Relativistic relative velocity of particles

  • #1
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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.
 

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Answers and Replies

  • #2
PeroK
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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?
 
  • #3
TSny
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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##?
 
  • #4
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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.
 
  • #5
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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?
 
  • #6
PeroK
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No. You need to use the velocity transformation formula, which is more than just ##\gamma## factor.
 
  • #7
TSny
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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.
 
  • #8
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Thanks for all your help. I managed to solve it.
 

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