Astroparticle Physics

42 3 Kinematics and Cross Sectionsdecay kinematicsπ 0 decayFor m µ = 105.658 369MeV and m π ± = 139.570 18MeV one gets Eµ kin = E µ − m µ = 4.09 MeV. Forthe two-body decay of the kaon, K + → µ + + ν µ ,(3.36) gives Eµ kin = E µ − m µ = 152.49 MeV (m K ± =493.677 MeV).Due to helicity conservation the decay π + → e + + ν eis strongly suppressed (see Fig. 2.10). Using (3.36) thepositron would get in this decay a kinetic energy of) 2= E e + −m e = m π2+ m2 e2m π−m e = m π1 − m em π ≈69.3 MeV, which is approximately half the pion mass.This is not a surprise, since the ‘heavy’ pion decays intotwo nearly massless particles.Example 8: The decay π 0 → γ + γE kine +2(The kinematics of the π 0 decay at rest is extremelysimple. Each decay photon gets as energy one half ofthe pion rest mass. In this example also the decay of aπ 0 in flight will be considered. If the photon is emittedin the direction of flight of the π 0 , it will get ahigher energy compared with the emission opposite tothe flight direction. The decay of a π 0 in flight (Lorentzfactor γ = E π 0/m π 0) yields a flat spectrum of photonsbetween a maximum and minimum energy. Fourmomentumconservationq π 0 = q γ1 + q γ2leads toq 2 π 0 = m 2 π 0 = q 2 γ 1+ q 2 γ 2+ 2q γ1q γ2. (3.37)Since the masses of real photons are zero, the kinematiclimits are obtained from the relation2q γ1q γ2= m 2 π 0 . (3.38)In the limit of maximum or minimum energy transfer tothe photons they are emitted parallel or antiparallel tothe direction of flight of the π 0 .Thisleadstop γ1 ‖−p γ2 . (3.39)Using this, (3.38) can be expressed as2(E γ1 E γ2 − p γ1 · p γ2 ) = 4E γ1 E γ2 = m 2 π 0 . (3.40)