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Background

I am aware that the colour ordered amplitudes are cyclically symmetric, for example, for a 4 point amplitude:

$$A_4(1,2,3,4)= A_4(2,3,4,1) \tag{1}$$

Problem

I am trying to prove the BCJ relation:

$$s_{14} A_4[1234] − s_{13} A_4[1243] = 0 \tag{2}$$

where $s_{ij} = (p_i + p_j )^2$ are Mandelstam invariants.

Doubt

To do this I started by introducing the photon-decoupling identity (for a 4 gluon case):

$$A(1,2,3,4) + A(2,1,3,4) +A(2,3,1,4)=0 \tag{3}$$

On New Relations for Gauge-Theory Amplitudes (on page 10) it is stated that:

$$ A_4(1, 2, 3, 4) +A_4(1, 3, 4, 2) + A_4(1, 4, 2, 3) = 0 \tag{4}$$

I don't understand how this last equation (4) respects cyclicity. How do get this from (3)?

How are the Mandesltam variables introduced in the equation?

In the notes mentioned above the following is stated but I don't seem to understand it:

...the decoupling identity (..) does not rely on the specific polarizations or space-time dimension, the cancellation is entirely due to the amplitude’s dependence on the Mandelstam variables. (..) it cannot rely on four-dimensional spinor identities.

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