From the post "Thought Experiment Gone Wrong" dated 17 Nov 2017,
\(x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}\)
So,
\(x^{''}=\cfrac{x^{'}-v^{'}t}{\sqrt{1-{v^{'}}^{2}}}\)
It is tempting to substitute one into the other, and get
\(x^{"}=\cfrac{\cfrac{x-vt}{\sqrt{1-v^2}}-v^{'}t}{\sqrt{1-{v^{'}}^{2}}}\)
in this case, \(v^{'}\) is the speed measured in \(x^{'}\) frame not \(x\). Every unit in \(x^{'}\) as measured in \(x\) is scaled by the factor \(\cfrac{1}{\sqrt{1-v^2}}\). Instead we should have,
\(x^{"}=\cfrac{{x-vt}-v^{'}t}{\sqrt{({1-v^2})(1-{v^{'}}^{2})}}=\cfrac{x-(v+v^{'})t}{\sqrt{({1-v^2})(1-{v^{'}}^{2})}}\)
where \(v^{'}\) is also scaled by \(\cfrac{1}{\sqrt{1-v^2}}\),
\(v^{'}\rightarrow \cfrac{v^{'}}{\sqrt{1-v^2}}\).
Do we scale \(v^{'}\) again in the denominator? No. The scaling factor from \(x^{"}\) to \(x^{'}\) does not get scaled again. The translation term get scaled.
Just for the fun of it...
