General velocity Lorentz transformation

  • Thread starter carllacan
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  • #1
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Homework Statement


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.


Homework 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]


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.
 

Answers and Replies

  • #2
TSny
Homework Helper
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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.
 
  • #3
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You might be able to simplify this a little by making use of the identity:
[tex]\frac{γ-1}{β^2}=\frac{γ^2}{γ+1}[/tex]
and then using a common denominator for the second two terms in the numerator.
 

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