1. The problem statement, all variables and given/known data
A particle's movement is described by [itex]\vec{r}[/itex] in the inertial system IS. Find the velocity of the particle [itex]\vec{\dot{r'}}[/itex] in the system IS', which is moving with arbitrary velocity [itex]v[/itex] from IS. Both inertial systems are arbitrary.
2. Relevant equations
For the position vector the Lorentz transformation is [itex]\vec{r'} = \vec{r} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{r})\vec{\beta} -\gamma\vec{\beta}ct[/itex], and for the time [itex]ct' = \gamma(ct-\vec{\beta}·\vec{r}))[/itex]
3. The attempt at a solution
Suppose that the Lorentz transformations are still valid when applied to differential quantities [itex]d\vec{r}[/itex] and [itex]dt[/itex]. Then:
[itex]\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})}[/itex]
Taking a [itex]dt[/itex] out of the denominator:
[itex]\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}[/itex]
And so:
[itex]\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}})}[/itex]
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.
A particle's movement is described by [itex]\vec{r}[/itex] in the inertial system IS. Find the velocity of the particle [itex]\vec{\dot{r'}}[/itex] in the system IS', which is moving with arbitrary velocity [itex]v[/itex] from IS. Both inertial systems are arbitrary.
2. Relevant equations
For the position vector the Lorentz transformation is [itex]\vec{r'} = \vec{r} + \frac{\gamma - 1}{\beta ^2}(\vec{\beta}·\vec{r})\vec{\beta} -\gamma\vec{\beta}ct[/itex], and for the time [itex]ct' = \gamma(ct-\vec{\beta}·\vec{r}))[/itex]
3. The attempt at a solution
Suppose that the Lorentz transformations are still valid when applied to differential quantities [itex]d\vec{r}[/itex] and [itex]dt[/itex]. Then:
[itex]\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})}[/itex]
Taking a [itex]dt[/itex] out of the denominator:
[itex]\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}[/itex]
And so:
[itex]\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}})}[/itex]
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.
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