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Proper Time For Photon on Null Geodesic

  1. Oct 26, 2008 #1
    I can understand the logic from some arguments as to why proper time in a photon's "frame of reference" is zero. I cannot understand how this follows from the argument that (SPACE)2 - (TIME) 2 = 0. This to me says that the SPACE-TIME interval for the photon is zero (null interval) and SPACE = TIME, but how does it follow that proper time in the null geodesic is zero? Thanks.
     
  2. jcsd
  3. Oct 26, 2008 #2

    atyy

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    Science Advisor

    If the worldline is a straight line, the proper time between two points on that worldline is the interval. We can get the proper time along an arbitrary wordline by dividing the worldline into many small straight line segments, and adding up the intervals for the segments. Since the interval is the same in all reference frames, the proper time along a worldline is the same in all reference frames. A photon travels along a path of zero proper time in all reference frames. Assuming we set up perpendicular coordinate axes, the proper time along a worldline is defined as the integral of (SPACE)2 - (TIME) 2, just like arc length along a path is Euclidean space is defined as the integral along the path of (SPACE)2.
     
  4. Oct 26, 2008 #3

    Mentz114

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    Gold Member

    Proper time and proper interval are one and the same thing, only a factor of c between them. The Minkowski metric may be written

    [tex]c^2d\tau^2=ds^2=c^2dt^2-dx^2-dy^2-dz^2[/tex]
     
  5. Oct 26, 2008 #4

    Dale

    Staff: Mentor

    I don't know about this, I think that the spacetime interval is only equal to the proper time for timelike intervals. I am not sure that it makes sense to talk about the proper time along a null worldline any more than it makes sense to talk about the proper time along a spacelike worldline.
     
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