Velocity addition

Consider a train moving through a station at speed $\vec v = v \e_x$ . A person in the train throws a ball forward and observes it moving at speed $u_T^x$ in the $+x$ direction. How fast does an observer standing at the station observe the ball to be moving?

On the train, in the time $\Delta t_T$ the ball moves $\Delta x_T$ , so

\begin{align*} u_T^x &= \frac{\Delta x_T}{\Delta t_T} \\ u_S^x &= \frac{\Delta x_S}{\Delta t_S} = \frac{\gamma (\Delta x_T + v \Delta t_T)}{\gamma (\Delta t_T + v \Delta x_T / c^2)} \\ &= \frac{\Delta x_T / \Delta t_T + v}{1 + \frac{v}{c^2} \frac{\Delta x_T}{\Delta t_T}} \\ &= \frac{u_T^x + v}{1 + u_T^x v / c^2} \end{align*}

This is the relativistic velocity addition formula. We see that the sum of velocities is scaled such that it never exceeds $c$ so long as neither velocity individually exceeds $c$ .

Importantly, this is only the formula if both velocities are in the same direction

Perpendicular velocities

If the person in the train instead throws the ball up with speed $u_T^y$ in the $y$ direction, then we have

\begin{align*} u_T^y &= \frac{\Delta y_T}{\delta t_T} \\ u_S^y &= \frac{\Delta y_S}{\delta t_S} = \frac{\Delta y_T}{\gamma(\Delta t_T + v\Delta x_T /c^2} \\ &= \frac{u_T^y}{\gamma(1 + u_T^x v/c^2)}. \end{align*}