# Relative velocity in special relativity

• Physicssssss
In summary, the conversation discusses the relative velocity of two aeroplanes observed from the ground, and how to express this velocity in terms of their individual velocities. Using the general Lorentz transformation and the equation for the square of the vector, the final solution is derived as an expression depending only on the velocities of the two planes. Further simplification can be done by expressing gamma in terms of the velocity u. f

## 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?

It is an answer that depends on v and u only, that is good. Now you can simplify it (don't forget to express gamma via u). Not nice, and there could be a shorter way, but it should work.

It works! Thanks