I have been trying to do this question for ages now:
A particle has a velocity v ={$v_x,v_y,v_z$} in S and a velocity $\bf v'$={$ v'_x, v'_y,v'_z$} in S'. Prove from velocity transformations that
$c^2-v^2 =\frac{c^2(c^2-v'^2)(c^2-V^2)}{(c^2+{Vv'_z})^2}$.
The obvious thing, in my opinion is that we have to use "inverse velocity transformations", basing on the denominator. The particle is moving along the z-axis, and I also know that the invariant line element
$(ds)^2/dt =(cdt/dt)^2-{(dx/dt)^2 +(dy/dt)^2+(dz/dt)^2}$ such that
$(ds/dt)^2=c^2-v^2 =\frac{1}{\gamma^2}$
My opinion is that i must show that
$(cdt/dt)^2-{(dx/dt)^2 +(dy/dt)^2+(dz/dt)^2}$=$\frac{c^2(c^2-v'^2)(c^2-V^2)}{(c^2+{Vv'_z})^2}$.
I have that
$v_z= \frac{v'_z+V}{1+\frac{Vv'_z}{c^2}}$
$v_y=\frac{v'_y}{\gamma(1+\frac{Vv'_z}{c^2})}$
$v_x=\frac{v'_x}{\gamma(1+\frac{Vv'_z}{c^2})}$