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How exactly does FTL travel cause reverse time travel?

  1. Sep 22, 2013 #1
    So I look at the Minowski diagrams, and I can understand that moving on the y means moving in time, and moving on the x means moving in space, but drawing some diagrams of my own, I find it impossible to see why it would imply FTL travel. All objects travel either directly on the x (instantaneous jump) or forwards. I can understand why you would have some weird visual effects (seeing yourself jump into hyperspace after the jump), but I cannot understand why it would imply time travel, and have given up all attempts to interpret the Minowski diagrams.

    Could anyone help by explaining exactly what would happen if a spaceship was equipped with an FTL jump drive?
     
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  3. Sep 22, 2013 #2

    HallsofIvy

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    No one can answer that because no one know exactly what an "FTL jump drive" would do!
    It is true that the Lorentz transformations include
    [tex]\frac{t- vx/c^2}{\sqrt{1- v^2/c^2}}[/tex]

    If we were to take v> 0, that denominator would be an imaginary number. What does that mean? Another way of looking at is that time appears to move more slowly for a frame as its speed (relative to the reference frame) approaches c. "In the limit", as v approaches c, time intervals go to 0 so one can imagine that, if you take v> c, time intervals would become negative, reversing time.

    Of course, this is using one aspect of relativityb (time goes more slowly as speed increases) while denying another (that nothing can move faster than light). The more sensible thing to do is to accept all of the consequences of relativity including tht nothing can move faster than light.
     
  4. Sep 23, 2013 #3
    Basically, what I meant by 'FTL jump drive' is that it disappears in one place and instantaneously reappears in another. It isn't affected by relativistic effects. If it were affected by relativistic effects, time runs sideways or something.
     
  5. Sep 23, 2013 #4
    The faster you travel, the slower time goes relative to something else. ##t\prime=t\sqrt{1-\frac{v^2}{c^2}}## and so when your velocity squared reaches the speed of light squared ##t\prime=t\sqrt{1-\frac{299,792,458m/s^2}{299,792,458m/s^2}}## you simply end up with ##t\prime=0## meaning that time relative to something else is 0. This means that you effectively enter a changeless state.

    The blood in your veins will stop flowing, you will stop breathing, you will not age, your body will not do anything, you enter a stage of eternal hibernation if you will... you could travel at the speed of light for a million years and then stop, the blood would resume flowing, you would start breathing again ect and as far as you're concerned, absolutely 0 time has passed. Nada.

    So the fun part, if you go faster than the speed of light, even by 1m/s ##t\prime=t\sqrt{1-\frac{299,792,459m/s^2}{299,792,458m/s^2}}## which equals ##-\frac{1}{299,792,458}## or ##-3.3356......\times10^{-9}##

    And that is why if you go faster than light, time will go backwards. Moreover though, with jump drives or more accurately warp drives, you're not going faster than light, you're manipulating space-time, making the distance smaller.

    A simple analogy... If two cars at equal speed race round a track, the first car goes round the perimetre, the second car cuts across the track, who will get to the finish line first? The car that took the shortcut of course, but it didn't go faster than the first car, it simply took a shorter route.

    I hope that clears things up.
     
  6. Sep 24, 2013 #5

    Borek

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    Nonsense. Square root of a negative number is not another negative number, but an imaginary number. HallsofIvy already wrote that.
     
  7. Sep 24, 2013 #6
    Note first, that I don't believe time travel is possible.

    However, if it was, you would get visual effects that can be modelled using sound. Sound travels at the speed of sound. What happens if something travels faster than sound?
    First of all, you would get Cherenkov radiation, which is analogous to the sonic boom.
    Secondly, you will loose casuality. Imagine a set of devices that communicate by sound. You can observe that they obey some laws regarding the order of their activation. When you exceed the speed of sound, you invalidate that laws.
    Third thing is that you get two "sonic images" for an object that travels faster than sound. Imagine an object that makes a beep, moves faster than sound and makes another beep. You will hear two beeps coming from two separate objects apparently. If the moving objects beeps constantly, you will hear it as two beeping objects, moving in opposite directions and finally meeting at the point where the object's path is closest to you. This was once a proposed interpretation of annihilation of matter and antimatter, but now it's known to be false.
     
  8. Sep 24, 2013 #7
    I've just realised that I've made a typographical error in my first post ("imply FTL travel" rather than "reverse time travel"), but all of you got the idea anyway.

    So let's say that Alice jumps away from Bob by about one light second, quickly snaps a picture of the moon and returns to him to give him the picture before the moon's light reaches him. I see no reverse time travelling there.

    @haael: I don't really get it. I imagine that all you would have would be sonic images that lagged behind the actual object.
     
  9. Sep 24, 2013 #8

    stevendaryl

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    First, let's prove a little theorem:

    Theorem 1: If FTL travel is possible in every frame, then instantaneous travel is possible.

    To see this, let an FTL rocket travel from the point [itex]x=0, t=0[/itex] to the point [itex]x=L, t=L/w[/itex], where [itex]w[/itex] is the speed of the FTL rocket. Now, switch to another frame using the Lorentz transforms.

    [itex](x=0, t=0) \rightarrow (x'=0, t'=0)[/itex]
    [itex](x=L, t=L/w) \rightarrow (x'=\gamma L (1-v/w), t' = \gamma L/w (1 - vw/c^2))[/itex]

    Now, if we choose [itex]v = c^2/w[/itex], then we have the rocket arriving at time
    [itex]t'=0[/itex] in the primed frame. So in the primed frame, the rocket arrives instantaneously.

    So FTL travel in one frame implies instantaneous travel in a different frame. If instantaneous travel is possible in one frame, then it is possible in every frame, by the relativity principle.

    Theorem 2: If instantaneous travel is possible in every frame, then back-in-time travel is possible.

    To see this, just pick any two frames, F and F' with a relative (slower-than-light) speed v.

    Send a rocket instantaneously according to frame F from
    [itex]x=0[/itex] at time [itex]t=0[/itex] to [itex]x=L[/itex] at time [itex]t=0[/itex].

    According to frame F', the rocket left at time [itex]t'=0[/itex] and arrived at time [itex]t'=- \gamma vL/c^2[/itex]. Now, send the rocket back to the point [itex]x=0[/itex] instantaneously according to frame F'. The rocket arrives at time [itex]t'=- \gamma vL/c^2[/itex]. Using the Lorentz transforms again, we see that:

    [itex]t' = \gamma (t - vx/c^2) = \gamma t[/itex]

    So [itex]t = t'/\gamma = - vL/c^2[/itex]

    So the rocket gets back at a time BEFORE time [itex]t=0[/itex].
     
  10. Sep 24, 2013 #9
    (snip basically correct reasoning)
    Non sequitur - the "relativity principle" itself is not confirmed here.
    Consider:
    Positive statements (something normally happens) are better founded in experiment than negative ones (something is impossible).

    For example parity violation - electromagnetic, gravitational and strong forces are completely symmetric against left and right.

    Yet weak interaction exists - and includes complete parity violation.

    There is also a CP violation... supposed to entail T violation.

    Or does it?

    It is claimed that a CPT violation would authomatically involve a Lorentz violation.

    Suppose it does. What next?

    Imagine that a hidden preferred frame exists. It has no interaction with the common physical processes - so time is dilated as if no preferred frame existed.

    Except that the preferred frame does exist - and plays a role in some rare interactions.

    Causing FTL.

    Pretty clearly, if FTL travel A-B were possible, due to relativity of simultaneity you can always choose a frame where arrival at B is before departure from A.

    But how about A-B-A, or A-B-C-A?

    Is it possible to have FTL where A-A always takes zero or positive time irrespective of where B or C may be, or what the frame of observer is?
     
  11. Sep 24, 2013 #10

    stevendaryl

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    I don't know what you mean by "not confirmed". If something is possible in one frame, but not in another, then that is a violation of the principle of relativity.

    The proof that FTL implies back-in-time travel is the reason people believe that nothing (or at least nothing that can carry information) can travel faster than light. So basically you're bringing up "nothing can travel faster than light" in the context of a proof that nothing can travel faster than light.

    That means that relativity is wrong. The argument here is that ASSUMING relativity, it follows that FTL implies backward-in-time travel (or the ability to send messages back in time).

    Absolutely. There is no problem with FTL if there is a preferred frame.
     
  12. Sep 24, 2013 #11

    stevendaryl

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    You can certainly cook up a theory in which
    • There is a preferred frame, F.
    • Ordinary objects moving at speed v relative to F are contracted by a factor of [itex]\sqrt{1-(v/c)^2}[/itex] in the direction of their motion.
    • Ordinary clocks moving at speed v relative to F are slowed by the same factor.
    • There is a special rocket (necessarily NOT made out of ordinary matter) that travels at speed [itex]w > c[/itex] relative to frame F.

    This theory would be equivalent to SR for experiments involving clocks and rulers made of ordinary matter, but would be different from SR for experiments involving unordinary matter.
     
  13. Sep 24, 2013 #12
    R7Wp89C.png
    As I understand it, when looking at it from the primed perspective, Point A and B shift down the y axis, implying reverse time travel?
     
  14. Sep 24, 2013 #13
     
  15. Sep 24, 2013 #14

    ghwellsjr

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    Here is a labeled version of your Minkowski drawing with a line going just from A to B (the markings along the top and right are for the yellow grid lines):

    attachment.php?attachmentid=62164&stc=1&d=1380063615.png

    There is nothing special about this worldline. It's just a stationary object located at x=5 with time going from 3 to 4 (according to the frame depicted with the grey grid lines). Here's another spacetime diagram showing just the rest frame of this object:

    attachment.php?attachmentid=62145&stc=1&d=1380046622.png

    Now I use the Lorentz Transformation process on points A and B to see what it looks like in a frame moving at 0.5c with respect to the first frame:

    attachment.php?attachmentid=62146&stc=1&d=1380046622.png

    Now you can see that the object is moving at -0.5c and its time and space coordinates are the same as the yellow coordinates in the first drawing.

    As dauto pointed out, the FTL occurs when the object instantly moves from its location near x=0 to Point A and then later from Point B back to near x=0.

    The bottom line is that the Minkowski diagram does not say anything about whether or not an object can actually travel at FTL as you can draw any lines going any way you want. Normally, people use a Minkowski diagram to depict the rest frames of two different objects traveling at different speeds with respect to each other but that speed cannot be equal or greater than c.

    By the way, what is the purpose of the blue line in your drawing?
     

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  16. Sep 24, 2013 #15

    DrGreg

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  17. Sep 24, 2013 #16

    QuantumPion

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    An interesting, related paradox involves merely sending messages FTL. If A sends a message to B FTL, and then B sends a response back to A FTL, it is possible for A to receive the response before sending the initial message. This leads to all sorts of logical paradoxes.
     
  18. Sep 24, 2013 #17

    QuantumPion

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    I totally wrote my post first but DrGreg got his in first using his tachyonic antitelephone. :p
     
  19. Sep 25, 2013 #18
    @ghwellsjr: The blue line is supposed to be a relativistic object, and the light blue lines its Lorentz-transformed frame of reference.
    Looking at your diagram, it seems that the origin of the path to Point A (from the frame of the relativistic object) would be roughly -1.5, 3.5, while point A itself is 4, 0.5. That means that the net movement in time is about -3 units. Still equipped with the jump drive, the ship would be able to jump to its past self and stop it from ever leaving in the first place! Out the window with casualty!
     
  20. Sep 25, 2013 #19

    ghwellsjr

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    I don't know how you arrived at that. Usually, when we talk about an object's frame of reference, we mean one in which it is at rest, meaning its worldline is aligned with the time axis or at least parallel with it so that there is no change in the x-axis as a function of time. You didn't draw the dark blue line aligned or parallel with the light blue ct' line so you need to explain what you mean.

    Where did you get these numbers from? I can't figure this out.

     
  21. Sep 25, 2013 #20
    Okay, I obviously am misunderstanding these diagrams somewhere with the reference frames. Ignore that.

    About the numbers, I assumed that the ship would be jumping from 0, 3 to 5, 3, then from 5, 4 to 0, 4. I then visually estimated the coordinates on the yellow grid, then applied them to the last diagram.

    I'm doing something wrong, right?
     
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