# Lorentz Transformation for S”

1. The problem statement, all variables and given/known data

Frame S’ travels at speed V1 along the x axis of frame S. Frame S” travels at speed V2 along the x’ axis of frame S’. Apply the Lorentz transformation twice to find the coordinates x”, y”, etc of any event in terms of x, y, z, t. Show that this is the same as the standard Lorentz transformation with velocity V given by the relativistic sum of V1 and V2

2. Relevant equations

$\x’=\gamma(x-Vt)$
$\t’=\gamma(t’-\frac{Vx}{c^{2}})$
$\v_{x}’=\frac{v_{x}-V}{1-\frac{v_{x}V}{c^{2}}}$

3. The attempt at a solution

No problem with the first part. I’ll just put my answer for the x” coordinate for reference:

$\x”=\gamma_{1}\gamma_{2}(x(1+\frac{V_{1}V_{2}}{c^{2}})-(V_{1}+V_{2})t)$

The different gamma factors come from transforming between S and S’ and then again between S’ and S”

For showing that this is equivalent to applying the Lorentz transformation using the relativistic sum:

$\x”=\gamma(x-Vt)=\gamma(x-\frac{V_{1}+V_{2}}{1+\frac{V_{1}V_{2}}{c^{2}}}t)$

This is almost in the correct form. If I kill the fraction I can get

$\x”(1+\frac{V_{1}V_{2}}{c^{2}})=\gamma(x(1+\frac{V_{1}V_{2}}{c^{2}})-(V_{1}+V_{2})t)$

The only differences between this and the expression I am looking for is the extra factor by x” and the fact that the factor gamma should be the product of two gamma factors. I figured I could write the gammas explicitly in terms of V1 and V2 in my first equation for x and then do the same in the second equation by substituting the relativistic sum for V and then rearrange things until I got the extra factor next to x in the second equation to cancel. I worked through the algebra for a little while, but it didn’t seem like it was going to happen. I just want to make sure I’m not overlooking something before I continue with the algebraic gymnastics. If Im on the right track please confirm and Ill continue to work at it.

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