Finding the energies of 2 photons from a decay of a travelling mass

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SUMMARY

The discussion focuses on calculating the energies of two photons resulting from the decay of a mass M traveling at v=c/3. The mass M is formed by a mass m traveling at v=3c/5 colliding with a stationary mass m, leading to M being defined as M=3m/√2. The key equations utilized include E=K + mc^2, p = gamma m v, and E^2 - p^2 c^2 = m^2 c^4. Participants emphasize the importance of conserving energy and momentum in the calculations to determine the energy ratio of the photons.

PREREQUISITES
  • Understanding of special relativity concepts, particularly energy-momentum relations.
  • Familiarity with the equations of motion for particles, including gamma factor calculations.
  • Knowledge of photon properties, including energy and momentum relationships.
  • Ability to solve simultaneous equations involving energy and momentum conservation.
NEXT STEPS
  • Study the derivation and application of the energy-momentum relation E^2 - p^2 c^2 = m^2 c^4.
  • Learn about the conservation of energy and momentum in relativistic collisions.
  • Explore the concept of gamma factor in special relativity and its implications for moving masses.
  • Investigate the properties of photons, including their energy-momentum relationship and implications in decay processes.
USEFUL FOR

Students of physics, particularly those focusing on special relativity, particle physics, and energy conservation principles. This discussion is beneficial for anyone studying photon interactions and decay processes in high-energy physics contexts.

Aaron7
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Homework Statement


Mass M traveling at v=c/3 decays into 2 photons in x and -x directions.

Mass M orginally formed by mass m traveling at v=3c/5 hitting a stationary mass m.
This gives M v=c/3 and M=3m/√2.

Homework Equations



E=K + mc^2

p = gamma m v
E = gamma m c^2

K = (gamma -1) mc^2

E^2 - p^2 c^2 = m^2 c^4

The Attempt at a Solution



I am confused how the photons form from a moving mass. How would I find out the ratio of the energies of the photons?
 
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Use your E2 - p2c2 = m2c4 formula to find the relationship of a given photon's momentum and its energy. Set m = 0. Then solve for p.

Then set up the following simultaneous equations (noting that the photons fly off in opposite directions on the x-axis):
  • (Energy of one photon) + (Energy of the other photon) = (The original energy of the system)
  • (|Momentum of one photon|) - (|Momentum of the other photon|) = (|Original momentum of the system|)

"Original energy of the system" above is the original total energy of the system; not just its kinetic energy. You have two equations and two unknowns. You can take it from there.
Aaron7 said:
Mass M orginally formed by mass m traveling at v=3c/5 hitting a stationary mass m.
This gives M v=c/3 and M=3m/√2.
Your above calculations kinda worry me. Is this intermediate step something you are explicitly told to calculate? If so, I don't think it's correct: According to my calculations, (total) energy and momentum are not conserved in your calculations (as they should be). If you are not required to perform this intermediate step, then I don't think it's necessary.
 
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