366
6 Special Theory of Relativity
6.118 T threshold = [(m p + m p + m p )
2
− (m p + m p )
2 ]/2m p by Eq. (6.53)
m p = 1,837 m e = 1,837 × 0.00.00051 GeV = 0.937 GeV
M = 273 × 0.00051 GeV = 0.137 GeV
Using the above values we find T threshold = 0.167 GeV
6.119 T threshold = (m k + m Λ )
2
− (m p + m p )
2
/2m p
m k = 0.498 GeV, m Λ = 1.115 GeV, m π = 0.140 GeV, m p = 0.938 GeV
Using these values, we find T threshold = 0.767 GeV = 767 MeV
Note that when pions are used as bombarding particles the threshold for
strange particle production is lowered then in N–N collisions. However, first
a beam of pions must be produced in N–N collisions.
6.120 Consider the reaction P + P → P + P + nπ
T threshold = [(m p + m p + nm π )
2
− (m p + m p )
2 ]/2m p
Simplifying we get the desired result
6.121 T threshold = [(m p + m π0 )
2
− (m p + 0)
2 ]/2m p
Using m p = 940 MeV and m π0 = 135 MeV, we
find T threshold = 145 MeV
Note that the threshold energy for pion production in collision with gamma
rays is only half of that for N–N collisions. But the cross-section is down by
two orders of magnitude as the interaction is electromagnetic.
6.122 T threshold = [(m Ξ− + m k + m k0 )
2
− (m π− + m p )
2 ]/2m p
= [(1,321 + 494 + 498)
2
− (140 + 938)
2 ]/2 × 938
= 2,233 MeV
Note that for Ξ production, the threshold is much higher than that for
−
production as it has to be produced in association with two other strange
particles (see Chaps.9 and 10).
6.123 Using the invariance, E
2
− |
p|
2
= E
∗2
− |
p
∗
|
2
At threshold: (E + M p )
2
− E ν
2
= (M p + M μ + M w )
2
− 0
(5 + 0.938)
2
− 5
2
= (0.938 + 0.106 + M w )
2
M w = 2.16 GeV
Since the reaction does not proceed, M w > 2.16 GeV
6.124 T thr = [(m p + m Λ + m k )
2
− (2m p )
2 ]/2m p
= [(0.938 + 1.115 + 0.494)
2
− (2 × 0.938)
2 ]/2 × 0.938
= 1.58 GeV
6 Special Theory of Relativity
6.118 T threshold = [(m p + m p + m p )
2
− (m p + m p )
2 ]/2m p by Eq. (6.53)
m p = 1,837 m e = 1,837 × 0.00.00051 GeV = 0.937 GeV
M = 273 × 0.00051 GeV = 0.137 GeV
Using the above values we find T threshold = 0.167 GeV
6.119 T threshold = (m k + m Λ )
2
− (m p + m p )
2
/2m p
m k = 0.498 GeV, m Λ = 1.115 GeV, m π = 0.140 GeV, m p = 0.938 GeV
Using these values, we find T threshold = 0.767 GeV = 767 MeV
Note that when pions are used as bombarding particles the threshold for
strange particle production is lowered then in N–N collisions. However, first
a beam of pions must be produced in N–N collisions.
6.120 Consider the reaction P + P → P + P + nπ
T threshold = [(m p + m p + nm π )
2
− (m p + m p )
2 ]/2m p
Simplifying we get the desired result
6.121 T threshold = [(m p + m π0 )
2
− (m p + 0)
2 ]/2m p
Using m p = 940 MeV and m π0 = 135 MeV, we
find T threshold = 145 MeV
Note that the threshold energy for pion production in collision with gamma
rays is only half of that for N–N collisions. But the cross-section is down by
two orders of magnitude as the interaction is electromagnetic.
6.122 T threshold = [(m Ξ− + m k + m k0 )
2
− (m π− + m p )
2 ]/2m p
= [(1,321 + 494 + 498)
2
− (140 + 938)
2 ]/2 × 938
= 2,233 MeV
Note that for Ξ production, the threshold is much higher than that for
−
production as it has to be produced in association with two other strange
particles (see Chaps.9 and 10).
6.123 Using the invariance, E
2
− |
p|
2
= E
∗2
− |
p
∗
|
2
At threshold: (E + M p )
2
− E ν
2
= (M p + M μ + M w )
2
− 0
(5 + 0.938)
2
− 5
2
= (0.938 + 0.106 + M w )
2
M w = 2.16 GeV
Since the reaction does not proceed, M w > 2.16 GeV
6.124 T thr = [(m p + m Λ + m k )
2
− (2m p )
2 ]/2m p
= [(0.938 + 1.115 + 0.494)
2
− (2 × 0.938)
2 ]/2 × 0.938
= 1.58 GeV