348
6 Special Theory of Relativity
6.64 (a) π
+
→ μ
+
+ ν μ
T = Q(Q + 2m ν )/2(m μ + m ν + Q) = (m π − m μ )
2
/2m π
(Q = m π − m μ − m ν = m π − m μ and m ν = 0)
m μ = 206.9m e = 206.9 × 0.511 = 105.7 MeV
Therefore,
T μ = 4.12 = (m π − 105.7)
2
/2m π
Solving for m π
m π = 141.39 MeV/c
2
= 141.39/0.511m e
= 276.7m e
(b) μ
+
→ e
+
+ ν e + ν μ
T max for electron is obtained when ν e and ν μ fly together in opposite direction. Thus, the three-body problem is reduced to a two-body one.
m μ c
2
= 206.9 × 0.511 = 105.72 MeV
Q = 105.72 − 0.51 = 105.21 MeV
T e (max) = Q
2
/2(m e + Q) = (105.72)
2
/2(0.51 + 105.72) = 52.6 MeV
6.65 Q = 938.2 + 1,875.5 − (939.5 + 135.0) − 2.2 = 1,737 MeV
T π = Q(Q + 2m n )/2(Q + m π + m n )
= 1,737(1,737 + 2 × 939.5)/2(1, 737 + 135 + 939.5) = 1, 117 MeV
Total energy E π = 1, 117 + 135 = 1, 252 MeV
6.66 M
2
= m 1
2
+ m 2
2
+ 2(E 1 E 2 − P 1 P 2 cos θ)
( 1 )
m 1 = 966 m e = 966 × 0.511 MeV = 493.6 MeV
(2)
m 2 = 273 m e = 273 × 0.511 MeV = 139.5 MeV
(3)
p 1 = 394 MeV, p 2 = 254 MeV
(4)
E 1 = ( p 1
2
+ m 1
2 )
1/2
= [(394)
2
+ (93.6)
2 ]
1/2
= 631.6
( 5 )
E 2 = ( p 2
2
+ m 2
2 )
1/2
= [(254)
2
+ (139.5)
2 ]
1/2
= 289.5
( 6 )
cos θ = cos 154
◦
= −0.898
(7)
Using (2) to (7) in (1), M = 899.4 MeV
6.67 Using the results of Problem 6.60, the angle between the outgoing particles
after the collision is given by
cos θ = T /(T + 4M)
Here T = 940 MeV = M
Therefore, cos θ = 0.2 → θ = 78.46
◦
6.68 Using the invariance, E
2
− |
p|
2
= E
∗2
E
∗2
= (E + E f )
2
− ( p
2
+ p
2
f + 2 pp f cos θ)
= (E + E f )
2
− (E
2
− M
2
+ E
2
f − M
2
+ 2 pp f cos θ
= 2M
2
+ 2(E E f − p.p f )
(a) For parallel momenta, θ = 0, p.p f = +pp f
(b) For anti-parallel momenta θ = π, p.p f = −pp f
(c) For orthogonal momenta θ = π/2, p.p f = 0
6.69 By problem 6.13 it is sufficient to show that tan θ tan ϕ = 2/(γ + 1)
p = γβm = (γ
2
− 1)
1/2 m