6.3 Solutions
361
6.103 The angle between the two γ -rays in the LS can be found fom the formula
m
2 c
4
= m 1
2 c
4
+ m 2
2 c
4
+ 2(E 1 E 2 − c
2 p 1 p 2 cos ϕ)
Putting m 1 = m 2 = 0, cp 1 = E 1 , cp 2 = E 2
m
2 c
4
= 2E 1 E 2 (1 − cos ϕ) = 4E 1 E 2 sin
2 (ϕ/2)
sin (ϕ/2) = mc
2
/2(E 1 E 2 )
1/2
(15)
6.104 For small angle ϕ,
ϕ = mc
2
/(E 1 E 2 )
1/2
(16)
Set E 2 = γ mc
2
− E 1
ϕ = mc
2
/[E 1 (γ mc
2
− E 1 )]
1/2
(17)
For minimum angle d ϕ/dE 1 = 0. This gives E 1 = γ mc
2
/2.
Using this value of E 1 in (17), we obtain
ϕ min = 2/γ
(18)
Measurement of ϕ min affords the determination of E π via γ .
ϕ min = 2mc
2
/E π
6.105 E 1 + E 2 = E (energy conservation)
(1)
p 1 + p 2 = p (momentum conservation)
(2)
Taking the scalar product
(p 1 + p 2 ).(p 1 + p 2 ) = p.p
or
p 1
2
+ p 2
2
+ 2 p 1 p 2 cos θ = p
2
(3)
Using c = 1, Eq. (3) becomes
E 1
2
+ E 2
2
+ 2E 1 E 2 cos θ = E
2
− m
2
(4)
Let E 2 /E 1 = D,
(5)
the disparity factor. Then (1) becomes
E 1 (D + 1) = E
(6)
Combining (4), (5) and (6)
2DE
2 (1 − cos θ) = m
2
Or sin θ/2 = [m/2E][
√
D + 1/
√
D]
For small θ ,
θ = [m/E][
√
D + 1
√
D]