Instead of one unit,
\(x^{'}cos(\theta)=x^{'}.\cfrac{\sqrt{c^2-v^2}}{c}=x^{'}.\sqrt{1-\cfrac{v^2}{c^2}}=x\)
So,
\(x^{'}=\cfrac{x}{\sqrt{1-\cfrac{v^2}{c^2}}}\) --- (*)
as the origin is still displaced by \(vt\),
\(x^{'}=x-vt\)
but the dilation needed is now given by (*), so,
\(x^{'}=\cfrac{x-vt}{\sqrt{1-\cfrac{v^2}{c^2}}}\)
and the rest of Lorentz transformations follows.
What does it mean when we use light speed this way? \(c\), instead of one unit? It means, \(v\) is in units of 299792458 ms-1 and \(x\) in units of 299792458 m. But it is OK because,
\(c.x^{'}=\cfrac{c.x-c.vt}{\sqrt{1-\cfrac{(c.v)^2}{c^2}}}\)
where \(x^{' }\), \(x\) and \(v\) are in units or unit meter and unit meter per unit time, respectively.
\(c.x^{'}=c.\cfrac{x-vt}{\sqrt{1-v^2}}\)
\(x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}\)
which is the same as before. But! when \(\gamma=\cfrac{1}{\sqrt{1-\cfrac{v^2}{c^2}}}\),
\(\cfrac{x}{|c|}\rightarrow x\)
\(\cfrac{x^{'}}{|c|}\rightarrow x^{'}\)
and
\(\cfrac{v}{c}\rightarrow v\)
we are using light speed as reference, everything is measured in 299792458 m or 299792458 ms-1.
Have a nice day.
