Momentum of W Bosons After Collision in Particle Physics Lab

AI Thread Summary
In a particle physics lab, an electron and positron collide, resulting in the production of W+ and W− bosons. The total energy before the collision is 200 GeV, with both particles having identical speeds and opposite momentum, leading to a total momentum of zero prior to the collision. Using the equation E^2 = (pc)^2 + (mc^2)^2, a calculated momentum of 3.17 x 10^-17 for each boson is derived, but there is disagreement on the correctness of this solution. Participants emphasize the importance of conservation of momentum and energy in the analysis. Clarification is sought on the potential errors in the calculations.
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Homework Statement


In a particle physics lab, an electron e− and a positron e+ collide, annihilate, and produce a W+ boson and a W− boson. Just before the collision, the electron and positron have a total energy of E = 100 GeV each, with velocities pointing along the +x-axis and -x-axis respectively.

  1. What is the momentum p of each of the W bosons after the collision?

Homework Equations


me− = me+ = 0.511 MeV/c^2, mW− = mW+ = 80.385 GeV/c^2, E = γmc^2, E^2 = (pc)^2 + (mc^2)^2.

The Attempt at a Solution


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Energy conserved so total energy is 200 GeV.

Since they are going in opposite directions and have opposite directions the γ of both electron and position must be identical since they both have the same mass and same total energy. Therefore their speeds must be identical.
Similarly this would suggest that the momentum of both particles must be equal and opposite and so the total momentum before collision is zero.

Using E ^2 = (pc)^2 + (mc^2)^2
Yields a momentum of 3.17 * 10^-17 of both Bosons (but both in different directions)


Could anyone please confirm whether my solutions are correct.
 
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Momentum and energy must be conserved, as you said, so the answer is correct
 
Your answer is wrong.
 
vela said:
Your answer is wrong.
Any more detAil as to where I went wrong?
 
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