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| − | <center><math>\frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4{\theta}}</math></center> | + | <center><math>\frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4=\frac{8E^{*4}\left(1-\cos^2{\theta}\right)}{p^{*4}\sin^4{\theta}}</math></center> |
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Revision as of 16:33, 1 January 2019
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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+2−2ust2−2st−2tsu2−2su+(u2+s2)t2)
Using the fine structure constant (with c=ℏ=ϵ0=1)
α≡e24π
dσdΩ=α22s((t2+s2)u2+2s2tu+2−2ust2−2st−2tsu2−2su+(u2+s2)t2)
In the center of mass frame the Mandelstam variables are given by:
s≡4E∗2
t≡−2p∗2(1−cosθ)
u≡−2p∗2(1+cosθ)
Calculating the parts to have common denominators:
2=2p∗4sin4θp∗4sin4θ
t2u2=4p∗2(1−cosθ)24p∗2(1+cosθ)2=tan4θ2=p∗4(1−cosθ)4p∗4sin4θ
s2u2=16E∗44p∗4(1+cosθ)2=E∗4sec4θ2p∗4=4E∗4p∗4(1+cosθ)2=4E∗4(1−cosθ)2p∗4(1+cosθ)2(1−cosθ)2=4E∗4(cos2θ−2cosθ+1)p∗4sin4θ
u2t2=4p∗2(1+cosθ)24p∗2(1−cosθ)2=cot4θ2=p∗4(1+cosθ)4p∗4sin4θ=p∗4(1+cosθ)4p∗4(1−cos4θ)
s2t2=16E∗44p∗4(1−cosθ)2=E∗4csc4θ2p∗4=4E∗4p∗4(1−cosθ)2=4E∗4(1+cosθ)2p∗4(1−cosθ)2(1+cosθ)2=4E∗4(cos2θ+2cosθ+1)p∗4sin4θ
\frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4=\frac{8E^{*4}\left(1-\cos^2{\theta}\right)}{p^{*4}\sin^4{\theta}}
−2st=8E∗22p∗2(1−cosθ)=4E∗2p∗2(1−cosθ)=4E∗2(1+cosθ)p∗2(1−cosθ)(1+cosθ)=4E∗2(1+cosθ)p∗2sin2θ=4E∗2p∗2sin2θ(1+cosθ)p∗4sin4θ
−2su=8E∗22p∗2(1+cosθ)=4E∗2p∗2(1+cosθ)=4E∗2(1−cosθ)p∗2(1+cosθ)(1−cosθ)=4E∗2(1−cosθ)p∗2sin2θ=4E∗2p∗2sin2θ(1−cosθ)p∗4sin4θ
−2tsu2=4p∗2(1−cosθ)4E∗24p∗2(1+cosθ)2=E∗2(1−cosθ)sec4θ2p∗2=E∗2p∗2(1−cosθ)p∗4(1+cosθ)2=E∗2p∗2(1−cosθ)(1−cosθ)2p∗4(1+cosθ)2(1−cosθ)2=E∗2p∗2(−cos3θ+3cos2θ−3cosθ+1)p∗4sin4θ
−2ust2=4p∗2(1+cosθ)4E∗24p∗2(1−cosθ)2=E∗2(1+cosθ)csc4θ2p∗2=E∗2p∗2(1+cosθ)p∗4(1−cosθ)2=E∗2p∗2(1+cosθ)(1+cosθ)2p∗4(1−cosθ)2(1+cosθ)2=E∗2p∗2(cos3θ+3cos2θ+3cosθ+1)p∗4sin4θ
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