
After they've crossed over, they are now each moving through the "downhill" part of the other's simultaneity; each is therefore simultaneous with the other's past. So the pastward link is only a problem (for these observers) when they are moving apart.

Grant Hutchison

Webbo
2010-Apr-08, 01:35 PM
I had a read of Einsteins description of The Relativity of Simultaneity and to me it just explains why one reference frame observes 2 events as simultaneous while another will observe as not because the second frame has moved since the light signal was transmitted thereby moving towards one event and away from the other. I still don't see why this create causality problems if you could travel instantly to the source of either signal as they have already been omitted and are in transit. He is only concerned with the observation of these events and is not trying to infer any paradoxes from his example.

Look at this quote;

"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment."

He is specifically avoiding saying whether the 2 events actually occured simultaneously, he is only saying the "rays of light emitted" meet at the midpoint. The whole thing is about the observation of the events and how it differs in different frames. That's all. Maybe others have progressed the definition since then but it seems as though they have made more complex issues from a very simple explanation.

grant hutchison
2010-Apr-08, 04:59 PM
Yes, Einstein used the passage of light signals across known distances to set up his definition of simultaneity. The constancy of the speed of light for all observers then leads to his result that observer in relative motion disagree about simultaneity. He does this by telling us how different observers synchronize their clocks using light signals.

This disagreement about simultaneity then creates the problem that observers disagree about the temporal order of spacially separated events, which in turn leads to the potential causality violations of FTL travel. It's all in there with the 1905 bricks of special relativity.

Grant Hutchison

Ken G
2010-Apr-08, 07:01 PM
Perhaps it has been said, but note that the causality problems are not just from going FTL, they are from a combination of that and an assumed symmetry principle that implies "instantaneous travel" has to mean the same thing to all observers in all frames (so either grant hutchison's blue or red observer can follow their own simultaneity lines, and if done in succession, could travel to their own past if they are moving apart as he explained). If one adopts instead the perspective of a preferred frame, say an aether frame that is unseen in special relativity because the Lorentz transformation conceals it (which is perfectly possible but we have no reason to include this view in our theory), then FTL travel in that preferred frame could be possible without causality problems (that's also why the OP could not find causality problems by considering instantaneous travel from only one frame). This tells us that if we ever did achieve FTL travel, yet we did not encounter causality problems, we could study the limitations on our FTL travel to detect the presence of such a preferred frame.

grant hutchison
2010-Apr-08, 07:46 PM
He is specifically avoiding saying whether the 2 events actually occured simultaneously, he is only saying the "rays of light emitted" meet at the midpoint. The whole thing is about the observation of the events and how it differs in different frames. That's all. Maybe others have progressed the definition since then but it seems as though they have made more complex issues from a very simple explanation.Going back to 1905 and On the electrodynamics of moving bodies (http://www.fourmilab.ch/etexts/einstein/specrel/www/), we find Einstein being quite specific about simultaneity, and how it is defined using light signals (Section 1, Definition of Simultaneity):
We have so far defined
only an “A time” and a “B time.” We have not defined a common “time” for
A and B, for the latter cannot be defined at all unless we establish by definition
that the “time” required by light to travel from A to B equals the “time” it
requires to travel from B to A. Let a ray of light start at the “A time” tA from
A towards B, let it at the “B time” tB be reflected at B in the direction of A,
and arrive again at A at the “A time” t'A.
In accordance with definition the two clocks synchronize if
tB − tA = t'A − tB.
We assume that this definition of synchronism is free from contradictions,
and possible for any number of points; and that the following relations are
universally valid:—
1. If the clock at B synchronizes with the clock at A, the clock at A syn-
chronizes with the clock at B.
2. If the clock at A synchronizes with the clock at B and also with the clock
at C, the clocks at B and C also synchronize with each other.
Thus with the help of certain imaginary physical experiments we have set-
tled what is to be understood by synchronous stationary clocks located at dif-
ferent places, and have evidently obtained a definition of “simultaneous,” or
“synchronous,” and of “time.”
Grant Hutchison

RAMS57
2010-Apr-09, 01:11 AM
Actually these type of suppositions always are conducive to the question of 'Time'. Since anything beyond 1LY falls apart in our own snail oike pace of time event management (put another way, we today would not tolerate taking 10 months to go from St. Louis To LA as they did 130 years ago, since we can now pick up our cell phone and instantly talk-see from one side of the earth to the other.)

Thus, you are missing nothing, actually, since it is the event of time that comes into play. Until we break the Light Speed Barrier with Einstien's Special Loophole in Relativity we are forever marooned in our Time-Place.

Webbo
2010-Apr-09, 12:39 PM
Yes, Einstein used the passage of light signals across known distances to set up his definition of simultaneity. The constancy of the speed of light for all observers then leads to his result that observer in relative motion disagree about simultaneity. He does this by telling us how different observers synchronize their clocks using light signals.

This disagreement about simultaneity then creates the problem that observers disagree about the temporal order of spacially separated events, which in turn leads to the potential causality violations of FTL travel. It's all in there with the 1905 bricks of special relativity.

Grant Hutchison

I dont see that just because the observers disagree about the order of events that it follows on that this can cause causality violations if travelling FTL. The observers are not witnessing the actual order of events, they are witnessing their personal observed order of events. This observation has no bearing on the reality of the situation. If, in the example, the observers were hearing the lightening strike, and they were able to travel to the source of the strike at 0.9999 C they would not arrive there before the sound left, even though they would have travelled exponentially faster then the sound wave.

Ken G
2010-Apr-09, 12:58 PM
I dont see that just because the observers disagree about the order of events that it follows on that this can cause causality violations if travelling FTL. The observers are not witnessing the actual order of events, they are witnessing their personal observed order of events. This observation has no bearing on the reality of the situation.
That's why I stressed the additional importance of an assumed symmetry principle that says if you can do something (like move instantaneously) in one frame, then you can also do it in another frame the same way. That is crucial for using FTL to get causality violations-- it never happens just from FTL travel in a single frame. So in Grant's picture, you take the case of two observers moving away from each other, and you move one to the other, instantaneously in the reckoning of the one (follow Grant's simultaneity lines). Then once with the other, the first does the same thing again to return to his/her original world line, except this time, the instantaneous movement is in regard to the second person's line of simultaneity. So now we don't just have lines of simultaneity (which are nothing but coordinates), we have actual FTL movement reckoned in two different frames (the symmetry principle), and the first person ends up in their own past. That's a causality violation, not just a coordinate issue.

The symmetry principle is called "relativity", and so sometimes the way this is put is:
Relativity, FTL, causality: choose two.

Webbo
2010-Apr-09, 01:17 PM
That's why I stressed the additional importance of an assumed symmetry principle that says if you can do something (like move instantaneously) in one frame, then you can also do it in another frame the same way. That is crucial for using FTL to get causality violations-- it never happens just from FTL travel in a single frame. So in Grant's picture, you take the case of two observers moving away from each other, and you move one to the other, instantaneously in the reckoning of the one (follow Grant's simultaneity lines). Then once with the other, the first does the same thing again to return to his/her original world line, except this time, the instantaneous movement is in regard to the second person's line of simultaneity. So now we don't just have lines of simultaneity (which are nothing but coordinates), we have actual FTL movement reckoned in two different frames (the symmetry principle), and the first person ends up in their own past. That's a causality violation, not just a coordinate issue.

The symmetry principle is called "relativity", and so sometimes the way this is put is:
Relativity, FTL, causality: choose two.

And what if you use sound waves to determine simultaneity? Grants diagram should operate in exactly the same way but as far as I am aware no causality issues would be experienced by travelling faster than sound towards either or both sources.

Webbo
2010-Apr-09, 01:35 PM
I think the key point I am making here is this;

Why, when travelling instantly, does the observer travel along the simultaneity line (after all it's just a path the light has taken between the 2 points)? Shouldn't he just move horizontaly accross the diagram? (ie distance between the observers at 0 time interval).

Ken G
2010-Apr-09, 01:58 PM
And what if you use sound waves to determine simultaneity? Grants diagram should operate in exactly the same way but as far as I am aware no causality issues would be experienced by travelling faster than sound towards either or both sources.That's because there's no analogous symmetry principle for sound waves-- there's no "relativity of sound waves" (sound has a medium, which establishes a clear preferred frame). This again underscores the point that causality problems with FTL are not just a coordinate effect, they are a symmetry effect (a "relativity" effect).

Ken G
2010-Apr-09, 02:00 PM
I think the key point I am making here is this;

Why, when travelling instantly, does the observer travel along the simultaneity line (after all it's just a path the light has taken between the 2 points)? Shouldn't he just move horizontaly accross the diagram? (ie distance between the observers at 0 time interval).That is indeed the key point. To move horizontally is to say that "instantaneous movement" (or anything FTL in general) is to be measured in a single preferred frame. You'll never get causality problems with that type of FTL because it lacks the symmetry principle that allows all observers to use the same prescription for defining what "instantaneous" means. That prescription has to be something they can follow with their own rulers and clocks, not some "absolute" scale like "horizontal in the diagram." This is not a technical detail, it is the beating heart of relativity. Grant's diagram is using this principle implicitly-- the lines of simultaneity are drawn according to a symmetric prescription that any observer can use, and it is only "FTL travel" defined in those terms that creates causality problems.

Webbo
2010-Apr-09, 02:15 PM
That is indeed the key point. To move horizontally is to say that "instantaneous movement" (or anything FTL in general) is to be measured in a single preferred frame. You'll never get causality problems with that type of FTL because it lacks the symmetry principle that allows all observers to use the same prescription for defining what "instantaneous" means. That prescription has to be something they can follow with their own rulers and clocks, not some "absolute" scale like "horizontal in the diagram." This is not a technical detail, it is the beating heart of relativity. Grant's diagram is using this principle implicitly-- the lines of simultaneity are drawn according to a symmetric prescription that any observer can use, and it is only "FTL travel" defined in those terms that creates causality problems.

I still don't see why you are required to travel along the simultaneity line. It's just the path that the light signal has taken. What is so special about it? As I have stated all through this, it all appears to be issues relating to the observation of the light signals and bears no relation to reality. It's almost as if the observer is stating that an event cannot have happened until they have seen it. In there own reference frame that may be true (or at least they cannot prove anything to be true until they have seen it) but that has no impact in the reality of the situation.

Webbo
2010-Apr-09, 02:20 PM
That's because there's no analogous symmetry principle for sound waves-- there's no "relativity of sound waves" (sound has a medium, which establishes a clear preferred frame). This again underscores the point that causality problems with FTL are not just a coordinate effect, they are a symmetry effect (a "relativity" effect).

Sound may have a medium but it would be the consistent throughout the example as it would be for light in a vacuum. Grants diagram is valid for any determination of simultaneity whether the signal is light or sound or any other consistently measurable indicator. However, travelling faster than sound does not mean you must travel along the sound simultaneity line.

Ken G
2010-Apr-09, 02:35 PM
I still don't see why you are required to travel along the simultaneity line. It's just the path that the light signal has taken. What is so special about it?What is special about it is revealed by experiment: using this approach to simultaneity, the symmetry principle I'm talking about actually works-- the laws of physics are the same in all frames (inertial frames, if we're doing SR). That is not true for sound waves-- if you use sound waves to establish simultaneity with people moving through air, you will not get that the laws of physics work the same for everyone. The people moving through air will notice decidedly different physics if they use your coordinatization of time. It is quite an interesting exercise to see how that plays out. Let's say you draw Grant's picture but using sound waves instead of light, and you define "instantaneous" travel to mean movement along a line of simultaneity. Let's also say that the blue observer is moving at speed v with respect to air, and the red observer is moving at speed -v. If they both use sound waves to draw their simultaneity lines, the picture will not look like Grant's, because a funny thing happens to the sound waves red and blue emit-- they do not move at the speed of sound relative to red and blue. If you work it out, you will find that lines of simultaneity are indeed horizontal if we use sound instead of light (ignoring corrections on the order of the square of the ratio of the speed of sound to the speed of light).

What this means is, the slanting lines of simultaneity that Grant drew are not just a matter of his coordinate instructions, they have to do with how those coordinate instructions interact with reality, and are seen in the "central frame" where blue and red are moving away from each other (the frame of the figure axes). That's also why the causality violations are real, if instantaneous travel in the red or blue frame means movement along their own simultaneity lines, and that holds for each frame. In the sound case, such travel will be horizontal in the figure-- and we're back to "instantaneous" meaning only one thing, rather than a bunch of different things in different frames.


As I have stated all through this, it all appears to be issues relating to the observation of the light signals and bears no relation to reality. That's because relativity is quite a bit more subtle than people normally imagine. It is not really a statement about what light does, it is a statement about our desire to have a certain symmetry principle among observers. We say "we want this symmetry", and reality says "OK, here's how you can have it", where that means a constant speed of light in all frames and a sense of simultaneity mediated by the action of light. Reality gives us those instructions for how to have the symmetry, using light, and we like that, so we use light.

Let me say it another way. Let's say you make a clock by bouncing a sound wave back and forth between two walls. Now a wind blows through your clock-- it keeps a different time. You can turn the clock perpendicular to the wind, or you can turn it into the wind, and it matters-- the time you keep is different each way. But if you make a clock by bouncing a light beam, this is not the case-- you can turn that clock any which way and the time it keeps is the same. We like this symmetry, so we make clocks out of the light waves, but not out of the sound waves.

If we had no such thing as light, we could not make clocks that way, but we would still encounter the symmetry. We could use any good clocks (clocks that exhibit the desired symmetry, which includes atomic oscillators), and turn them any which way, and they would still keep the same time, and using those clocks, we would still find that the laws of physics work the same for any inertial observer if we apply those laws to the readings of clocks and rulers, and we transform those readings between frames using the Lorentz transformation. Since we have laws that work the same for all inertial observers, we then assume that FTL signals would work the same for all inertial observers, and it follows immediately that we could send signals into our own past, with the help of a symmetric observer moving away from us. None of this has anything to do with what the observers see happening at some distance from them-- if the symmetry principle holds, the FTL signal really shows up in their own past, right where they are standing.

JMV
2010-Apr-09, 02:48 PM
(after all it's just a path the light has taken between the 2 points)?
No, it is not. The path that light takes is called light cone in space-time diagrams and Grant hasn't included them into either of the two diagrams in this thread, ie. none of the lines in Grant's diagrams in this thread represent the path that light takes to travel between two points. Simultaneity lines connect the events that actually are simultaneous in the respective rest frames that the simultaneity lines belong to, not events that visually appear to be simultaneous. The slanting of these simultaneity lines is the necessary result of the postulate that speed of light is invariant.


Shouldn't he just move horizontaly accross the diagram? (ie distance between the observers at 0 time interval).
He does move horizontally in a diagram that represents his own rest frame. In other words, the simultaneity lines are horizontal in their respective rest frames. Any events in Grant's diagrams that are situated along a horizontal line are simultaneous only in the frame of the outside observer that the diagram represents.

Webbo
2010-Apr-09, 03:36 PM
No, it is not. The path that light takes is called light cone in space-time diagrams and Grant hasn't included them into either of the two diagrams in this thread, ie. none of the lines in Grant's diagrams in this thread represent the path that light takes to travel between two points. Simultaneity lines connect the events that actually are simultaneous in the respective rest frames that the simultaneity lines belong to, not events that visually appear to be simultaneous. The slanting of these simultaneity lines is the necessary result of the postulate that speed of light is invariant.


He does move horizontally in a diagram that represents his own rest frame. In other words, the simultaneity lines are horizontal in their respective rest frames. Any events in Grant's diagrams that are situated along a horizontal line are simultaneous only in the frame of the outside observer that the diagram represents.

Sorry but I just cant follow this. If you are stating that the simultaneity lines connects events that actually are simultaneous then observer A would determine that the observer B's past is simultaneous with their own present. This is clearly wrong. In addition Einstein clearly states in his example that the observers are not witnessing the timing of the actual events, they are just witnessing the point where the light signals meet.

"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment."

Also, Grants description of his diagram was from the perspective of each observer so it it those observers experiences that I am that I am querying about the simultaneity lines. Why do they travel along the simultaneity lines as he stated?

Ken G
2010-Apr-09, 03:47 PM
Sorry but I just cant follow this. If you are stating that the simultaneity lines connects events that actually are simultaneous then observer A would determine that the observer B's past is simultaneous with their own present. He was making sure you recognized that the lines in Grant's diagram are lines of simultaneity, in Einstein coordinates (which respect the desired symmetry I've been talking about), not the paths of light rays. At one point you made a comment where it sounded like you thought those were intended to be light paths.

Also, Grants description of his diagram was from the perspective of each observer so it it those observers experiences that I am that I am querying about the simultaneity lines. Why do they travel along the simultaneity lines as he stated?JMV is saying that the figure itself is taken from the perspective of an observer in a frame "between" blue and red, i.e., blue moves at v and red at -v. It is still intended to talk about events occuring to blue and red, it is merely the perspective of the figure that is intermediate. The figure does not tell you how things look to blue and red, in a literal sense, it tells you how they reckon the events, using their own Einstein coordinates. The coordinates don't really matter-- what matters is the symmetry principle, and the invariance of the causality violation that results.

Webbo
2010-Apr-09, 04:03 PM
He was making sure you recognized that the lines in Grant's diagram are lines of simultaneity, in Einstein coordinates (which respect the desired symmetry I've been talking about), not the paths of light rays. At one point you made a comment where it sounded like you thought those were intended to be light paths.
Then what do they represent? They can't represent actual simultaneous events as we know that the 2 observers met earlier and the same amount of time and distance has progressed for each. How can observer A's present be simultaneous with observer B's past and vice versa?


JMV is saying that the figure itself is taken from the perspective of an observer in a frame "between" blue and red, i.e., blue moves at v and red at -v. It is still intended to talk about events occuring to blue and red, it is merely the perspective of the figure that is intermediate. The figure does not tell you how things look to blue and red, in a literal sense, it tells you how they reckon the events, using their own Einstein coordinates. The coordinates don't really matter-- what matters is the symmetry principle, and the invariance of the causality violation that results.

From his description Grant talks about this from the perspective of each observer not the third party reading the diagram.

Ken G
2010-Apr-09, 04:47 PM
Then what do they represent? They can't represent actual simultaneous events as we know that the 2 observers met earlier and the same amount of time and distance has progressed for each. How can observer A's present be simultaneous with observer B's past and vice versa?The blue (or red) parallel lines represent all the events that blue (or red) will reckon as simultaneous, in a spacetime diagram built in the coordinates of a third observer "between" red and blue. It seems there are some basic concepts in relativity that you need to get first:

In SR, using Einstein coordinates (setting clocks using the prescription Grant quoted), a symmetry appears where any inertial observer can use their own Einstein coordinates and tell a consistent "tale" about what is happening around them, where by "consistent" I mean that it's all the same equations of physics, and the outcomes of their measurements can be translated back and forth between observer frames using the Lorentz transformation. What is not present in that symmetry is that two obsevers whose clocks were synchronized on passing will continue to think their clocks are synchronized (each will think the other's clock is lagging theirs, or more correctly, is lagging their Einstein coordinates at the location of the other's clock, but lagging in a symmetric way). Also, what is not present in the symmetry is that distant events that one observer reckons (not sees) as simultaneous (indicated by Grant's lines), the other will not so reckon. That is called "relativity of simultaneity", and is what Grant's figures shows-- the way red and blue reckon simultaneity is different.

Note that reckoning simultaneity just means matching up events in spacetime, and labeling them as "simultaneous". It is a rather arbitrary process, and quite coordinate dependent, but the causality attribute that is preserved is that if one observer reckons two events as simultaneous, another may not, but the other will never reckon the two events as being causally connected either. So the actual meaning of "simultaneity" is pretty unimportant, what matters is "spacelike separated" (acausal) versus "timelike separated" (causal), and those classes are not mixed by the different observers (unless FTL travel is possible in a symmetric way, as we've discussed).



From his description Grant talks about this from the perspective of each observer not the third party reading the diagram.Again, the diagram itself takes the perspective of an "intermediate" third observer, that's what defines the vertical (time for that observer) and horizontal (space for that observer) in the diagram. Nevertheless, use of that third observer is just a device-- what is aimed at is understanding the events as they happen to red and blue, merely from a more "fair" intermediate perspective that sees red and blue in a symmetric way. The same diagram could be done from either red or blue's perspective, and then their own simultaneity lines would be horizontal, but I think Grant made a good choice by underscoring the symmetry between red and blue, as that is the crux of the whole causality violation issue.

Webbo
2010-Apr-09, 05:06 PM
OK. I think I am destined to never understand what to make of this diagram. Just to be clear on one thing though, if the observers were neither moving apart or towards each other, then in the diagram the observers would now have vertical lines pointing upwards and the lines of simultaneity would be horizontal lines straight accross. Am I therefore correct in thinking that in this scenario instant travel would not produce any causality violations?

grant hutchison
2010-Apr-09, 05:18 PM
OK. I think I am destined to never understand what to make of this diagram. Just to be clear on one thing though, if the observers were neither moving apart or towards each other, then in the diagram the observers would now have vertical lines pointing upwards and the lines of simultaneity would be horizontal lines straight accross. Am I therefore correct in thinking that in this scenario instant travel would not produce any causality violations?In that case the observers would share a common rest frame, and would agree about simultaneity. So they couldn't get into each other's past as I've been describing. But some third observer, moving relative to this pair, would then be able to observe ambiguity in the temporal order of these instantaneous journeys (arriving before they departed, for instance). So as soon as you have someone performing instantaneous journeys, you are in principle able to find a moving reference frame from which someone else could observe causality violations.

And just to confirm what others have told you, the simultaneity lines in my diagrams connect events which are actually simultaneous, not events which are seen simultaneously. There are no light rays in those diagrams. Observers in relative motion slice spacetime differently: what is simultaneous for you is not simultaneous for me.

Grant Hutchison

grav
2010-Apr-09, 06:57 PM
OK. I think I am destined to never understand what to make of this diagram. Okay then, let's try the math. I'll keep it simple. All observers measure the speed of light to be c in all directions. So if we have one observer A measure the speed of light to be c in both directions along the x axis and now we'll add another observer B travelling in the positive x direction at a relative speed to A of v, then if B were to measure the speed of light ballistically, then since A measures c in both directions along the length of B's ship, then B would measure c-v for the light when travelling from the back of the ship to the front and c+v for the light travelling from the front of the ship to the back. However, stellar aberration and experiments show that light is always measured to travel at the same speed to an observer regardless of the motion of the source, so B must also measure the speed of light as the same in both directions along the length of B's ship also. In order to do this, then according to A's perspective, the clock at the back of B's ship must be set further ahead than the clock in the front of B's ship. That way, B will measure the same time for the light to travel from the back to the front as from the front to the back using the difference in the times when a pulse passes each point, both over the same distance, so measures the speed of light as the same in both directions. So observer A will see a simultaneity shift between the clocks at the front and back of B's ship but observer B says they are synchronized because the light takes the same time to travel in either direction according to the difference in times that pass upon B's own clocks. The way SR works out, according to A, B's clocks are also time dilating and the length of B's ship is contracted in such a way that B also measures the speed of light as c in all directions, and this goes for all observers, so none can claim a preferred frame of reference because they all measure the same things of light and of each other.

So if B says B's own clocks are synchronized, while A does not, then according to A, if a traveller could be sent fast enough from the back of B's ship to the front, where the clocks at the front of the ship read a lesser time than at the back, then the traveller would be travelling into B's past as far as B is concerned, because he is departing the back of B's ship at a greater time than he is arriving at the front. This in itself won't produce a causality violation, when only jumping once or many times in respect to the same frame, but if the traveller were to then jump faster than light in respect to another frame, then the traveller could arrive back at the departure point before he left. For instance, the way simultaneity works out with SR, A sees the clock at the back of B's ship set to a greater reading of tl = d v / c^2 than the clock at the front of B's ship, where d is the length of B's ship as B measures it. So if from the perspective of A, the traveller were to instantly jump from the back of B's ship to the front, then the traveller would be travelling back in time according to B's frame by a time of d v / c^2. This also requires some distance to be travelled in B's frame as well, though, so the traveller could never get back to the back of the ship before he left by travelling less than light speed (actually less than c^2 / v even) to B's frame or any positive speed from A's frame. However, since the traveller is now in B's past according to B's frame, if the traveller were to now instantly jump to the back of the ship as would be measured in B's frame, then the traveller would now arrive again at the back of the ship, but at a time of d v / c^2 before he left according to the same clock at the back of B's ship.

Ken G
2010-Apr-09, 07:35 PM
In that case the observers would share a common rest frame, and would agree about simultaneity. So they couldn't get into each other's past as I've been describing. But some third observer, moving relative to this pair, would then be able to observe ambiguity in the temporal order of these instantaneous journeys (arriving before they departed, for instance). So as soon as you have someone performing instantaneous journeys, you are in principle able to find a moving reference frame from which someone else could observe causality violations.I don't think that is actually correct. You can never get causality violations just from instantaneous travel in a single frame, it has to be able to happen in multiple frames. In other words, there has to be "relativity". This is an important point, because we don't really know we have relativity, the observations only tell us we are allowed to imagine we have relativity, so we do imagine it (that's physics). But if there were "really" an absolute frame that instantaneous travel could occur in, and only in that frame (other frames would be subservient to the concept of FTL travel in the preferred frame, and no "faster than instantaneous" travel was allowed in the preferred frame), we could still have all the same observations we now have, if the Lorentz transformation still applied (for whatever reason, no longer because of relativity). What's more, there would never be any causality violations. There would be inertial observers who think signals went backward in coordinate time (but we have that already with noninertial observers in SR), but causality violations could be avoided by avoiding them in the "preferred" frame. That's why to get causality violations, you need the symmetry principle to hold between at least two frames-- you need to reckon your FTL travel (most easily, instantaneous travel) in a symmetric way from two different frames (and as you showed, they even have to be diverging symmetric observers, which I think is really interesting and trying to tell us something we haven't yet gleaned).

grant hutchison
2010-Apr-09, 07:53 PM
I don't think that is actually correct.I'm considering this as a situation in which the two ends of the instantaneous journey can be clearly labelled "cause" and "effect", for instance because the traveller experiences some proper time during the journey. And I used the phrase "causality violation" to label the situation of the observer who measures "cause" and "effect" to be temporally reversed. Are we at cross-purposes because of terminology? Or are you telling me something else?

Grant Hutchison

Ken G
2010-Apr-09, 08:43 PM
I'm considering this as a situation in which the two ends of the instantaneous journey can be clearly labelled "cause" and "effect", for instance because the traveller experiences some proper time during the journey. And I used the phrase "causality violation" to label the situation of the observer who measures "cause" and "effect" to be temporally reversed. Are we at cross-purposes because of terminology? Or are you telling me something else?
I think I'm telling you something else. It is true that in your scenario, there will be inertial observers who reckon causes as preceding effects, but that is just a coordinate time re-ordering problem, it's not a causality violation (don't reify...). A true causality violation is when a cause creates an effect which can then be a cause of the original cause! That won't happen if there is only instantaneous travel allowed in one special frame (i.e., no symmetry principle), as long as we assert that any process takes some time (so a process cannot be its own effect and thus its own cause!).

The way to prove this is simply to assert that non-coordinate types of causality violations must be invariants, they must be coordinate independent. As such, they can be reckoned in any frame, including the preferred frame. So as long as causality is protected in that one frame, it will never be violated (in the "real" way, not the "coordinate" time re-ordering) in any other frame either. Since instantaneous travel in the preferred frame will protect causality in that frame, it is actually faster-than-instantaneous (FTI) travel that is needed to get real causality violations. That requires two frames, with symmetric instantaneous travel in succession, as seen in your diagram. In other words, symmetric FTL in two frames can yield FTI travel in one frame, and only that is a "real" causality violation.

If the distinction between "real" and "coordinate" time-ordering violations is not convincing, simply consider non-inertial observers in our own universe, and the "Andromeda paradox." Einstein coordinates, reckoned from changing inertial frames, yields time-ordering violations, but not causality violations.

grant hutchison
2010-Apr-09, 08:51 PM
I think I'm telling you something else. ... A true causality violation is when a cause creates an effect which can then be a cause of the original cause!OK ... then I think you're telling me that I got careless with my terminology. Up to now I've been fairly careful to distinguish "temporal ambiguity" from "causality violation". It seems I lost control of that distinction in the post that prompted your comment, and in my reply to your comment.

Grant Hutchison

Ken G
2010-Apr-09, 09:04 PM
Yes, that may well be the issue. It occurs to me that what I'm saying is something of an "aha"-- it was never FTL travel that generated causality violations, it was always FTI travel (faster than instantaneous-- where I restrict to isotropic coordinatizations of each frame or else even FTI is just a coordinate issue too). It's just that when you combine FTL travel with a symmetry principle of handling said FTL travel in the same way for two diverging observers, that you can generate FTI travel in some third frame. It is always that FTI travel that creates the causality problem, never the FTL by itself. (I think this is also the OPer's problem with seeing it.) That also means that you can have FTL, or even instantaneous, motion in a single preferred frame, but simply disallow it to exist in the same way in any other frame, because you above all disallow FTI travel in the preferred frame. As long as there is one frame where FTI travel is strictly forbidden, you can have any kind of FTL travel you like, and still never get a true causality violation in any frame. So far we have not accomplished FTL travel at all, so we cannot explore whether or not it respects causality in a single preferred frame.

Ken G
2010-Apr-09, 09:29 PM
I'm not quite happy with the "FTI" terminology, because it is really still just coordinate language-- a signal arrives at a retarded time coordinate from when it was sent out. That's still no big deal, it's just a way the clock was set. The actual causality violation is if there can be FTI travel in a frame (a "preferred" frame, if you will) in which the laws of the physics are isotropic, so that FTI in one direction can be followed by FTI travel in the reverse direction. Showing up in the past on the same world line is the causality violation, so there will always have to be two-way travel to get it (barring some GR topology business). It's just that we can still have a single frame that has isotropic laws of physics, and instantaneous travel can happen in all directions in that frame, without causality violations as long as all other frames have nonisotropic physics that prevents FTI travel in the isotropic frame. If there are no isotropic frames at all, then even FTI travel doesn't insure causality violations, it could just be a manifestation of the nonisotropy in how the coordinate prescription interacts with reality.

Now, someone might point out that the Einstein prescription sounds isotropic, but we only know that the prescription is thus, we don't know that its interaction with reality is still isotropic (i.e., we don't know that the speed of light is isotropic). We only know that by assuming it is, we get a symmetry we want (isotropy within frames and frame-to-frame symmetry), so we go with it as long as we can. So this means that we don't really know that the success of relativity tells us that FTL is impossible, relativity could succeed for other reasons, there could still be a preferred frame where causality is preserved, and FTL could be possible. It would just mean that the isotropic Einstein prescription interacts with a non-isotropic reality in a way we have not yet noticed.

BioSci
2010-Apr-09, 11:33 PM
Yes, that may well be the issue. It occurs to me that what I'm saying is something of an "aha"-- it was never FTL travel that generated causality violations, it was always FTI travel (faster than instantaneous-- where I restrict to isotropic coordinatizations of each frame or else even FTI is just a coordinate issue too). It's just that when you combine FTL travel with a symmetry principle of handling said FTL travel in the same way for two diverging observers, that you can generate FTI travel in some third frame. It is always that FTI travel that creates the causality problem, never the FTL by itself. (I think this is also the OPer's problem with seeing it.) That also means that you can have FTL, or even instantaneous, motion in a single preferred frame, but simply disallow it to exist in the same way in any other frame, because you above all disallow FTI travel in the preferred frame. As long as there is one frame where FTI travel is strictly forbidden, you can have any kind of FTL travel you like, and still never get a true causality violation in any frame. So far we have not accomplished FTL travel at all, so we cannot explore whether or not it respects causality in a single preferred frame.

Hmm, that sounds a bit to me like the "no true Scotsman" routine :)

The problem is the combination of: Relativity, FTL, and Causality - you can have any two, just not all three.

A single FTL trip results in another frame (in relative motion) being able to observe a result before the cause - it is just not as obvious a case of a "causality violation" as when a second FTL trip is combined with the moving frame to show up back in the past of the initial frame - but the physical paradox is already present at the first instance of FTL + relativity.

Another nice diagram of this FTL+relativity+Causality is here: http://www.theculture.org/rich/sharpblue/archives/000089.html

Ken G
2010-Apr-10, 01:05 AM
The problem is the combination of: Relativity, FTL, and Causality - you can have any two, just not all three.Right. So then the further question is, what is the aspect of relativity that is combining with FTL to blow the causality? If we simply adopt relativity as true, the question does not come up, but relativity is a set of assumptions. We can ask, which of those assumptions would need to be relaxed, and in what way, for FTL to exist without causality problems? In other words, is there a modification to relativity that allows FTL and does not contradict any experiment that has been done? What I'm trying to say is that the answer is yes-- the assumptions of relativity that would need to be relaxed, yet without contradicting any existing experiment, have to do with the assumption of isotropic laws of physics in every inertial frame. The simplest way to relax that is to allow isotropic laws in only one frame, an aether frame if you will, and have Lorentz transformations to all other frames. Then FTL can occur in the aether frame without any causality violations, and without contradicting any experiment that has ever been done. All that is lost is the philosophy of relativity, and the fact that this philosophy represents our best, simplest, most symmetric, and most appealing theory.


A single FTL trip results in another frame (in relative motion) being able to observe a result before the cause - it is just not as obvious a case of a "causality violation" as when a second FTL trip is combined with the moving frame to show up back in the past of the initial frame - but the physical paradox is already present at the first instance of FTL + relativity.What I am trying to show is that this is not a causality violation at all, not just a not-as-obvious one-- it only becomes a causality violation when we further add additional assumptions about the laws of physics being isotropic in every inertial frame-- i.e., when we add additional aspects of relativity to the "single FTL trip". I'm exposing these implicit assumptions that are required to yield actual causality violations. You're still right that relativity+FTL= causality violation, but it is important to recognize where the relativity parts come in-- not just in the Einstein coordinates, but in additional assumptions about what they mean (that they represent true symmetries of nature, and not just a way to coordinatize spacetime).

In other words, relativity is not Einstein coordinates, so showing spacetime diagrams is just not enough-- there also has to be stated the principle of relativity itself, which is a symmetry between observers. We could discover FTL tomorrow, and it might not yield any causality violations, and no past observation would need to be wrong. But we wouldn't have relativity any more, so no one is banking on that.

Webbo
2010-Apr-10, 01:31 AM
And just to confirm what others have told you, the simultaneity lines in my diagrams connect events which are actually simultaneous, not events which are seen simultaneously.

I dont think I am able to get past this statement. Effectively I think you are saying that observer A's present is simultaneous with observer B's past, while at the same time observer B's present is simultaneous with observers A's past. That seems to be a clear paradox and break down of logic.

I didn't follow most of what Ken G was stating but the part that I could agree with is that faster than instantaneous travel will indeed create causality problems. That makes obvious sense to me.

grant hutchison
2010-Apr-10, 01:48 AM
I dont think I am able to get past this statement. Effectively I think you are saying that observer A's present is simultaneous with observer B's past, while at the same time observer B's present is simultaneous with observers A's past. That seems to be a clear paradox and break down of logic.Nevertheless, that's the relativity of simultaneity. We don't find this sort of coordinate rotation at all surprising when it applies to space alone. If two observers face one another, no-one is surprised that one's left is the other's right. But when one's space becomes another's time (as it does under SR), then intuition fails.

Grant Hutchison

Ken G
2010-Apr-10, 01:53 AM
I dont think I am able to get past this statement. Effectively I think you are saying that observer A's present is simultaneous with observer B's past, while at the same time observer B's present is simultaneous with observers A's past. That seems to be a clear paradox and break down of logic.Actually, there is really no physical way to give meaning to the statement you just made. What is A's past? According to whom, and when? You can't even ask A what his/her own past is, because the answer will depend on when you ask them! Ask me right now, I have a past. And I have a concept of what your past is. But how do I compare that to your concept of what your past is? It depends on when I ask you! And I can't ask you "now", because that simply begs the issue-- the very question we are asking is what is now for you! Is it when you read this, or when I wrote it, or when you understand what it means? There are lots of "nows", the word doesn't mean a whole lot.

However, once we choose a coordinatization, we can start matching up the "nows". That's all a coordinatization is, it's a way to match up "nows", and it is totally arbitrary. But we get to choose it the way we like, and in SR, Einstein gives us a nice coordinatization to choose. That's what Grant uses in his figure. It has nice properties, that's all we can say about it-- it is not the "actual nows", there's no such thing. Matching nows is like matching socks when there are no obvious pairs-- you do it any way that appeals to you.

So the first lesson about "simultaneity" is that all it is is matching up instants between two distant observers, and labeling them as instantaneous. That's it, that's all it is, there's zero physical meaning in it until we dig further into what is going on. The first gem we dig up is the concept of causality, and how if we use Einstein coordinates, we have found experimentally that all inertial observers agree on the causal connection between every two events (that's what the "light cone" is really all about). We like that, point for Einstein's sock-matching. The second gem we dig up is symmetry, wherein all inertial observers see the same laws of physics (and if A thinks B's clock is lagging A's Einstein coordinates, then B must think A's clock is lagging B's Einstein coordinates, by that symmetry). We really like that, game to Einstein.

But there's a price-- we cannot find any absolute concept of simultaneity that all observers agree on. Instead, the concept fragments into a million little pieces-- every observer gets their own concept of how the socks should be matched. You say that's a paradox, but there are no contradictions-- the socks aren't actually paired in any physical way, it's just coordinatization.

Webbo
2010-Apr-10, 02:17 AM
What we do know is that at some point in the past observer A and observer B were at the same location and sychronised clocks. Since then they have travelled the same distance over the same period of time so each must know that, for example, a year has passed and they have travelled 1 billion miles, so it would be acceptable to assume that a year has passed on the other ship and they have also travelled 1 billion miles. However, when drawing the diagram they will discover that in fact their present is simultaneous with say, only 6 months passing on the other ship which has only travelled 0.5 billion miles, and vice versa. So, the only way I can see this working is if it's the observation of these simultaneous events that they are representing on the diagram, not their actual simultaneity.

Incidently on the page provided earlier one of the posters provided a diagram and suggested that FLT just widens the space-time cone and does not create causality issues.

http://s180.photobucket.com/albums/x58/riidii/?action=view&current=Sonic-FTLDiagram.png

Webbo
2010-Apr-10, 02:28 AM
Nevertheless, that's the relativity of simultaneity. We don't find this sort of coordinate rotation at all surprising when it applies to space alone. If two observers face one another, no-one is surprised that one's left is the other's right. But when one's space becomes another's time (as it does under SR), then intuition fails.

Grant Hutchison

But there was nothing paradoxical in Einsteins description of the simultaneity of relativity. He quite clearly stated that it's the light signals that are determining the order or simltaneity of events and only their apparent contradictions. Nothing in his description implies that each observer it witnessing the actual events happening. How/where is it that this progressess to to the paradoxical situation in your diagram?

Ken G
2010-Apr-10, 03:39 AM
What we do know is that at some point in the past observer A and observer B were at the same location and sychronised clocks. Since then they have travelled the same distance over the same period of time so each must know that, for example, a year has passed and they have travelled 1 billion miles, so it would be acceptable to assume that a year has passed on the other ship and they have also travelled 1 billion miles. But when you match your own 100-year sock with their 1-year sock, you do not think they've gone 100 light years, you think they've gone 1 light year (approximately)! What's more, they think they've gone 100 light years, but you attribute that to their rulers having shrunk by that same factor of 100! You are looking for a symmetry, but the symmetry is completely maintained if when A has lived 100 years, A reckons B as living 1, as long as when B has lived 100 years, B reckons A as living 1. That's perfect symmetry. There's absolutely no requirement that A has to think B has lived 100 years when A has lived 100 years. Why would there be? There's no further connection between A and B after they part company, so there's no need for them to match the 100-year sock with the 100-year sock, and it certainly is not what you get with the Einstein convention.

If we want to use Einstein coordinates, and take advantage of all the nice properties I mentioned, then the times do not match up in the way you (and everyone else) expected they would. What's more, any coordinate system which matched the socks the way you want to, would lose the symmetry between all observers-- you would in effect have a single frame designated as the absolute timekeeper, and no other frame could use their own clocks to do physics. It would be too high a price to pay for the coordinates you want to impose. That's the point here, you get to choose any coordinates you want, and any simultaneity convention-- but then you turn it over to nature, and see if nature makes a hash out of your choices. When we turn the Einstein coordinates over to nature, nature says, "OK, then you lose absolute simultaneity, but you get the same physics for all inertial observers." And we say, "deal".


However, when drawing the diagram they will discover that in fact their present is simultaneous with say, only 6 months passing on the other ship which has only travelled 0.5 billion miles, and vice versa. So, the only way I can see this working is if it's the observation of these simultaneous events that they are representing on the diagram, not their actual simultaneity.No, it is their "actual" simultaneity (which just means, as defined by the simultaneity convention chosen-- the socks matching chosen), Grant's figure has nothing to do with what anyone "sees". It's a figure about Einstein coordinates, and we supplement that with the statement that reality respects those coordinates enough to annoint them with very desirable symmetry properties.

Incidently on the page provided earlier one of the posters provided a diagram and suggested that FLT just widens the space-time cone and does not create causality issues. I've no doubt they are simply forgetting about the implicit symmetry assumptions that go into relativity that I have been going on about. If you don't make those assumptions, you can have FTL in a single frame without any causality problems, but you are violating the assumptions of relativity unless you allow the symmetry to extend to the "return trip" of the signal (as seen from the other frame, to impose the symmetry)-- and that's when you get the causality violations.

Van Rijn
2010-Apr-10, 05:29 AM
But there was nothing paradoxical in Einsteins description of the simultaneity of relativity.


. . . as long as there was no faster than light communication or travel. The issue here has been about what could happen per relativity assuming there was FTL communication.

grant hutchison
2010-Apr-10, 12:29 PM
I dont think I am able to get past this statement. Effectively I think you are saying that observer A's present is simultaneous with observer B's past, while at the same time observer B's present is simultaneous with observers A's past. That seems to be a clear paradox and break down of logic.Actually, there is really no physical way to give meaning to the statement you just made.I think Webbo was implicitly using the time and simultaneity axes of the reference frame in my diagrams: an observer who sees the red and blue observers diverge symmetrically, and who plots their simultaneity lines with a "pastward" slope towards each other.

Grant Hutchison

grant hutchison
2010-Apr-10, 01:17 PM
But there was nothing paradoxical in Einsteins description of the simultaneity of relativity. He quite clearly stated that it's the light signals that are determining the order or simltaneity of events and only their apparent contradictions. Nothing in his description implies that each observer it witnessing the actual events happening. How/where is it that this progressess to to the paradoxical situation in your diagram?The process by which he establishes simultaneity (synchronizing clocks using light signals) leads directly to the slope of the simultaneity lines in my diagram. Einstein is quite clear that moving observers don't share the same simultaneity, both in the 1905 paper I quoted earlier, and the 1920 book from which you took your quote (the famous "train and embankment" section). From Chapter 9 of Relativity: The Special and General Theory (http://www.bartleby.com/173/9.html), a little after the section you quoted:
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.I know you've been taking this to be a statement about the arrival of light signals at observers, but it's about the actual time and space that different observers experience. Having established clock synchronization with light signals, Einstein uses synchronized clocks as his standard for simultaneous events: no more light signals are required. And observers in relative motion disagree about whether clocks are synchronized or not. That is, they disagree about simultaneity at a fundamental level, separate from the effects of light signal delay.

Einstein's result wouldn't have been remotely world-shaking if all he'd said was "we see distant stuff later than we see nearby stuff".

Grant Hutchison

Ken G
2010-Apr-10, 01:47 PM
I think Webbo was implicitly using the time and simultaneity axes of the reference frame in my diagrams: an observer who sees the red and blue observers diverge symmetrically, and who plots their simultaneity lines with a "pastward" slope towards each other.Yes, I think you're right about what he was saying, but even in that frame the statement is meaningless without putting in additional words that expose its core problems. What he would have to say instead is "If we contrast A's concept of B's present with C's concept of both A and B's present, A's concept of B's present falls in the past of C's concept of B's present." So not only do we need to identify a reference frame in the statement (as you well know), we also have to mix reference frames in the same statement-- there's no statement like "the present moment for A falls in B's past" that just uses a single reference frame. Mixing reference frames like that is an incoherent thing to do-- in any one frame (even C's), the present for A is the present for B-- as these presents are individual events. Or we can use global versions of their "present", and say that "A's global version of the present is not the same as B's global version of the present", but no past/future comparison can be made without mixing frames.

Webbo
2010-Apr-10, 04:12 PM
I know you've been taking this to be a statement about the arrival of light signals at observers, but it's about the actual time and space that different observers experience.

So what exactly is it that "different observers experience"? Is it something other than the light signals?

Also, why does Einstein specifically refer to the light signals in his experiment?

"When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment."

What is the point of that statement? He appears here, to me, to be deliberatly pointing out that this is purely an observer phenomenon.

Take this summary sentence - "Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). "

He qualfied what simultaneous meant above. It's not a statement of actual events, just when the observer perceived them.


Einstein's result wouldn't have been remotely world-shaking if all he'd said was "we see distant stuff later than we see nearby stuff".


I think that's exacly what he's saying. Why read more into his very clear statements. He was just pointing out in different frames events are viewed in a different order.

Webbo
2010-Apr-10, 04:23 PM
Yes, I think you're right about what he was saying, but even in that frame the statement is meaningless without putting in additional words that expose its core problems. What he would have to say instead is "If we contrast A's concept of B's present with C's concept of both A and B's present, A's concept of B's present falls in the past of C's concept of B's present." So not only do we need to identify a reference frame in the statement (as you well know), we also have to mix reference frames in the same statement-- there's no statement like "the present moment for A falls in B's past" that just uses a single reference frame. Mixing reference frames like that is an incoherent thing to do-- in any one frame (even C's), the present for A is the present for B-- as these presents are individual events. Or we can use global versions of their "present", and say that "A's global version of the present is not the same as B's global version of the present", but no past/future comparison can be made without mixing frames.

Then maybe we should use 1 global frame to avoid these problems.

Say I am observer A and I can see observer B is 1 light year away and his clock is also 1 light year behind. If I am able to travel instantly to observer B I would just arrive at the time I observe him now + the distance I travel instantly. Therefore I would arrive at the same time as my clock is now.

grant hutchison
2010-Apr-10, 08:35 PM
So what exactly is it that "different observers experience"? Is it something other than the light signals?It is. It's the experience of working out which events are simultaneous. If a clock one light-second distant appears to be running one second slow when observed through a telescope, then our observers allow for light-travel time and deduce that the distant clock is synchronized with their local clock. That is the standard of simultaneity Einstein describes.


Also, why does Einstein specifically refer to the light signals in his experiment?Because the method I describe above is Einstein's definition of simultaneity. We exchange light signals with distant clocks, allow for the time lag due to the light signals, and judge simultaneity accordingly.


Take this summary sentence - "Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). "

He qualfied what simultaneous meant above. It's not a statement of actual events, just when the observer perceived them.No, it's a statement of actual events. I can only suggest you read Einstein's words more completely, and more carefully.

The important consideration here is that each observer must measure the speed of light to be the same, whatever their state of motion. Once you accept that, then the result for the train and the embankment becomes mysterious and deeply significant. Observers must disagree about what is actually simultaneous, rather than having some trivial difference about when light signals were received. Relativistic time dilation and length contraction fall out of that disagreement, and the predictions from that theory have been exquisitely well-confirmed.

Grant Hutchison

Ken G
2010-Apr-11, 03:35 AM
Then maybe we should use 1 global frame to avoid these problems.Yes, choosing a frame and sticking to it is a good way to avoid confusion.


Say I am observer A and I can see observer B is 1 light year away and his clock is also 1 light year behind. If I am able to travel instantly to observer B I would just arrive at the time I observe him now + the distance I travel instantly. Therefore I would arrive at the same time as my clock is now.There's no need to consider a time that you "observe" now, you have a reckoning of the situation and that's all you need-- if you reckon his time as being 1 light year behind (I presume you mean 1 year behind, a light year is not a time), and you reckon that you travel "instantly", then you must arrive when his clock is 1 year behind-- that's what you reckon "instantly" means, and what you reckon being "1 year behind" means as well. You see also why I avoid saying you are moving into B's "past", you are simply moving to B's world line at a point where his clock, which was once synchronized to yours, is now 1 year behind yours. Some would say at this point that "his clock must be slow compared to yours", but I'd rather say his clock is fine, you just took a different path than he did to the point of the second meeting, and there is a different time elapsed over that different path.

Note there is no causality violation yet-- that will appear if you are allowed to now use B's concept of "instantly" to travel back instantly to your own original world line (which would be "faster than instantaneous" travel if we stick to the same original frame). As in Grant's diagram, you will arrive in your own past, and that's a problem-- but it comes not just from traveling instantly, but from traveling instantly twice, and from the point of view of two diverging world lines. For that, you had to invoke a symmetry that "instant travel" works the same in all frames, and that's why it is FTL plus relativity that leads to causality violations.

Webbo
2010-Apr-12, 01:19 PM
It is. It's the experience of working out which events are simultaneous. If a clock one light-second distant appears to be running one second slow when observed through a telescope, then our observers allow for light-travel time and deduce that the distant clock is synchronized with their local clock. That is the standard of simultaneity Einstein describes.
OK. So if I have worked out that our clocks are synchronised based on the fact that they are one second behind and one light-second distant, if I am then able to instantly travel to their location, our clocks should surely be synchronised and I wouldn’t travel back in time. I have calculated the information I am receiving now was sent 1 second ago. How can I arrive before it left.

Because the method I describe above is Einstein's definition of simultaneity. We exchange light signals with distant clocks, allow for the time lag due to the light signals, and judge simultaneity accordingly.
But this is exactly what I am saying. The simultaneity lines are judged based on observations of light signals. However, previously I kept being told there are no light signals in the diagram. And where on the diagram are the lagging adjustments? What would they look like without lagging adjustments?

No, it's a statement of actual events. I can only suggest you read Einstein's words more completely, and more carefully.
I think I have. He qualified at the beginning what simultaneous meant. If they are a statement of events then he would be stating a paradox which he isn’t. He clearly points out the reason for the apparent paradox and it’s because the observer in the middle of the train has moved towards one signal and away from another hence their order of events has altered.
“Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

The important consideration here is that each observer must measure the speed of light to be the same, whatever their state of motion. Once you accept that, then the result for the train and the embankment becomes mysterious and deeply significant. Observers must disagree about what is actually simultaneous, rather than having some trivial difference about when light signals were received. Relativistic time dilation and length contraction fall out of that disagreement, and the predictions from that theory have been exquisitely well-confirmed.
I don’t see anything mysterious. They don’t disagree about what is actually simultaneous. Einstein just describes the events as observed by each and gives the reason; light signal delay. And referring back to the first point the observer on the train could easily calculate the simultaneity and agree with the other observer. When the signal is sent both observers are say 10 ms away from each strike. The observer on the embankment, when receiving the signals simultaneously, would work out that each travelleved 10 ms and they are 10 ms old before reaching them so would deduce they happened simultaneously. The observer on the train has moved say 2 ms toward B and he would see that signal as being sent 8 ms ago. Next he would see A and that would have been sent 12 ms ago. He knows that the train has moved 2 ms toward B so simple math could work out that the events in reality happened simultaneously. Only the observations are in disagreement.

Webbo
2010-Apr-12, 01:31 PM
Yes, choosing a frame and sticking to it is a good way to avoid confusion.
There's no need to consider a time that you "observe" now, you have a reckoning of the situation and that's all you need-- if you reckon his time as being 1 light year behind (I presume you mean 1 year behind, a light year is not a time), and you reckon that you travel "instantly", then you must arrive when his clock is 1 year behind-- that's what you reckon "instantly" means, and what you reckon being "1 year behind" means as well. You see also why I avoid saying you are moving into B's "past", you are simply moving to B's world line at a point where his clock, which was once synchronized to yours, is now 1 year behind yours. Some would say at this point that "his clock must be slow compared to yours", but I'd rather say his clock is fine, you just took a different path than he did to the point of the second meeting, and there is a different time elapsed over that different path..
Where did I state that instantly means I would travel to the when the light signal was sent? I know that the light signal has been in transit for 1 year. My instant travel is from location A to B with no time in between so if something is one light year away I need to add 1 year of time to what I see now. I know I am not observeing an actual event. It just a light signal.

Ken G
2010-Apr-12, 01:44 PM
Where did I state that instantly means I would travel to the when the light signal was sent? Neither Grant nor I are talking about when any signals were sent, we are both talking about how you reckon simultaneity (a la Einstein) using those signals-- which is totally different. You don't even have to use light signals in Einstein's approach, an alternative scheme involves "very slow movement" to separate synchronized clocks, but it's clunky and impractical so we just use the light standard. But either way, the time the light was emitted is still unimportant-- what matters is how you use the light to reckon simultaneity.

Judging from your last post to Grant, you still are not getting this. If you think we are wrong, ask someone else. If you think we are right but you are not understanding something crucial that we are saying, I don't think either of us understand what is the missing piece, other than what we've already said: forget about the time the light was emitted, simultaneity involves the calculation you do with what you observe, not what you observe itself.

Also, in your last post to Grant, you repeated that you don't get a causality violation just by moving instantly one time. I have confirmed that many times now-- to get a causality violation, you have to move instantly forward and then backward, and what's more, the "instantaneity" must be reckoned as such from the two different frames. That's exactly what Grant's diagram does, it's why there are two different colors there. So pointing out that you don't get causality violations with just one instantaneous transit is not contradicting what we are telling you.

Webbo
2010-Apr-12, 02:35 PM
Neither Grant nor I are talking about when any signals were sent, we are both talking about how you reckon simultaneity (a la Einstein) using those signals-- which is totally different. You don't even have to use light signals in Einstein's approach, an alternative scheme involves "very slow movement" to separate synchronized clocks, but it's clunky and impractical so we just use the light standard. But either way, the time the light was emitted is still unimportant-- what matters is how you use the light to reckon simultaneity.

Judging from your last post to Grant, you still are not getting this. If you think we are wrong, ask someone else. If you think we are right but you are not understanding something crucial that we are saying, I don't think either of us understand what is the missing piece, other than what we've already said: forget about the time the light was emitted, simultaneity involves the calculation you do with what you observe, not what you observe itself.

Also, in your last post to Grant, you repeated that you don't get a causality violation just by moving instantly one time. I have confirmed that many times now-- to get a causality violation, you have to move instantly forward and then backward, and what's more, the "instantaneity" must be reckoned as such from the two different frames. That's exactly what Grant's diagram does, it's why there are two different colors there. So pointing out that you don't get causality violations with just one instantaneous transit is not contradicting what we are telling you.

I still don't follow where I said that my reckoning is that if I look at an object 1 light year away and therefore also 1 year old, I would travel to instantly to what I am observing. I have specifically stated repeatedly that I believe I would travel instantly to what I observe now (1 year ago) plus the distance/time (1 light-year/1 year) and arrive at exactly the same time at the destination that I have on my wristwatch. If I turn around and observe where I came from, I would see that it is also 1 year ago (because its obviously now 1 light-year distant) so if I were to travel back instantly, I would also arrive at the time I now observe from my original location (1 year ago) plus the distance/time (1 light-year/1 year) to again arrive at the time I have on my wristwatch now. Ergo, no moving back (or forward) in time in either direction. I'm just racing light signals.

grant hutchison
2010-Apr-12, 03:58 PM
OK. So if I have worked out that our clocks are synchronised based on the fact that they are one second behind and one light-second distant, if I am then able to instantly travel to their location, our clocks should surely be synchronised and I wouldn’t travel back in time. I have calculated the information I am receiving now was sent 1 second ago. How can I arrive before it left.Single journey doesn't do it for you. But someone else, moving relative to you, will measure (that is, after allowing for light lag) that you did arrive before you left. The event "arrival" will really be before the event "departure" for them, once they have done the necessary sums, or seeded your route with the necessary clocks synchronized by them. And if, after your arrival, you change your state of motion and jump instantaneously back to your departure point, then you can arrange to arrive before you left. I've shown how that works.


But this is exactly what I am saying. The simultaneity lines are judged based on observations of light signals. However, previously I kept being told there are no light signals in the diagram. And where on the diagram are the lagging adjustments? What would they look like without lagging adjustments?There are no light signals in the diagram. They don't matter. We're talking about actual simultaneity, as calculated by different observers, not the simultaneous arrival of light signals. I can draw you some light signals if you want, but they don't change the outcome.


I think I have. He qualified at the beginning what simultaneous meant. If they are a statement of events then he would be stating a paradox which he isn’t. He clearly points out the reason for the apparent paradox and it’s because the observer in the middle of the train has moved towards one signal and away from another hence their order of events has altered.You need to read more. I'd suggest going back to the 1905 paper I linked to earlier, and working through the first section.


I don’t see anything mysterious. They don’t disagree about what is actually simultaneous. Einstein just describes the events as observed by each and gives the reason; light signal delay. And referring back to the first point the observer on the train could easily calculate the simultaneity and agree with the other observer. When the signal is sent both observers are say 10 ms away from each strike. The observer on the embankment, when receiving the signals simultaneously, would work out that each travelleved 10 ms and they are 10 ms old before reaching them so would deduce they happened simultaneously. The observer on the train has moved say 2 ms toward B and he would see that signal as being sent 8 ms ago. Next he would see A and that would have been sent 12 ms ago. He knows that the train has moved 2 ms toward B so simple math could work out that the events in reality happened simultaneously. Only the observations are in disagreement.But each observer measures the speed of light as being equal to c, relative to his own standard of rest. The observer on the train finds that a light signal from the rear of the train arrived later than one from the front, both signals travelling the same distance along the train, and both moving at the same speed relative to the train. So he calculates that the front flash happened before the rear flash. For him, using Einstein's mechanism for synchronizing clocks, that's really true. Things that are simultaneous on the embankment aren't simultaneous on the train.
This is absolutely fundamental to Special Relativity, and if you don't get it, then you're missing the whole point.

Grant Hutchison

Webbo
2010-Apr-12, 05:03 PM
But each observer measures the speed of light as being equal to c, relative to his own standard of rest. The observer on the train finds that a light signal from the rear of the train arrived later than one from the front, both signals travelling the same distance along the train, and both moving at the same speed relative to the train. So he calculates that the front flash happened before the rear flash. For him, using Einstein's mechanism for synchronizing clocks, that's really true. Things that are simultaneous on the embankment aren't simultaneous on the train.
This is absolutely fundamental to Special Relativity, and if you don't get it, then you're missing the whole point.

Grant Hutchison

Einsteins observer on the train does not calculate that B arrives before A. He observes B arriving before A.

"Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

If the observer does a calculation as I did previously he will calculate they were simultaneous.

I know that events observed as simultaneous on the train, will not be simultaneous as observed from the embankment. I have quoted it often enough!

grant hutchison
2010-Apr-12, 05:43 PM
If the observer does a calculation as I did previously he will calculate they were simultaneous.The train observer will not calculate that they were simultaneous, for reasons which I've explained, and which Einstein explains in the first eight chapters of the book you're quoting from, as well as in the 1905 paper I have commended to your attention.

Grant Hutchison

Webbo
2010-Apr-12, 05:52 PM
What do the observers witness rather than calculate?

M sees the flashes simultaneously. Are you saying that M1 also sees the flashes simultaneously?

grant hutchison
2010-Apr-12, 06:01 PM
What do the observers witness rather than calculate?

M sees the flashes simultaneously. Are you saying that M1 also sees the flashes simultaneously?No. He sees the front flash before the back flash. And because he sees the front flash before the back flash, and knows that he is in the centre of the train, and knows that the speed of light is measured as c relative to the train, he is compelled to deduce that the front flash was emitted before the back flash. The relativity of simultaneity is about the emission, not the reception.

Grant Hutchison

Ken G
2010-Apr-12, 06:18 PM
I still don't follow where I said that my reckoning is that if I look at an object 1 light year away and therefore also 1 year old, I would travel to instantly to what I am observing. I have specifically stated repeatedly that I believe I would travel instantly to what I observe now (1 year ago) plus the distance/time (1 light-year/1 year) and arrive at exactly the same time at the destination that I have on my wristwatch. If I turn around and observe where I came from, I would see that it is also 1 year ago (because its obviously now 1 light-year distant) so if I were to travel back instantly, I would also arrive at the time I now observe from my original location (1 year ago) plus the distance/time (1 light-year/1 year) to again arrive at the time I have on my wristwatch now. Ergo, no moving back (or forward) in time in either direction. I'm just racing light signals.And that would all be true if the other location was not moving-- then both instantaneous trips would be reckoned as such in the same frame, and that never leads to causality violations. If you do the problem again, except let the distant point be moving away from you, then you will get Grant's figure. The numbers you are using are not the correct numbers unless the destination is not moving (the error is in the text I bolded-- watch out, in relativity nothing is "obvious"), so the other observer moves out a large distance and stops, and your calculation proceeds from there. Then there is indeed no causality violation. When there is a violation is when the other observer is still moving away, so that when you reckon the "instantaneous" trip home, you reckon it from his frame (i.e., you change from a blue path to a red path in Grant's figure), which is what will put you in your own past.

pzkpfw
2010-Apr-12, 07:30 PM
I know that events observed as simultaneous on the train, will not be simultaneous as observed from the embankment. I have quoted it often enough!

Two more observers on that train, one near the front and one near the back, would observe the two flashes at different times than that observer in the exact middle of the train - but they would agree with that middle observer about when the lightening strikes occured, because they have a common frame of reference and can account for the distances travelled by light at c.

This is quite different to the train observer observing the flashes at different times than the embankment observer; the diffence being the relative motion while both train and embankment observer are entitled to consider themselves as at rest and the other as moving.

Webbo
2010-Apr-13, 02:43 PM
No. He sees the front flash before the back flash. And because he sees the front flash before the back flash, and knows that he is in the centre of the train, and knows that the speed of light is measured as c relative to the train, he is compelled to deduce that the front flash was emitted before the back flash. The relativity of simultaneity is about the emission, not the reception.

Grant Hutchison

I think then that we fundamentally disagree with what Einstien is trying to show in his example. I believe that he is trying to show that the reason for the apparent paradox is purely because of the delay in light signal and the movement of M1 towards one light signal and away from the other. However your interpretation is that each observer has witnessed a different reality and a therefore there is a paradox. Einstein doesn't mention any paradoxes here as IMO he is not implying one. I don't see a paradox because M1 has moved towards B and shortened the distance from emission, and away from A and lengthened the distance from emission, so if light was emitted from each of these locations at the same time (in M's frame), I would expect them to be received at different times in M1's frame as the distances from emission are now different. However, as I know the distances are different I would deduce that emissions were simultaneous even though I observed a delay.

Webbo
2010-Apr-13, 03:01 PM
And that would all be true if the other location was not moving-- then both instantaneous trips would be reckoned as such in the same frame, and that never leads to causality violations. If you do the problem again, except let the distant point be moving away from you, then you will get Grant's figure. The numbers you are using are not the correct numbers unless the destination is not moving (the error is in the text I bolded-- watch out, in relativity nothing is "obvious"), so the other observer moves out a large distance and stops, and your calculation proceeds from there. Then there is indeed no causality violation. When there is a violation is when the other observer is still moving away, so that when you reckon the "instantaneous" trip home, you reckon it from his frame (i.e., you change from a blue path to a red path in Grant's figure), which is what will put you in your own past.

Sorry, I should have stated my example assumed there was no moving away, so at least we agree that FTL even in both directions would not create causality problems. As for the moving scenario, I still think it would be OK because taking it from my example, if the second observer (whom I have now joined) was moving away (say a further light-month), when I looked back to my origin I would observe that the time is 13 months ago (the new distance), so if I travelled back instantly I would arrive at the time observed (13 months ago) plus the new distance (1 light year + 1 light month) which still equals the time now according to my wristwatch. However, rather than go around in more circles on this point I think we should just agree to disagree.

grant hutchison
2010-Apr-13, 03:06 PM
I think then that we fundamentally disagree with what Einstien is trying to show in his example.Yes. Your interpretation is incorrect, and I've been over the missing elements for you several times. So if you're genuinely interested in what Special Relativity does (and I'm not really getting a sense that you are), I can only suggest you stop trying to decode a couple of paragraphs and read a bit more.


However your interpretation is that each observer has witnessed a different reality and a therefore there is a paradox. Einstein doesn't mention any paradoxes here as IMO he is not implying one.There is no paradox in Special Relativity, which is why Einstein doesn't mention one. Both observers witness the same reality, they just disagree about which events are simultaneous. This is not paradoxical in the slightest.
Paradoxes only arise when we add faster-than-light travel to Special Relativity, and require causality to be preserved. As has been said several times already, you can't have SR, FTL and intact causality.

Grant Hutchison

Webbo
2010-Apr-13, 03:16 PM
Two more observers on that train, one near the front and one near the back, would observe the two flashes at different times than that observer in the exact middle of the train - but they would agree with that middle observer about when the lightening strikes occured, because they have a common frame of reference and can account for the distances travelled by light at c.

This is quite different to the train observer observing the flashes at different times than the embankment observer; the diffence being the relative motion while both train and embankment observer are entitled to consider themselves as at rest and the other as moving.

Not sure exactly how they will agree. According to this common frame which you considered at rest, Observer X at the back of the train will see flash A at 0 time and flash B 10 ms later (using my numbers earlier) and observer Y at the front will see flash B at 0 time and flash A 10 ms later. I can see that they could easily agree with each other that they were simultaneous strikes when taking the distance from each strike into consideration. However, when discussing this with observer M1 he will state that flash B was first. Your common frame should make M1 see them as simultaneous which Grant explicitly states they are not.

Webbo
2010-Apr-13, 03:26 PM
Both observers witness the same reality, they just disagree about which events are simultaneous. This is not paradoxical in the slightest.

I dont understand how you can state this. According to you M says the flashes were emitted simultaneously and M1 says the flashes were emitted one before the other. Paradox.

grant hutchison
2010-Apr-13, 04:32 PM
I dont understand how you can state this. According to you M says the flashes were emitted simultaneously and M1 says the flashes were emitted one before the other. Paradox.This isn't according to me, it's according to Einstein. That's what he's telling you. And it's not a paradox: it's simply the nature of reality according to special relativity. It's a coordinate thing, like the reversal of left and right when two observers face each other. It's harder to understand because it involves time, and you're surprised by it because it doesn't accord with your intuition about time. So it took someone of Einstein's abilities to see past the surprise to the deeper significance.

Grant Hutchison

Ken G
2010-Apr-13, 05:14 PM
Sorry, I should have stated my example assumed there was no moving away, so at least we agree that FTL even in both directions would not create causality problems.Yes, causality problems are from FTL with relativity, but it turns out that even the "relativity" piece does not become relevant to causality violation until it involves FTL seen from two observers on diverging paths.
As for the moving scenario, I still think it would be OK because taking it from my example, if the second observer (whom I have now joined) was moving away (say a further light-month), when I looked back to my origin I would observe that the time is 13 months ago (the new distance), so if I travelled back instantly I would arrive at the time observed (13 months ago) plus the new distance (1 light year + 1 light month) which still equals the time now according to my wristwatch. However, rather than go around in more circles on this point I think we should just agree to disagree.What we can agree on is that you are not doing the mathematics correctly. Use the Lorentz transformation, and you will see. (That's what Grant's figure does-- so if you think his figure is wrong, then you should be trying to show why it does not obey the Lorentz transformation, instead of just making up numbers.) I can do a little better and tell you that what is wrong with your numbers is that you think there is a meaningful distance between the two frames without keeping track of the different notions of simultaneity in the two frames, and you can't do that, because to get a distance from something that is moving, you also have to specify a time that applies to the moving object-- that's what you are not doing according to the Einstein simultaneity prescription.

What's more, everything Grant is telling you is correct. I teach this stuff, Grant knows it quite well. The only real question here is if you want to cling to the misconceptions you now possess, or if you want to understand Einstein's relativity. But to clarify what seems to be a key issue, I want to underscore what Grant just said-- we all agree there is not a paradox in special relativity when all signals are limited by the speed of light. That's why the combination of relativity with respect for causality requires a theorem that no information can be propagated faster than c. That theorem is generally accepted in the physics community, not just by Grant and myself, and you have the wherewithall at this point to know why-- if you choose to.

Jeff Root
2010-Apr-13, 05:19 PM
It is a paradox to Webbo, Grant, even if it isn't a paradox to you or me.
Paradoxes are relative.

-- Jeff, in Minneapolis

Webbo
2010-Apr-13, 05:24 PM
This isn't according to me, it's according to Einstein. That's what he's telling you. And it's not a paradox: it's simply the nature of reality according to special relativity. It's a coordinate thing, like the reversal of left and right when two observers face each other. It's harder to understand because it involves time, and you're surprised by it because it doesn't accord with your intuition about time. So it took someone of Einstein's abilities to see past the surprise to the deeper significance.

Grant Hutchison

Alternatively, when refering to observers, rays of light and hastening toward beams of light, he was discussing the relative observations of this phenomena from the signals received, not the reality of the emissions. As I see no reference to this deeper meaning nor do I see any surprises within the example, I'm happy with my interpretation.

Ken G
2010-Apr-13, 05:25 PM
It is a paradox to Webbo, Grant, even if it isn't a paradox to you or me.
Paradoxes are relative.But that isn't the issue, of whether or not Webbo wants to call it a paradox. Webbo is doing the math wrong, and reaching the conclusion that what Grant sees as no paradox is something that Einstein would have seen as a paradox, and would therefore have rejected. Webbo is making an error, pure and simple, and it is the error of not using Einstein coordinates even as he speaks about the kind of reality that Einstein was envisioning. He has not seen that the coordinates he is using do not respect the correct symmetries of special relativity, and are not the ones for which the laws of physics will work as formulated by Einstein.

Webbo
2010-Apr-13, 05:25 PM
It is a paradox to Webbo, Grant, even if it isn't a paradox to you or me.
Paradoxes are relative.

-- Jeff, in Minneapolis

Excellent. Then we can both be correct.

Ken G
2010-Apr-13, 05:32 PM
Alternatively, when refering to observers, rays of light and hastening toward beams of light, he was discussing the relative observations of this phenomena from the signals received, not the reality of the emissions. As I see no reference to this deeper meaning nor do I see any surprises within the example, I'm happy with my interpretation.All the same, your interpretation is quite incorrect, because it will not agree with observations. You are missing the key elements of relativity, which expressed in coordinate form, are:
1) in Einstein coordinates, the speed of light is the same in all inertial frames
2) the laws of physics work the same in those coordinates for all inertial observers
3) the coordinates are globally arbitrary, but locally correspond to actual measurements we can make on rulers and clocks.
Your approach to the problem is actually using a different coordinatization (you are using an absolute concept of time), and it has the value to you of resolving a "paradox" that the rest of us simply do not view as a paradox (the "relativity of simultaneity", which we view as surprising and counterintuitive but not paradoxical), but its disadvantages you do not recognize: it would not have any of the above three properties, and so physics would simply not work in the coordinates you imagine. As such, you have missed Einstein's major contribution-- do you really think that all Einstein did was notice that light had a speed? That was well known for centuries beforehand.

grant hutchison
2010-Apr-13, 05:48 PM
As I see no reference to this deeper meaning nor do I see any surprises within the example, I'm happy with my interpretation.And so long as you happily restrict yourself to an eccentric interpretation of a couple of paragraphs lifted out of the context of Einstein's work, you're utterly missing the point. It's a really good and exciting point, it tells us something about how the Universe is actually observed to work, and it gives rise to all that other stuff about length contraction and time dilation.
It's up to you, though.

Grant Hutchison

Webbo
2010-Apr-13, 05:49 PM
1) in Einstein coordinates, the speed of light is the same in all inertial frames


Just concentrating on this for a second. If this is true then why does the light not reach M1 simultaneously? We are given the "train as a rigid reference-body (co-ordinate system)" and are also told that at the begining of the example that A & B have simultaneous strikes and are an equal distance from M1.

Also what possible meaning could there be to Einstein stating that "he is hastening towards the beam of light" or "riding on ahead of the beam of light". These can have no possible meaning in respect of statement 1 as he is not going anywhere in his reference frame.

Ken G
2010-Apr-13, 05:56 PM
Just concentrating on this for a second. If this is true then why does the light not reach M1 simultaneously? We are given the "train as a rigid reference-body (co-ordinate system)" and are also told that at the begining of the example that A & B have simultaneous strikes and are an equal distance from M1.The answer to your question is straightforward: the only frame in which we are given that the strikes are simultaneous is the frame where M1 is moving, and that's why, in that frame, the strikes are not seen simultaneously by M1. Einstein then asks, as should we all, can we allow M1 to do a calculation in which he can conclude the strikes were simulatanous? The answer is no-- not in a coordinate system with the attributes I mentioned above (i.e., a coordinate system that exhibits the principle of relativity). That was Einstein's contribution-- the arguments you are giving any of us could have done centuries before Einstein's day! The really key thing you need to get is that in the frame where M1 is stationary (the train frame), the strikes are not simultaneous, in the Einstein coordinatization (which is the means being used to breathe meaning into the term "simultaneous", because of its attractive properties I listed above).

Jeff Root
2010-Apr-13, 06:04 PM
But that isn't the issue ...
Whether it's the issue or not isn't the issue, either. Yet you said it, anyway.

Webbo thinks something is a paradox. If he thinks it's a paradox, it is.
The math and the theory and the physics are all side issues.

That is, they are side issues to my point that Webbo sees a paradox.
They are, of course, the central issues of the thread, so I guess what is
central and what is off to the side is relative.

-- Jeff, in Minneapolis

pzkpfw
2010-Apr-13, 07:30 PM
Not sure exactly how they will agree. According to this common frame which you considered at rest, Observer X at the back of the train will see flash A at 0 time and flash B 10 ms later (using my numbers earlier) and observer Y at the front will see flash B at 0 time and flash A 10 ms later. I can see that they could easily agree with each other that they were simultaneous strikes when taking the distance from each strike into consideration. However, when discussing this with observer M1 he will state that flash B was first. Your common frame should make M1 see them as simultaneous which Grant explicitly states they are not.

No! I didn't exactly say they'd agree with M' that the events were simultaneous, I said they'd agree about when they occured. If M' doesn't think the events were simultaneous, neither would these two observers. But if M' did then they would too. (So they'd have the same difference of opinion with M as M' has).


Your common frame should make M1 see them as simultaneous which Grant explicitly states they are not.

The common frame of the train means all observers on the train will agree about when the events occured, regardless of when they see the events. It does not at all force the events to be simultaneous. The events will or won't have been simultaneous with respect to the train, all train observers will agree with that.


According to this common frame which you considered at rest, ...

Do you not agree that observer M' can consider him/herself as at rest and the embankment moving?

If you don't, then there's no surprise you disagree with the mainstream interpretation, because you are essentially making the embankment frame a preferred frame that the train "really is" moving relative to - and not the converse.

grav
2010-Apr-13, 07:35 PM
1) in Einstein coordinates, the speed of light is the same in all inertial framesJust concentrating on this for a second. If this is true then why does the light not reach M1 simultaneously? We are given the "train as a rigid reference-body (co-ordinate system)" and are also told that at the begining of the example that A & B have simultaneous strikes and are an equal distance from M1.

Also what possible meaning could there be to Einstein stating that "he is hastening towards the beam of light" or "riding on ahead of the beam of light". These can have no possible meaning in respect of statement 1 as he is not going anywhere in his reference frame.One of the two postulates of SR is that observers in any frame will measure the speed of light as c in any direction regardless of the motion of the source. That means that from the embankment frame, if two pulses of light are emitted simultaneously at A and B, then the pulse of light travelling at c from B toward M1 while M1 is also travelling toward that pulse at v will reach M1 in a lesser time than the pulse that is travelling from A while M1 travels away from that pulse at v. In order for the train frame to measure the speed of light the same in both directions, then the time that the train frame measures for the pulses to travel over the same distance in either direction must also be the same. That means that according to the embankment frame, the train frame's clocks must be set differently at the front and back of the train, in order for the train frame to measure the same difference between the reading on the clocks when the pulse is emitted at either end and at M1 when it arrives. We find that if the length of the train is contracted and the train clocks are time dilating as the embankment frame measures them, both by a factor the same factor, and the train's clocks are set to greater readings upon clocks toward the back of the train, then the train frame will measure the speed of light at c in all directions.

Although the embankment frame observes the train's length as contracted in the line of motion, the train's rulers are also contracted to the same degree, so the train observers measure a normal length of the train using their own rulers. Although the embankment frame says the train's clocks are ticking slower, all physical processes upon the train occur more slowly also, so the train observers notice no difference. And although the embankment frame reads a simultaneity difference upon the train's clocks from the front to the back of the train, the train observers measure the same speed of light in either direction so say their clocks are synchronized. However, with the clocks unsynchronized as the embankment frame views them, if a traveller could be instantly transported from the back of the train to the front relative to the embankment frame, although it occurs over zero time according to the embankment frame's clocks, since the clock at the front of the train reads a lesser time than the clock at the back of the train as the embankment frame views them, then the traveller will have travelled back in time according to the train frame, arriving at the front of the train when a clock there reads a lesser time than the clock at the back of the train read when he departed.

pzkpfw
2010-Apr-13, 07:43 PM
...because M1 has moved towards B and shortened the distance from emission, and away from A and lengthened the distance from emission, so if light was emitted from each of these locations at the same time (in M's frame), I would expect them to be received at different times in M1's frame as the distances from emission are now different. However, as I know the distances are different I would deduce that emissions were simultaneous even though I observed a delay.

But what about the frame of M'?

He or she can consider themselves as at rest, and for them the lightening strikes occured at the front and back of the train. (Look at fig 1 of the page linked to in post #1 of Grimbles ATM thread and read the text above it).

[ http://www.bartleby.com/173/9.html ]

Since the distance from the front and back of the train to the observer in the exact middle is equal, and the speed of light is always c, then that observer must see the flashes at the same time, if the strikes occured at the same time (in the frame of that observer).

The frame of the train is no more or less important than the frame of the embankment.

The train observer is perfectly welcome, even if "moving" (according to an observer on the embankment), to have an occurance of lightening strikes at each end of his or her train, that they consider simultaneous. (But if that's the case, the embankment observer won't be agreeing that those strikes were simultaneous.)

Ken G
2010-Apr-13, 07:44 PM
Webbo thinks something is a paradox. If he thinks it's a paradox, it is.But he does not think there is a paradox, because he has both the wrong coordinates, and the wrong concept of what would be a paradox, to understand relativity. Ergo, none of this is about what he or I or Grant classifies as paradoxical-- that is all a complete red herring. What does matter here is that Webbo is not getting the correct understanding of relativity, and both Grant and I have tried hard to rectify that. Saying that he is actually not wrong, and it's all a question of labeling things as paradoxical or not, is quite counterproductive toward that end, because it encourages Webbo to think there is nothing to correct, robbing him of the opportunity to correct it.

That is, they are side issues to my point that Webbo sees a paradox.
They are, of course, the central issues of the thread, so I guess what is
central and what is off to the side is relative.
It certainly may be relative what Webbo's personal interest is in asking the question, but it is our job to see that the question is answered correctly anyway. In other words, if someone asks a question, gives an answer they like, and says they don't care what the right answer is, they only care about the answer they like, it then becomes our job to show them why they should not like the wrong answer.

grant hutchison
2010-Apr-13, 08:23 PM
It is a paradox to Webbo, Grant, even if it isn't a paradox to you or me.
Paradoxes are relative.And if my aunt had wheels, she'd be a tea-trolley.
Tea-trolleys are relative. :whistle:

Grant Hutchison

Jeff Root
2010-Apr-13, 08:26 PM
But he does not think there is a paradox ...
Yes he does. Post 215.

-- Jeff, in Minneapolis

Ken G
2010-Apr-13, 08:31 PM
Yes he does. Post 215.
You are not following his argument. Pay more attention to his key phrase "according to you" (Grant). He is using the "paradox" in post 215 to object to Grant's interpretation of Einstein. His own belief is that there is no paradox, and he knows that Einstein did not see a paradox. His mistake is incorrectly interpreting both Grant and Einstein, who are in mutual agreement. Perhaps you are entering the discussion late...

Jeff Root
2010-Apr-14, 09:07 AM
You are not following his argument. Pay more attention to his key phrase
"according to you" (Grant). He is using the "paradox" in post 215 to object
to Grant's interpretation of Einstein.
Yes. That is where he sees a paradox, even though you don't. He sees a
paradox, so for him, there is a paradox. It doesn't matter what you believe
about relativity, nor what Grant believes about relativity, nor what Einstein
believed or said about relativity.



His own belief is that there is no paradox, and he knows that Einstein did not
see a paradox. His mistake is incorrectly interpreting both Grant and Einstein,
who are in mutual agreement. Perhaps you are entering the discussion late...
I understand that. He interprets Einstein so as to avoid the paradox Grant posed.

Hey, if there is one thing that makes relativity fascinating, it is the paradoxes.

-- Jeff, in Minneapolis

grant hutchison
2010-Apr-14, 09:43 AM
Might I suggest that a prolonged discussion of the (alleged) subjective nature of paradoxes is a topic for another thread?

Grant Hutchison

Ken G
2010-Apr-14, 12:29 PM
Excellent idea. You and I are both interested in helping Webbo see that his current understanding of special relativity does not go past the idea that light has a speed, and he needs to take it to the place where he sees that the real heart of SR, in coordinate terms, is that there exists a coordinatization in which the speed of light is the same in all frames, and which all inertial observers see the same physics, and which the meaning of x and t locally (in that coordinatization) is what rulers and clocks give. Further, when you put those pieces together, FTL communication in two directions between (or travel by) two diverging observers can produce a causality violation.

Then, if he wants to get really into it, he will be ready for the concept of coordinate-free physics based on invariants. Maybe if SR was taught that way from the outset, and just toss the whole coordinate and simultaneity business, it would work better, but I suspect the usual wisdom that such an approach is mathematically overloaded might be correct.

Have we worked out the combination of relative speed and distance between the two observers, and the speed of the FTL motion relative to them, required to get a causality violation?

grant hutchison
2010-Apr-14, 02:21 PM
Have we worked out the combination of relative speed and distance between the two observers, and the speed of the FTL motion relative to them, required to get a causality violation?I think the threshold FTL velocity for causality violation is dependent only on their relative velocity, and not on their separation. Their separation comes into play in determining how far into the past a given FTL speed (above threshold) will take them.
The symmetry of my little diagram shows this, I think. Draw a horizontal line connecting the two worldlines. By symmetry, this represents the same FTL speed for both observers. So an observer making the double journey at that speed (with instantaneous velocity-matching and turnaround) will arrive back at the instant he departed. Anything faster will allow travel into the past of his own worldline.

Grant Hutchison

Webbo
2010-Apr-14, 02:49 PM
The answer to your question is straightforward: the only frame in which we are given that the strikes are simultaneous is the frame where M1 is moving, and that's why, in that frame, the strikes are not seen simultaneously by M1. Einstein then asks, as should we all, can we allow M1 to do a calculation in which he can conclude the strikes were simulatanous? The answer is no-- not in a coordinate system with the attributes I mentioned above (i.e., a coordinate system that exhibits the principle of relativity). That was Einstein's contribution-- the arguments you are giving any of us could have done centuries before Einstein's day! The really key thing you need to get is that in the frame where M1 is stationary (the train frame), the strikes are not simultaneous, in the Einstein coordinatization (which is the means being used to breathe meaning into the term "simultaneous", because of its attractive properties I listed above).

How can a fixed coordinate system move. It can only move relative to another frame. M1's experience is within that fixed frame so SR should be telling us that as he is an equal distance from A & B and stays that way, and as c is constant in his frame, he should see the the flashes simultaneously. As Einstein states within the example "If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously". So if the train suddenly stopped M1 would be see the flashes simultaneously. This should not change reality just because they move. c is supposed to be constant from A & B whether moving or not as within the frame and the fixed elements of it, movement is meaningless.

Webbo
2010-Apr-14, 02:49 PM
Deleted as duplicate of above

Webbo
2010-Apr-14, 03:20 PM
No! I didn't exactly say they'd agree with M' that the events were simultaneous, I said they'd agree about when they occured. If M' doesn't think the events were simultaneous, neither would these two observers. But if M' did then they would too. (So they'd have the same difference of opinion with M as M' has).

OK. M1 says that B was emitted before A, so therefore my observers X and Y at the back and front of the train respectively will also both say B emitted before A. Is that what you are stating?



The common frame of the train means all observers on the train will agree about when the events occured, regardless of when they see the events. It does not at all force the events to be simultaneous. The events will or won't have been simultaneous with respect to the train, all train observers will agree with that.

And how will they determine the difference between what is seen and what occurred? Light signal delay?

Also say 2 more observers were stationed on the embankment at A & B and fixed to M’s frame. Would they agree the flashes were simultaneous? If so, how? Light signal delay?



Do you not agree that observer M' can consider him/herself as at rest and the embankment moving?

Yes, according to SR. But if so, then when A & B strike simultaneously (remember at this point in the experiment the 2 frames are exactly the same point in space) the embankment is moving to the right so M1 being an equal distance from A & B should see the strikes simultaneously and M should therefore see A before B. This is exactly the same as the experiment by Einstein but having the embankment move (which should be irrelevant) instead of the train. Why the different result then? [QUOTE]

Webbo
2010-Apr-14, 03:41 PM
One of the two postulates of SR is that observers in any frame will measure the speed of light as c in any direction regardless of the motion of the source. That means that from the embankment frame, if two pulses of light are emitted simultaneously at A and B, then the pulse of light travelling at c from B toward M1 while M1 is also travelling toward that pulse at v will reach M1 in a lesser time than the pulse that is travelling from A while M1 travels away from that pulse at v. In order for the train frame to measure the speed of light the same in both directions, then the time that the train frame measures for the pulses to travel over the same distance in either direction must also be the same. That means that according to the embankment frame, the train frame's clocks must be set differently at the front and back of the train, in order for the train frame to measure the same difference between the reading on the clocks when the pulse is emitted at either end and at M1 when it arrives. We find that if the length of the train is contracted and the train clocks are time dilating as the embankment frame measures them, both by a factor the same factor, and the train's clocks are set to greater readings upon clocks toward the back of the train, then the train frame will measure the speed of light at c in all directions.

Although the embankment frame observes the train's length as contracted in the line of motion, the train's rulers are also contracted to the same degree, so the train observers measure a normal length of the train using their own rulers. Although the embankment frame says the train's clocks are ticking slower, all physical processes upon the train occur more slowly also, so the train observers notice no difference. And although the embankment frame reads a simultaneity difference upon the train's clocks from the front to the back of the train, the train observers measure the same speed of light in either direction so say their clocks are synchronized. However, with the clocks unsynchronized as the embankment frame views them, if a traveller could be instantly transported from the back of the train to the front relative to the embankment frame, although it occurs over zero time according to the embankment frame's clocks, since the clock at the front of the train reads a lesser time than the clock at the back of the train as the embankment frame views them, then the traveller will have travelled back in time according to the train frame, arriving at the front of the train when a clock there reads a lesser time than the clock at the back of the train read when he departed.

And from the perspective of the train observer why it is not he who sees the flashes simultaneously and the embankment moving past that requires the above corrections for time dilation and length contraction, and the the clocks on points A & B on the embankment that require adjusting?

Webbo
2010-Apr-14, 03:51 PM
But what about the frame of M'?

He or she can consider themselves as at rest, and for them the lightening strikes occured at the front and back of the train. (Look at fig 1 of the page linked to in post #1 of Grimbles ATM thread and read the text above it).

[ http://www.bartleby.com/173/9.html ]

Since the distance from the front and back of the train to the observer in the exact middle is equal, and the speed of light is always c, then that observer must see the flashes at the same time, if the strikes occured at the same time (in the frame of that observer).

The frame of the train is no more or less important than the frame of the embankment.

The train observer is perfectly welcome, even if "moving" (according to an observer on the embankment), to have an occurance of lightening strikes at each end of his or her train, that they consider simultaneous. (But if that's the case, the embankment observer won't be agreeing that those strikes were simultaneous.)

There is only one point A & B and this is fixed with respect to the embankment. The pointson the train that used to correspond with A & B (lets say A1 & B1) have also moved to the right so observer M1 is central to A1 & B1 but not central to A & B by the time the light signals arrive.

grant hutchison
2010-Apr-14, 03:58 PM
I think the threshold FTL velocity for causality violation is dependent only on their relative velocity, and not on their separation. Their separation comes into play in determining how far into the past a given FTL speed (above threshold) will take them.
The symmetry of my little diagram shows this, I think. Draw a horizontal line connecting the two worldlines. By symmetry, this represents the same FTL speed for both observers. So an observer making the double journey at that speed (with instantaneous velocity-matching and turnaround) will arrive back at the instant he departed. Anything faster will allow travel into the past of his own worldline.Further to this, it looks as if the reference frame in my diagram (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity4.jpg) (the frame represented by orthogonal black spacetime axes) provides a useful reference for the red and blue observers. Each will find that the threshold FTL speed for causality violation is just the reciprocal of their velocity relative to the reference frame: if they're moving at 0.5c relative to that frame, then the causality-violation threshold is 2c, for instance.
If the red and blue observers can't measure their velocity directly against the reference frame, then the sums get a little more complicated. If all they have is their velocity relative to each other, they have to come at the problem via rapidity, or by the relativistic velocity-addition formula.

Grant Hutchison

Webbo
2010-Apr-14, 04:02 PM
I think the threshold FTL velocity for causality violation is dependent only on their relative velocity, and not on their separation. Their separation comes into play in determining how far into the past a given FTL speed (above threshold) will take them.
The symmetry of my little diagram shows this, I think. Draw a horizontal line connecting the two worldlines. By symmetry, this represents the same FTL speed for both observers. So an observer making the double journey at that speed (with instantaneous velocity-matching and turnaround) will arrive back at the instant he departed. Anything faster will allow travel into the past of his own worldline.

Grant Hutchison

Isn't this what I stated pages ago. And as you said before, faster than instant would create causality problems and I certainly agree with that.

grant hutchison
2010-Apr-14, 04:19 PM
Isn't this what I stated pages ago. And as you said before, faster than instant would create causality problems and I certainly agree with that.No, it isn't what you stated pages ago, and it isn't Ken's "faster than instant" travel, either (except in a round-about way). It's just faster-than-light (but slower than instantaneous) travel, first in the red frame and then in the blue.

Honestly, you're not going to get how that works until you've cleared up your misunderstanding about the train and embankment.

Grant Hutchison

Webbo
2010-Apr-14, 04:30 PM
One of the two postulates of SR is that observers in any frame will measure the speed of light as c in any direction regardless of the motion of the source.

I think this is a rewording of the postulate that changes its meaning entirely. IMO that is not what it says. To quote;

"...light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body."

IMO that has an entirely different meaning to your rewording. To me that states that always light propagates at c relative to empty fixed space. If light is measured at c from a body in a moving frame then c would be dependent on that moving body but it should be independent of it.

In addition, why did we even need this second postulate if it means, under your interpretation, exactly the same as the first;

"...the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good."

The reason is because it is stating something different to the second as Einstein states himself;

"...and also introduce another postulate, which is only apparently irreconcilable with the former..."

So he is fully aware they are essentially contradictory postualtes on the surface.

Further, would you say that material objects are covered under SR? If I am sitting on a moving train and I throw an apple towards the front or back of the train with the same force, it will travel at the same speed and go exactly the same distance. That is exactly how you would describe the speed of c from a flashlight pointed forwards or backwards on the train, but isn't light supposed to behave differently and be independent of the moving body?

Webbo
2010-Apr-14, 04:34 PM
No, it isn't what you stated pages ago, and it isn't Ken's "faster than instant" travel, either (except in a round-about way). It's just faster-than-light (but slower than instantaneous) travel, first in the red frame and then in the blue.

Honestly, you're not going to get how that works until you've cleared up your misunderstanding about the train and embankment.

Grant Hutchison

My apologies, I though you posted the FTI comments. At least you are both partially accepting that FTL can be possible without causation issues.

grant hutchison
2010-Apr-14, 05:02 PM
My apologies, I though you posted the FTI comments. At least you are both partially accepting that FTL can be possible without causation issues.I've no recollection of "partially accepting"anything quite so vague. I've explained why FTL under SR immediately raises the potential for causality violation, and I've discussed specific examples in which causality violation does or doesn't occur.

Grant Hutchison

Ken G
2010-Apr-14, 06:24 PM
I think the threshold FTL velocity for causality violation is dependent only on their relative velocity, and not on their separation. Their separation comes into play in determining how far into the past a given FTL speed (above threshold) will take them.
The symmetry of my little diagram shows this, I think. Draw a horizontal line connecting the two worldlines. By symmetry, this represents the same FTL speed for both observers. So an observer making the double journey at that speed (with instantaneous velocity-matching and turnaround) will arrive back at the instant he departed. Anything faster will allow travel into the past of his own worldline.Yes, quite right-- your figure gives the instant-speed limit of a general theorem that to get causality violations via symmetric FTL communication between two diverging observers (by which I mean communication with an FTL signal sent out and returned at the same speed in each frame), that signal must travel "faster than instantaneous" in the intermediate frame. If the observers' speed in that intermediate frame is v, then the signal in the frame it is sent must be sent at a speed u that obeys both u > v, and uv > c2, and it doesn't matter how close to the synchronization point the signal transfer occurs. Cute.

Ken G
2010-Apr-14, 06:43 PM
M1's experience is within that fixed frame so SR should be telling us that as he is an equal distance from A & B and stays that way, and as c is constant in his frame, he should see the the flashes simultaneously.What we are telling you is that your philosophically-based conclusion is experimentally incorrect. It was a big surprise that what you are saying should happen is not what happens, and Einstein's contribution was figuring out how to describe the way the experiment actually does come out. But what you need to get, and indeed can make no progress until you get, is that if the embankment assumes the speed of light is c, and does a calculation using that, and concludes that the strikes at A and B are simultaneous, then the observer at M1, also doing a calculation assuming light moves at c, would be forced to conclude the strikes at A and B were not simultaneous (and as a result, M1 would also have to conclude that the strikes did not occur the same distance apart that the embankment-observer measures by looking at the blackened spots). That argument was done long before Einstein, it's just that before Einstein, the general conclusion from the bizarre consequence of assuming c was constant, was that coordinates (and a concomitant meaning of simultaneity) should be used such that light does not move at c in both frames (i.e., Maxwell's equations are only correct in one frame, the "aether" frame), such that M1's calculation should yield that the strikes actually were simultaneous (anything else being seen as paradoxical). Einstein's contribution was to point out that the actual resolution that allows us to apply the laws of physics the same in all inertial frames is to use a different set of coordinates (and simultaneity), described early in this thread, for which the speed of light was indeed c in all frames (so Maxwell's equations work in all frames, being a law of physics), but for which the strikes are just plain not simultaneous for M1, and this is not a paradox.

As Grant also pointed out, until you get the above paragraph, you are not going to understand Einstein's relativity, nor are you going to get correct predictions to experiments by using the recognized forms of the laws of physics (like Maxwell's equations) applied to the "naive" coordinates you are imagining. I have stressed that there is nothing wrong with the coordinates you use to imagine the situation (and define simultaneity), coordinates are just language-- but they are not Einstein's coordinates, they won't work with the forms of the laws of physics you'll find in textbooks, and they won't exhibit the beautiful symmetries of Einstein's coordinates. In short, they just won't be the standard language used in SR, and you will get confused if you mistake them for SR coordinates. That confusion will lead you to miss why symmetric FTL signals (i.e., reckoned in symmetric Einstein coordinates) will indeed result in concrete causality violations.


So if the train suddenly stopped M1 would be see the flashes simultaneously.That is certainly clear, long before Einstein.

This should not change reality just because they move.I'm not sure what this is meant to mean, you are describing two different realities and asserting that the reality should not be changed, despite changing it. But that is not your core problem here, it is your unwillingness to part with an absolute sense of simultaneity. If you want to use Einstein coordinates and his concept of simultaneity, and thereby understand the language he developed, the very first thing you will need to do is understand why you have to part with your concept of absolute simultaneity in favor of "relativity of simultaneity." Google that term for guidance, or consider the attibutes of the "Lorentz transformation."


c is supposed to be constant from A & B whether moving or not as within the frame and the fixed elements of it, movement is meaningless.It is true that M1 can call his own frame, and his train, stationary, but the presence or absence of relative motion in regard to the embankment represents two different realities that must not be treated the same.
At least you are both partially accepting that FTL can be possible without causation issues.Well, I would say that we are both completely accepting that, in a case where a signal is sent back and forth at speed u in the frame of two observers each diverging at speed v < c in the intermediate frame, FTL only causes causation issues when uv > c2, which is also the requirement for "faster than instantaneous" signals in the intermediate frame.