1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length contraction

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A rod of length L_0 moves with a speed v along the horizontal direction. The rod makes an angle of θ_0 with respect to the x'-axis.

    a. Show that the length of the rod as measured by a stationary observer is given by L = L_o [1 - (v^2 / c^2) cos^2 (θ_o) ]^(.5)


    2. Relevant equations

    L = L_p/ gamma

    3. The attempt at a solution

    I have a few questions:

    Is L proper L_o (the S frame?)
    When I am trying to find the length of the rod as measured by a stationary observer do I refer to the graph to the right?

    Horizontal length is all I need to worry about, correct?

    L = L_o / gamma

    x' = L cos θ_o

    x' = x ( 1 - v^2/c^2 )^(-1/2)

    L cos θ_o = x ( 1 - v^2/c^2 )^(-1/2)

    L = x / [ (1 - v^2/c^2 )^(1/2) cos θ_o ] where x = L_o

    L = L_o / [ (1 - v^2/c^2 )^(1/2) cos θ_o ]

    Am I on the right track? I can't seem to make it look like:
    L = L_o [1 - (v^2 / c^2) cos^2 (θ_o) ]^(.5)
     

    Attached Files:

  2. jcsd
  3. Jan 29, 2012 #2
    Wait, huh, is the picture you attached one given by the problem or something you made? The picture is really different from how I read the problem.

    The way you're approaching the problem seems fine. But if you're going to find the x coordinates, you need to find two x coordinates because length is x2-x1. Looks like you're doing a strange mixing of length contraction and x coordinates. Either approach is okay, but you'll probably confuse yourself going between them (confused me at least).
     
  4. Jan 29, 2012 #3
    Yes I think that is my problem. I don't know how to draw the diagram. Can you explain me what you understood? I have read it several times and I just don't get it
     
  5. Jan 29, 2012 #4
    So you have an S' that moving relative to the S frame; their x axis is collinear. In S' we have a proper length L_o, and the length makes an angle with respect to the x' axis. So the part that will be contracted is L_o*cosØ'. L_o*sinØ' will stay the same in both frames. S even sees a different angle than S' does.
     
  6. Jan 29, 2012 #5
    Thank you, I figured out the problem. I was making it way too complicated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Length contraction
  1. Length contraction (Replies: 1)

  2. Length contraction (Replies: 3)

  3. Length contraction. (Replies: 6)

  4. Length Contraction (Replies: 3)

  5. Length contraction? (Replies: 11)

Loading...