Electron/Positron Annihilation

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SUMMARY

In the discussion on Electron/Positron Annihilation, participants clarified that when a positron collides with an electron, two photons are produced, each carrying energy equal to the sum of the rest mass energy and kinetic energy of the colliding particles. The equation E = m_o*c^2 + K accurately represents the energy of each photon, but it does not imply that each photon contains double the energy of a single particle. The conservation of momentum and energy necessitates the creation of two photons to maintain balance during the annihilation process.

PREREQUISITES
  • Understanding of basic particle physics concepts, specifically electron and positron interactions.
  • Familiarity with the principles of energy conservation in particle collisions.
  • Knowledge of Einstein's mass-energy equivalence principle (E = mc²).
  • Basic understanding of photon properties and behavior in quantum mechanics.
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  • Study the principles of particle-antiparticle annihilation in detail.
  • Learn about the conservation laws in particle physics, focusing on momentum and energy conservation.
  • Explore the implications of Einstein's mass-energy equivalence in various particle interactions.
  • Investigate the properties of photons and their role in quantum field theory.
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Students and professionals in physics, particularly those focused on particle physics, quantum mechanics, and energy conservation principles. This discussion is beneficial for anyone seeking to deepen their understanding of electron-positron interactions and photon production.

Von Neumann
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I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?
 
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Von Neumann said:
I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?

The total energy of the two photons is the same as the sum of the mass and kinetic energy of the colliding electron positron pair. Two photons, two particles. What's this 2:1 ratio of which you speak?
 
The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.
 
Von Neumann said:
The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.

One photon could not be created. There is zero momentum total if they collide head on. It couldn't conserve momentum and energy. You need two. I think "each photon" doesn't mean each single photon created by the collision. It means each of the two photons created by the collision.
 
Last edited:
Oh right! Thanks!
 

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