Pi meson decay (relativistic momentum)

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SUMMARY

The discussion focuses on the decay of a charged π meson (rest mass = 273me) into a neutrino (zero rest mass) and a μ meson (rest mass = 207me). The key equations utilized include E = moγc² and K = mo(γ-1)c², which are essential for calculating kinetic energies. The initial and final energy equations were established in both the rest frame of the π meson and an alternative frame where the μ meson is at rest. The conservation of momentum principle was applied, leading to a quadratic equation for the speed of the μ meson, which was solved using the quadratic formula.

PREREQUISITES
  • Understanding of relativistic momentum and energy equations
  • Familiarity with the concepts of rest mass and gamma factor (γ)
  • Knowledge of conservation laws in particle physics
  • Ability to solve quadratic equations
NEXT STEPS
  • Study the derivation of the Lorentz factor (γ) in special relativity
  • Learn about conservation of momentum in particle decay processes
  • Explore the implications of mass-energy equivalence in particle physics
  • Practice solving problems involving relativistic kinetic energy and momentum
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying particle physics, as well as educators and researchers interested in relativistic decay processes and energy calculations.

RyanP
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Homework Statement


A charged π meson (rest mass = 273me) decays into a neutrino (zero rest mass) and a μ meson (rest mass = 207me). Find the kinetic energies of the neutrino and the mu meson.

Homework Equations


E = moγc2
K = mo(γ-1)c2
v = pc2/E
p = moγv

The Attempt at a Solution


In the rest frame of the pi meson (S frame),
Ei = 273mec2
Ef = 207meγc2 + K where γ is the gamma factor of the mu meson and K is the energy (kinetic) of the neutrino.

Since the neutrino has no rest mass, it travels at c, so pv = Ev/c = K/c.

I couldn't figure out how to carry on in this frame, so I switched to an S' frame where the mu meson is at rest.

In this frame,
p'i = -273meγv where v is the speed of the pi meson initially (equal to the speed of the mu meson in the pi rest frame).
p'f = -E'nu/c since the neutrino is the only thing moving after the collision in this frame.

E'i = 273meγc2
E'f = 207me + E'nu

How do I proceed from here?
 
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Go back to the rest frame of the pion. Since the pion is initially at rest, the initial momentum in this frame is zero. Since momentum is conserved, the final momentum is zero also. So what do you know about the momenta of the neutrino and the muon?
 
phyzguy said:
Go back to the rest frame of the pion. Since the pion is initially at rest, the initial momentum in this frame is zero. Since momentum is conserved, the final momentum is zero also. So what do you know about the momenta of the neutrino and the muon?
The momentum of the neutrino is equal and opposite to the momentum of the muon - should I solve for the speed of the muon?
 
You should be able to write four equations for the four unknowns, Enu, Emu, pnu, pmu.
 
Ended up with a quadratic equation for v (speed of the muon), in terms of both masses and c. Brute-forced it with the quadratic formula and got the right answer.
 

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