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I have a question regarding the relative velocities of two objects in $3$D space. If objects $A$ and $B$'s velocities are being observed by stationary observer $C$, what is the velocity vector $A$ will see $B$ with or vice versa. This question also assumes $A$ and $B$ are also going at relativistic speeds.

$$\mathbf{v}_A=(a,b,c)$$ $$\mathbf{v}_B=(d,e,f)$$

If we assume all the vector components are being multiplied by $c$ (the speed of light) and can't be greater than $1$, what would $B$'s velocity vector be from $A$'s perspective?

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  • $\begingroup$ Could you describe the directions A and B are traveling relative to each other? I'm having difficulty visualizing your situation. $\endgroup$
    – Jake Watrous
    Commented Feb 8, 2017 at 3:00
  • $\begingroup$ In this scenario, A and B could have velocities pointing in similar directions or opposing directions. I mainly want the mathematical formula to do the calculation with. $\endgroup$
    – Laff70
    Commented Feb 8, 2017 at 3:48
  • $\begingroup$ You're looking to find relative velocities, then? $\endgroup$
    – Jake Watrous
    Commented Feb 8, 2017 at 3:50
  • $\begingroup$ In this example the vectors for velocity are broken into x, y, and z vectors. $\endgroup$
    – Laff70
    Commented Feb 8, 2017 at 3:52

1 Answer 1

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Let $C$ be rest frame $S$ (or lab frame, earth frame, etc) and $A$ be the moving frame $S'$.

Transformation of velocity

Velocity of $B$ observed by $A$ is $$\mathbf{v}'= \frac{\mathbf{u}+ \left( \dfrac{\gamma_{v}-1}{v^2}\mathbf{u}\cdot \mathbf{v}-\gamma_{v} \right)\mathbf{v}} {\gamma_{v} \left( 1-\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2} \right)}$$

where $\gamma_{v}=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$, $\mathbf{u}=(d,e,f)$ is the velocity of $B$ observed by $C$ and $\mathbf{v}=(a,b,c)$ is the velocity of $A$ observed by $C$

Please refer to the derivation in pp. 33-4 of the following notes.

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  • $\begingroup$ Thank you for your help. This should help greatly with my research. $\endgroup$
    – Laff70
    Commented Feb 8, 2017 at 19:21
  • $\begingroup$ @Ng Chung Tak I would be interested in a derivation of this formula. Where could it be found? $\endgroup$
    – Quaerendo
    Commented Jan 31, 2020 at 16:48
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    $\begingroup$ @Quaerendo Please refer to pp. 33-4 of the following notes. $\endgroup$
    – Ng Chung Tak
    Commented Jan 31, 2020 at 22:30

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