1000 Solved Problems in Modern Physics

360 6 Special Theory of Relativity

This shows that small emission angles of photons in the lab system are

favored.

6.102 In Fig. 6.16, AB and AD represent the momentum vectors of the two photons
in the LS. BC is drawn parallel to AD so that ABC forms the momentum
triangle, that is

AB+BC=AC

Fig. 6.16Locus of the tip of
the momentum vector of
γ-rays fromπ^0 decay is an
ellipse

From energy conservation

hν 1 +hν 2 =γmc^2 /2(1+βcosθ∗)+(γmc^2 /2)(1+βcos(π−θ∗))

=γmc^2 =const

where we have used the fact that the angles of emission of the two photons

in the rest frame ofπ^0 , are supplementary.

Since momentum is given byp=h/c,itfollowsthatAB+BC=constant,

which means that the locus of the tip of the momentum vector is an ellipse.

E=mc^2 /^2 γ(1−βcosθ) (14)

Compare this with the standard equation for ellipse

r=a(1−ε^2 )/(1−εcosθ)

We findε=β

a(1−β^2 )=mc^2 /^2 γ

Ora=γmc^2 / 2

The larger the velocity ofπ^0 , the greater will be the eccentricity,ε.