Astroparticle Physics

358 17 SolutionsN 1 (> 2GeV) = a∫ ∞2ε −2.7 dε =−(a/1.7)ε −1.7∣ ∣∞= 0.181 a.With full field on, only particles with E > 〈E〉 = 12 GeV will reach sea level forlatitudes between 0 ◦ and 50 ◦ :N 2 (> 12 GeV) ≈ 0.0086 a.These results have to be combined with the surface of the Earth that is affected. Assumingisotropic incidence and zero field, the rate of cosmic rays at sea level is proportionalto the surface of the Earth,Φ 1 = AN 1 = 4πR 2 N 1 .With full field on, only the part of the surface of Earth is affected for which 0 ◦ ≤ λ ≤50 ◦ . The relevant surfaces can be calculated from elementary geometry yielding a fluxΦ 2 = A(50 ◦ –90 ◦ )N 1 + A(0 ◦ –50 ◦ )N 2 ,whereA(50 ◦ –90 ◦ ) = 2π[(R cos 50 ◦ ) 2 + (R(1 − sin 50 ◦ )) 2]andA(0 ◦ –50 ◦ ) = 4πR 2 − A(50 ◦ –90 ◦ ).This crude estimate leads to Φ 1 /Φ 2 = 3.7, i.e., in periods of transition when the magneticfield went through zero, the radiation load due to cosmic rays was higher by abouta factor of 4.8. In principle, neutrons are excellent candidates. Because they are neutral, they wouldpoint back to the sources of cosmic rays. Their deflection by inhomogeneous magneticfields is negligible. The only problem is their lifetime, τ 0 = 885.7 seconds. At 10 20 eVthe Lorentz factor extends this lifetime considerably toτ = γτ 0 = 1020 eVm n c 2 τ 0 = 9.4 × 10 13 s ̂= 2.99 × 10 6 light-years = 0.916 Mpc .Still the sources would have to be very near (compare λ γp ≈ 10 Mpc), and there is noevidence for point sources of this energy in the close vicinity of the Milky Way.217.8 Chapter 81. Due to the relative motion of the galaxy away from the observer, the energy of the photonappears decreased. The energies of emitted photons and observed photons are related bythe Lorentz transformation,E em = γE obs + γβcp ‖obs = γ(1 + β)E obs .