Moller Differential Cross Section
Using the equation from [1]
dσdΩ′1=e48E∗2{1+cos4(θ∗2)sin4(θ∗2)+1+sin4(θ∗2)cos4(θ∗2)+2sin2(θ∗2)cos2(θ∗2)}
where α=e2ℏcwithℏ=c=1 and θ∗=θ∗1=θ∗2
This can be simplified to the form
dσdΩ′1=α24E∗2(3+cos2θ∗)2sin4θ∗
Plugging in the values expected for 2 scattering electrons:
α2=5.3279×10−5
E∗≈106.031MeV
Using unit analysis on the term outside the parantheses, we find that the differential cross section for an electron at this momentum should be around
5.3279×10−54×1.124×1016eV2=1.18×10−21eV−2=1.18×10−211eV2×1×10181×1018=.0012GeV2
Using the conversion of
11GeV2=.3894mb
.00121GeV2=.0012111GeV2=.0012×.3894mb=.467×10−3mb
We find that the differential cross section scale is dσdΩ≈.5×10−3mb=.5μb
Different p21 Values
Using the conversion of
11GeV2=.3894mb
σ=∫dσ=∫dσdΩ′2dΩ
The range of the detector is considered to be .10≤θ≤.87,−π≤ϕ≤π
σ=∫2.531.611∫π−πdσdΩ′2sinθdθdϕ
σ=2π∫2.531.611dσdΩ′2sinθdθ
σ=2π(1.638)dσdΩ′2
σ=(10.294)dσdΩ′2
Differential Cross Section Scale for Different p21 Values
| p′2(MeV)
|
\frac{d\sigma}{d\Omega_{2}^'}(eV^{-2})
|
\frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2})
|
\frac{d\sigma}{d\Omega_{2}^'}(mb)
|
\frac{d\sigma}{d\Omega_{2}^'}(b)
|
σ(b)
|
| 10000
|
9.357×10−11
|
9.357×107
|
3.644×107
|
3.644×104
|
3.751×105
|
| 5000
|
3.743×10−10
|
3.743×108
|
1.458×108
|
1.458×105
|
1.501×106
|
| 1000
|
9.357×10−9
|
9.357×109
|
3.644×109
|
3.644×106
|
3.751×107
|
| 500
|
3.743×10−8
|
3.743×1010
|
1.458×1010
|
1.458×107
|
1.501×108
|
CM to Lab Frame
We can substitute in for θ
dσdΩ′1=α24E∗2(3+cos2θ∗)2sin4θ∗
dσdΩ′1=α24E∗2(3+cos2θ∗)2sin(θ∗)sin(θ∗)sin(θ∗)sin(θ∗)
Using,
sin(θ∗)=sin(θ∗2)=p′2p∗2 sin(θ′2)
| ∂2σ(E,θ,ϕ)∂E∂Ω=∂2σ∗(E∗,θ∗,ϕ∗)∂E∗∂Ω∗pp∗
|
dσdΩ′1=α24E∗2(3+cos2θ∗)2p′2p∗2 sin(θ′2)p′2p∗2 sin(θ′2)p′2p∗2 sin(θ′2)p′2p∗2 sin(θ′2)
dσdΩ′1=α2p∗424E∗2p′42(3+cos2θ∗)2sin4(θ′2)
Now, using the trigometric identity,
sin2t+cos2t=1⟹cos2(θ∗)=1−sin2(θ∗)
dσdΩ′1=α2p∗424E∗2p′42(3+1−sin2(θ∗))2sin4(θ′2)
dσdΩ′1=α2p∗424E∗2p′42(4−sin(θ∗)sin(θ∗))2sin4(θ′2)
dσdΩ′1=α2p∗424E∗2p′42(4−p′2p∗2 sin(θ′2)p′2p∗2 sin(θ′2))2sin4(θ′2)
dσdΩ′1=α2p∗424E∗2p′42(4−p′22p∗22 sin2(θ′2))2sin4(θ′2)
dσdΩ′1=α2p∗424E∗2p′42(16−8p′22p∗22 sin2(θ′2)+p′42p∗42 sin4(θ′2))sin4(θ′2)
Substituting,
p∗2=√E∗22−m2
dσdΩ′1=α2(√E∗22−m2)44E∗2p′42(16−8p′22p∗22 sin2(θ′2)+p′42p∗42 sin4(θ′2))sin4(θ′2)
dσdΩ′1=α2(E∗22−m2)24E∗2p′42(16−8p′22p∗22 sin2(θ′2)+p′42p∗42 sin4(θ′2))sin4(θ′2)
Substituting in for m, E2*,and E*
α2=5.3279×10−5
dσdΩ′1=(5.3279×10−5(((53.015MeV)2−(.511MeV)2)24×(106.031MeV)2p′42(16−8p′22p∗22 sin2(θ′2)+p′42p∗42 sin4(θ′2))sin4(θ′2)
dσdΩ′1=9.357×109eV2p′42(16−8p′22p∗22 sin2(θ′2)+p′42p∗42 sin4(θ′2))sin4(θ′2)