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Relativity of simultaneity through lorentz transformation

  1. Apr 3, 2013 #1
    Hello friend,

    Can you give me an example that shows simultaneous events in one reference frame not simultaneous in other reference frame with the help of lorentz Transformation?
     
  2. jcsd
  3. Apr 3, 2013 #2

    ghwellsjr

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    Sure. I'll use the simplified LT where c=1, and where y and z are zero. First we calculate gamma as a function of the relative speed beta (the ratio of the speed to the speed of light) between the two frames:

    γ = 1/√(1-β2)

    Next we have the two equations for the distance, x', and time, t', coordinates in the new frame as a function of the distance, x, and time, t, in the original frame along with gamma and beta:

    x' = γ(x-βt)
    t' = γ(t-βx)

    So let's say our two events are at t1=100, x1=100 and t2=100, x2=200. Note that these two events in our original frame are simultaneous because their time coordinates are the same (t1 = t2 = 100).

    And let's say beta = 0.6.

    Gamma is calculated as:

    γ = 1/√(1-β2) = 1/√(1-0.62) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

    Now we want to calculate the coordinates in the new frame for the first event:

    x1' = γ(x1-βt1) = 1.25(100-0.6*100) = 1.25(100-60) = 1.25(40) = 50
    t1' = γ(t1-βx1) = 1.25(100-0.6*100) = 1.25(100-60) = 1.25(40) = 50

    Finally the coordinates in the new frame for the second event:

    x2' = γ(x2-βt2) = 1.25(200-0.6*100) = 1.25(200-60) = 1.25(140) = 175
    t2' = γ(t2-βx2) = 1.25(100-0.6*200) = 1.25(100-120) = 1.25(-20) = -25

    Actually, I didn't really need to calculate the new distance coordinates because all you wanted was to show that the new time coordinates would not be the same (t1' = 50 ≠ t2' = -25) but it doesn't hurt to see both the new time and distance coordinates.
     
  4. Apr 3, 2013 #3
    thank you!
     
  5. Apr 3, 2013 #4

    ghwellsjr

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    You're welcome.

    And now that you are an expert on doing Lorentz Transforms, take a look at the diagrams I drew that I pointed you to on another thread and see if you can confirm that every event in the original frame transforms to the correct coordinates in the other diagrams. My second diagram is moving at -0.6 with respect to the first one and the third one is moving at 0.6c with respect to the first one.
     
  6. Apr 3, 2013 #5

    Dale

    Staff: Mentor

    Here is a little widget I wrote. You specify the relative velocity and the x and t coordinates then it gives you the x' and t' coordinates along with the coordinate lines. Hopefully it helps you get a visual feel for how the transforms work. The relativity of simultaneity is demonstrated by the fact that the lines of constant t are not parallel to the lines of constant t'.

    You may need to download the free player from Mathematica: http://www.wolfram.com/cdf-player/

    ghwellsjr, btw, I fixed the number formatting issue you pointed out.
     

    Attached Files:

  7. Apr 3, 2013 #6

    ghwellsjr

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    Got it. Problem fixed. Thanks.
     
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