1000 Solved Problems in Modern Physics

340 6 Special Theory of Relativity

Soct 1 =d+βct 1

Solving fort 1 ,

t 1 =d/c(1−β)

Forβ= 0 .6,γ= 1. 25

t 1 = 5 × 108 × 103 / 3 × 108 × 0. 4 =4167 s

6.30 From Lorentz transformations we get
Δt′=γΔt=γd 1 /c= 1. 25 × 5 × 108 × 103 / 3 × 108
= 2 ,083 s

6.31 τ=d/vγ=d/γβc=d/c(γ^2 −1)^1 /^2 (1)
γ=E/m= 1 +(T/m)= 1 +(100/140)= 1. 714
D= 4 .88 m,c= 3 × 108 ms−^1
Using these values in (1), we getτ= 1. 17 × 10 −^8 s

6.32 I=I 0 e−t/τ=I 0 e−γd/cβτ
γ= 1 / 0. 14 = 7. 143 ,d=10 m,c= 3 × 108 ms−^1
β=(1− 1 / 7. 1432 )^1 /^2 = 0. 99 ,τ= 2. 56 × 10 −^8 s
I 0 = 106
Using the above values, we findI= 83

6.33 (γ−1)M=M
Orγ= 2
β=(γ^2 −1)^1 /^2 /γ=(2^2 −1)^1 /^2 / 2 =

√

3 / 2

The dilated timeT=γT 0 = 2 × 2. 5 × 10 −^8 = 5 × 10 −^8 s

The distance traveled before decaying is

d=vT=βcT=

√

3 / 2 × 3 × 108 × 5 × 10 −^8 =13 m

6.34 Timet=d/v= 300 / 3 × 108 = 1. 0 × 10 −^6 s
Asv≈cat ultrarelativistic velocity
The proper lifetime is dilated
τ=τ 0 γ=τ 0 E/m= 2. 6 × 10 −^8 ×(200× 103 +140)/ 140
= 3. 71 × 10 −^3 s
Fractionfof pions decaying is given by the radioactive law
f= 1 −exp(−T/τ)
= 1 −exp(− 0 .0269)
= 0. 027
The pions and muons are subsequently stopped in thick walls of steel and
concrete, pions through their nuclear interactions and muons through absorp-
tion by ionization. The neutrinos being stable, neutral and weakly interacting
will survive.

6.35 Assuming that the pions decay exponentially (the law of radioactivity), then
after timetthey travel a distanced, with velocityv=βcso thatt=d/βc
and their mean lifetime is lengthened by the Lorentz factorγ.
(a) The intensity at counter B will be

IB=IAexp (−d/γβcτ)(1)