Positive pion decay and kinetic energy

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SUMMARY

The discussion centers on the decay of a positive pion into a positive muon and a neutrino within a nucleus containing Z protons. The key equations utilized include Coulomb's law, F = k (q1q2/r^2), and conservation of energy and momentum principles. The muon's kinetic energy increases due to electrostatic repulsion as it moves away from the nucleus. The consensus is that the initial kinetic energy (Ki) of the muon equals its final kinetic energy (Kf) after accounting for the energy gained from the electrostatic interaction.

PREREQUISITES
  • Understanding of particle physics concepts, specifically pion decay.
  • Familiarity with Coulomb's law and electrostatic forces.
  • Knowledge of conservation of energy and momentum in particle interactions.
  • Basic understanding of 4-vectors in relativistic physics.
NEXT STEPS
  • Study the principles of pion decay and its implications in particle physics.
  • Learn about the application of Coulomb's law in particle interactions.
  • Explore conservation laws in particle physics, focusing on energy and momentum conservation.
  • Investigate the use of 4-vectors for calculating energy and momentum in relativistic contexts.
USEFUL FOR

This discussion is beneficial for physics students, particle physicists, and anyone interested in understanding the dynamics of particle decay and interactions within atomic nuclei.

justtram13
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Homework Statement


A nucleus contains Z protons that on average are uniformly distributed throughout a tiny sphere of radiues R.
Suppose that in an accelerator experiment a positive pion is produced at rest at the center of a nucleus containing Z protons. The pion decays into a positive muon (essentially a heavy positron) and a neutrino. The muon has initial kinetic energy Ki.
How much kinetic energy does the muon have by the time it has been repelled very far away from the nucleus? (The muon interacts with the nucleus only through Coulomb's law and is unaffected by nuclear forces. The massive nucleus hardly moves and gets negligible kinetic energy.)


Homework Equations


F = k (q1q2/r^2)




The Attempt at a Solution


Neutrino product -> E = pc
Since pion is at rest, its energy equals its mass
E(sub∏) = m(sub∏)

E(sub∏) = E(subμ) + E(subv)
E(sub∏) = E(subμ) + 0
E(sub∏) = E(subμ)

I have no idea where to go from here. As of now, I'm assuming that Ki = Kf, but I don't think that that's right. Any suggestions on where to go from here?
 
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F = k (q1q2/r^2)
Only for point-charges, or spherical charge distributions with no overlap. You can use this once the muon left the nucleus.

Do you have to calculate Ki? In that case: Energy and momentum are conserved in the pion decay and you can neglect the neutrino mass. 4-vectors are the quickest way to calculate the muon energy.

I'm assuming that Ki = Kf
The electrostatic repulsion will increase the muon energy.
 

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