# Composition of velocities

### Galilean result

Suppose we have three frames - unprimed, primed and double-primed - where the primed frame is moving at speed $v$ relative to the unprimed frame, and the double-primed frame is moving at speed $u$ relative to the primed frame.

We would like to know the speed of the double-primed frame, $w$ say, relative to the unprimed frame.

For Galileo and everyone else before Einstein, this simply would have been the sum of the two velocities, $$w = u + v$$

For example, if $$v = u = \frac{3}{4}$$ then according to Galileo, $$w = \frac{3}{4} + \frac{3}{4} = \frac{3}{2} \gt 1$$

But we know that this cannot be correct because nothing can travel faster than the speed of light, which we have taken to be $c \equiv 1$, as usual.

### Relativistic result

The correct value is $$w = \frac{u+v}{1+uv}$$ which we can compute by compounding the matrices representing the two boosts corresponding to $v$ and $u$.

These are \begin{align*}\begin{bmatrix}t'\\x'\end{bmatrix} &= \gamma_v\begin{bmatrix}1&-v\\-v&1\end{bmatrix} \begin{bmatrix}t\\x\end{bmatrix}\\\begin{bmatrix}t''\\x''\end{bmatrix} &= \gamma_u\begin{bmatrix}1&-u\\-u&1\end{bmatrix} \begin{bmatrix}t'\\x'\end{bmatrix}\end{align*} where we have dropped the irrelevant $(y, z)$ components, and written \begin{align*}\gamma_v &= \frac{1}{\sqrt{1 - v^2}}\\\gamma_u &= \frac{1}{\sqrt{1 - u^2}}\end{align*}

Since we do know that the double-primed frame is moving at a constant speed relative to the unprimed frame, we expect $w$ to satisfy $$\begin{bmatrix}t''\\x''\end{bmatrix} = \gamma_w\begin{bmatrix}1&-w\\-w&1\end{bmatrix} \begin{bmatrix}t\\x\end{bmatrix}$$ where $$\gamma_w = \frac{1}{\sqrt{1 - w^2}}$$

Now, \begin{align*}\begin{bmatrix}t''\\x''\end{bmatrix} &= \gamma_u\begin{bmatrix}1&-u\\-u&1\end{bmatrix} \begin{bmatrix}t'\\x'\end{bmatrix}\\&= \gamma_u \gamma_v \begin{bmatrix}1&-u\\-u&1\end{bmatrix}\begin{bmatrix}1&-v\\-v&1\end{bmatrix} \begin{bmatrix}t\\x\end{bmatrix}\\&= \gamma_u \gamma_v \begin{bmatrix}1+uv&-(u+v)\\-(u+v)&1+uv\end{bmatrix}\begin{bmatrix}t\\x\end{bmatrix}\\&= \left(1+uv\right)\gamma_u \gamma_v\begin{bmatrix}1&-\frac{u+v}{1+uv}\\-\frac{u+v}{1+uv}&1\end{bmatrix}\begin{bmatrix}t\\x\end{bmatrix}\end{align*}

We would like to take $$w = \frac{u+v}{1+uv}$$ but that would require that $$\gamma_w = \left(1+uv\right)\gamma_u \gamma_v$$

But, \begin{align*}\left(1+uv\right)\gamma_u \gamma_v &= \frac{1+uv}{\sqrt{\left(1 - v^2\right)\left(1 - u^2\right)}}\\&= \frac{1+uv}{\sqrt{\left(1 + uv\right)^2 - \left(u + v\right)^2}}\\&= \frac{1}{\sqrt{1 - \left(\frac{u+v}{1+uv}\right)^2}}\\&= \gamma_w\end{align*}

### Justifications

If we try our example again, where $$v = u = \frac{3}{4}$$ then $$w = \frac{3/4 + 3/4}{1+9/16} = \frac{3/2}{25/16} = \frac{24}{25} \lt 1$$

More generally if $v, u \lt 1$, so that $1 - v \lt 1$ as well, then \begin{align*} & u\left(1 - v\right) \lt 1 - v\\\Rightarrow & u + v \lt 1 + uv\\\Rightarrow & \frac{u + v}{1 + uv} \lt 1\end{align*}

In the case where one of $v,u = 1$, then we get $$\frac{u + 1}{1 + u\cdot1} = 1$$ and similarly, if both $v = u = 1$ then $$\frac{1 + 1}{1 + 1\cdot1} = 1$$

Finally, if we put back in the speed of light, $c \neq 1$, so that $$w = \frac{u + v}{1 + uv/c^2}$$ and if $u,v$ are small compared to $c$, then we recover the intuitive, Galilean result, $w = u + v$.