6.3 Solutions 359
Note that if theπ^0 s were to decay at rest (γ=1) then the rectangle would
have reduced to a spike atE= 67 .5 MeV, half of rest energy ofπ^0.6.101 Theγ-rays of intensityI(θ∗) which are emitted in the solid angle dΩ∗in
the CMS will appear in the solid angle dΩ∗in the LS with intensityI(θ).
Therefore
I(θ)dΩ=I(θ∗)dΩ∗
or
I(θ)=I(θ∗)sinθ∗dθ∗/sinθdθ (10)Fig. 6.15γ-ray energy
spectrum fromπ^0 decay in
cosmic ray events
From the Lorentz transformation
E∗=γE(1−βcosθ)
=γE∗γ(1+cosθ∗)(1−cosθ)
Where we have used (2)
1 /γ^2 (1−βcosθ)= 1 +βcosθ∗Differentiating
−βsinθdθ/γ^2 (1−βcosθ)^2 =−βsinθ∗dθ∗
Therefore sinθ∗dθ∗/sinθdθ= 1 /γ^2 (1−βcosθ)^2 (11)
AlsoI(θ∗)= 1 / 4 π (12)
because of assumption of isotropy of photons in the rest frame ofπ^0
Combining (10), (11) and (12)
I(θ)= 1 / 4 πγ^2 (1−βcosθ)^2 (13)