6.3 Solutions
347
Fig. 6.7 Symmetrical elastic
collision between identical
particles
6.61 (γ − 1)mc
2
= mc
2
γ = 2
β = (1 − 1/γ
2 )
1/2
=
√
3/2
p e = mγβc
cp e = mc
2
γβ = mc
2
× 2 ×
√
3/2 = mc
2
×
√
3 MeV
p γ = p e =
√
3E 0 MeV/c
6.62 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
Δ ABC.
180
◦
− θ = 60
◦
Therefore, θ = 120
◦
Thus the paths of any two leptons are equally inclined to 120
◦
Fig. 6.8 Decay of a muon at
rest into three leptons whose
masses are neglected
6.63 By Problem 6.53, T = 2mc
2
β
2 cos
2
θ/[1 − β
2 cos
2
θ ]
( 1 )
T = 10
9 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
Using mc
2
= 0.511 MeV, β = 0.875 and θ = 3
◦ in (1),
We find T = 3.3 MeV
347
Fig. 6.7 Symmetrical elastic
collision between identical
particles
6.61 (γ − 1)mc
2
= mc
2
γ = 2
β = (1 − 1/γ
2 )
1/2
=
√
3/2
p e = mγβc
cp e = mc
2
γβ = mc
2
× 2 ×
√
3/2 = mc
2
×
√
3 MeV
p γ = p e =
√
3E 0 MeV/c
6.62 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
Δ ABC.
180
◦
− θ = 60
◦
Therefore, θ = 120
◦
Thus the paths of any two leptons are equally inclined to 120
◦
Fig. 6.8 Decay of a muon at
rest into three leptons whose
masses are neglected
6.63 By Problem 6.53, T = 2mc
2
β
2 cos
2
θ/[1 − β
2 cos
2
θ ]
( 1 )
T = 10
9 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
Using mc
2
= 0.511 MeV, β = 0.875 and θ = 3
◦ in (1),
We find T = 3.3 MeV