Momentum of W Bosons After Collision in Particle Physics Lab

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SUMMARY

In a particle physics lab, an electron (e−) and a positron (e+) collide to produce W+ and W− bosons, each with a mass of 80.385 GeV/c². The total energy before the collision is 200 GeV, leading to a total momentum of zero due to the equal and opposite velocities of the colliding particles. The calculated momentum of each W boson is 3.17 x 10^-17, indicating that while energy is conserved, the momentum calculation requires further verification as there are discrepancies in the solution provided by the participants.

PREREQUISITES
  • Understanding of particle physics concepts such as annihilation and boson production.
  • Familiarity with relativistic energy-momentum equations, specifically E^2 = (pc)^2 + (mc^2)^2.
  • Knowledge of the properties of electrons and positrons, including their mass (0.511 MeV/c²).
  • Basic grasp of conservation laws in physics, particularly conservation of momentum and energy.
NEXT STEPS
  • Review the derivation of relativistic momentum and energy conservation in particle collisions.
  • Study the properties and interactions of W bosons in particle physics.
  • Explore advanced topics in particle physics, such as Feynman diagrams and decay processes.
  • Learn about experimental techniques used to measure particle momenta in collider experiments.
USEFUL FOR

Students and researchers in particle physics, physicists involved in collider experiments, and anyone interested in the dynamics of particle collisions and boson interactions.

David0709
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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


[/B]
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.
 
Last edited by a moderator:
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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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