362
6 Special Theory of Relativity
The minimum angle is found by setting dθ/dD = 0
This gives us D = 1, that is E 1 = E 2 = E/2.
θ min = 2mc
2
/E
The Lorentz transformation of angles gives us the relation
E = [m γ /2][1 + β cos θ
∗ )
We need to consider one of the photons in the forward hemisphere. The
fraction of photons emitted in the CMS (rest frame of π
0 ) within the angle θ
∗
is (1 − cos θ
∗ ). This fraction is 1/2 for θ
∗
= 60
0 , that is cos θ
∗
= 1/2. When
one photon goes at θ
∗
= 60
0 , the other photon will go at θ
∗
= 120
0 with the
direction of flight of π
0 . Hence cos θ
∗
= −1/2 for the second photon. The
disparity factor
D = E 2 /E 1 = (1 + β/2)/(1 − β/2) = (2 + β)/(2 − β).
For relativistic pions β = 1 Hence D > 1.
A quarter of pions will be emitted within an angle θ
∗
= 41.4
0 , that is
cos θ
∗
= 0.75. In this case
D = (1 + 3β/4)/(1 − 3β/4)
And the previous argument gives us D > 7
6.106 First find E
∗ the total energy available in the CMS
E
∗2
= (E π + m p )
2
− P
2
π ≈ (P π + m p )
2
− P π
2
(Because E π m π )
E
∗
= 4.436 GeV
Total energy carried by K
∗ in the CMS
E k
∗
= (E
∗2
+ m k
∗2
− m γ 0 )/2E
∗
= 1.942 GeV
γ K
∗
= E K
∗
/m K
∗
= 1.942/0.89 = 2.18
β K
∗
= 0.8888
γ c = (γ + ν)/(1 + 2γ ν + ν
2 )
1/2
γ = 10/0.14 = 71.4
ν = m 2 /m 1 = 0.940/0.140 = 6.71
γ c = 2.466, β c = 0.9141
tan θ = sin θ
∗
/(cos θ
∗
+ β c /β
∗ )
( 1 )
Differentiate with respect to θ
∗ and set
∂ tan θ/∂θ
∗
= 0.This gives cos θ
∗
= −β
∗
/β c
cos θ
∗
= −0.8888/0.9141 = −0.9723
θ
∗
= 166.5
0
Using the values of θ
∗
, γ c , and the ratio β c /β
∗ in (1) we find θ m = 59.3
◦