1000 Solved Problems in Modern Physics

6.3 Solutions 357

From (1) we have

E′^2 =(E 0 −E+m)^2 (4)

Comparing (3) and (4) and simplifying

cosφ= 1 −m(E 0 −E−m)/E 0 E≈ 1 −m(E 0 −E)/E 0 E (5)

where we have neglectedmin comparison withE 0 −E.

For small angle,(5) becomesφ=[2m(E 0 −E)/E 0 E]^1 /^2 (6)

ForE 0 =2GeV,E= 0 .5GeV,m= 0. 51 × 10 −^3 GeV

φ= 0 .039 radians= 2. 24 ◦

6.97 Let the mass of the primary particle beM, and that of secondary particlesm 1

andm 2. Let the total energy of the secondary particles in the LS beE 1 and

E 2 , and momentap 1 andp 2. Using the invariance of (total energy)^2 −(total

momentum)^2

M^2 =(E 1 +E 2 )^2 −(p 12 +p 22 + 2 p 1 p 2 cosθ)

=E 12 −p 12 +E 22 −p 22 +2(E^21 −p 12 cosθ)

=m 12 +m 22 +2(E 12 −p 12 cosθ) (SinceE 1 =E 2 ,p 1 =p 2 )

= 2 m 12 +2(m 12 +p 12 −p 12 cosθ)

= 4 m 12 + 4 p 12 sin^2 θ/ 2

=4(140)^2 +4(300)^2 sin^235 ◦

= 196 , 836

M=444 MeV/c^2

6.98 Consider one of the twoγ-rays. From Lorentz transformation

cpx=γc(cpx∗+βcE∗)

where the energy and momentum refer to one of the twoγ-rays and the

subscriptCrefers toπ^0. Starred quantities refer to the rest system ofπ^0.

cpcosθ=γ(cp∗cosθ∗+βE∗)

where we have dropped off the subscriptC.Butforγ-rayscp∗=E∗and

cp=E, and because the twoγ-rays share equal energy in the CMS,E∗=

mc^2 /2, wheremis the rest mass ofπ◦.

Thereforecpcosθ=(γmc^2 /2)(β+cosθ∗)(1)

Also,E=γ(E∗+βcpx∗)=γ(E∗+βcp∗cosθ∗)

or

cp=E=(γmc^2 /2)(1+βcosθ∗)(2)

Whenθ∗= 0

Emax=

1

2

Eπ◦(1+β)(3)

Whenθ∗=π

Emin=

1

2

Eπ◦(1−β)(4)