6.3 Solutions
337
E 1
∗
= E 2
∗
= (1/4)[(E +
2
− p +
2
+ E −
2
− p −
2
+ 2(E + E − − p + p − cos θ )]
E 1
∗
= E 2
∗
= (1/4)[m
2
+ m
2
+ 2(E + E − − p + p − cos θ )]
= (1/2)(m
2
+ E + E − − p + p − cos θ)
or E 1
∗
= E 2
∗
= [1/2(m
2
+ E + E − − p + p − cos θ )]
1/2
6.13 tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β 1
∗ ) = (1/γ c ) tan θ
∗
/2
( 1 )
(Because β c = β 1
∗ )
Also
tan ϕ = (1/γ c ) tan θ
∗
/2
( 2 )
Where θ
∗ and ϕ
∗ are the corresponding angles in the CM system
Multiply (1) and (2)
tan θ tan ϕ = (1/γ c
2 ). tan θ
∗
/2. tan ϕ
∗
/2
But ϕ
∗
= π − θ
∗ and for m 1 = m 2 , γ c = ((γ + 1)/2)
1/2
tan θ tan ϕ = 2/(γ + 1)
In the classical limit, γ → 1 and tan θ tan ϕ = 1
But since tan(θ + ϕ) = (tan θ + tan ϕ)/(1 − tan θ tan ϕ)
tan(ϕ + θ ) → ∞
i.e. θ + ϕ = π/2
For γ > 1, tan θ tan ϕ < 1. For tan(θ + ϕ) to be finite (θ + ϕ) < π/2
Hence, for γ > 1; θ + ϕ < π/2
6.14 The total energy carried by π
+ in the rest frame of K
+ can be calculated from
E π + = (m K
2
+ m π
2
− m ν
2 )/2m k = 265 MeV
(1)
γ c = γ
∗
π + = 265/139.5 = 1.9
and β c = β π
∗
= (γ
∗2
π + − 1)
1/2
/γ
∗
π +
In the rest frame of pion, the total energy of muon is obtained again by (1)
E μ+
2
= (m μ + 2 + m π + 2 − 0)/2m π + = 110 MeV
γ μ
∗
= 110/106 = 1.0377
β μ
∗
= (γ
∗2
μ − 1)
1/2
/γ μ
∗
= 0.267
The Lorentz factor γ μ for the muon in the LS is obtained from γ μ = γ c γ
∗
μ (1 +
β
∗
μ β c cos θ
∗
μ ) where θ
∗
μ is the emission angle of the muon in the rest frame of
pion. Put θ
∗
μ = 0 to obtain γ μ (max) and θ
∗
μ = 180
◦ to obtain γ μ (min). The
maximum and minimum kinetic energies are 150 and 55.5 MeV.
6.15 First we find the momenta of three pions by using the formula
P 1 = (E
2
1 − m
2 )
1/2 etc, where the total energy, E 1 = T 1 + m, m = 140 MeV is
the pion mass. We need to find the total momentum of the three particles. Take
the direction of the middle particle as x-axis. Calculate these components as
below:
T 1 (MeV) = 190, E 1 = 330(MeV), P 1 = 299 MeV/c,
p 1 (x) = 276.4 MeV/c, p 1 (y) = −114 MeV/c
T 2 = 321, E 2 = 461, P 2 = 439, p 2 (x) = 439, p 2 (y) = 0
T 3 = 58, E 3 = 198, P 3 = 140, p 3 (x) = 137, p 3 (y) = 227
E = 989 MeV,
p(x) = 852.4 MeV/c
P(y) = −84 MeV/c
337
E 1
∗
= E 2
∗
= (1/4)[(E +
2
− p +
2
+ E −
2
− p −
2
+ 2(E + E − − p + p − cos θ )]
E 1
∗
= E 2
∗
= (1/4)[m
2
+ m
2
+ 2(E + E − − p + p − cos θ )]
= (1/2)(m
2
+ E + E − − p + p − cos θ)
or E 1
∗
= E 2
∗
= [1/2(m
2
+ E + E − − p + p − cos θ )]
1/2
6.13 tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β 1
∗ ) = (1/γ c ) tan θ
∗
/2
( 1 )
(Because β c = β 1
∗ )
Also
tan ϕ = (1/γ c ) tan θ
∗
/2
( 2 )
Where θ
∗ and ϕ
∗ are the corresponding angles in the CM system
Multiply (1) and (2)
tan θ tan ϕ = (1/γ c
2 ). tan θ
∗
/2. tan ϕ
∗
/2
But ϕ
∗
= π − θ
∗ and for m 1 = m 2 , γ c = ((γ + 1)/2)
1/2
tan θ tan ϕ = 2/(γ + 1)
In the classical limit, γ → 1 and tan θ tan ϕ = 1
But since tan(θ + ϕ) = (tan θ + tan ϕ)/(1 − tan θ tan ϕ)
tan(ϕ + θ ) → ∞
i.e. θ + ϕ = π/2
For γ > 1, tan θ tan ϕ < 1. For tan(θ + ϕ) to be finite (θ + ϕ) < π/2
Hence, for γ > 1; θ + ϕ < π/2
6.14 The total energy carried by π
+ in the rest frame of K
+ can be calculated from
E π + = (m K
2
+ m π
2
− m ν
2 )/2m k = 265 MeV
(1)
γ c = γ
∗
π + = 265/139.5 = 1.9
and β c = β π
∗
= (γ
∗2
π + − 1)
1/2
/γ
∗
π +
In the rest frame of pion, the total energy of muon is obtained again by (1)
E μ+
2
= (m μ + 2 + m π + 2 − 0)/2m π + = 110 MeV
γ μ
∗
= 110/106 = 1.0377
β μ
∗
= (γ
∗2
μ − 1)
1/2
/γ μ
∗
= 0.267
The Lorentz factor γ μ for the muon in the LS is obtained from γ μ = γ c γ
∗
μ (1 +
β
∗
μ β c cos θ
∗
μ ) where θ
∗
μ is the emission angle of the muon in the rest frame of
pion. Put θ
∗
μ = 0 to obtain γ μ (max) and θ
∗
μ = 180
◦ to obtain γ μ (min). The
maximum and minimum kinetic energies are 150 and 55.5 MeV.
6.15 First we find the momenta of three pions by using the formula
P 1 = (E
2
1 − m
2 )
1/2 etc, where the total energy, E 1 = T 1 + m, m = 140 MeV is
the pion mass. We need to find the total momentum of the three particles. Take
the direction of the middle particle as x-axis. Calculate these components as
below:
T 1 (MeV) = 190, E 1 = 330(MeV), P 1 = 299 MeV/c,
p 1 (x) = 276.4 MeV/c, p 1 (y) = −114 MeV/c
T 2 = 321, E 2 = 461, P 2 = 439, p 2 (x) = 439, p 2 (y) = 0
T 3 = 58, E 3 = 198, P 3 = 140, p 3 (x) = 137, p 3 (y) = 227
E = 989 MeV,
p(x) = 852.4 MeV/c
P(y) = −84 MeV/c