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FTL and time travel

  1. Jan 19, 2012 #1
    Hi 'body,
    ok, this is a well known issue, but i still can't find a valid explanation of why FTL implies time travel.
    I've read lot of explanations, and each one ends up with a nonsense telling something like "if you drive a FTL car when you'll arrive you'll see yourself opening the car door and entering it, so you've travelled in the past". Of course I'll see myself getting in the car, because the light is slower than me and it will reach me after I've stopped. Another example is the "tachyon pistols" http://sheol.org/throopw/tachyon-pistols.html which is clearly an optical illusion due to the light speed "slowness".
    Is there any better explanation of this FTL and time travel connection, or I have just to believe it is true because someone tell it is true?
  2. jcsd
  3. Jan 19, 2012 #2
    I've found a some better explanation, here, but I think that it turns out that what he call "trick" is just an error: Lorentz Transformation can't be applied since the FTL speed is NOT a relative velocity, and this means that the Lorentz transformation gamma can't be calculated that way.
  4. Jan 19, 2012 #3
    For the example they give the relative velocity between Bob and Alice is only 3/5 c, so you can certainly calculate γ.
  5. Jan 19, 2012 #4


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    frakie, obviously, with or without FLT, time travel and the ensuing paradoxes are not possible. However, I do not think that the explanations you mentioned are flawed in the sense you point out. Maybe this other link, which puts things very simply, helps you better understand.

    I will make a try at an explanation myself. In the end it is a simple triangle, like this:


    Imagine a duel on a Train (red frame), the same duel that Brian Green describes in The Fabric of Cosmos. Duelers (Back and Front) shoot when they receive light signals from the mid-point of the wagon. Now, when Back is warned by the corresponding signal, a guy by him on the Ground (blue frame), shoots at Front (event P). He shoots a superluminal bullet (let us imagine, for simplicity, an instantaneous one) and wounds Front... when? If you follow with your finger the simultaneity blue line of the Ground frame (horizontal), you see that Front would be wounded at event Q. In the Ground frame, that means that Front is hurt before receiving its own signal. That is already unfair enough. Now Front Fires back the same bullet, which is supposed to travel also instantaneously. When does it hit the aggressor? If you follow with your finger the red simultaneity line of the Train frame (descending), you will find out that the aggressor is punished before he committed his own felony. So the bullet in question has travelled back in time and this has created an inconsistency, a paradox.

    All this looks like an orthodox application of SR math and geometry, except that it is combined with FLT, which in principle SR rules out, but what if experiment re-introduces it?

    However, as I said, it is quite obvious that, even if FLT were possible, those paradoxes would not happen. If some people fear they could, it is because they misinterpret the concepts. If I had to put it briefly, I would say that the error is mixing up the old absolute simultaneity concept with the relative simultaneity concept introduced by SR.
  6. Jan 20, 2012 #5
    Yes you're right, I was wrong because I thought in
    gamma = 1 / root(1-(v/c)^2)

    v was the FTL velocity.
  7. Jan 20, 2012 #6
    I see your point and I also see there are many contradictions in your example:
    the very first contradiction is the use of relativistic simultaneity concept; this concept lies on an axiom which states that every event spreads through the spacetime at velocity c. This is in contradiction with our axiom "exist one object which can travel at FTL speed".
    In your example this contradicion leads you to tell that simultaneity (simultaneity based on FTL speed) from P to Q is not the same from Q to R (because you here use simultaneity based on c speed). In fact your image shows that the P-Q arrow goes back in time from time tP to tQ(=t0) in the red X line, which is obviously wrong.
    Furthermore your example is complicated because the train is moving from the ground point of view and this fact complicates the maths.

    It seems to me that the relativistic simultaneity concept can't cope with the reality we can see: if you say that the relativistic simultaneity concept is true, then you must say that the stars we see in the sky are NOW exactly how we see them, because what we see is simultaneus to our "now-here". And we already know it is untrue.
    Last edited: Jan 20, 2012
  8. Jan 20, 2012 #7


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    I am not sure I follow all your comments. But for example the one I quoted seems wrong: Special relativity (the relativistic concept of simultaneity) does not state that things, say a star, are now how we SEE them. The relativistic observer is an intelligent one who accounts for the time that light takes to travel from its origin. So if you see a star exploding, you do not say it is exploding now, you say it exploded at time now - minus light's trip time.
    Last edited: Jan 21, 2012
  9. Jan 21, 2012 #8
    According to special relativity an observer must say two events A and B are simultaneous when he receives - instantly - the two informations "A is happening" and "B is happening". Those informations travel at c speed. So if i receive a supernova light right now i have to tell that the event "the supernova is exploding" is simultaneous with "I'm here now".
  10. Jan 21, 2012 #9


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    No, you should check your sources, SR does not say so at all. Regard a frame of reference as a network of clocks at rest with each other. You are in frame S and there is a clock belonging to your frame when and where the supernova explodes at time t (event 1). When that happens, that time t1 is transmitted to you through a radio signal, which travels in parallel with the light of the supernova. At time t2 = t1 + distance * c (event 2), you will receive the two notices: visual notice of the explosion and the radio notice of the time t1 when the star exploded, which is earlier than your now t2.
  11. Jan 21, 2012 #10


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    No, if you receive light, now, from a supernova that is 100 light-years away, you will say the supernova occurred 100 years ago.

    On the other hand, if I am flying past Earth in a very high speed spaceship at the same time, I may say the supernova is 80 light-years away (length contraction) and it occurred 80 years ago.
  12. Jan 21, 2012 #11
    If that clock belongs to my frame,then how is it supposed to be synchronized with my clock and be brought next to the supernova? I think that travel would break the synchronization. Nevertheless you're right,I need to check some better sources. Have you got any suggestion?
  13. Jan 21, 2012 #12
    ros.gif This image explains what relativity of simultaneity is. And since "simultaneous" means "happening at exactly the same time" it is clear and obvious to me that the consequence is what i wrote in my previous comment.
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