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Traveling Backwards in Time

  1. Mar 3, 2014 #1
    The equation for time dilation goes something like

    t' = t/ (√(1-v2/c2)

    I have heard that if a particle can travel at a speed such that v>c, then the particle will go backwards in time.

    But how is this true? According to this equation, if v>c, this implies that t' is an imaginary number...not a real valued negative number.
  2. jcsd
  3. Mar 3, 2014 #2


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    Since particles CAN'T travel faster than c, your question has no meaning other than playing games with math.
  4. Mar 3, 2014 #3


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    It's not true, and you're right.
  5. Mar 3, 2014 #4


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    Just to add a little, if a particle goes faster than light in one frame, then in other frames of reference, it will arrive before it left. Furthermore, under plausible assumptions, if particles can go faster than light, you can send a message and get an answer before you sent it. See:


    So, it is true that the particle won't go back in time in a frame in which it is moving faster than light; and it is true the mass is imaginary not negative. However, faster than light particles do readily lead to time travel 'paradoxes'.
  6. Mar 4, 2014 #5


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    I know very little quantum field theory. But as far as I know, when there is a field with an imaginary mass term, it means that we have to do a perturbation about some value of the field such that the resulting particle has real positive mass. For example Goldstone bosons (and Higgs bosons, I think). They have imaginary mass terms, but the associated particle(s) have real mass, because we must do the perturbation at some non-zero value of the field. It is not possible to do a perturbation around values of the field which would give 'imaginary mass' particles. Or, I think we would get a condensate at these values, which is because we cannot do a perturbation to just get single particles.

    So anyway, I think that in quantum field theory, this idea of imaginary mass particles is kinda nonsensical. But I don't know for sure, I don't know much about quantum field theory, so don't take me too seriously.
  7. Mar 5, 2014 #6


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    The claim about time travel is not (directly, anyway) about time dilation. It's about the Lorentz transformations. To get an extreme case, let's suppose that you have a teleportation device, which from any frame allows you to teleport instantaneously (according to that frame) to any other location in the universe.

    Consider two frames, [itex]F[/itex] and [itex]F'[/itex], where objects at rest in [itex]F'[/itex] are moving at speed [itex]v[/itex] in the x-direction, as measured in frame [itex]F[/itex]. Then what we do is this: We have a teleporter at rest in frame [itex]F[/itex]. You teleport from the point
    [itex]x=0, t=0[/itex]

    to the point
    [itex]x=L,\ \ t=0[/itex]

    Now, switch to frame [itex]F'[/itex]. In its coordinates, the second event is:
    [itex]x'=\gamma L,\ \ t' = -\gamma \dfrac{v L}{c^2}[/itex]

    Now, hop in a teleporter that is at rest in frame [itex]F'[/itex]. Teleport back to the point [itex]x=0[/itex] where you came from. The coordinates of this point in frame F' is:

    [itex]x'=\gamma \dfrac{v^2}{c^2} L,\ \ t' = -\gamma \dfrac{vL}{c^2}[/itex]

    The coordinates in frame F are:

    [itex]x=0,\ \ t= - \dfrac{vL}{c^2}[/itex]

    So the round-trip brought you back to where you started (in space), but to an earlier time.

    You might think this is an artifact of assuming instantaneous travel. But another exercise with Lorentz transformations shows that if travel is faster-than-light in one frame, then there is a second frame in which it is instantaneous. So a faster-than-light rocket, together with the ability to change frames (which only takes a slower-than-light rocket) will give you instantaneous travel, which can be used for back-in-time travel.

    None of this analysis uses the concept of applying the Lorentz transformations to "frames" traveling faster than light.
  8. Mar 5, 2014 #7


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    I see that PAllen already said this more succinctly.
  9. Mar 5, 2014 #8


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    This equation is valid for v<c only.
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