Photon energy, momentum, and velocity

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SUMMARY

The discussion centers on calculating the total energy and mass of a system comprising two photons with energies of 200 MeV and 100 MeV, respectively. The total energy of the system is confirmed to be 300 MeV. To find the mass of a single particle with the same energy and momentum, the equations p = E/c for photons and p = γmu for massive particles are utilized. The correct velocity of the particle is determined to be 0.74c, emphasizing the importance of vector addition in momentum calculations.

PREREQUISITES
  • Understanding of photon energy and momentum equations
  • Familiarity with relativistic mass-energy equivalence
  • Knowledge of vector addition in physics
  • Basic concepts of special relativity
NEXT STEPS
  • Study the implications of relativistic momentum in particle physics
  • Learn about the Lorentz factor (γ) and its applications
  • Explore the concept of energy-momentum conservation in collisions
  • Investigate the differences between massless and massive particles in relativity
USEFUL FOR

Students and educators in physics, particularly those focusing on particle physics and special relativity, will benefit from this discussion.

warfreak131
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Homework Statement



Suppose one photon has an energy of 200 MeV and it is traveling along the x-axis. Suppose another has an energy of 100 MeV and is traveling along the y-axis. (A) What is the total energy of this system. (The answer is 300 MeV) (B)If a single particle has this same energy and momentum, what would be its mass?

The answer is 200 MeV, I am having trouble reaching that answer.

Homework Equations



[tex]p=\frac{E}{c}[/tex] for a photon
[tex]p={\gamma}mu[/tex] for a particle with mass
[tex]E=cp[/tex] for a photon
[tex]E={\gamma}mc^{2}[/tex] for a moving particle with mass.

The Attempt at a Solution



Energy of the photon = energy of the moving particle:
[tex]cp={\gamma}mc^{2}[/tex]
[tex]300 MeV={\gamma}mc^{2}[/tex]

Momentum of the photon = momentum of the moving particle:
[tex]\frac{E}{c}={\gamma}mu[/tex]
[tex]\frac{300 MeV}{c}={\gamma}mu[/tex]

This is all I can do so far, after this I get stuck. If I continue, I get the result that u=c. The correct answer is u=.74c
 
Last edited:
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The total momentum of the system of the two photons is not 300 MeV. Momentum is a vector so you have to add the vectors in the correct way.
 

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