How Fast Are Proton and Antiproton Moving Towards Each Other?

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SUMMARY

A proton and an antiproton, each moving at a speed of 0.8c towards a collision point, have been analyzed for their relative speed using the Einstein velocity addition formula. The calculation shows that the speed of the antiproton in the proton's frame of reference is 0.976c. The approach taken involved considering the collision point as moving towards the stationary proton, which is a valid method for this scenario. The final result confirms the correct application of relativistic principles in determining relative velocities.

PREREQUISITES
  • Understanding of Einstein's theory of relativity
  • Familiarity with the Einstein velocity addition formula
  • Knowledge of relativistic speeds (e.g., speeds approaching the speed of light)
  • Basic concepts of particle physics, specifically protons and antiprotons
NEXT STEPS
  • Study the implications of relativistic speeds on mass and energy
  • Learn more about the center of mass frame in particle collisions
  • Explore advanced applications of the Einstein velocity addition formula
  • Investigate the behavior of particles in high-energy physics experiments
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Students and educators in physics, particularly those focusing on relativity and particle physics, as well as researchers analyzing high-energy particle collisions.

beecher
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Homework Statement



A proton and an antiproton are moving toward each other in a head-on collision. If each has a speed of 0.8c with respect to the collision point, how fast are they moving with respect to each other?


Homework Equations



Ux = [U'x + V] / [1 + (v/c^2) Ux']

The Attempt at a Solution



I believe I have solved it correctly, but I am unsure of one assumption which I made, and want to make sure that it is alright. Instead of looking at it as the collision point being stationary, and the proton moving towards it at 0.8c, I assume that the proton is stationary, and the collision point moves towards it at 0.8c. Then the anti-proton moves towards the collision point at 0.8c as measured in the frame of the collision point.
This gives me Ux = the speed of the antiproton in the proton frame
Ux' = the speed of the anti-proton in the collision point frame (0.8c)
V = Speed of collision point in the proton frame (0.8c)

Thus, Ux = [0.8c + 0.8c] / [1 + (0.8c/c^2) 0.8c] = 1.6c / 1.64 = 0.976c

Does this look correct?

Thanks
 
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beecher said:
I believe I have solved it correctly, but I am unsure of one assumption which I made, and want to make sure that it is alright. Instead of looking at it as the collision point being stationary, and the proton moving towards it at 0.8c, I assume that the proton is stationary, and the collision point moves towards it at 0.8c. Then the anti-proton moves towards the collision point at 0.8c as measured in the frame of the collision point.
This gives me Ux = the speed of the antiproton in the proton frame
Ux' = the speed of the anti-proton in the collision point frame (0.8c)
V = Speed of collision point in the proton frame (0.8c)

Thus, Ux = [0.8c + 0.8c] / [1 + (0.8c/c^2) 0.8c] = 1.6c / 1.64 = 0.976c

Does this look correct?
Rather than thinking of the collision point, think of the centre of mass frame of reference (the frame in which the velocities are initially measured). You wish to determine the velocity of the anti-proton in the proton's frame of reference. To do this, you apply the Einstein velocity addition to translate the velocity of the anti-proton in the centre of mass frame to the proton frame, which you have done correctly.

AM
 
Thanks for the confirmation
 

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