Elong-13-05-01

Cross-Section Issue - Fixed

From looking at the dilution factor, f, I found a problem with how the cross-sections were calculated. We know that, naively,
<math>f=\frac{N_{D}}{(N_{D}+N_{N}+N_{He})} = \frac{t\cdot R_{D}}{(t\cdot R_{D}+t\cdot R_{N}+t\cdot R_{He})} = \frac{R_{D}}{(R_{D}+R_{N}+R_{He})} \approx \frac{N_{D}}{N_{D}+N_{N}}\approx \frac{3\sigma_{D}}{3\sigma_{D}+\sigma{N}}\approx \frac{3\cdot 2}{3\cdot 2 + 14}\approx 6/20=0.3</math>.

Originally, I was (wrongly) using <math>\frac{d^2\sigma^u}{d\Omega dE'} = \left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}} \left[ \frac{2\cdot \left(\frac{F_1^{X}}{A_X} \right)}{m_{X}}\tan^2\left( \frac{\theta_{e'}}{2} \right) + \frac{\left( \frac{F_2^X}{A_X}\right) }{\nu} \right]</math>

where X = nucleus and <math>\left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}}=\frac{Z^2 \alpha^2 \hbar^2 c^2}{4E^2\sin^4\left( \frac{\theta}{2} \right)}\cos^2\left( \frac{\theta}{2} \right)</math>.
This gave <math>f \approx 0.062</math>.

The problem came from using a nuclear Mott cross-section but a nucleon structure cross-section. To fix this, I first found the Mott cross-section for a proton,
<math>\left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}}=\frac{1^2 \alpha^2 \hbar^2 c^2}{4E^2\sin^4\left( \frac{\theta}{2} \right)}\cos^2\left( \frac{\theta}{2} \right)</math>,

and then multiplied that by the structure cross-section at the total number of nucleons,

<math>\frac{d^2\sigma^u}{d\Omega dE'} = A_X \left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}} \left[ \frac{2\cdot \left(\frac{F_1^{X}}{A_X} \right)}{m_{X}}\tan^2\left( \frac{\theta_{e'}}{2} \right) + \frac{\left( \frac{F_2^X}{A_X}\right) }{\nu} \right]</math>.

This gave <math>f \approx \frac{3\sigma_{D}}{3\sigma_{D}+\sigma{N}} \approx 0.29</math>, as expected.

Plugging all this into the rates code gave