Obtaining an angular distribution from the Klein-Nishina formula

Question

I'm having trouble on how one might obtain an angular distribution from the Klein-Nishina formula for compton scattering of polarised photons, more specifically on how this report is obtaining it. (http://geant4.cern.ch/G4UsersDocuments/UsersGuides/PhysicsReferenceManual/html/node56.html)

The experiment concerns gamma ray compton polarimetry and detects correlations of two scattered back to back entangled photons by having two detectors perpendicular to the beam and rotating one about the azimuth, centered on the scatterers.

They obtain an equation for the scattered intensity relative to incident flux (equation 22) which I am fine with, however the transition from equation 22 to equation 44 and the end result, equation 45 seems ambiguous to me. Could anyone shed some light on how $\psi$ (Detector separation angle) is introduced here?

In short, the report states:

$I = \int_{\phi_{min}}^{\phi_{max}} \int_{\delta_{min}}^{\delta_{max}}\frac{d\sigma}{d\Omega} \,d\delta d\phi$

where $I$ is a beam intensity relative to the incident beam, $\delta$ is the polar scattering angle and $\phi$ is a polarisation angle about the beam axis.

the Klein-Nishina formula for compton scattering of polarised radiation is:

$\frac{d\sigma}{d\Omega} = \frac{1}{2} r_o^2 \frac{\frac{1}{2-\cos{\delta}} + 2 - \cos{\delta} - 2\sin^2{\delta} \cos^2{\phi}}{2 - \cos^2{\delta}}$

and then an equation is obtained in the form $I = a - bcos^2(\psi)$

where $\psi$ is the detector separation angle.

How has this distribution dependent on the detector separation angle $\psi$ resulted from the previous equations?

Sorry if this is obvious, as it seems mysterious to me. The relevant derivations and diagrams are present from appendix A, page 13 of the report.

Thanks for your time.

This post imported from StackExchange Physics at 2016-01-23 21:11 (UTC), posted by SE-user Bolly Vonning