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The cross section for a b $\longrightarrow $ R in the medium

Now we want to consider a resonance with a potential. Our philosphy will be to evaluate the mass $\mu$ of the produced resonance as described above in equation (1.1), and to treat hereafter the process like a vacuum collision at $s=\mu$. Therefore we exchange in a first step $s$ by $\mu$

\begin{eqnarray*}
\sigma_{a\,b \rightarrow R}(s)\sim
\frac{1}{p^{2}_{ab}(\mu(s))...
...{\left(\mu(s)^2-M_R^2\right)^2+\mu(s)\Gamma_{tot}^2(\mu(s))} \ .
\end{eqnarray*}

As a next step we need to consider a correction due to the relative velocities of the colliding particles which was introduced by Effenberger [Eff99]1.1. The flux-factor in the cross section is proportional to $\frac{1}{v_{rel}}$. Therefore we correct by setting

\begin{displaymath}
\sigma(s)=\frac{v_{rel}\left(\mu(s)\right)}{v_{rel}(s)}\, \sigma\left(\mu(s)\right) \ .
\end{displaymath}

Hence the cross section in the case of a resonance being produced in a potential is given by
$\displaystyle \sigma_{a\,b \rightarrow R}(s)=$   $\displaystyle \frac{v^{ab}_{rel}\left(\mu(s)\right)}{v^{ab}_{rel}(s)} F_I \frac{2J_R+1}{\left(2J_a+1\right)\left(2J_b+1\right)}
\frac{1}{\mathcal{S}_{ab}} \,$  
    $\displaystyle \frac{4\pi}{p^{2}_{ab}(\mu(s))} \,
\frac{\mu(s)\, \Gamma_{R\right...
..._{tot}(\mu(s))}{\left(\mu(s)^2-M_R^2\right)^2+\mu(s)\Gamma_{tot}^2(\mu(s))} \ .$ (1.4)


next up previous contents
Next: Reactions of the type Up: Resonance production: Previous: The cross section for   Contents
Oliver Buss 2005-03-16