# Relativistic relative velocity

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1. Jan 9, 2015

### scottJH

1. The problem statement, all variables and given/known data
In a given inertial frame two particles are shot out simultaneously from a given point with equal speeds u at an angle of 60 degrees with respect to each other. Using the concept of 4-velocity or otherwise, show that the relative speed of the particles is given by

$u_R = u(1-3u^2/4c^2)/(1-u^2/2c^2)$

I have tried this a number of ways but always end up getting

$u_R = u(1-3u^2/4c^2)$

So I guess my question is which answer is correct?

2. Relevant equations

$u'_x = (ux-v)/(1-v*ux/c^2)$ and $u'_y = uy/\gamma(1-v*ux/c^2)$

3. The attempt at a solution

I set S' as the stationary state of particle A, moving at a velocity u in the x direction. This meant that particle A was at rest in this frame. Therefore the velocity of B in the S' frame is the relative velocity of the two particles.

For particle B I obtained:

$u'_x = -u/2$ and $u'_y = usin60/\gamma$

Then using those values calculated the relative velocity to be

$u_R = u(1-3u^2/4c^2)$

I apologise for the layout of the equations because it is my first time posting.

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Last edited: Jan 9, 2015
2. Jan 9, 2015

### PeroK

You got off on the wrong foot. You need to work out the velocity of particle B in the initial frame, where the angle is valid.

7. Jan 9, 2015

### TSny

ux and uy are the x and y components of the velocity of B in the original unprimed frame.

8. Jan 9, 2015

### scottJH

Thanks for all your help. I managed to solve it.