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Question on lorentz transformation equations

  1. Aug 16, 2010 #1
    i am reading Lillian R. Lieber's book on the einstein theory of relativity and i am a bit confused on page 65. she wants to take the equations:
    x=x'cosθ - y'sinθ
    y=x'sinθ + y'cosθ

    and compare them to:
    x'=β(x-vt)
    t'=β(t-vx/c2)

    she takes c as one so:
    x'=β(x-vt)
    t'=β(t-vx)

    she solves for x and t and gets:
    x=β(x'+vt')
    t=β(t'+vx')

    then she replaces t with iτ and t' with iτ' and she gets:
    x=β(x'+vt')
    iτ = iβτ' + βvx'

    x=β(x'+vt')
    τ=βτ' + iβvx'

    next is the part i am confused about. she sets β = cosθ and -iβv = sinθ. this nicely turns the Lorentz equations into:
    x=x'cosθ - τ'sinθ
    τ=x'sinθ + τ'cosθ

    what i don't understand is how did she choose -iβv = sinθ? it works out all nicely in the end but how did she know that sinθ had to equal -iβv? was it arbitrary as a result of trial and error or did she use β = cosθ in order to figure out that -iβv = sinθ?
     
  2. jcsd
  3. Aug 17, 2010 #2
    This can't be, are you sure that the author is not setting:

    [tex]\beta=cosh(\theta)[/tex] and [tex]\beta v= sinh (\theta)[/tex]. Note the use of hyperbolic trigonometry as in https://www.physicsforums.com/blog.php?b=1911 [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Aug 17, 2010 #3

    jcsd

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    Science Advisor
    Gold Member

    But you can use the substitutions sinh x = -isin ix and cosh x = cos ix.

    Anyway the answer to demonelite, thoguh I haven't been looked at myself, is that your question should be answered by considering what θ is.
     
    Last edited by a moderator: May 4, 2017
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