... energy is Eph1max = Eoe/ (1 – k). (6.48)If, for example, the incident total positron energy is 30 MeV, and Ee0 = 0. 511 MeVthen Eph1max = 0. 511 / [1 – (29.489/30. 511 ) 1/ 2] MeV = 30.25 ... E20). (6.33)Using T 1 = γ 1 E 1 0 – E 1 0, where γ 1 = (1 – β 1 2) 1/ 2 and β 1 = v 1 /c, we have E30 2 = E 1 0 2 + E20 2 + 2γ 1 E 1 0E20 . (6.34)If two identical particles ... invariant.Now, sc2 = E 1 2 + 2E 1 E2 + E22 – (p 1 2 + 2p 1 ⋅p2 + p22)c2 = E 1 0 2 + E20 2 + 2E 1 E2 – 2p 1 ⋅p2c2 = E 1 0 2 + E20 2 + 2(E 1 , p 1 c)[E2, –p2c]....