6.3 Solutions 351
Because of (2) and (5)
p′e=p=E/c
(b)β=v/c=(1− 1 /γ^2 )^1 /^2 =(1−m^2 /Ee^2 )^1 /^2
=(1−m^2 /(pe^2 +m^2 ))^1 /^2 =(pe^2 /(pe^2 +m^2 ))^1 /^2
=(1+(mc^2 /E)^2 )−^1 /^2
wherePe=Eandc=1.6.3.4 Invariance Principle ................................
6.76 Referring to Fig. 7.8, letPγandPγ′be the four-momentum vectors of photon
before and after the scattering, respectively,PeandPe′the four vectors of
electron before and after scattering respectively. Form the scalar product of
the four-vector of the photon and the four-vector of photon+electron. Since
this scalar product is invariant.
Pγ.(Pγ+Pe)=Pγ′.(Pγ′+Pe′)(1)Further, total four-momentum is conserved.Pγ+Pe=Pγ′+Pe′ (2)
Now,Pγ=[hν, 0 , 0 ,ihν](3)
Pe=[0, 0 , 0 ,imc] (4)
Pγ′=[hν′cosθ,hν′sinθ, 0 ,ihν]wheremis the rest mass of the electron.
Using (1) to (4)
[hν, 0 , 0 ,ihν].[hν, 0 , 0 ,i(hν+mc^2 )]
=[hν′cosθ,hν′sinθ, 0 ,ihν].[hν, 0 , 0 ,i(hν+mc^2 )]
Therefore,h^2 ν^2 −hν(hν+mc^2 )=h^2 νν′cosθ−hν′(hν+mc^2 )
Simplifying
hν′ν(1−cosθ)=mc^2 (ν−ν′)
Orh/mc(1−cosθ)=c(1/ν′− 1 /ν)=λ′−λ
OrΔλ=λ′−λ=(h/mc)(1−cosθ)
This is the well known formula for Compton shift in wavelength (formula
7.37).6.77 Let the initial and final four momenta of the electron bePi=(Ei/c,pi) and
Pf=(Ef,pf), respectively. The squared four-momentum transfer is defined by
Q^2 =(Pi−Pf)^2 =− 2 m^2 c^2 +2 EiEf
c^2− 2 Pi·Pf
However,Ei=Ef=Eand|pi|=|pf|=E/c; so neglecting the electron
mass
Q^2 = 2 E^2 (1−cosθ)/c^2