356 6 Special Theory of Relativity
6.91 (a)λ/λ′=[(1+β)/(1−β)]^1 /^2 = 656 / 434 = 1. 5115
v=βc= 1. 17 × 108 ms−^1
(b) the nebula is receding
6.92λ/λ′=[(1+β)/(1−β)]^1 /^2 =(1+β)^1 /^2 (1−β)−^1 /^2
≈(1+β/ 2 +...)(1+β/ 2 −...)
= 1 +β+...(neglecting higher order terms)
Δλ/λ′=(λ/λ′)− 1 =β=v/c6.93λ/λ′=[(1+β)/(1−β)]^1 /^2 = 670 / 525
β= 0. 239
v=βc= 0. 239 × 3 × 108 = 7. 17 × 107 ms−^1
= 7. 17 × 104 kms−^1
This speed exceeds the escape velocity. Hence the explanation is not valid.
6.94λ′=λ[(1−β)/(1+β)]^1 /^2 = 589 .3[(1− 0 .21)/(1+ 0 .21)]^1 /^2 = 476 .2nm
The color is blue6.95 The source velocity is perpendicular to the line of sight.θ= 90 ◦,
ν′=νγ
λ=γλ′
γ= 1 /(1−β^2 )^1 /^2 = 1 /(1− 0. 052 )^1 /^2 = 1. 00125
Δλ=λ−λ′=λ′(γ−1)=589(1. 00125 −1)
= 0 .736 nm= 7. 36 A ̊6.96 Let the electron recoil at angleφwith momentump, and neutrino get scat-
tered with energyE′and momentump′.
Energy conservation gives
E 0 +m=E+E′ (1)
From the momentum triangle
p′^2 =p 02 +p^2 − 2 p 0 pcosφ (2)
We can writep′=E′,p=E,p 0 =E 0 , so that (2) becomes
E′^2 =E 02 +E^2 − 2 E 0 Ecosφ (3)Fig. 6.13Collision of an energetic neutrino with a stationary electron