# Relativistic relative velocity of particles

## Homework Statement

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?

## Homework Equations

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

## The Attempt at a Solution

[/B]
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.

#### Attachments

• 816 bytes Views: 406
• 596 bytes Views: 374
• 2.6 KB Views: 399
• 854 bytes Views: 415
Last edited:

Related Advanced Physics Homework Help News on Phys.org
PeroK
Homework Helper
Gold Member
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.

Also, should the answer you're given have $2c^2# in the denominator? TSny Homework Helper Gold Member Hello, and welcome to PF! ## Homework Statement show that the relative speed of the particles is given by$u_R = u(1-3u^2/4c^2)/(1-u^2/c^2)$I believe there is a misprint in this expression. Should the expression in parentheses in the numerator be raised to the 1/2 power? ## Homework Equations$u'_x = (ux-v)/(1-v*ux/c^2)$and$u'_y = uy/\gamma(1-v*ux/c^2)$## The Attempt at a Solution [/B] I set S' as the stationary state of particle A, moving at a velocity u in the x direction. This meant that particle A was at rest in this frame. Therefore the velocity of B in the S' frame is the relative velocity of the two particles. For particle B I obtained:$u'_x = -u/2$and$u'_y = usin60/\gamma$I do not get these results using your relevant equations. In particular, what happened to the denominators of$u'_x$and$u'_y$? Hello, and welcome to PF! I believe there is a misprint in this expression. Should the expression in parentheses in the numerator be raised to the 1/2 power? You are correct, the parentheses should be raised to the 1/2 power. I do not get these results using your relevant equations. In particular, what happened to the denominators of$u'_x$and$u'_y$? For particle B I set the velocity in the x-direction to be$ucos60$and in the y-direction to be$usin60$. I then just substituted those values into the equations for u'_x and u'_y. I obtained those values for u'_x and u'_y because both u_x and u_y are zero since the we are considering the rest frame of particle A. 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. Also, should the answer you're given have$2c^2# in the denominator?
Is the angle still not valid if we consider the frame where particle A is stationary? Surely the relative velocity of particle B in that frame it the overall relative velocity?

PeroK
Homework Helper
Gold Member
No. You need to use the velocity transformation formula, which is more than just $\gamma$ factor.

TSny
Homework Helper
Gold Member
For particle B I set the velocity in the x-direction to be $ucos60$ and in the y-direction to be $usin60$. I then just substituted those values into the equations for u'_x and u'_y. I obtained those values for u'_x and u'_y because both u_x and u_y are zero since the we are considering the rest frame of particle A.
ux and uy are the x and y components of the velocity of B in the original unprimed frame.

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