Astroparticle Physics

342 17 Solutionsor2m e (E e + + m e ) ≥ m 2 ZE e + ≥ m2 Z2m e− m e ≈ 8.1 × 10 15 eV = 8.1PeV.17.3 Chapter 31. Similarly to Problem 2.4, for the reaction γ + γ → µ + + µ − the threshold condition is(q γ1 + q γ2 ) 2 ≥ (2m µ ) 2 .For photons qγ 2 1= qγ 2 2= 0. Then(q γ1 + q γ2 ) 2 = 2(E γ1 E γ2 − p γ1 · p γ2 ) = 4E γ1 E γ2for the angle π between the photon 3-momenta. Finally,E γ1 ≥ (m µc 2 ) 2≈ 1.1 × 10 19 eV .E γ22. The number of collisions is obtained by subtracting the number of unaffected particlesfrom the initial number of particles:N = N 0 − N = N 0 (1 − e −x/λ ).For thin targets x/λ ≪ 1 and the expansion in the Taylor series givesN = N 0 (1 − (1 − x/λ +···)) = N 0xλ = N 0N A σ N x.For the numerical example one getsN = 10 8 × 6.022 × 10 23 g −1 × 10 −24 cm 2 × 0.1g/cm 2 ≈ 6 × 10 6 .3. The threshold condition is(q¯νe + q p ) 2 ≥ (m n + m e ) 2 .Taking into account that q 2¯ν e= 0andq p = (m p , 0), one obtainsandm 2 p + 2E¯ν em p ≥ (m n + m e ) 2E¯νe ≥ (m n + m e ) 2 − m 2 p2m p≈ 1.8MeV,where the following values of the particles involved were used: m n = 939.565 MeV,m p = 938.272 MeV, m e = 0.511 MeV.