How Fast Are Proton and Antiproton Moving Towards Each Other?

AI Thread Summary
A proton and an antiproton, each moving at 0.8c towards a collision point, need their relative speed calculated. The user proposes a method by assuming the proton is stationary and the collision point moves towards it at 0.8c, while the antiproton also moves at 0.8c towards the collision point. Using the Einstein velocity addition formula, the user calculates the relative speed as approximately 0.976c. Another participant confirms the approach and emphasizes the importance of using the center of mass frame for accurate calculations. The discussion highlights the correct application of relativistic principles in determining relative velocities.
beecher
Messages
15
Reaction score
0

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
 
Physics news on Phys.org
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
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top