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Deriving Lorentz transformation by light cone coordinates

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Derive the Lorentz Transformation using light cone coordinates defined by

    ##x^±=t±x##

    ##x^+ x^-~## is left invariant if we multiply ##~e^φ~## to ##~x^+~## and ##~e^{-φ}~## to ##~x^-~##, that is ##~x'^+ x'^-=x^+ x^-##

    2. Relevant equations
    ##t'^2 - x'^2 = t^2 - x^2 ~~~(invariance~of~spacetime~separation)##
    ##~x'^+ = e^φ x^+~## and ##~x'^- = e^{-φ} x^-##
    Set ##~c = 1##
    3. The attempt at a solution
    Since ##t'^2 - x'^2 = t^2 - x^2~##, we can set ##~x'^±=t'±x'~##

    So that ##~t' = ½( x'^+ + x'^-) = ½( e^φ x^+ + e^{-φ} x^-) = t(coshφ) + x(sinhφ)~##
    Similarly ##~x' = t(sinhφ) + x(coshφ)##.

    Suppose ##~x' = 0~##, this implies ##~x = -ut~## then ##~u = \frac{sinhφ}{coshφ} = tanhφ##
    From the identity ##~cosh^2φ - sinh^2φ = 1~##, we have ##~coshφ = \frac{1}{(1-u^2)^½}~## and ##~sinhφ = \frac{u}{(1-u^2)^½}~##.
    Therefore,
    ##t' = \frac{t + ux}{(1-u^2)^½}##
    ##x' = \frac{x + ut}{(1-u^2)^½}##

    Now to derive the inverse Lorentz transformation,
    ##~x^+ = e^{-φ} x'^+~## and ##~x^- = e^φ x'^-## (transferred ##e^φ## to the other side)

    By using ##~x^±=t±x~##, we have ##~t = ½( x^+ + x^-) = ½( e^{-φ} x'^+ + e^φ x'^-) = ½( t'(e^{φ} + e^{-φ}) - x'(e^{φ} - e^{-φ})) = t'(coshφ) - x'(sinhφ)~##
    Similarly ##~x = -t'(sinhφ) + x'(coshφ)~##
    This implies that ##φ → -φ##, so
    ##~cosh(-φ) = \frac{1}{(1-u^2)^½}~## and ##~sinh(-φ) = \frac{u}{(1-u^2)^½}~## implies ##~coshφ = \frac{1}{(1-u^2)^½}~## and ##~sinh(φ) = \frac{-u}{(1-u^2)^½}~##.
    Therefore,
    ##t = \frac{t' - ux'}{(1-u^2)^½}##
    ##x = \frac{x' - ut'}{(1-u^2)^½}##

    Is my argument ##φ → -φ## correct? Such that I plugged it in the hyperbolic functions then produced the crucial negative ##~u~## in the numerator of ##sinhφ##?
     
  2. jcsd
  3. Mar 13, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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