Verifying Conservation of Energy and Momentum for γ + p → Δ+ → πo + p

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SUMMARY

The discussion focuses on verifying the conservation of energy and momentum in the reaction γ + p → Δ+ → πo + p, where an incident photon with energy 0.34 GeV excites a proton into a Δ+ baryon that subsequently decays into a neutral pion and a proton. The Δ+ mass is established at 1.23 GeV, and the conservation laws are applied to demonstrate that the peak photon energy aligns with this mass. Key equations include the four-momentum [E,p] and the relationship between momentum and energy in relativistic physics.

PREREQUISITES
  • Understanding of relativistic energy-momentum relations
  • Familiarity with the concept of baryon decay
  • Knowledge of the rest mass energy calculation (E=Mc²)
  • Basic principles of conservation laws in physics
NEXT STEPS
  • Study the derivation of four-momentum in relativistic physics
  • Learn about baryon decay processes and their implications
  • Explore the concept of invariant mass in particle physics
  • Investigate the role of conservation laws in high-energy particle collisions
USEFUL FOR

This discussion is beneficial for physics students, particle physicists, and educators focusing on energy and momentum conservation in particle interactions.

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Homework Statement



γ + p → Δ+ → πo + p

incident photon with energy 0.34 GeV excites a photon into Δ+ which decays into πo + p

Write down and apply the laws of conservation of energy and momentum to show taht the peak photon energy of 0.34 GeV is consistent with Δ+ mass of 1.23 GeV

Homework Equations



four momentum [E,p]

The Attempt at a Solution



just wondering what to start with exactly after writing down the equations.

M = 938.727MeV/c^2
so the rest energy is just Mc^2 with 0 momentum for the intial proton

do I assume that all the momentum from the photon is transferred to the Δ+ because the photon does not show up in the final state?

In the rest frame the pion and proton are ejected at 180 degrees from each other and have the same momentum, but not energy or velocity?

so

p' = 0.34/2c = γMv where gamma here is 1/sqrt(1-β2) and v is the velocity of the proton
 
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in the lab frame?then E' = sqrt(M2c4 + p'2c2) = 1.23GeVbut then how do I show that the peak photon energy of 0.34 GeV is consistent with this?
 

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