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$\mathbf{\pi N \longrightarrow \eta \Delta}$

This cross section is given by

\begin{eqnarray*} \frac{d\sigma_{\pi N \rightarrow \eta \Delta}}{dm_{\Delta}}=\l... ...lta})}{s p^{initial}_{cm}} \mathcal{A}_{\Delta}(m_{\Delta}) . \end{eqnarray*}

Assuming a constant matrix element $\left\vert\mathcal{M}_{\pi N \rightarrow \eta \Delta} \right\vert^{2}=7 \mathrm{ mB GeV^2}$ we can integrate over the mass of the $\Delta$ resonance and get

\begin{eqnarray*} \sigma_{\pi N \rightarrow \eta \Delta}=\left\langle 1 \frac{... ...cm}(m_{\Delta})\mathcal{A}_{\Delta}(m_{\Delta}) dm_{\Delta} . \end{eqnarray*}


Oliver Buss 2005-03-16