With the knowledge that the speed of light is constant and the same for every reference frame, and that no object can travel at the speed of light or faster than the speed of light in any reference frame, we ask ourselves the question: what happens when two spaceships leave from a space station in opposite directions and both reach a constant speed of 0.95c (with c the speed of light) relative to that space station? In the space station’s reference frame, the two ships travel at 2 \cdot 0.95c = 1.9c relative to each other. This doesn’t violate relativity, as *neither spaceship* is actually going at or faster than the speed of light. But, ignoring the effects of relativity, in the reference frame of either spaceship, the other ship would appear to travel at 1.9c. This *would* violate relativity!

Applying the equations we found before, we will find out what actually happens.

Let’s call the spaceship that travels to the “left” of the space station ship *A*, and the other ship that travels to the “right”, ship *B*. As pointed out, from the point of view of an observer at the space station the ships separate with a speed of 1.9c. This is not what an observer at either ship sees.

We choose our coordinate systems such that the ships travel along the X-axis. We take two reference frames, one for the space station and one for ship *A.* Seen from the reference frame of the space station, ship *A* moves with velocity v_{A} = -0.95c (0.95c in the negative X-direction).

We create a coordinate system k_A with coordinates (x’, t’) corresponding to ship *A*‘s reference frame. In k_A, ship *A* is at x’ = 0 and at rest. We create a second coordinate system k_S with coordinates (x, t) that corresponds to the space station’s reference frame. In k_S, the space station is at x = 0 and at rest. Assume that at t = 0, both ships are at x = 0 (i.e., the space station and both ships are at the exact same position).

As seen in this post, we have:

\begin{aligned}

t’ &= \gamma (t – v_{A} x / c^2) \\

x’ &= \gamma (x – v_{A} t)

\end{aligned}

With \gamma = \frac{1}{\sqrt{1 – v_{A}^2 / c^2}}, the Lorentz factor.

We know that ship *B* is traveling with velocity v_{B} = 0.95c in system k_S. So, we have x = v_{B} t. Required is the velocity of ship *B* in system k_A. We have:

\begin{aligned}

t’ &= \gamma (t – v_{A} x / c^2) = \gamma (t – v_{A} v_{B} t / c^2) \\

x’ &= \gamma (x – v_{A} t) = \gamma(v_{B} t – v_{A} t)

\end{aligned}

Then, the velocity v’ of ship *B* in system k_A is the derivative of *B*‘s position x’ as function of time t’ in k_A:

\begin{aligned}

v’ &= \frac{d x’}{d t’} = \frac{d \gamma(v_{B} t – v_{A} t)}{d \gamma (t – v_{A} v_{B} t / c^2)} \\

&= \frac{d (v_{B} t – v_{A} t)}{d (t – v_{A} v_{B} t / c^2)} \\

&= \frac{v_{B} – v_{A}}{1 – v_{A} v_{B} / c^2}

\end{aligned}

So, the velocity v’ of ship *B* as seen from ship *A* is:

\begin{aligned}

v’ &= \frac{v_{B} – v_{A}}{1 – v_{A} v_{B} / c^2} \\

&= \frac{0.95c – – 0.95c}{1 – -0.95c \cdot 0.95c / c^2} \\

&\approx 0.9987 c

\end{aligned}

This is slower than the speed of light!

In these equations, we have v_{A} < 0, as it was moving to the left of the space station. If we take v = -v_{A} and u = v_{B}, we get:

s = \frac{v + u}{1 + v u / c^2}

Which is the general velocity-addition formula in special relativity. v and u have to be in a straight line (collinear) and the formula assumes they are in opposite direction. Often it is described as the velocity s an observer *A* sees an object *C* move at, when *C* is seen to move at velocity u by observer *B*, and *B* is seen to move at velocity v by *A*. Below is a plot of the equation, with s, v and u in fractions of the speed of light.