General velocity Lorentz transformation

[carllacan]

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]

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.

 

[Chestermiller]

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.