# Relativistic relative velocity of particles

scottJH

## 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

Last edited:

Homework Helper
Gold Member
2021 Award
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?

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##?

scottJH
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.

scottJH
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?

Homework Helper
Gold Member
2021 Award
No. You need to use the velocity transformation formula, which is more than just ##\gamma## factor.

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.

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