What Is the Minimum Beam Energy for Proton-Proton Collisions?

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SUMMARY

The minimum beam energy required for proton-proton collisions to initiate the p+p→p+p+n0 reaction is calculated to be 1.212 GeV. This value is derived from the rest energy of the neutral pion (n0) at 547.3 MeV and the rest energy of two protons, each at 938.3 MeV. The calculation confirms that the total energy must account for both the rest energy of the protons and the additional energy required to produce the neutral pion. The method used involves conservation of energy principles in particle physics.

PREREQUISITES
  • Understanding of particle physics and conservation of energy principles
  • Familiarity with relativistic energy equations
  • Knowledge of rest mass energies of protons and neutral pions
  • Basic proficiency in high-energy physics calculations
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  • Study the principles of relativistic energy and momentum conservation in particle collisions
  • Learn about the rest mass energies of various subatomic particles
  • Explore the concept of center of mass frames in high-energy physics
  • Investigate the implications of beam energy in collider experiments
USEFUL FOR

This discussion is beneficial for physics students, particle physicists, and researchers involved in high-energy collider experiments, particularly those focusing on proton-proton interactions and particle production mechanisms.

J_M_R
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Homework Statement



Calculate the minimum beam energy in a proton-proton collider to initiate the p+p→p+p+n0 reaction. The rest energy of the no is 547.3MeV.

Homework Equations



For a head on collision between particles a and b, from conservation of total energy: Ex = Ep + Ep' = 2Ep

so that mx^2c^4 = Ex^2 - px^2c^2 = (2Ep)^2

The Attempt at a Solution


[/B]
Rearranging the above, gives mx = 2Ep/c^2

So using the values I have been provided: Ep = (547.3MeV)/2 = 0.2737Gev

I am not sure if my method is correct?
 
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I guess "beam energy" means the total energy, including the rest energy of the protons.
For the additional energy: yes it is correct. The lab frame is also the center of mass frame, so both protons simply contribute half of the n0 energy plus their rest energy.
 
So to get Ep, I am also required to add the rest energy of the two protons?:

Ep = (547.3MeV + 2(938.3MeV))/2 = 1.212GeV

If rest energy of a proton is 938.3Mev.
 
Right.
 
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