6.3 Solutions
359
Note that if the π
0 s were to decay at rest (γ = 1) then the rectangle would
have reduced to a spike at E = 67.5 MeV, half of rest energy of π
0 .
6.101 The γ -rays of intensity I (θ
∗ ) which are emitted in the solid angle d Ω
∗ in
the CMS will appear in the solid angle dΩ
∗ in the LS with intensity I (θ ).
Therefore
I (θ )d Ω = I (θ
∗ )d Ω
∗
or
I (θ ) = I (θ
∗ ) sin θ
∗ dθ
∗
/ sin θdθ
(10)
Fig. 6.15 γ -ray energy
spectrum from π
0 decay in
cosmic ray events
From the Lorentz transformation
E
∗
= γ E(1 − β cos θ)
= γ E
∗
γ (1 + cos θ
∗ )(1 − cos θ)
Where we have used (2)
1/γ
2 (1 − β cos θ) = 1 + β cos θ
∗
Differentiating
− β sin θ dθ/γ
2 (1 − β cos θ)
2
= −β sin θ
∗ dθ
∗
Therefore sin θ
∗ dθ
∗
/ sin θdθ = 1/γ
2 (1 − β cos θ )
2
(11)
Also I (θ
∗ ) = 1/4π
(12)
because of assumption of isotropy of photons in the rest frame of π
0
Combining (10), (11) and (12)
I (θ ) = 1/4πγ
2 (1 − β cos θ)
2
(13)