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Special Relativity and direction of travel

  1. Jun 17, 2011 #1
    As I have read, a clock in a spaceship approaching the speed of light will appear to slow down in time from the point of view of an outside stationary observer. Does the direction of travel of the space ship relative to the observer have an effect? If it is travelling towards the observer it seems to me like it should speed up in time, and slow down when moving away? Furthermore, what would the time dilation be to the observer if the space ship was moving in a circle around the observer? Thanks
     
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  3. Jun 17, 2011 #2

    bcrowell

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    The standard way of talking about this is that time dilation is measured by comparing clocks while the ship is next to the observer, and then comparing later when it's next to the observer again. By this definition, the time dilation is given by the factor [itex]\gamma=1/\sqrt{1-v^2/c^2}[/itex], which is independent of the direction of the velocity v.

    If you do it instead by measuring the frequency of a radio wave sent from the spaceship to the observer, then you get the relativistic Doppler shift http://en.wikipedia.org/wiki/Relativistic_Doppler_effect , which does depend on the direction of motion.

    -Ben
     
  4. Jun 17, 2011 #3

    tiny-tim

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    welcome to pf!

    hi hrob64! welcome to pf! :smile:
    ah, you're confusing how it appears to how we measure it

    yes, if the speed of the clock is v = c times tanh(x),

    then looking through a telescope, the clock's rate will be multiplied by ex (ie faster) if the clock is approaching the observer, and by e-x (ie slower) if the clock is going away from the observer,

    but the observer knows that he is always looking at the past, and when he accounts for that, he finds the clock (in both cases) is going slower by 1/cosh(x), = √(1 - v2/c2)
    the same, √(1 - v2/c2)
     
  5. Jun 17, 2011 #4
    That would involve General Relativity. Special relativity only treats straight motion at constant velocity.
     
  6. Jun 17, 2011 #5

    WannabeNewton

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    As long as we are dealing with a manifold with the property of global flatness then circular motion doesn't require GR. The same lorentz factor would apply and I think you would just have to evaluate it at each point of the circular path.
     
  7. Jun 17, 2011 #6
    I see. And that would be homologous with looking out at a circle of infinite diameter?
     
  8. Jun 18, 2011 #7

    tiny-tim

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    no, special relativity is about the Lorentz transformation , and that converts any description of motion in (x,y,z,t) coordinates to a description in different (possibly instantaneous) (x'.y',z',t') coordinates

    as WannabeNewton :smile: says …
    uhh? :redface: do you mean a straight line? :confused:
     
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