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Help with time-space diagram equations.

  1. May 17, 2013 #1
    Consider a situation where we want to combine spacetime diagrams of Alice and Bob, where Bob is moving at a speed of 0.4c to the right (positive x direction). If we draw Alice’s xA axis as horizontal and tA axis as vertical, answer the following questions.

    What is the equation, written in terms of xA and tA, that describes Bob’s tB axis (the world line where xB = 0) on Alice’s diagram?

    My options are the following:

    1. tA = γ xA/(0.4c), where γ is the Lorentz factor
    2. tA = xA/(0.4c)
    3. tA = (0.4c)(xA)
    4. tA = γ (0.4c)(xA), where γ is the Lorentz factor

    Can someone explain, step by step, how to derive the correct equation?


    p.s. this is not homework, it`s just something for fun I found online that I want to understand...
     
  2. jcsd
  3. May 17, 2013 #2

    ghwellsjr

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    Step 1: Since speed is distance divided by time, 0.4c = xA/tA.

    Step 2: Rearrange by multiplying both sides of the equation by tA and dividing by 0.4c to get answer 2.
     
  4. May 17, 2013 #3

    pervect

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    let ##t_a, x_a## be Alice's coordinates
    let ##t_b, x_b## be Bob's coordinates
    let ## \beta = v/c \quad \gamma = 1 / \sqrt{1-\beta^2}##

    then write down the Lorentz tranform:

    ##x_b = 0 = \gamma \left( x_a - \beta t_a \right) ##

    From which you conclude the equation is ##x_a = \beta t_a ##
    which is option 2
     
  5. May 18, 2013 #4

    Fredrik

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    The ##t_B## axis is the image of Bob's world line (the set of all events in spacetime that he will pass through) in Alice's coordinates. So for all points ##(t_A,x_A)## on that line, ##x_A## and ##t_A## are (according to Alice) respectively the distance he has traveled since the (0,0) event, and the time that has passed since the (0,0) event. So his velocity is ##v=x_A/t_A##.

    This means that for any point ##(t_A,x_A)## on that line, we have ##x_A=0.4c t_A##.

    Hm, maybe I should have looked more closely at the previous answers before I wrote this. I thought that no one had pointed out that velocity=distance/time, but it's right there in ghwellsjr's post.
     
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