The t quantity is known as the square of the 4-momentum transfer
t≡(P∗1−P′∗1)2=(P∗2−P′∗2)2
In the CM Frame
P∗1=−P∗2
P′∗1=−P′∗2
E∗1=E′∗1=E∗2=E′∗2
|→p∗1|=|→p′∗1|=|→p∗2|=|→p′∗2|
t=(P∗1−P′∗1)2=(P∗2−P′∗2)2
t=P∗21+P′∗21−2P∗1P′∗1=P∗22+P′∗22−2P∗2P′∗2
t=2m2−2E∗1E′∗1+2→p∗1→p′∗1=2m2−2E∗2E′∗2+2p∗2p′∗2
t=2m2−2E∗21+2|p∗21|cosθ1 1′=2m2−2E∗22+2|p∗22|cosθ2 2′
where θ1 1′ and θ2 2′is the angle between the before and after momentum in the CM frame
Using the relativistic relation E2=m2+p2 this reduces to
t=−2p∗21+2|p∗21|cosθ1 1′=−2p∗22+2|p∗22|cosθ2 2′
t=−2p∗21(1−cosθ1 1′)=−2p∗22(1−cosθ2 2′)
The maximum momentum is transfered at 90 degrees, i.e. cos90circ=0
t=−2p∗21
This can be rewritten again using the relativistic energy relation E2=m2+p2
t=−2(m2−E∗21)=−2(m2−E∗22)
In the Lab Frame
t=P21+P′21−2P1P′1=P22+P′22−2P2P′2
t=2m2−2E1E′1+2→p1→p′1=2m2−2E2E′2+2p2p′2
with p2=0
and E2=m
t=2m2−2mE′2=2(m2−E′2m)
Between Frames
t=2(m2−E′2m)=2(m2−E∗22)
→E′2=E∗21m
withE∗2≈53 MeV for E1=11000 MeV
The Moller electron has a maximum energy possible of:
E′2=5500 MeV
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