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Sure this is easy for you but I dont understand

  1. Jun 4, 2005 #1
    I have one question, could you help me please...

    Suppose one event is emission of light from some point A, and another event is its arrival to some point B. There is one observer who is watching these events
    So and if we write the interval (rewrite it in the way to separate the time term-dx00) we get the quadratic equation with respect to dx00, solving this equation we have to roots (dx00)1, (dx00)2, in SR they differ with the sign only, (it means forward-backward spread of light(from A to B and from B to A)But generally in GR these are two different solutions and in that case does it mean that external observer will say that light takes (dx00)1 time to go from A to B and it will take (dx00)2 time to go backward (from B to A)?
     
  2. jcsd
  3. Jun 4, 2005 #2
    Neitrino,

    I don't understand your notation, but if the observer is stationary with respect to both location A and location B, then the the time he measures for light to go from A to B or from B to A will be equal.
     
  4. Jun 4, 2005 #3

    pervect

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    I think your interpretation of the math may be off. One way of interpreting the math is to say that ring laser gyroscopes can and do work in GR. Another way of interpreting it is that if you have a time varying system, geodesics into the future are not the same into the past.

    The metric coefficients will have dx*dt terms only when the system is rotating, or when the system is otherwise evolving in time so that it is not symmetrical with time (as in the metric of a moving mass).

    What you are doing is taking a fixed (or perhaps zero) interval of proper time, and computing the coordinates. What you are then noting is that the coordiantes (dx,dt) for an object following some sort of geodesic are different going into the past and into the future. But this isn't surprising when the system itself is varying with time.
     
    Last edited: Jun 4, 2005
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