Astroparticle Physics

36 3 Kinematics and Cross SectionsE 2 − p 2 c 2 = m2 0 c41 − β 2 (1 − β2 ) = m 2 0 c4 . (3.6)This result shows that E 2 −p 2 c 2 is a Lorentz-invariant quan-tity. This quantity is the same in all systems and it equals thesquare of the rest energy. Consequently, the total energy ofa relativistic particle can be expressed by√E = c p 2 + m 2 0 c2 . (3.7)This equation holds for all particles. For massless particlesor, more precisely, particles with rest mass zero, one obtainsE = cp . (3.8)Particles of total energy E without rest mass are also subjectto gravitation, because they acquire a mass according tom = E/c 2 . (3.9)The transition from relativistic kinematics to classical(Newtonian) mechanics (p ≪ m 0 c) can also be derivedfrom (3.7) by series expansion. The kinetic energy of a particleis obtained to√E kin = E − m 0 c 2 = c p 2 + m 2 0 c2 − m 0 c 2invariant massmass equivalentclassical approximationE 2 − p 2 c 2 = γ 2 m 2 0 c4 − γ 2 m 2 0 β2 c 4can be written as( ) p 2= m 0 c√1 2 + − m 0 c 2m 0 c(≈ m 0 c 2 1 + 1 ( ) )p2− m 0 c 22 m 0 c= p22m 0= 1 2 m 0v 2 , (3.10)in accordance with classical mechanics. Using (3.4) and(3.5), the velocity can be expressed byorv =p = c2 pγm 0 Eβ = cp E . (3.11)In relativistic kinematics it is usual to set c = 1. This simplifiesall formulae. If, however, numerical quantities haveto be calculated, the actual value of the velocity of light hasto be considered.