6.3 Solutions
365
γ = L x /L x
= tan θ/ tan θ = tan 45
◦
/ tan 30
◦
=
√
3
β = (γ
2
− 1)
1/2
/γ =
2
3
= 0.816
The speed at which the rod is moving is v = βc = 0.816 c
6.3.6 Threshold of Particle Production
6.113 For the production reaction
m 1 + m 2 → m 3 + m 4
The threshold energy for m 1 when m 2 is at rest is
T 1 = [(m 3 + m 4 )
2
− (m 1 + m 2 )
2 ]/2m 2
In the given reaction we can put
m 3 + m 4 = 4M, m 1 = m 2 = M
T 1 = 6M or T 1 = 6Mc
2
6.114 Here m 1 = 0, m 2 = m, (m 3 + m 4 ) = 3m
T 1 = 4mc
2
6.115 If m 1 is the projectile mass, m 2 target mass, and m 3 + m 4 + m 5 , the mass of
product particles. The threshold is given by formula
T 1 = [(m 3 + m 4 + m 5 )
2
− (m 1 + m 2 )
2 ]/2m 2
= [(940 + 940 + 140)
2
− (940 + 940)
2 ]/2 × 940
= 290.4 MeV
The threshold energy is thus slightly greater than twice the rest-mass
energy of pion (140 MeV). Non-relativistically, the result would be 280 MeV,
that is double the rest mass energy of Pion. The extra energy of 10 MeV is to
be regarded as relativistic correction
6.116 Use the invariance of E
2
− P
2
= E
∗2
− P
∗2
= E
∗2
− 0 = E
∗2
E
∗2
= (4m)
2
= (T + m + m + 0.025)
2
− (P 1 − 0.218)
2
Putting m = 0.938, P 1 = (T
2
+ 2T m)
1/2 and
solving for T , we find that T (threshold) = 4.3 GeV
6.117 T threshold = [(m 3 + m 4 )
2
− (m 1 + m 2 )
2 ]/2m 2
= [(0.89 + 1.11)
2
− (0 + 0.94)
2 ]/2 × 0.94
= 3.12 GeV