Velocity-addition formula

Remarks

Proof idea: apply the inverse Lorentz transformation $x = \gamma (x' + v t')$, $t = \gamma (t' + v x' / c^2)$, $y = y'$, $z = z'$ to the differentials along the world line and divide by $dt$. Example: $u' = 0.6 c$ and $v = 0.6 c$ give $u = 1.2 c / 1.36 = 15 c / 17 \approx 0.88 c$, below $c$. If $\abs{u'} < c$ and $\abs{v} < c$ then $\abs{u} < c$, and $u' = c$ gives $u = c$, so the speed of light is the same in both frames. In terms of rapidities, collinear velocities add: $\operatorname{artanh}(u/c) = \operatorname{artanh}(u'/c) + \operatorname{artanh}(v/c)$, which is the addition formula of $\tanh$. For $\abs{u'} \ll c$ and $\abs{v} \ll c$ the formula reduces to the Galilean rule $u = u' + v$.

Going deeper

For a student

The formula is a direct computation from the Lorentz transformation, and it shows that boosts along one axis form a group isomorphic to $(\R, +)$ via rapidity. The transverse components get the extra factor $1/\gamma$ because of time dilation. For non-collinear velocities, the composition of two boosts is not a pure boost but a boost followed by a rotation, the Wigner rotation.