Elong-14-05-23

Clarifying Azz

In the steps mentioned below, following in the footsteps of Frankfurt & Strikman, everything is defined in terms of ratios of the polarized cross-section over the unpolarized cross sections. This leads to

$r(0,k) = \frac{\sigma_m^0}{\sigma_u}$

$r(+1,k) = \frac{\sigma_m^+}{\sigma_u}$

$r(-1,k) = \frac{\sigma_m^-}{\sigma_u}$

To go from these into our Azz observable, which is just the tensor-polarized cross section over the unpolarized cross section, we need $\sigma_u = \frac{1}{3}(\sigma_m^0 + \sigma_m^+ + \sigma_m^-)$ and $\sigma_p^0 = \frac{1}{3}(\sigma_m^+ + \sigma_m^- - 2\sigma_m^0 )$ such that

$A_{zz} = \frac{\sigma_p^0 - \sigma_u}{\sigma_u} $

$A_{zz} = \frac{\frac{1}{3}(\sigma_m^+ + \sigma_m^- - 2\sigma_m^0 ) - \frac{1}{3}(\sigma_m^0 + \sigma_m^+ + \sigma_m^-)}{\sigma_u}$

$A_{zz} = \frac{1}{3}\frac{(\sigma_m^+ - \sigma_m^+) + (\sigma_m^- - \sigma_m^-) + (-2\sigma_m^0 - \sigma_m^0)}{\sigma_u}$

$A_{zz} = \frac{-3}{3}\frac{\sigma_m^0}{\sigma_u}$

$A_{zz} = -\frac{\sigma_m^0}{\sigma_u}$

$A_{zz} = -r(0,k) \frac{}{}$