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Differential Cross-Section
dσdΩ=164π2spfinalpinitial|M|2
Working in the center of mass frame
pfinal=pinitial
Determining the scattering amplitude in the center of mass frame
M=e2(u−st+t−su)
M2=e4(u−st+t−su)(u−st+t−su)
M2=e4((u−s)2t2+(t−s)2u2+2(u−s)t(t−s)u)
M2=e4((u2−2us+s2)t2+(t2−2ts+s2)u2+2(ut−st+s2−us)tu)
M2=e4((t2+s2)u2−2s2tu+(u2+s2)t2)
Using the fine structure constant (with c=ℏ=ϵ0=1)
α≡e24π
dσdΩ=α22s((t2+s2)u2−2s2tu+(u2+s2)t2)
In the center of mass frame the Mandelstam variables are given by:
s≡4E∗2
t≡−2p∗2(1−cosθ)=−2p∗2(1−2cos2θ2+1)=−4p∗2(1−2cos2θ2)=−4p∗2sin2θ2
u≡−2p∗2(1+cosθ)=−2p∗2(1+2cos2θ2−1)=−4p∗2cos2θ2
Where the using the relationship
cosθ=−1+cosθ2
dσdΩ=α28E∗2(16p∗4sin4θ2+16E∗416p∗4cos4θ2−32E∗44p∗2sin2θ24p∗2cos2θ2+16p∗4cos4θ2+16E∗416p∗4sin4θ2)
dσdΩ=α28E∗2(16p∗4sin4θ2+16E∗416p∗4cos4θ2−32E∗44p∗2(sin2θ2+cos2θ2)+16p∗4cos4θ2+16E∗416p∗4sin4θ2)
\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left( \frac{16p^{*4}\sin^4{\frac{\theta}{2}}}{16p^{*4}\cos^4{\frac{\theta}{2}}}+\frac{16E^{*4}}{16p^{*4}\cos^4{\frac{\theta}{2}}}-\frac{32E^{*4}}{4p^{*2}}+\frac{16p^{*4}\cos^4{\frac{\theta}{2}}{16p^{*4}\sin^4{\frac{\theta}{2}}}+\frac{16E^{*4}}{16p^{*4}\sin^4{\frac{\theta}{2}}}\right )
∴E2≡m2+p2
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