Variables Used in Elastic Scattering

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Lorentz Invariant Quantities

Total 4-Momentums

As was shown earlier the scalar product of a 4-Momentum vector with itself ,

P1P1=E1E1p1p1=m21=s

,

and the length of a 4-Momentum vector composed of 4-Momentum vectors,

P2=(P1+P2)2=(E1+E2)2(p1+p2)2=(m1+m2)2=s

,

are invariant quantities.

It was further shown that

P2=P2


where P=(P1+P2)2 represents the 4-Momentum Vector in the CM frame


and P=(P1+P2)2 represents the 4-Momentum Vector in the initial Lab frame

which can be expanded to

{\mathbf P^*}^2={\mathbf P^{'*}}^2={\mathbf P}^2={\mathbf P^'}^2


where {\mathbf P^'}=({\mathbf P_1^'}+{\mathbf P_2^'})^2 represents the 4-Momentum Vector in the final Lab frame


and P=(P1+P2)2 represents the 4-Momentum Vector in the final CM frame

New 4-Momentum Quantities

Working in just the CM frame, we can form new 4-Momentum Vectors comprised of 4-Momenta in this frame, with

P1P1=(E1E1p1(x)p1(x)p1(y)p1(y)p1(z)p1(z))=Pa


P1P2=(E1E2p1(x)p2(x)p1(y)p2(y)p1(z)p2(z))=Pb


P2P1=(E2E1p2(x)p1(x)p2(y)p1(y)p2(z)p1(z))=Pc


P2P2=(E2E2p2(x)p2(x)p2(y)p2(y)p2(z)p2(z))=Pd

Using the algebraic fact

(ab)2=(ba)2


and the fact that the length of these 4-Momentum Vectors are invariant,

(P1P1)2=(P122P1P1+P1)=(E1E1p1(x)p1(x)p1(y)p1(y)p1(z)p1(z))2=(Pa)2


(P1P2)2=(P122P1P2+P2)=(E1E2p1(x)p2(x)p1(y)p2(y)p1(z)p2(z))2=(Pb)2


(P2P1)2=(P222P2P1+P1)=(E2E1p2(x)p1(x)p2(y)p1(y)p2(z)p1(z))2=(Pc)2


(P2P2)2=(P222P2P2+P2)=(E2E2p2(x)p2(x)p2(y)p2(y)p2(z)p2(z))2=(Pd)2

Using the fact that the scalar product of a 4-momenta with itself is invariant,


P1P1=E1E1p1p1=m21


We can simiplify the expressions

(P1P1)2=(m212P1P1+m21)=(Pa)2


(P1P2)2=(m212P1P2+m22)=(Pb)2


(P2P1)2=(m222P2P1+m21)=(Pc)2


(P2P2)2=(m222P2P2+m22)=(Pd)2

Finding the cross terms,

P1P=(E1p1(x)p1(y)p1(z))(1000010000100001)(Ep1(x)p1(y)p1(z))=E1E1p1p1


(P1P1)2=(m212E1E12p1p1+m21)=(Pa)2


(P1P2)2=(m212E1E22p1p2+m22)=(Pb)2


(P2P1)2=(m222E2E12p2p1+m21)=(Pc)2


(P2P2)2=(m222E2E22p2p2+m22)=(Pd)2


Using the fact that in the CM frame,

CM.png


p1=p2
p1=p2

We can further simplify

(P1P1)2=(m212E1E12p2p2+m21)=(Pa)2


(P1P2)2=(m212E1E22p2p1+m22)=(Pb)2


(P2P1)2=(m222E2E12p2p1+m21)=(Pc)2


(P2P2)2=(m222E2E22p2p2+m22)=(Pd)2

Mandelstam Representation

Mandelstam.png