1000 Solved Problems in Modern Physics

6.3 Solutions 365

γ=Lx/Lx′=tanθ/tanθ=tan 45◦/tan 30◦=

√

3

β=(γ^2 −1)^1 /^2 /γ=

√

2

3

= 0. 816

The speed at which the rod is moving isv=βc= 0. 816 c

6.3.6 Threshold of Particle Production .....................

6.113 For the production reaction

m 1 +m 2 →m 3 +m 4

The threshold energy form 1 whenm 2 is at rest is

T 1 =[(m 3 +m 4 )^2 −(m 1 +m 2 )^2 ]/ 2 m 2

In the given reaction we can put

m 3 +m 4 = 4 M,m 1 =m 2 =M

T 1 = 6 MorT 1 = 6 Mc^2

6.114 Herem 1 =0,m 2 =m,(m 3 +m 4 )= 3 m

T 1 =^4 mc^2

6.115 Ifm 1 is the projectile mass,m 2 target mass, andm 3 +m 4 +m 5 , the mass of
product particles. The threshold is given by formula
T 1 =[(m 3 +m 4 +m 5 )^2 −(m 1 +m 2 )^2 ]/ 2 m 2
=[(940+ 940 +140)^2 −(940+940)^2 ]/ 2 × 940
= 290 .4MeV

The threshold energy is thus slightly greater than twice the rest-mass

energy of pion (140 MeV). Non-relativistically, the result would be 280 MeV,

that is double the rest mass energy of Pion. The extra energy of 10 MeV is to

be regarded as relativistic correction

6.116 Use the invariance ofE^2 −P^2 =E∗^2 −P∗^2 =E∗^2 − 0 =E∗^2

E∗^2 =(4m)^2 =(T+m+m+ 0 .025)^2 −(P 1 − 0 .218)^2

Puttingm= 0 .938,P 1 =(T^2 + 2 Tm)^1 /^2 and

solving forT, we find thatT(threshold)=4.3 GeV

6.117 Tthreshold=[(m 3 +m 4 )

(^2) −(m
1 +m 2 )
(^2) ]/ 2 m
2
=[(0. 89 + 1 .11)^2 −(0+ 0 .94)^2 ]/ 2 × 0. 94
= 3 .12 GeV