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− | <center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}p^4\sin^4{\theta}}\left( p^4\left(\cos^4{\theta}+6\cos^2{\theta}+1 \right)-4E^{*4}p^2\sin^4{\theta}-4E^{*4}\sin^2{\theta}+68E^{*4} \right)</math></center> | + | <center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}p^4\sin^4{\theta}}\left( p^4\left(\cos^4{\theta}+6\cos^2{\theta}+1 \right)-4E^{*4}p^2\sin^4{\theta}-4E^{*4}\sin^2{\theta}+8E^{*4} \right)</math></center> |
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Revision as of 20:28, 30 December 2018
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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
Simplifying 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)
dσdΩ=α28E∗2(16p∗4sin4θ216p∗4cos4θ2+16E∗416p∗4cos4θ2−32E∗44p∗2+16p∗4cos4θ216p∗4sin4θ2+16E∗416p∗4sin4θ2)
dσdΩ=α28E∗2(16p∗4sin4θ216p∗4cos4θ2+16E∗4sin4θ216p∗4cos4θ2sin4θ2−32E∗44p∗2+16p∗4cos4θ216p∗4sin4θ2+16E∗4cos4θ216p∗4sin4θ2cos4θ2)
dσdΩ=α28E∗2(tan4θ2+16E∗4sin4θ2p∗4sin4θ−32E∗44p∗2+cot4θ2+16E∗4cos4θ2p∗4sin4θ)
dσdΩ=α28E∗2(tan4θ2+cot4θ2−32E∗44p∗2+16E∗4cos4θ2p∗4sin4θ+16E∗4sin4θ2p∗4sin4θ)
dσdΩ=α28E∗2(tan4θ2+cot4θ2−32E∗44p∗2+4E∗4(cos2θ+3)p∗4sin4θ)
dσdΩ=α28E∗2(tan4θ2+cot4θ2−32E∗44p∗2+4E∗4cos2θp∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2(tan4θ2+cot4θ2−8E∗4p∗2+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2(sin4θ(cosθ+1)4+sin4θ(cosθ−1)4−8E∗4p∗2+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2(sin4θ(cosθ−1)4(cosθ+1)4(cosθ−1)4+sin4θ(cosθ+1)4(cosθ−1)4(cosθ+1)4−8E∗4p∗2+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2(sin4θ(cosθ−1)4sin8θ+sin4θ(cosθ+1)4sin8θ−8E∗4p∗2+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2((cosθ−1)4sin4θ+(cosθ+1)4sin4θ−8E∗4p∗2+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α28E∗2((2cos4θ+12cos2θ+2)sin4θ−8E∗4sin4θp∗2sin4θ+4E∗4(cos2θ−sin2θ)p∗4sin4θ+12E∗4p∗4sin4θ)
dσdΩ=α24E∗2sin4θ((cos4θ+6cos2θ+1)1−4E∗4sin4θp∗2+2E∗4(cos2θ−sin2θ)p∗4+6E∗4p∗4)
dσdΩ=α24E∗2sin4θ(p4(cos4θ+6cos2θ+1)p4−4E∗4p2sin4θp∗4+2E∗4(cos2θ−sin2θ)p∗4+6E∗4p∗4)
dσdΩ=α24E∗2p4sin4θ(p4(cos4θ+6cos2θ+1)−4E∗4p2sin4θ+2E∗4(cos2θ−sin2θ)+6E∗4)
dσdΩ=α24E∗2p4sin4θ(p4(cos4θ+6cos2θ+1)−4E∗4p2sin4θ+2E∗4(1−2sin2θ)+6E∗4)
dσdΩ=α24E∗2p4sin4θ(p4(cos4θ+6cos2θ+1)−4E∗4p2sin4θ−4E∗4sin2θ+8E∗4)
∴E2≡m2+p2
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