1000 Solved Problems in Modern Physics

342 6 Special Theory of Relativity

6.38 (a)β=v/c=

√

3

2

γ=(1−β^2 )−^1 /^2 =(1− 3 /4)−^1 /^2 = 2

Total energy of the particle

E=γMc^2 = 2 Mc^2

(b) Distance traveled on an average

d=γβct 0 = 2 ×

√ 3

2 cτ=

√

3 cτ

(c) The sphere will shrink in the direction of motion but will not in the trans-

verse direction. Consequently, its shape would appear as that of a spheroid

as shown in Fig. 6.5

Fig. 6.5

6.39 ν′= 1 −νu−ν/uc 2
putv=c
ν′= 1 −cuc−u/c 2 = 1 c−−uu/c=c

6.3.3 Mass,Momentum,Energy ..........................

6.40 (a)m=m 0 γ

γ=τ/τ 0 = 6. 6 × 10 −^6 / 2. 2 × 10 −^6 = 3

m= 3 × 207 =621 me

(b)T=(γ−1)m 0 c^2 =(3−1)(207× 0 .51)

=211 MeV

(c) Total energyE=mc^2 = 621 mec^2 = 621 × 0. 511 = 3 .173 MeV

p=βE/c

β=(γ^2 −1)^1 /^2 /γ=(3^2 −1)^1 /^2 / 3 = 0. 9428

p=(0.9428)(317.3)/c

=299 MeV/c

6.41 E=m 0 c^2 =(1×^10 −^3 kg)(3×^108 )^2 =^9 ×^1013 J

6.42 T=(γ−1)m=m

γ= 2

β=(γ^2 −1)^1 /^2 /γ= 0. 866

The result is independent of the mass of the particle.