Relativistic energy/momentum, massless particles

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Homework Help Overview

The discussion revolves around a problem in relativistic energy and momentum, specifically concerning the decay of a pion at rest into a muon and an antineutrino. Participants are examining the energies and momenta of the resulting particles, given the masses of the pion and muon, while noting the massless nature of the antineutrino.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of energy and momentum in the decay process, with initial attempts to apply different energy equations. There is a question regarding the appropriateness of using the rest mass energy equation for the muon due to its momentum.

Discussion Status

Some participants have provided guidance on the correct equations to use for the muon’s energy, suggesting the need to apply the relativistic energy-momentum relation. Others have proposed a method for solving for momentum and energy, leading to specific values that appear to satisfy conservation laws.

Contextual Notes

There is an acknowledgment of the complexities involved in applying the equations correctly, particularly in relation to the muon's motion and energy. The discussion reflects a mix of interpretations and approaches without reaching a definitive conclusion.

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Homework Statement



A pion at rest decays into a muon and an antineutrino. The mass of the antineutrino is zero, find the energies and momenta of the muon and antineutrino. Mass of the pion is 139.57 MeV/c^2 and the mass of the muon is 105.66 MeV/c^2

Homework Equations



pion -> muon + antineutrino

(1) E=mc^2
(2) E=pc
(3) E^2 = (pc)^2 + (mc^2)^2

The Attempt at a Solution



Conservation of energy: E(pion) = E(muon) + E(antineutrino)
Using equation 1 and the given masses:
E(pion) = 139.57 MeV
E(muon) = 105.66 MeV (?)
so E(antineutrino) = E(pion) - E(muon) = 33.91 MeV
Use equation 2 for the massless antineutrino and get p = 33.91 MeV/c

since momentum is conserved and p(pion) = 0 (at rest),
p(muon) = -p(antineutrino) = -33.91 MeV/c

I don't think I'm right because if the muon has momentum it is moving and thus I can't find its energy by simply plugging its mass into equation 1.

Thanks for any help.
 
Last edited:
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You are correct in that you can't use Equation 1 for the energy of the muon. You have to use Equation 3. Otherwise you are on target.
 
How about this:

p(muon) = p(antineutrino) = p
E(antineutrino) = pc
E(muon) = sqrt[(pc)^2+(mc^2)^2]

plug these into conservation of E and solve for p
after some math I get

p = 29.79 MeV/c
so E(antineutrino) = 29.79 MeV
E(muon) = 109.78 MeV

The two E's add up to equal E(pion), 139.57
 
Looks good to me.
 

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