carllacan
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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.