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Special relativity train problem

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm having a tough time with relativity, and I thought I'd try this problem to see if I understood the topic very well.

    Let's consider a train moving at 0.65c along a straight track. Its headlight produces a beam with an angular spread of 60 degrees. If you astanding alongside the track (rails are 1.5 m apart), how far from you is the train when you see its approaching headlight suddenly disappear?


    2. Relevant equations



    3. The attempt at a solution
    I drew a picture, but I really need a hint on how to get started. I'm thinking about using length contraction somehow, but I just can't get my head around what I need to do.

    Here's the best I've got right now: From trigonometry I have x= tan(60)*.75, x = 1.3, and I'm pretty sure this isn't right.
    Any help would be greatly appreciated.
     
    Last edited: Sep 12, 2009
  2. jcsd
  3. Sep 13, 2009 #2

    tiny-tim

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    Hi bcjochim07! :smile:

    .75 (or .76) is γ, but apart from that I don't understand what you're doing, or why. :confused:

    What is x? What equation or diagram are you using?
     
  4. Sep 13, 2009 #3
    Actually I'm going to take a second shot at this problem.


    The angular spread of the headlight in the frame of the train is 60 degrees, so the observer along tracks sees this angular spread:

    theta = arccos((cos(60) + 0.76)/(1+0.76cos(60)) = 24.07 degrees.

    First of all, I'm picturing a moving cone of light centered on the train tracks and an observer at the side
    From my drawing, I think that the headlight should disappear when you are at the point where one of the edges of the cones intersects the tracks. (On second thought, maybe this is not right?)

    Anyway, I've drawn a triangle with legs x (distance of observer from train in x direction) and 0.75 (half the width between the tracks) and the angle opposite to the 0.75 side is 12.035 degrees. So I get:

    0.75/x = tan(12.035) , x = 3.5 m. My book actually gives 3.0 m, so I'm not doing something right.
     
  5. Sep 14, 2009 #4

    tiny-tim

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    Well, that can't possibly be right, because if the train is stationary, then you replace 0.76 by 1, and the result is arccos(1).
    Sorry, I've still no idea what you're doing. :confused:

    The question is asking, if a ray of light is at 60º as measured on the train, then what is the angle as measured on the ground?

    So use equations

    put x = (√3/2)t, y = (1/2)t …

    then what are x' and y' ? :smile:
     
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