6.3 Solutions
353
But γ = 10/0.14
Therefore, ϕ min = 0.028 rad or 1.6
◦
6.83 Rest mass energy of ω
0
= Total available energy – (total kinetic energy +
mass energy of π
+ and π
− )
m ω c
2
= 2.29 − (1.22 + 0.14 + 0.14) = 0.79 GeV
6.84 Energy conservation gives
m 1 γ 1 + m 2 = M γ
(1)
Momentum conservation gives
m 1 γ 1 β 1 = M γ β
(2)
Squaring (1)
m 1
2
γ 1
2
+ m 2
2
+ 2γ 1 m 1 m 2 = M
2
γ
2
(3)
Squaring (2)
m 1
2
γ 1
2
β 1
2
= M
2
γ
2
β
2
(4)
Using β 1 = (1 − 1/γ 1
2 )
1/2 and β = (1 − 1/γ
2 )
1/2
(4) becomes
m 1
2 (γ 1
2
− 1) = M
2 (γ
2
− 1)
(5)
Subtracting (5) from (3)
m 1
2
+ m 2
2
+ 2m 1 m 2 /(1 − v
2
/c
2 )
1/2
= M
2
6.85 E 0 = E p + E π (energy conservation)
(1)
Q = m 0 − (m p + m π )
( 2 )
P 0
2
= P p
2
+ P π
2
+ 2P p P π cos θ
(3)
Or E 0
2
− m 0
2
= E p
2
− m p
2
+ E π
2
− m π
2
+ 2P p P π cos θ
(4)
Using (1) in (4) and simplifying
2E p E π − 2P p P π cos θ + m p
2
+ m π
2
= m 0
2
= (Q + m p + m π )
2
(5)
Or Q = (m p
2
+ m π
2
+ 2E p E π − 2P p P π cos θ )
1/2
− (m p + m π )
( 6 )