Something I don't understand about FTL travel


by AndromedaRXJ
Tags: travel
AndromedaRXJ
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#1
Jan20-13, 01:42 AM
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Okay first of all, I know it's impossible to travel faster than light. But they say that IF.... IF you could, then you would arrive at your destination before you left.

That's what I don't understand.

Like for example, it takes light 8 minutes to get to us from the Sun.

So if a person left Earth at 6:00 and arrived at the Sun at 6:07, he traveled faster than light. But that doesn't sound like he arrived there before leaving. It just sounds like he arrived 7 minutes LATER. Faster than expected of light, but still LATER.

Can someone explain this?
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Matterwave
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Jan20-13, 01:45 AM
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In some valid reference frame which is in motion with respect to the Earth-Sun system, it will be that you arrived before you left because any two events which are space-like separated (you leaving and you arriving would be 2 events which are space-like separated) can be found to change order by making a lorentz boost. I will let someone else show the math because I'm lazy.
stevendaryl
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Jan20-13, 08:10 AM
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Quote Quote by AndromedaRXJ View Post
Okay first of all, I know it's impossible to travel faster than light. But they say that IF.... IF you could, then you would arrive at your destination before you left.
It has to do with the way time transforms in Special Relativity. Suppose you have some faster-than-light ship that travels at some speed [itex]U > c[/itex]. So pick two frames F and F' with relative speed v. Have your FTL ship travel from [itex]x_1=0, t_1=0[/itex] to some point [itex]x_2=UT, t_2 = T[/itex]. Under Lorentz transforms, we can compute the coordinates in frame F' as follows:

[itex]x_1' = 0[/itex]
[itex]t_1' = 0[/itex]
[itex]x_2' = \gamma (x_2 - v t_2) = \gamma (U-v)T[/itex]
[itex]t_2' = \gamma (t_2 - \dfrac{v}{c^2} x_2) = \gamma (1-\dfrac{Uv}{c^2})T[/itex]

Note: if [itex]U = \dfrac{c^2}{v}[/itex], then [itex]t_2' = 0[/itex]. So if [itex]U > c[/itex] in one frame, then travel time can be instantaneous in another frame. But it gets weirder than that. If [itex]U > \dfrac{c^2}{v}[/itex], then [itex]t_2' < 0[/itex]. So in frame F', the faster-than-light ship arrives Before it leaves.

So if faster-than-light travel is possible in one frame, then there are frames in which back-in-time travel is possible.

Fredrik
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Jan20-13, 12:47 PM
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Something I don't understand about FTL travel


Quote Quote by AndromedaRXJ View Post
Okay first of all, I know it's impossible to travel faster than light. But they say that IF.... IF you could, then you would arrive at your destination before you left.

That's what I don't understand.

Like for example, it takes light 8 minutes to get to us from the Sun.

So if a person left Earth at 6:00 and arrived at the Sun at 6:07, he traveled faster than light. But that doesn't sound like he arrived there before leaving. It just sounds like he arrived 7 minutes LATER. Faster than expected of light, but still LATER.

Can someone explain this?
You're right. However, in a coordinate system that's comoving with someone going fast (but <c) in the direction towards the sun, the same sequence of events would be described very differently. I think a spacetime diagram is the best way to explain it. I drew this one in paint, so it's ugly (and not 100% accurate, because the lines don't have the right slopes), but it illustrates the main point well enough.



The t and x axes are the ones used by the observer on Earth. The points on a line that's parallel to the x axis all have the same t coordinate. The t' and x' axes are the ones used by an observer whose motion is described by the t' axis. The points on a line that's parallel to the x' axis all have the same t' coordinate. Note that it's possible to draw a horizontal line in the diagram below the Sun 6:01 event and above the Earth 6:00 event. This means that the Sun 6:01 event has a later t coordinate than the Earth 6:00 event. But the Earth 6:00 event is above the x' axis, while the Sun 6:01 event is below the x' axis. This means that the former has a later t' coordinate than the latter.

So in the "primed" coordinate system, the thing you think of as going from the Earth to the sun is described as going from the Sun to the Earth. Alternatively, it can be considered going from the Earth to the sun, but arriving before departing (i.e. going back in time).

Edit: I have to add that it doesn't make sense to use a person in the example, because a person moving FTL contradicts SR. So SR plus your assumption is logically inconsistent, and in an inconsistent system, every statement is true (including things like 7=3). It never makes sense to ask what would happen if something that contradicts the theory we're supposed to use to answer the question is true. But it's fine to ask what happens if some weird-*** particle makes this trip, because that doesn't immediately contradict SR. (Apparently we can't say "***". I didn't know that).
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AndromedaRXJ
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#5
Jan22-13, 11:03 AM
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Thanks everyone for answering. I've read all your posts.

I'm not that great in understanding mathematics, however. So in other words, not all observers will see the ships arrival before it's departure? Only the one moving towards the Sun?

Is the way I described it in my first post assumes that time and simultaneity are absolute and not relative? Or is there at least one observer who would see it that way?

Quote Quote by stevendaryl View Post
It has to do with the way time transforms in Special Relativity. Suppose you have some faster-than-light ship that travels at some speed [itex]U > c[/itex]. So pick two frames F and F' with relative speed v. Have your FTL ship travel from [itex]x_1=0, t_1=0[/itex] to some point [itex]x_2=UT, t_2 = T[/itex]. Under Lorentz transforms, we can compute the coordinates in frame F' as follows:

[itex]x_1' = 0[/itex]
[itex]t_1' = 0[/itex]
[itex]x_2' = \gamma (x_2 - v t_2) = \gamma (U-v)T[/itex]
[itex]t_2' = \gamma (t_2 - \dfrac{v}{c^2} x_2) = \gamma (1-\dfrac{Uv}{c^2})T[/itex]

Note: if [itex]U = \dfrac{c^2}{v}[/itex], then [itex]t_2' = 0[/itex]. So if [itex]U > c[/itex] in one frame, then travel time can be instantaneous in another frame. But it gets weirder than that. If [itex]U > \dfrac{c^2}{v}[/itex], then [itex]t_2' < 0[/itex]. So in frame F', the faster-than-light ship arrives Before it leaves.

So if faster-than-light travel is possible in one frame, then there are frames in which back-in-time travel is possible.
Also, what is the capital T in these equations?
Fredrik
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Jan22-13, 12:06 PM
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Quote Quote by AndromedaRXJ View Post
I'm not that great in understanding mathematics, however.
You should focus on spacetime diagrams then.

Quote Quote by AndromedaRXJ View Post
So in other words, not all observers will see the ships arrival before it's departure? Only the one moving towards the Sun?
Correct.

Quote Quote by AndromedaRXJ View Post
Is the way I described it in my first post assumes that time and simultaneity are absolute and not relative? Or is there at least one observer who would see it that way?
The Earth observer sees it that way. A thing is only "absolute" if all observers agree about it.

Quote Quote by AndromedaRXJ View Post
Also, what is the capital T in these equations?
It's the time coordinate of the arrival event in the coordinate system comoving with Earth. Since the departure event has time coordinate 0, it's also the travel time in that same coordinate system.
Whovian
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#7
Jan22-13, 01:10 PM
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Someone in a certain reference frame will see you arrive before you left, but they'll see you arrive as a flying purple telephone made of strawberry sausage since the journey took you 5i years.
georgir
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#8
Jan24-13, 04:35 AM
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Quote Quote by Whovian View Post
Someone in a certain reference frame will see you arrive before you left, but they'll see you arrive as a flying purple telephone made of strawberry sausage since the journey took you 5i years.
No, imaginery durations or coordinates are the result of the observer himself moving at above c, but in this case we are talking about a subluminal observer observing superluminal-separated (spacelike separated) events. He will just see a reversed order of events.
georgir
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Jan24-13, 05:32 AM
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I think the distinction between spacelike-separated, timelike-separated and null-separated events is good to point out here.

If a velocity under the speed of light is required to visit both of a pair of events in one reference frame, then that is also true for all other inertial reference frames. The order of the events is also always the same in all reference frames. Such events are called "timelike separated", and because of the invariance of their relative order, causal relation between them is possible.

If a velocity above the speed of light is required to visit both of a pair of events in one reference frame, then that is also true for all other inertial reference frames. The order of the events is not the same in all reference frames. In some frames they are simultaneous, in others one preceeds the other, and in others yet their order is reversed. Such events are called "spacelike separated" and because of the undefiniteness of their relative order, causal relation between them is considered impossible. In other words, they can not belong on a single "worldline", or be events concerning a single object.

If a velocity exactly equal the speed of light is required to visit both of a pair of events in one reference frame, then that is also true for all other inertial reference frames. The order of the events is also always the same in all reference frames. Such events are called "null separated", causal relation between them is possible, but only as electromagnetic or gravity field changes, not massive particle travel.
vemvare
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Jan29-13, 06:54 PM
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Do I read Fredrik's diagram correctly if I conclude that the fast-moving observer will see the "FTL" ship moving backwards, but due to the distances between Earth and sun and the observer moving slower than light it could never interfere, fr. ex by first seeing the ship leaving (backwards) the sun and then sending a message to tell the ship not to start the journey in the first place.

Oh, this is hard to put to words, but my point is, is there some kind of difference between a scenario leading to a paradox where an observer would observe an effect leading to a cause (=time going backwards) but not being able to interfere due to the limits of lightspeed, and an observer being able to either contact his own past, or convey a message from the sun to Earth backwards in time?

Are both scenarios "equally impossible?"
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Jan30-13, 03:43 AM
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Quote Quote by vemvare View Post
Do I read Fredrik's diagram correctly if I conclude that the fast-moving observer will see the "FTL" ship moving backwards, but due to the distances between Earth and sun and the observer moving slower than light it could never interfere, fr. ex by first seeing the ship leaving (backwards) the sun and then sending a message to tell the ship not to start the journey in the first place.
The world line of the "fast-moving" observer is the t' axis. As you can see, it intersects the world line of the FTL object. So they actually meet at that event, and can do things to each other. Suppose e.g. that a short time before that event, the "fast-moving" ship drops an empty fuel tank or something, that collides with and destroys the FTL object. This would eliminate the part of the FTL object's world line that's to the right of the explosion event, not the part that's to the left. From the fast-moving ship's point of view, this is what happened: It dropped a fuel tank, which subsequently exploded, and out of the explosion came an object that was moving FTL towards Earth.

Quote Quote by vemvare View Post
Oh, this is hard to put to words, but my point is, is there some kind of difference between a scenario leading to a paradox where an observer would observe an effect leading to a cause (=time going backwards) but not being able to interfere due to the limits of lightspeed, and an observer being able to either contact his own past, or convey a message from the sun to Earth backwards in time?
To get an actual contradiction, you must consider two devices that are moving at different velocities, and are both capable of receiving and emitting FTL particles. See this post to find out how the ability to send messages that have infinite speed in your own rest frame leads to a contradiction.
stevendaryl
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Jan30-13, 07:26 AM
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Quote Quote by vemvare View Post
Do I read Fredrik's diagram correctly if I conclude that the fast-moving observer will see the "FTL" ship moving backwards, but due to the distances between Earth and sun and the observer moving slower than light it could never interfere, fr. ex by first seeing the ship leaving (backwards) the sun and then sending a message to tell the ship not to start the journey in the first place.

Oh, this is hard to put to words, but my point is, is there some kind of difference between a scenario leading to a paradox where an observer would observe an effect leading to a cause (=time going backwards) but not being able to interfere due to the limits of lightspeed, and an observer being able to either contact his own past, or convey a message from the sun to Earth backwards in time?

Are both scenarios "equally impossible?"
What I think you might be getting at is that travel back in time doesn't necessarily lead to a paradox. What is paradoxical is a closed time loop. If I can send a message to my own younger self, then that would lead to a paradox, because I can tell myself "Don't send this message".

But if FTL is possible, and all frames are equivalent (the relativity principle), then it is possible to send a message to your own younger self. Send a back-in-time message from yourself to a colleague, and have the colleague send a back-in-time message back to you.
Fredrik
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Jan30-13, 07:58 AM
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Quote Quote by stevendaryl View Post
But FTL is possible, and all frames are equivalent (the relativity principle), then it is possible to send a message to your own younger self. Send a back-in-time message from yourself to a colleague, and have the colleague send a back-in-time message back to you.
This is a bit of an oversimplification. I would describe it like this: Send a FTL message to a colleague who is moving away from you at a speed close to c, and have him send a FTL reply back to you. If the speeds and distances involved are great enough, then you will receive the reply before you sent the message.
Dadface
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Jan30-13, 08:34 AM
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What am I missing here? Andromeda RXJ seems to be asking .........What would the outcome be if you did something that's impossible to do?
stevendaryl
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Jan30-13, 08:59 AM
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Quote Quote by Fredrik View Post
This is a bit of an oversimplification. I would describe it like this: Send a FTL message to a colleague who is moving away from you at a speed close to c, and have him send a FTL reply back to you. If the speeds and distances involved are great enough, then you will receive the reply before you sent the message.
But you can prove that if it is possible to send FTL messages between two observers, then it is possible to send back-in-time messages between the same two observers.
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Jan30-13, 08:59 AM
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Quote Quote by Dadface View Post
What am I missing here? Andromeda RXJ seems to be asking .........What would the outcome be if you did something that's impossible to do?
It's not possible to do something that is impossible.
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Jan30-13, 09:04 AM
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Quote Quote by stevendaryl View Post
It's not possible to do something that is impossible.
That's exactly my point. If FTL is impossible then why discuss the consequences of moving at FTL speeds? The question is contradictory and doesn't make sense.
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Jan30-13, 09:12 AM
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Quote Quote by stevendaryl View Post
But you can prove that if it is possible to send FTL messages between two observers, then it is possible to send back-in-time messages between the same two observers.
To produce a message that's going back in time in all inertial coordinate systems, you need two devices that can send and receive FTL messages, and there must be a large velocity difference or large distance between them.


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