= Solution
For a standard collinear <Lorentz transformation>, put $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$, with $|v|<c$. Then $x'=\gamma(x-vt)$ and $ct'=\gamma(ct-\beta x)$. Define the <rapidity> $\phi=\operatorname{artanh}\beta$. The identities $\cosh^2\phi-\sinh^2\phi=1$ and $\tanh\phi=\beta$, with $\cosh\phi>0$, give $\cosh\phi=\gamma$ and $\sinh\phi=\gamma\beta$. Substitution proves the desired hyperbolic form.
Adding and subtracting those transformed coordinates uses $\cosh\phi\mp\sinh\phi=e^{\mp\phi}$ and yields
$$
\boxed{x'+ct'=e^{-\phi}(x+ct),\qquad x'-ct'=e^{\phi}(x-ct).}
$$
A second collinear <Lorentz transformation> of <rapidity> $\phi'$ multiplies these two null coordinates by $e^{-\phi'}$ and $e^{\phi'}$. Their total factors are therefore $e^{-(\phi+\phi')}$ and $e^{\phi+\phi'}$, proving
$$
\boxed{\phi''=\phi+\phi'.}
$$
The hyperbolic tangent addition formula now gives the velocity composition law
$$
\boxed{v''=c\tanh(\phi+\phi')=\frac{v+v'}{1+vv'/c^2}.}
$$
All velocities here are signed along the same axis; the speed is the absolute value if either frame moves in the opposite direction.
Back to article page
