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Lorentz Length Contraction Derivation

  1. Dec 30, 2011 #1
    In this video:

    Special Relativity: Length contraction in more detail
    http://www.youtube.com/watch?v=s3ZqBt8KWVQ&feature=mfu_in_order&list=UL

    for the derivation of Lorentz contraction equation at 2.48 the guy says, "I won't bore you with the details".

    I'd appreciate is someone would please "bore" me with the details, as I'm struggling to do the algebra which he has not shown here in order to go to the next line.

    Thank you
     
  2. jcsd
  3. Dec 30, 2011 #2

    BruceW

    User Avatar
    Homework Helper

    If you look at the uploader's comments, he made a mistake in the video at that bit. The equation should be:
    [tex]\frac{L'}{v+c} + \frac{L'}{-v+c} = \frac{2L}{c} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} [/tex]

    From here, you need to rearrange.
     
  4. Dec 30, 2011 #3
    Well spotted BruceW, I hadn't noticed that.

    I'm trying to arrange the formula you have above so it's equal to L'

    I can't seem to manage it.

    Concerning the LHS, I was thinking of multiplying (-v + c) with L', and multiplying (v + c) with L', and multiplying (v + c) with (c - v) to obtain (c^2 - v^2) as the denominator.
     
  5. Dec 30, 2011 #4
    Yes, so on the LHS you want a common denominator so you can group the L' terms, then you can rearrange to get L' alone.
     
  6. Dec 30, 2011 #5
    I know denominator I have (c^2 - v^2) can be rearranged to become (c-v)(c+v) and I have both of these individually multiplied with an L' on the numerator.

    Am I going about this the right way? I'm not sure what do do next.
     
  7. Dec 30, 2011 #6
    What have you got so far? Have you grouped the L' terms?
     
  8. Jan 4, 2012 #7
    Is L' equal to

    [L(c-v)(c+v)] / [c^2 * (1-v^2/c^2)^0.5]
     
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