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Two body final states : $\mathbf{X \rightarrow c d}$ in the vacuum

Here we need to utilize a Monte-Carlo description to choose the final state. In equation (1.6) we have seen that the two-particle final state depends both on the masses of the outgoing particles and their directions of motion. The Mandelstam $s$ and $p_{ab}$ are determined by the initial state. So we choose the masses $\mu_c$ and $\mu_d$ at the same time as we choose $\Omega$. First we utilize a variable transformation

\begin{displaymath}
y_i(\mu_i)=2\,\arctan\left[2\,\frac{\mu_i-M_i^0}{\Gamma^0_i}\right]; i\in\left\lbrace c,d\right\rbrace
\end{displaymath}

with $M_{i}^0$ and $\Gamma_{i}^0$ being the values at the pole position in the vacuum. Hence we can rewrite equation (1.6) as
$\displaystyle \frac{d\sigma_{a b \rightarrow c d}}{dy_c\, dy_d\, d\Omega}(s)$ $\textstyle =$ $\displaystyle \frac{1}{64\pi^2\, s} \frac{p_{cd}}{p_{ab}} \mathcal{A}_c(\mu_{c},p_c(\mu_c,\Omega)) \mathcal{A}_d(\mu_{d},p_d(\mu_d,\Omega))$  
    $\displaystyle \left\vert\mathcal{M}_{a b \rightarrow c d } (s)\right\vert^2 \times \frac{d\mu_c}{dy_c} \times \frac{d\mu_d}{dy_d}$ (2.14)

with

\begin{eqnarray*}
\frac{dy_i}{d\mu_i}&=&2 \times 2\, \frac{1}{\Gamma^0_i} \times...
...rac{\left(\Gamma^0_i\right)^2}{4}+\left(\mu_i-M_i^0\right)^2}\ .
\end{eqnarray*}

Now we choose $y_c, y_d$ and $\Omega$ according to a flat distribution. The probability that a random ensemble $(y_c,y_d,\Omega)$ will be accepted is therefore given by :

\begin{displaymath}
p_{accept}(y_c,y_d,\Omega)= \frac{p_{cd} \mathcal{A}_c(\mu_{...
...A}_d \frac{d\mu_c}{dy_c} \frac{d\mu_d}{dy_d}\right)_{max}} \ .
\end{displaymath}

The denoted by $\left(p_{cd} \mathcal{A}_c \mathcal{A}_d\frac{d\mu_c}{dy_c} \frac{d\mu_d}{dy_d}\right)_{max}$ is actually hard to find. We parametrize this maximal value by

\begin{displaymath}
\left(p_{cd} \mathcal{A}_c \mathcal{A}_d\frac{d\mu_c}{dy_c} ...
...{d\mu_d}{dy_d}\right)_{max}=Q_{cd} \times \max\{p^{vac}_{cd}\}
\end{displaymath}

The dimensionless factor $Q$ is of the order of $10$ and depends on the outgoing particles. It has to be readjusted if one establishes new in-medium effects.
next up previous contents
Next: Three body final states Up: Final State Decisions Previous: Resonance Production   Contents
Oliver Buss 2005-03-16