1000 Solved Problems in Modern Physics

6.3 Solutions 337

E 1 ∗=E 2 ∗=(1/4)[(E+^2 −p+^2 +E−^2 −p−^2 +2(E+E−−p+p−cosθ)]

E 1 ∗=E 2 ∗=(1/4)[m^2 +m^2 +2(E+E−−p+p−cosθ)]

=(1/2)(m^2 +E+E−−p+p−cosθ)

orE 1 ∗=E 2 ∗=[1/2(m^2 +E+E−−p+p−cosθ)]^1 /^2

6.13 tanθ=sinθ∗/γc(cosθ∗+βc/β 1 ∗)=(1/γc)tanθ∗/2(1)

(Becauseβc=β 1 ∗)

Also

tanφ=(1/γc)tanθ∗/2(2)

Whereθ∗andφ∗are the corresponding angles in the CM system

Multiply (1) and (2)

tanθtanφ=(1/γc^2 ).tanθ∗/ 2 .tanφ∗/ 2

Butφ∗=π−θ∗and form 1 =m 2 ,γc=((γ+1)/2)^1 /^2

tanθtanφ= 2 /(γ+1)

In the classical limit,γ→1 and tanθtanφ= 1

But since tan(θ+φ)=(tanθ+tanφ)/(1−tanθtanφ)

tan(φ+θ)→∞

i.e.θ+φ=π/ 2

Forγ>1, tanθtanφ<1. For tan(θ+φ) to be finite (θ+φ)<π/ 2

Hence, forγ>1;θ+φ<π/ 2

6.14 The total energy carried byπ+in the rest frame ofK+can be calculated from
Eπ+=(mK^2 +mπ^2 −mν^2 )/ 2 mk=265 MeV (1)
γc=γπ∗+= 265 / 139. 5 = 1. 9
andβc=βπ∗=(γπ∗^2 +−1)^1 /^2 /γπ∗+
In the rest frame of pion, the total energy of muon is obtained again by (1)
Eμ+^2 =(mμ+^2 +mπ+^2 −0)/ 2 mπ+=110 MeV
γμ∗= 110 / 106 = 1. 0377
βμ∗=(γμ∗^2 −1)^1 /^2 /γμ∗= 0. 267
The Lorentz factorγμfor the muon in the LS is obtained fromγμ=γcγμ∗(1+
βμ∗βccosθμ∗) whereθμ∗is the emission angle of the muon in the rest frame of
pion. Putθμ∗=0 to obtainγμ(max) andθμ∗= 180 ◦to obtainγμ(min). The
maximum and minimum kinetic energies are 150 and 55.5 MeV.

6.15 First we find the momenta of three pions by using the formula
P 1 =(E^21 −m^2 )^1 /^2 etc, where the total energy,E 1 =T 1 +m,m=140 MeV is
the pion mass. We need to find the total momentum of the three particles. Take
the direction of the middle particle asx-axis. Calculate these components as
below:
T 1 (MeV)=190,E 1 =330(MeV),P 1 =299 MeV/c,
p 1 (x)= 276 .4MeV/c,p 1 (y)=−114 MeV/c
T 2 =321,E 2 =461,P 2 =439,p 2 (x)=439,p 2 (y)= 0
∑T^3 =58,E^3 =198,P^3 =140,p^3 (x)=137,p^3 (y)=^227
E=989 MeV,

∑

∑ p(x)=^852 .4MeV/c

P(y)=−84 MeV/c