Relativistic Proton-Photon Collision and GZK Cutoff · USAPhO 2017 #A4
Physics
Problem
Relativistic particles obey the mass energy relation
E2=(pc)2+(mc2)2
where E is the relativistic energy of the particle, p is the relativistic momentum, m is the mass, and c is the speed of light.
A proton with mass mp and energy Ep collides head on with a photon which is massless and has energy Eb. The two combine and form a new particle with mass m∆ called Δ, or "delta". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.
a. Determine Ep in terms of mp, m∆, and Eb. You may assume that Eb is small.
b. In this case, the photon energy Eb is that of the cosmic background radiation, which is an EM wave with wavelength 1.06 mm. Determine the energy of the photons, writing your answer in electron volts.
c. Assuming this value for Eb, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by mpc² = 938 MeV and the mass of the Δ is given by m∆c² = 1232 MeV.
The following relationships may be useful in solving this problem:
velocity parameter β = v/c
Lorentz factor γ = 1/√(1−β²)
relativistic momentum p = γβmc
relativistic energy E = γmc²
relativistic doppler shift f/f₀ = √((1−β)/(1+β))