Difference between revisions of "Differential Cross-Section"

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<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left( \frac{\sin^4{\frac{\theta}{2}}+1}{\cos^4{\frac{\theta}{2}}}-\frac{2}{\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}}+\frac{\cos^4{\frac{\theta}{2}}+1}{\sin^4{\frac{\theta}{2}}}\right )</math></center>
 
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left( \frac{\sin^4{\frac{\theta}{2}}+1}{\cos^4{\frac{\theta}{2}}}-\frac{2}{\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}}+\frac{\cos^4{\frac{\theta}{2}}+1}{\sin^4{\frac{\theta}{2}}}\right )</math></center>
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[[File:DiffXSect_Solution.pdf]]

Revision as of 14:38, 28 March 2018

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(ust+tsu)


M2=e4(ust+tsu)(ust+tsu)


M2=e4((us)2t2+(ts)2u2+2(us)t(ts)u)


M2=e4((u22us+s2)t2+(t22ts+s2)u2+2(utst+s2us)tu)


M2=e4((t2+s2)u22s2tu+(u2+s2)t2)


Using the fine structure constant (with c==ϵ0=1)

αe24π


dσdΩ=α22s((t2+s2)u22s2tu+(u2+s2)t2)


In the center of mass frame the Mandelstam variables are given by:

s4E2


Using the relationship

cosθ=1+cosθ2



In the ultra-relativistic limit, the electron mass is small enough compared to the energy such that it can be neglected when compared to the momentum


mp


E2m2+p2E2p2



t2E2(1cosθ)=2E2(12cos2θ2+1)=4E2(12cos2θ2)=4E2sin2θ2


u2E2(1+cosθ)=2E2(1+2cos2θ21)=4E2cos2θ2


dσdΩ=α28E2(16E4sin4θ2+16E416E4cos4θ232E44E2sin2θ24E2cos2θ2+16E4cos4θ2+16E416E4sin4θ2)


dσdΩ=α28E2(sin4θ2+1cos4θ22sin2θ2cos2θ2+cos4θ2+1sin4θ2)


File:DiffXSect Solution.pdf