# Relative velocity in special relativity

## Homework Statement

Imagine we are observing two aeroplaes from the ground and let their velocities be $\mathbf{u}$ and $\mathbf{v}$ respectively. Assume that the first plane has radar equipment permitting a measurement of the speed of the other plane relative to itself. The velocity so measured will be the relative velocity of our definition. We must express this relative velocity in terms of the components of the velocities $\mathbf{u}$ and $\mathbf{v}$ of the two planes, as observed from the ground. The velocity of the second plane measured from the ground is $$\mathbf{v}=\frac{d\mathbf{r}}{dt}$$ while its velocity measured from the other plane is $$\mathbf{v}^{\prime}=\frac{d\mathbf{r}^{\prime}}{dt^{\prime}}.$$Using the general Lorentz transformation we have $$\mathbf{v}^{\prime}=\frac{\mathbf{v}-\mathbf{u}+\left(\gamma-1\right)\left(\frac{\mathbf{u}}{u^{2}}\right)\left\{ \left(\mathbf{u}\cdot\mathbf{v}\right)-u^{2}\right\} }{\gamma\left(1-\mathbf{u}\cdot\mathbf{v}\right)}$$where$$\gamma=\frac{1}{\sqrt{1-u^{2}}}.$$Calculate the square of the vector $\mathbf{v}^{\prime}$.

## Homework Equations

$$\left(\mathbf{u}\times\mathbf{v}\right)^{2}=u^{2}v^{2}-\left(\mathbf{u}\cdot\mathbf{v}\right)^{2}$$

## The Attempt at a Solution

The solutions is $$v^{\prime2}=1-\frac{\left(1-u^{2}\right)\left(1-v^{2}\right)}{\left(1-\mathbf{u}\cdot\mathbf{v}\right)^{2}}=\frac{\left(\mathbf{u}-\mathbf{v}\right)^{2}-\left(\mathbf{u}\times\mathbf{v}\right)^{2}}{\left(1-\mathbf{u}\mathbf{\cdot}\mathbf{v}\right)^{2}}.$$Taking the square of the vector $\mathbf{v}^{\prime}$ I have $$v^{\prime2}=\frac{1}{\gamma^{2}\left(1-\mathbf{u}\cdot\mathbf{v}\right)^{2}}\left\{ \left(\mathbf{v}-\mathbf{u}\right)^{2}+\frac{1}{u^{2}}\left(\gamma-1\right)^{2}\left(\mathbf{u}\cdot\mathbf{v}-u^{2}\right)^{2}+\frac{2}{u^{2}}\mathbf{u}\cdot\mathbf{v}\left(\gamma-1\right)\left(\mathbf{u}\cdot\mathbf{v}-u^{2}\right)-2\left(\gamma-1\right)\left(\mathbf{u}\cdot\mathbf{v}-u^{2}\right)\right\}.$$ Can someone help me?