General velocity Lorentz transformation

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SUMMARY

The discussion focuses on the application of the Lorentz transformation to determine the velocity of a particle in two inertial systems, IS and IS', where IS' moves with an arbitrary velocity v relative to IS. The relevant equations include the position transformation \(\vec{r'} = \vec{r} + \frac{\gamma - 1}{\beta^2}(\vec{\beta}·\vec{r})\vec{\beta} - \gamma\vec{\beta}ct\) and the time transformation \(ct' = \gamma(ct - \vec{\beta}·\vec{r})\). The user successfully derives the expression for the transformed velocity \(\frac{d\vec{r'}}{dt'}\) and seeks confirmation on its correctness. The discussion concludes that the approach is valid, and suggests further simplification using the identity \(\frac{\gamma - 1}{\beta^2} = \frac{\gamma^2}{\gamma + 1}\).

PREREQUISITES
  • Understanding of Lorentz transformations in special relativity
  • Familiarity with the concepts of velocity and acceleration in inertial frames
  • Knowledge of the Lorentz factor \(\gamma\) and its calculation
  • Ability to manipulate vector equations and perform vector dot products
NEXT STEPS
  • Study the derivation of the Lorentz transformation equations in detail
  • Learn how to apply Lorentz transformations to different physical scenarios
  • Explore the implications of relativistic velocity addition
  • Investigate the identity \(\frac{\gamma - 1}{\beta^2} = \frac{\gamma^2}{\gamma + 1}\) and its applications
USEFUL FOR

Students and professionals in physics, particularly those focused on special relativity, as well as educators seeking to clarify the application of Lorentz transformations in various inertial frames.

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.
 
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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.
 
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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.
 
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