6.3 Solutions 347
Fig. 6.7Symmetrical elastic
collision between identical
particles
6.61 (γ−1)mc^2 =mc^2
γ= 2
β=(1− 1 /γ^2 )^1 /^2 =
√
3 / 2
pe=mγβc
cpe=mc^2 γβ=mc^2 × 2 ×√
3 / 2 =mc^2 ×√
3MeV
pγ=pe=√
3 E 0 MeV/c6.62 For the three particles the energies (total) are equal.
E 1 =E 2 =E 3 (3)
(masses are neglected)
The magnitude of momenta are also equal
p 1 =p 2 =p 3
The momenta represented by the three vectors AC, CB and BA form the closed
ΔABC.
180 ◦−θ= 60 ◦
Therefore,θ= 120 ◦
Thus the paths of any two leptons are equally inclined to 120◦
Fig. 6.8Decay of a muon at
rest into three leptons whose
masses are neglected
6.63 By Problem 6.53,T= 2 mc^2 β^2 cos^2 θ/1−β^2 cos^2 θ
T= 109 eV= 1 ,000 MeV
γ= 1 +T/M= 1 + 1 , 000 / 940 = 2. 0638
β=[1−(1/γ^2 )]^1 /^2 =[1−(1/ 2 .0638)^2 ]^1 /^2 = 0. 875
Usingmc^2 = 0 .511 MeV,β= 0 .875 andθ= 3 ◦in (1),
We findT= 3 .3MeV