Special Relativity: Photoproduction of Mesons?

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SUMMARY

The discussion focuses on the photoproduction of K+ mesons through the reaction γ + p → K+ + Λ, specifically analyzing the conditions under which either the K+ or the Λ can be at rest in the lab frame. The conservation laws of energy and momentum are central to solving this problem, leading to the equations E^2 = p^2c^2 + m^2c^4 and p = mv. The use of momentum 4-vectors is suggested as an alternative method to simplify calculations, particularly when dealing with multiple momenta.

PREREQUISITES
  • Understanding of conservation laws in physics
  • Familiarity with the equations of special relativity, specifically E^2 = p^2c^2 + m^2c^4
  • Knowledge of momentum and energy conservation principles
  • Basic algebraic manipulation skills for solving equations
NEXT STEPS
  • Study the application of momentum 4-vectors in particle physics
  • Explore detailed examples of energy and momentum conservation in particle reactions
  • Learn about the properties and interactions of K+ mesons and Λ baryons
  • Review advanced algebra techniques for solving systems of equations in physics
USEFUL FOR

Students and educators in physics, particularly those focusing on particle physics and special relativity, as well as researchers interested in meson production and conservation laws.

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Homework Statement


12-16 K + mesons can be photoproduced by the reaction γ + p → K+ + Λ where the initial proton is at rest in the lab. From the conservation laws, discover if it is possible for either the K+ or the Λ to be at rest in the lab, and for what photon energy (in terms of the particle masses) this could happen.

Homework Equations


E^2 = p^2c^2 + m^2c^4
p = mv
E = mc^2

The Attempt at a Solution


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I'm really stuck on this problem. It's probably more of an algebraic answer, because I believe I'm supposed to somehow get the two equations of energy and momentum conservation to have like terms and then combine them somehow, but I'm at a loss as to what to do. Any help is appreciated.
 
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I think you already have the equations.
Conservation of Momentum: \displaystyle{p_\gamma =p_\Lambda \Rightarrow E_\gamma =p_\Lambda c\Rightarrow E_\gamma ^2=E_\Lambda ^2-m_\Lambda ^2c^4}
If you combine this with the conservation of energy you have two equations and two uknowns, \displaystyle{E_\gamma } and \displaystyle{E_\Lambda}.

Another way to solve such problems without algebraic difficulties is by using the momentum 4-vector (also known as 4-momentum). In this problem the calculations are not so tough, because there are only two momenta, so the 4-momentum solution is not much different. But I think it's worth reading about this method, because sometimes it's really helpful.
 

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