130_notes.dvi

We solve the problem using only conservation of energy and momentum. Lets work inunits in
whichc= 1 for now. We’ll put thecback in at the end. Assume the photon is initially moving in
thezdirection with energy E and that it scatters in theyzplane so thatpx= 0.Conservation of
momentumgives
E=E′cosθ+pecosφ

and
E′sinθ=pesinφ.

Conservation of energygives

E+m=E′+

√

p^2 e+m^2

Our goal is to solve forE′in terms of cosθso lets make sure we eliminate theφ. Continuing from
the energy equation
E−E′+m=

√

p^2 e+m^2

squaring and calculatingp^2 efrom the components

E^2 +E′^2 +m^2 − 2 EE′+ 2mE− 2 mE′= (E−E′cosθ)^2 + (E′sinθ)^2 +m^2

and writing out the squares on the right side

E^2 +E′^2 +m^2 − 2 EE′+ 2mE− 2 mE′=E^2 +E′^2 − 2 EE′cosθ+m^2

and removing things that appear on both sides

− 2 EE′+ 2mE− 2 mE′=− 2 EE′cosθ

and grouping

m(E−E′) = EE′(1−cosθ)

(E−E′)

EE′

=

(1−cosθ)

m

1

E′

−

1

E

=

(1−cosθ)

m