Astroparticle Physics

3.2 Four-Vectors 43Because of E γ2 = E π 0−E γ1 (3.40) leads to the quadraticequationEγ 2 1− E γ1 E π 0 + m2 π 0= 0 (3.41)4with the symmetric solutionsEγ max1= 1 2 (E π 0 + p π 0),Eγ min1= 1 (3.42)2 (E π 0 − p π 0).photon spectrumfrom π 0 decayBecause of E π 0 = γm π 0 and p π 0 = γm π 0β (3.42) canalso be expressed asE maxγ 1E minγ 1= 1 2 γm π 0(1 + β) = 1 2 m π 0 √1 + β1 − β ,= 1 2 γm π 0(1 − β) = 1 2 m π 0 √1 − β1 + β . (3.43)In the relativistic limit (γ ≫ 1, β ≈ 1) a photon emittedin the direction of flight of the π 0 gets the energyEγmax = E π 0 = γm π 0 and the energy of the backwardemittedphoton is zero.From (3.43) it is clear that for any energy of a neutralpion, a range of possible photon energies containsm π 0/2. If one has a spectrum of neutral pions, the energyspectra of the decay photons are superimposed insuch a way that the resulting spectrum has a maximumat half the π 0 mass.Much more difficult is the treatment of a three-body decay.Such a process is going to be explained for the exampleof the muon decay:three-body decayµ − → e − +¯ν e + ν µ . (3.44)Let us assume that the muon is originally at rest (E µ = m µ ).Four-momentum conservationq µ = q e + q¯νe + q νµ (3.45)can be rephrased as(q µ − q e ) 2 = (q¯νe + q νµ ) 2 ,