This question is a continuation of this previous question of mine and I am continuing with the same notation.
One claims that one can actually split this n-gluon amplitude such that there is just a single gluon propagating between two n−point amplitudes and the pn−1(z)− and pn(z) are on two sides. Define qi,n−1(z)=pi+pi+1+...+pn−1(z) and define h to be the helicity of the gluon when propagating out of the left amplitude. This is summarizied in saying that the following expression holds,
A(1,2,..,n,z)=∑n−3i=1∑h=±1AL(pi,pi+1,..,pn−1(z),qhi,n−1(z))1qi,n−1(z)2AR(pn(z),p1,p2,...,pi−1,q−hi,n−1(z))
Is there a "quick" explanation for the above split and why the propagating gluon has to flip helicity? (..it seems to be way of putting in the helicity conservation at high-energies but I can't make it very precise..)
In the above split shouldn't the sum be from i=2 since one can't get lower than 3-gluon vertices on either side?
Now one can apparently write the momentum squared of the propagator in the following way, qi,n−1(z)2=q2i,n−1−z[pn−1|γμqμi,n−1|pn>, where qi,n−1(0)=qi,n−1 and then apparently using the previous expression of A(1,2,..,n,z)=∑iRi(z−zi) one can re-write the amplitude as,
A(1,2,..,n)=∑n−3i=1∑h=±1AL(pi,pi+1,..,pn−1(zi),qhi,n−1(zi))1q2i,n−1AR(pn(zi),p1,p2,...,pi−1,−q−hi,n−1(zi))
where zi is such that qi,n−1(zi)2=0
- I would like to know how the above expression for A(1,2,..,n) was obtained. (..it looks like Cauchy's residue theorem but I can't make it completely precise..)
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