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$\mathbf{\pi N \longrightarrow K \bar{K} N}$

This cross section is choosen according to Sibirtsev et al. [SCK97]. Details concerning the Clebsch-Gordan coefficients an be found in [Eff99]. The result is given by

\begin{eqnarray*} \begin{array}{lll} \sigma_{\pi^- p\rightarrow n K^0 \bar{K}^0}... ...ar{K}^0}=\frac{1}{2} \ \sigma^{\mathrm{Sibirtsev}} && \end{array}\end{eqnarray*}

where

\begin{displaymath} \sigma^{\mathrm{Sibirtsev}}=1.121 \left(1-\frac{s_0}{s}\right)^{1.86} \left(\frac{s_0}{s}\right)^2 \mathrm{mB} \end{displaymath}

with $s_{0}=\left(m_{N}+2m_{K}\right)^2$ All channels with an incoming neutron are given by charge conjugation.


Oliver Buss 2005-03-16