340
6 Special Theory of Relativity
So ct 1 = d + βct 1
Solving for t 1 ,
t 1 = d/c(1 − β)
For β = 0.6, γ = 1.25
t 1 = 5 × 10
8
× 10
3
/3 × 10
8
× 0.4 = 4167 s
6.30 From Lorentz transformations we get
Δt
= γ Δt = γ d 1 /c = 1.25 × 5 × 10
8
× 10
3
/3 × 10
8
= 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 × 10
8 ms
−1
Using these values in (1), we get τ = 1.17 × 10
−8 s
6.32 I = I 0 e
−t/τ
= I 0 e
−γ d/cβτ
γ = 1/0.14 = 7.143, d = 10 m, c = 3 × 10
8 ms
−1
β = (1 − 1/7.143
2 )
1/2
= 0.99, τ = 2.56 × 10
−8 s
I 0 = 10
6
Using the above values, we find I = 83
6.33 (γ − 1)M = M
Or γ = 2
β = (γ
2
− 1)
1/2
/γ = (2
2
− 1)
1/2
/2 =
√
3/2
The dilated time T = γ T 0 = 2 × 2.5 × 10
−8
= 5 × 10
−8 s
The distance traveled before decaying is
d = vT = β cT =
√
3/2 × 3 × 10
8
× 5 × 10
−8
= 13 m
6.34 Time t = d/v = 300/3 × 10
8
= 1.0 × 10
−6 s
As v ≈ c at ultrarelativistic velocity
The proper lifetime is dilated
τ = τ 0 γ = τ 0 E/m = 2.6 × 10
−8
× (200 × 10
3
+ 140)/140
= 3.71 × 10
−3 s
Fraction f of 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 absorption 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 time t they travel a distance d, with velocity v = βc so that t = d/βc
and their mean lifetime is lengthened by the Lorentz factor γ .
(a) The intensity at counter B will be
I B = I A exp (−d/γβcτ )
( 1 )
6 Special Theory of Relativity
So ct 1 = d + βct 1
Solving for t 1 ,
t 1 = d/c(1 − β)
For β = 0.6, γ = 1.25
t 1 = 5 × 10
8
× 10
3
/3 × 10
8
× 0.4 = 4167 s
6.30 From Lorentz transformations we get
Δt
= γ Δt = γ d 1 /c = 1.25 × 5 × 10
8
× 10
3
/3 × 10
8
= 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 × 10
8 ms
−1
Using these values in (1), we get τ = 1.17 × 10
−8 s
6.32 I = I 0 e
−t/τ
= I 0 e
−γ d/cβτ
γ = 1/0.14 = 7.143, d = 10 m, c = 3 × 10
8 ms
−1
β = (1 − 1/7.143
2 )
1/2
= 0.99, τ = 2.56 × 10
−8 s
I 0 = 10
6
Using the above values, we find I = 83
6.33 (γ − 1)M = M
Or γ = 2
β = (γ
2
− 1)
1/2
/γ = (2
2
− 1)
1/2
/2 =
√
3/2
The dilated time T = γ T 0 = 2 × 2.5 × 10
−8
= 5 × 10
−8 s
The distance traveled before decaying is
d = vT = β cT =
√
3/2 × 3 × 10
8
× 5 × 10
−8
= 13 m
6.34 Time t = d/v = 300/3 × 10
8
= 1.0 × 10
−6 s
As v ≈ c at ultrarelativistic velocity
The proper lifetime is dilated
τ = τ 0 γ = τ 0 E/m = 2.6 × 10
−8
× (200 × 10
3
+ 140)/140
= 3.71 × 10
−3 s
Fraction f of 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 absorption 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 time t they travel a distance d, with velocity v = βc so that t = d/βc
and their mean lifetime is lengthened by the Lorentz factor γ .
(a) The intensity at counter B will be
I B = I A exp (−d/γβcτ )
( 1 )