Relativistic Differential Cross-section

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Relativistic Differential Cross-section

dσ=1F|M|2dQ

dQ is the invariant Lorentz phase space factor


dQ=(2\pi)^4\delta^4(\vec p_1 +\vec p_2 - \vec p_1^' -\vec p_2^')\frac{d^3 \vec p_1^'}{(2\pi)^3 2E_1^'}\frac{d^3 \vec p_2^'}{(2\pi)^3 2E_2^'}


and F is the flux of incoming particles


F=2E12E2|v1v2|=4|E1E2v21|


where v21 is the relative velocity between the particles in the frame where particle 1 is at rest


P1P2=E1E2(p1p2)=E1E2


Using the relativistic definition of energy

E2p2+m2=m2


P1P2=mE2


Letting E21E2 be the energy of particle 2 wiith respect to particle 1, the relativistic energy equation can be rewritten such that


|p221|=E221m2=(P1P2)2m2m2=(P1P2)2m4m2

where similarly p21 is defined as the momentum of particle 2 with respect to particle 1.



The relative velocity can be expressed as


v21=|p21|E21


F=2E12E2|v1v2|=4|mE21v12|=4|mE21|p21|E21|=4m|p21|


The invariant form of F is

F=4(P1P2)2m4


s2m2+2P1P2P1P2=s2m22


F=4(s2m22)2m4


F=4(s24sm2+4m44)m4=4(s24sm24)


where

sCM=4(m2+p1 2)=(2E1)2


F=4((4(m2+p1 2))24sm24)


F=2((4(m2+p1 2))24sm2)


Fcms=4pis


dσ=14pis|M|2dQ


d^3 \vec p_1^'=\vec p^{'3}_1 d \vec p^'  d\Omega


(E_1^')^2=(\vec p_1^')^2+(m_1)^2


E_1^' d E_1^'= \vec p_1^' d \vec p_1^'


dQ=\frac{1}{(4\pi)^2}\delta (E_1+E_2-E_1^'-E_2^')\frac{\vec p_1^'dE_1^'}{E_2^'}d\Omega<\center>


W_i \equiv E_1+E_2  \qquad \qquad W_f \equiv E_1^'+E_2^'


dW_f=dE_1^'+dE_2^'=\frac{\vec p_1^' d \vec p_1^'}{E_1^'}+\frac{p_2^' dp_2^'}{E_2^'}


In the center of mass frame

|\vec p_1^'|=|\vec p_2^'|=|\vec p_f^'| \rightarrow |\vec p_1^' d \vec p_1^'|=|\vec p_2^' d \vec p_2^'|=|\vec p_f^' d \vec p_f^'|


dW_f=\frac{W_f}{E_2^'}dE_1^'


dQcms=1(4π)2δ(WiWf)pfdWfWfdΩ


dQcms=1(4π)2pfsdΩ


dσdΩ=164π2spfpi|M|2