6.2 Problems
329
6.97 Assume the decay K
0
→ π
+
+π
− . Calculate the mass of the primary particle
if the momentum of each of the secondary particles is 3 × 10
8 eV and the
angle between the tracks is 70
◦
[University of Durham 1960]
6.98 Neutral pions of fixed energy decay in flight into two γ -rays. Show that the
velocity of pion is given by
β = (E max − E min )/(E max + E min )
where E is the γ -ray energy in the laboratory
6.99 In Problem 6.98 show that the rest mass energy of π
0 is given by mc
2
=
2(E max E min )
1/2
6.100 In Problem 6.98 show that the energy distribution of γ -rays in the laboratory
is uniform under the assumption that γ -rays are emitted isotropically in the
rest system of π
0
6.101 In Problem 6.98 show that the angular distribution of γ -rays in the laboratory
is given by
I (θ ) = 1/4πγ
2 (1 − β cos θ)
2
6.102 In Problem 6.98 show that the locus of the tip of the momentum vector is an
ellipse
6.103 In Problem 6.98 show that in a given decay the angle φ between two γ -rays
is given by
sin(φ/2) = mc
2
/2(E 1 E 2 )
1/2
6.104 In Problem 6.98 show that the minimum angle between the two γ -rays is
given by
φ min = 2mc
2
/E π
6.105 In Problem 6.98 find an expression for the disparity D (the ratio of energies)
of the γ -rays and show that D > 3 in half the decays and D > 7 in one
quarter of them
6.106 In the interaction π
−
+ p → K
∗ (890) + Y 0
∗ (1, 800) at pion momentum
10 GeV/c , K
∗ is produced at an angle θ in the lab system. Calculate the
maximum value θ m , given m π = 0.140 GeV/c
2 and m p = 0.940 GeV/c
2
6.107 A particle of mass m 1 travelling with a velocity v = βc collides elastically
with the particle m 2 at rest. The scattering angles of m 1 in the LS and CMS
are θ and θ
∗ . Show that
(a) γ c = (γ + ν)/(1 + 2γ ν + ν
2 )
1/2
(b) γ
∗
= (γ + 1/γ )/
√
(1 + 2γ /ν + 1/ν
2 )
(c) tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
(d) tan θ
∗
= sin θ/γ c (cos θ − β c /β
∗ )
where β c is the CMS velocity, β
∗ c is the velocity of m 1 in CMS,
γ c = (1 − β c
2 )
−1/2
, γ
∗
= (1 − β
∗2 )
−1/2
, ν = m 2 /m 1
329
6.97 Assume the decay K
0
→ π
+
+π
− . Calculate the mass of the primary particle
if the momentum of each of the secondary particles is 3 × 10
8 eV and the
angle between the tracks is 70
◦
[University of Durham 1960]
6.98 Neutral pions of fixed energy decay in flight into two γ -rays. Show that the
velocity of pion is given by
β = (E max − E min )/(E max + E min )
where E is the γ -ray energy in the laboratory
6.99 In Problem 6.98 show that the rest mass energy of π
0 is given by mc
2
=
2(E max E min )
1/2
6.100 In Problem 6.98 show that the energy distribution of γ -rays in the laboratory
is uniform under the assumption that γ -rays are emitted isotropically in the
rest system of π
0
6.101 In Problem 6.98 show that the angular distribution of γ -rays in the laboratory
is given by
I (θ ) = 1/4πγ
2 (1 − β cos θ)
2
6.102 In Problem 6.98 show that the locus of the tip of the momentum vector is an
ellipse
6.103 In Problem 6.98 show that in a given decay the angle φ between two γ -rays
is given by
sin(φ/2) = mc
2
/2(E 1 E 2 )
1/2
6.104 In Problem 6.98 show that the minimum angle between the two γ -rays is
given by
φ min = 2mc
2
/E π
6.105 In Problem 6.98 find an expression for the disparity D (the ratio of energies)
of the γ -rays and show that D > 3 in half the decays and D > 7 in one
quarter of them
6.106 In the interaction π
−
+ p → K
∗ (890) + Y 0
∗ (1, 800) at pion momentum
10 GeV/c , K
∗ is produced at an angle θ in the lab system. Calculate the
maximum value θ m , given m π = 0.140 GeV/c
2 and m p = 0.940 GeV/c
2
6.107 A particle of mass m 1 travelling with a velocity v = βc collides elastically
with the particle m 2 at rest. The scattering angles of m 1 in the LS and CMS
are θ and θ
∗ . Show that
(a) γ c = (γ + ν)/(1 + 2γ ν + ν
2 )
1/2
(b) γ
∗
= (γ + 1/γ )/
√
(1 + 2γ /ν + 1/ν
2 )
(c) tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
(d) tan θ
∗
= sin θ/γ c (cos θ − β c /β
∗ )
where β c is the CMS velocity, β
∗ c is the velocity of m 1 in CMS,
γ c = (1 − β c
2 )
−1/2
, γ
∗
= (1 − β
∗2 )
−1/2
, ν = m 2 /m 1