358 6 Special Theory of Relativity
From (3) and (4),(Emax−Emin)/(Emax+Emin)=β (5a)From the measurement ofEmaxandEmin, the velocity ofπ^0 can be deter-
mined.
6.99 In the solution of Problem 6.98 multiply (3) and (4) and writeE=γmπc^2
mπc^2 =2(EmaxEmin)^1 /^2 (5b)
From the measurement ofEmaxandEmin,massofπ^0 can be determined.
ItEmax=75 MeV andEmin=60 MeV, thenmc^2 = 2 ×(75×60)^1 /^2 =
134 .16 MeV
Hence the mass ofπ^0 is 134. 16 / 0. 51 = 262. 5 me
6.100dN/dE=(dN/dΩ∗).dΩ∗/dE=(1/ 4 π).2sinθ∗dθ∗/dE
=(1/2)d cosθ∗/dE (6)
where we have put dΩ∗= 2 πsinθ∗dθ∗for the element of solid angle and
dN/dΩ∗=^1 /^4 πunder the assumption of isotropy.
Differentiating (2) with respect to cosθ∗
dE/d cosθ∗=γβmc^2 / 2
or d cosθ∗/2dE= 1 /γβmc^2 (7)
Combining (6) and (7), the normalized distribution is
dN/dE= 1 /γβmc^2 =constant (8)
This implies that the energy spectrum is rectangular or uniform. It extends
from a minimum to maximum, Fig. 6.14.
From (3) and (4),
Emax−Emin=βEπ=γβmc^2 (9)
Note that the area of the rectangle is height×length
(dN/dE)×(Emax−Emin)=(1/γβmc^2 )×γβmc^2 = 1
That is, the distribution is normalized as it should.
The higher theπ^0 energy the larger is the spread in theγ-ray energy spec-
trum. For mono-energetic source ofπ^0 s, we will have a rectangular distribu-
tion ofγ-ray energy as in Fig. 6.14. But if theγ-rays are observed fromπ^0 s,
of varying energy, as in cosmic ray events the rectangular distributions may
be superimposed so that the resultant distributions may look like the solid
curve, shown in Fig. 6.15.Fig. 6.14γ-ray energy
spectrum fromπ^0 decay at
fixed energy