Deriving Lorentz transformation by light cone coordinates

[Whitehole]

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 $sinhφ$?

 

[mark726]

Yes, your argument is correct. When you plug in $φ → -φ$, you are essentially taking the inverse transformation, so your result is correct. This is a common technique used when deriving the Lorentz transformation in terms of hyperbolic functions.