6.3 Solutions 341
and at the counter C
Ic=IAexp (− 2 d/γβcτ)(2)∴Ic=I^2 B
IA
=
(470)^2
1000
= 221
(b) Take logarithm on both sides of (1) and simplify.
τ=d
γβcln(IA/IB)(3)
γ= 1 +T/mc^2 = 1 + 140 / 140 = 2. 0
β=(γ^2 −1)^1 /^2 /γ= 0. 866
d=10 m;c= 3 × 108 m/s
ln(IA/IB)=ln(1000/470)= 0. 755
Substituting the above values in (3),
τ= 2. 55 × 10 −^8 s
The accepted value is 2. 6 × 10 −^8 s6.36 The stationary object will appear to move with velocity−βctoward the
observer. The object moving with velocityαctoward the stationary object
would appear to have velocity
(αc−βc)/(1−αβ), as seen by the observer. If these two velocities are to be
equal then (αc−βc)/(1−αβ)=βc
Cross multiplying and simplifying we get the quadratic equation whose
solution isβ=[1−(1−α^2 )^1 /^2 ]/α
6.37
(i) t=
L
βc(1)
N 2 =N 1 exp[−t/γτ]=N 1 exp[
−
L
γβcτ]
(2)
Therefore (2) becomes
exp[
L
γβcτ]
=N 1 /N 2
Take logarithm on both sides
L
γβcτ=ln(
N 1
N 2)
Butγβ=√
γ^2 − 1
Thereforeτ= L
ln(N
N^1
2)√
γ^2 − 1 c(ii) γ= 1 /(1−β^2 )^1 /^2 = 1 /(1− 8 /9)^1 /^2 = 3
τ=200
ln(
10 , 000
8 , 983)√
32 − 1 × 3 × 108
= 2. 2 × 10 −^6 s.