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Proper time in Lorentz Space

  1. Jun 26, 2012 #1
    We know, when m=0, the schwarzschild space time becomes lorentz space time. Then, the proper time taken by one twin (A) to travel around the massive body in lorentz space while the other twin at rest can not be defined or it will be infinite. Is that true???
     
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  3. Jun 26, 2012 #2

    PeterDonis

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    If m = 0, there is no massive body. I'm not sure what scenario you are imagining.
     
  4. Jun 26, 2012 #3

    Nugatory

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    Do you mean "when r is very large, far enough away from the mass that spacetime is effectively flat"? Otherwise I can't make sense of the question.

    If that's what you mean, the proper time for a twin travelling in a giant circle is perfectly well defined and it's not zero - it's whatever the traveler's wristwatch records on the trip. It can calculated by computing the integral of ds along the path (and remember that the path is a closed curve in three-dimensional space but not in four-dimensional spacetime).
     
  5. Jun 26, 2012 #4
    Then, in de sitter space, also m=0. But, we can have some possible values for the proper time.
     
  6. Jun 26, 2012 #5

    PeterDonis

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    I think you need to be more explicit about exactly what types of trajectories you are talking about in which spacetimes.
     
  7. Jun 26, 2012 #6
    Yes, thanks, I can understand.
     
  8. Jun 26, 2012 #7
    No, it is not true, whhere did you get that? In the absence of any gravitational body (your [itex]m=0[/itex]), a circular path traversed by the "traveling twin" takes the amount of time [itex]\tau=\frac{2 \pi R}{v}\sqrt{1-(v/c)^2}[/itex]. The "stay at home" twin , measures on his clock the time [itex]t=\frac{2 \pi R}{v}[/itex]. Here [itex]R[/itex] is the radius of the circular path and [itex]v[/itex] is the speed of the "traveling" twin wrt the "stay at home" twin. The calculations are very simple.
     
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