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Special Relativity Angles

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A meter stick is at rest in the laboratory frame. It lies in the x-y plane, making an angle of 30° with the x-axis. What angle does this meter stick make with the x'-axis of a reference frame moving at V=0.8c in the x-direction.

    Laboratory frame is the rest frame.
    Variables: θ=30°, θ', V=0.8c

    2. Relevant equations
    Length Contraction
    x = x'*√1-V2/c2


    3. The attempt at a solution
    I tried treating the meter stick as a unit vector so that x=(1)cosθ and x'=L'cos(θ'). y=y' obviously.
    cos(θ') = cos(30)/√1-0.82
    = 0.866/0.6
    = 1.44
    θ'=cos-1(1.44)=domain error

    I feel like I'm making a stupid mistake and I just can't see it
     
  2. jcsd
  3. Feb 16, 2012 #2

    vela

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    In the moving frame, you expect x' < x because of length contraction, so it must be that x' = x/γ. You're using x' = γx.
     
  4. Feb 16, 2012 #3
    ahh so it would be tangent instead of cosine, right?

    In that case,
    tan(θ')=tan(30)/.6
    θ'= 43.9°

    Thanks!
     
  5. Feb 16, 2012 #4

    vela

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    No, I don't think that's right. How did you get tangent? That's a gamma, not a y. Also, remember the meter stick is no longer 1 meter long in the moving frame.
     
  6. Feb 16, 2012 #5
    Oh, I thought it was a y. Regardless, you can just say

    tanθ=y/x tanθ'=y'/x'

    (tanθ)/(tanθ')=yx'/xy'

    but since y=y'

    (tanθ)/(tanθ')=x'/x

    xtanθ=x'tanθ'

    so then use

    tanθ'=tanθ/√1-V2/c2

    plugging in the numbers gives me θ'=43.9, which is the answer in the book.

    please let me know if my reasoning is wrong, I want a better understanding of this.
     
  7. Feb 16, 2012 #6

    vela

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    That's right. Just wanted to make sure your reasoning for going from cos to tan was correct.
     
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