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1\subsection{The longitudinal and kinky strings are produced
2in hadronic collisions.}
3
4\subsection{The number of the cut Pomerons.}
5
6\hspace{1.0em}In the case of a nondiffractive interaction we can
7determine $n$ of cut Pomerons or $2n$ produced strings according to
8probability \cite{AGK74}
9\begin{equation}
10\label{SLKS1}
11 p^{(n)}_{ij}(\vec{ b}_i-\vec{ b}_j,s)
12=c^{-1}\exp{\{-2u(b_{ij}^2,s)\}}
13 \frac{[2u(b_{ij}^2,s)]^{n}}{n!}.
14\end{equation}
15
16\subsubsection{Separation of the longitudinal soft strings from
17 the kinky hard strings.}
18 
19\hspace{1.0em}We assume  \cite{ASGABP95}, \cite{WDFHO97} that each cut
20Pomeron can be substituted either by the two longitudinal soft strings
21or by the two kinky hard strings.
22
23At the moment it is not completely clear how to choose which cut
24pomeron should be
25substituted by longitudinal and which one should be
26substituted by kinky strings.
27
28One recipe is based on the eikonal model \cite{RCT94}, \cite{WDFHO97}
29\begin{equation}
30\label{SLKS2}u(b_{ij}^2,s)=u_{soft}(b_{ij}^2,s) + u_{hard}(b_{ij}^2,s).
31\end{equation}
32The soft eikonal part is defined as
33\begin{equation}
34\label{SLKS3}u_{soft}(b_{ij}^2,s) =
35\frac{\gamma _{soft}}{\lambda_{soft} (s)}(s/s_0)^{\Delta_{soft}} 
36\exp (-b_{ij}^2/4\lambda_{soft} (s)).
37\end{equation}
38The hard part is calculated according to
39\begin{equation}
40\label{SLKS4}u_{hard}(b_{ij}^2,s)=
41\frac{\sigma_{jet}}{8\pi\lambda_{hard}(s)}
42(s/s_0)^{\Delta_{hard}} 
43\exp (-b_{ij}^2/4\lambda_{hard} (s)).
44\end{equation} 
45The $\sigma_{jet}=0.027$ mbarn and $\Delta_{hard}=0.47$ were found from the fit of the two--jet
46experimental cross section \cite{UA1}. Then from the global fit of
47the total and
48elastic cross sections for $pp$ collisions the values of $\gamma_{soft} = 35.5$ mbarn, $\Delta_{soft}=
490.07$ and $R^2_{hard} = R^2_{soft} = 3.56$ GeV$^{-2}$ were
50found.
51
52Thus we can examine each cut Pomeron and substitute it by two kinky
53strings with probability
54\begin{equation}
55\label{SLKS5}
56P_{hard}(b^2_{ij},s) = \frac{u_{hard}(b_{ij}^2,s)}{u_{soft}(b_{ij}^2,s)+
57u_{hard}(b_{ij}^2,s)}.
58\end{equation}
59
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