Momentum transfer perpendicular to velocity of proton?

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SUMMARY

The discussion centers on the assumption that momentum transfer is perpendicular to the velocity of a proton when deducing the Bethe equation. It references Allison & Cobb (1980), which indicates that the first two terms of the Photoabsorption Ionization model are derived from the magnetic vector potential in the Coulomb gauge, where the electric field is transverse to the 3-momentum transfer. The conversation raises questions about the necessity of this perpendicularity assumption in the context of momentum transfer and its implications for understanding particle interactions.

PREREQUISITES
  • Understanding of the Bethe equation in particle physics
  • Familiarity with the Photoabsorption Ionization model
  • Knowledge of magnetic vector potential and its units
  • Concept of momentum transfer in particle interactions
NEXT STEPS
  • Research the derivation of the Bethe equation and its assumptions
  • Study the Photoabsorption Ionization model as outlined in Allison & Cobb (1980)
  • Explore the implications of the Coulomb gauge in electromagnetic theory
  • Investigate the role of momentum transfer in various particle collision scenarios
USEFUL FOR

Physicists, researchers in particle physics, and students studying electromagnetic interactions and momentum transfer in particle collisions will benefit from this discussion.

rjseen
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An assumption in deducing the Bethe equation is that the momentum transfer is perpendicular to the direction of the particle.

ALSO

In Allison & Cobb (1980), just after eq. 28, they state that the first two terms of the Photoabsorption Ionization model arise from the magnetic vector potential in the Coulomb gauge, for which the electric field is transverse to the direction of 3-momentum transfer.

I know that the magnetic vector potential A is in units of momentum per charge, so ignoring hard collisions, how does the justification of this look?
 
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