1000 Solved Problems in Modern Physics

362 6 Special Theory of Relativity

The minimum angle is found by setting dθ/dD= 0

This gives usD=1, that isE 1 =E 2 =E/2.

θmin= 2 mc^2 /E

The Lorentz transformation of angles gives us the relation

E=[mγ/2][1+βcosθ∗)

We need to consider one of the photons in the forward hemisphere. The

fraction of photons emitted in the CMS (rest frame ofπ^0 ) within the angleθ∗

is (1−cosθ∗). This fraction is 1/2 forθ∗= 600 , that is cosθ∗= 1 /2. When

one photon goes atθ∗= 600 , the other photon will go atθ∗= 1200 with the

direction of flight ofπ^0. Hence cosθ∗=− 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 pionsβ=1 HenceD>1.

A quarter of pions will be emitted within an angleθ∗= 41. 40 , that is

cosθ∗= 0 .75. In this case

D=(1+ 3 β/4)/(1− 3 β/4)

And the previous argument gives usD> 7

6.106 First findE∗the total energy available in the CMS

E∗^2 =(Eπ+mp)^2 −Pπ^2 ≈(Pπ+mp)^2 −Pπ^2 (BecauseEπmπ)

E∗= 4 .436 GeV

Total energy carried byK∗in the CMS

Ek∗=(E∗^2 +mk∗^2 −mγ^0 )/^2 E∗=^1 .942 GeV

γK∗=EK∗/mK∗= 1. 942 / 0. 89 = 2. 18

βK∗= 0. 8888

γc=(γ+ν)/(1+ 2 γν+ν^2 )^1 /^2

γ= 10 / 0. 14 = 71. 4

ν=m 2 /m 1 = 0. 940 / 0. 140 = 6. 71

γc= 2. 466 ,βc= 0. 9141

tanθ=sinθ∗/(cosθ∗+βc/β∗)(1)

Differentiate with respect toθ∗and set

∂tanθ/∂θ∗= 0 .This gives cosθ∗=−β∗/βc

cosθ∗=− 0. 8888 / 0. 9141 =− 0. 9723

θ∗= 166. 50

Using the values ofθ∗,γc, and the ratioβc/β∗in (1) we findθm= 59. 3 ◦