Energy of Kaon in Lab Frame: Relativistic Collision

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SUMMARY

The discussion focuses on calculating the energy of a kaon in the lab frame, specifically a kaon with a mass of 498 MeV/c² decaying into two pions, each with a mass of 137 MeV/c². One pion is produced at rest, leading to the application of conservation of energy and momentum principles. The participants emphasize the use of four-vectors to analyze the momentum and energy relationships, particularly noting that the stationary pion has a three-momentum of zero. The solution involves equating the total energy of the kaon to the sum of the energies of the two pions, factoring in the kaon's kinetic energy.

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  • Understanding of relativistic energy-momentum relations
  • Familiarity with four-vectors in physics
  • Knowledge of conservation laws in particle physics
  • Basic concepts of particle decay and rest mass energy
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  • Learn about conservation of energy and momentum in particle decay scenarios
  • Explore detailed examples of kaon decay processes
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Homework Statement


kaon of ss 498 MeV/c^2 traveling through laboratory decays into two pions each of mass 137MeV/c^2. Onee of the pions is produced at rest in the lab frame. What is the energy of the kaon in the lab frame?


Homework Equations

I think you need to consider 4 vectors for the momentum. Since one pion is stationary its 3 momentum is 0 right?
Not really sure how to work this out!
Do you equates
P = p1 +p2 where these are 4 vectors
gamma mkaon c^2 = gamma2 mpion c^2 + 0

Any suggestions of how to continue would be appreciated!

the energy of the second pion would be it's rest mass plus the deltamass times c^2 plus a factor which equals the kaon's kinetic energy. I think...



The Attempt at a Solution

 
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Just need to use conservation of energy and momentum. Then use the common expression involving energy and momentum.
 
Last edited:

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