# General velocity Lorentz transformation

## Homework Statement

A particle's movement is described by $\vec{r}$ in the inertial system IS. Find the velocity of the particle $\vec{\dot{r'}}$ in the system IS', which is moving with arbitrary velocity $v$ from IS. Both inertial systems are arbitrary.

## Homework Equations

For the position vector the Lorentz transformation is $\vec{r'} = \vec{r} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{r})\vec{\beta} -\gamma\vec{\beta}ct$, and for the time $ct' = \gamma(ct-\vec{\beta}·\vec{r}))$

## The Attempt at a Solution

Suppose that the Lorentz transformations are still valid when applied to differential quantities $d\vec{r}$ and $dt$. Then:

$\frac{d\vec{r'}}{d't} = c\frac{\vec{dr} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{dr})\vec{\beta} -\gamma\vec{\beta}cdt}{\gamma(cdt-\vec{\beta}·\vec{dr})}$

Taking a $dt$ out of the denominator:

$\frac{d\vec{r'}}{dt'} = c\frac{\vec{dr} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{dr})\vec{\beta} -\gamma\vec{\beta}cdt}{\gamma(c-\vec{\beta}·\vec{\frac{dr}{dt}})dt}$
And so:
$\frac{d\vec{r'}}{dt'} = c\frac{\vec{\dot{r}} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{\dot{r}})\vec{\beta} -\gamma\vec{\beta}c}{\gamma(c-\vec{\beta}·\vec{\dot{r}})}$
And then we can reorder that more nicely.

Is this right? I've been looking around but, surprisingly, haven't been able to find the answer to this.

TSny
Homework Helper
Gold Member
Yes, that all looks good. To check with standard formulas, find the components of the primed velocity that are parallel and perpendicular to the relative velocity of the two frames.

• 1 person
Chestermiller
Mentor
You might be able to simplify this a little by making use of the identity:
$$\frac{γ-1}{β^2}=\frac{γ^2}{γ+1}$$
and then using a common denominator for the second two terms in the numerator.

• 1 person