Deriving Lorentz transformation by light cone coordinates

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Whitehole
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Homework Statement


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^-##

Homework 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##

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##?
 
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