 Solution

For a standard collinear Lorentz transformation, put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, with $|v|<c$. Then $x^{\prime }=\gamma (x-vt)$ and $c{t}^{\prime }=\gamma (ct-\beta x)$. Define the rapidity $\varphi =\mathrm{artanh}\beta$. The identities ${\cosh }^2\varphi -{\sinh }^2\varphi =1$ and $\tanh \varphi =\beta$, with $\cosh \varphi >0$, give $\cosh \varphi =\gamma$ and $\sinh \varphi =\gamma \beta$. Substitution proves the desired hyperbolic form.

Adding and subtracting those transformed coordinates uses $\cosh \varphi \mp \sinh \varphi =e^{\mp \varphi }$ and yieldsA second collinear Lorentz transformation of rapidity ${\varphi }^{\prime }$ multiplies these two null coordinates by $e^{-{\varphi }^{\prime }}$ and $e^{{\varphi }^{\prime }}$. Their total factors are therefore $e^{-(\varphi +{\varphi }^{\prime })}$ and $e^{\varphi +{\varphi }^{\prime }}$, provingThe hyperbolic tangent addition formula now gives the velocity composition lawAll velocities here are signed along the same axis; the speed is the absolute value if either frame moves in the opposite direction.