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Spacelike interval

  1. Aug 25, 2007 #1
    I am given the four-vector of two events in a reference frame S and I found that they are spacelike. I want to then find the reference frame in which these two events are simultaneous. I am still a bit confused about how this is even possible.

    There is that equation:

    gamma * delta t' = delta t

    where t is the time interval in the original reference frame and t' is the coordinate system in the new reference frame. If the delta t' is 0, how is this equations even solvable?
     
  2. jcsd
  3. Aug 25, 2007 #2

    learningphysics

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    This is only true when in the frame where t' is measured, both events are at the same position. ie:delta x' = 0

    The times in the new reference frame will depend on the x-values. what is t1 in the new frame (depends on time and position in the old frame), what is t2 in the new frame (depends on time and position in the old frame)?
     
  4. Aug 25, 2007 #3
    Use the Lorentz transformation for time. If the interval between two events is spacelike, then there is one frame (i.e. frame with a particular velocity) where the two events are simultaneous. Similarly, if the interval is timelike, there is one frame where the events happen at the same point in space.
     
    Last edited: Aug 25, 2007
  5. Aug 26, 2007 #4
    Or when the direction of motion between the two reference is perpendicular to the direction of motion between the two events.
     
  6. Aug 26, 2007 #5
    So if I assume that the boost is in the x-direction, I have to solve


    [tex] x_0 = \gamma x_1 -5 *\gamma * \beta [/tex] which poses a very difficult challange unless you take v = 0 which does not make sense physically... [/tex]
     
  7. Aug 26, 2007 #6

    learningphysics

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    Can you describe how you get this equation?

    The way I understand the problem... there are two events (ct0,x0,0,0) and (ct1,x1,0,0) in frame S... and in frame S':

    [tex]ct_{0'} = \gamma{ct_{0}}-\beta\gamma{x_{0}}[/tex] and
    [tex]ct_{1'} = \gamma{ct_{1}}-\beta\gamma{x_{1}}[/tex]

    And you set [tex]ct_{0'} = ct_{1'}[/tex] and solve for v...

    Note, how I used x0 and x1... may not be the way you've used it... sorry for any confusion.
     
  8. Aug 26, 2007 #7

    robphy

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    If you think using [Minkowski] geometry, the answer is rather simple.

    You seek the [future-pointing] unit timelike vector that is orthogonal to the spacelike vector connecting your two events. In a Euclidean plane, how do you find the vector perpendicular to another vector? (There is more than one method.) Once you find that vector, you probably want the slope of that vector. What is the Minkowskian analogue?
     
  9. Aug 26, 2007 #8
    Nope. I meant this: [tex]c\Delta t^\prime = \gamma\left(c\Delta t - \beta\Delta x\right)[/tex]

    You know what [itex]c\Delta t'[/itex] is.
     
  10. Aug 27, 2007 #9
    I see. So if the two events are (ct_0, x_0, x_1, x_2) = (1,2,0,0) and (3,5,0,0) in S then they are simulaneous in an S' with beta = 2/3, right?
     
  11. Aug 27, 2007 #10
    Yes, that's right.
     
  12. Aug 27, 2007 #11

    robphy

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    Note the slope of the spacelike segment joining (1,2,0,0) and (3,5,0,0) is (5-2)/(3-1)=3/2.
    Note that this slope (3/2 > 1) multiplied by the slope [that you found] of the timelike segment perpendicular to it (2/3) is 1.
    That is, in Minkowski space, a timelike line is perpendicular to a spacelike line if the product of their slopes is 1 [i.e., the slopes are reciprocals of each other].
    In Euclidean space, the corresponding product is -1 [i.e. m2 = -1/m1 ].
    You can see this is true using the appropriate dot-products.
     
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