[1211] | 1 | \section{Longitudinal string decay.} |
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| 2 | |
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| 3 | \subsection{Hadron production by string fragmentation.} |
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| 4 | \hspace{1.0em}A string is stretched between flying away constituents: |
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| 5 | quark and antiquark or quark and diquark or diquark and antidiquark or |
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| 6 | antiquark and antidiquark. From knowledge of the constituents |
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| 7 | longitudinal $p_{3i}=p_{zi}$ and transversal $p_{1i}=p_{xi}$, |
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| 8 | $p_{2i}=p_{yi}$ momenta as well as their energies $p_{0i} = E_{i}$, |
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| 9 | where $i=1,2$, we can calculate string mass squared: |
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| 10 | \begin{equation} |
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| 11 | \label{LSD1} M^2_S=p^{\mu}p_{\mu}=p^2_0-p^2_1-p^2_2-p^2_3, |
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| 12 | \end{equation} |
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| 13 | where $p_{\mu} = p_{\mu 1}+p_{\mu 2}$ is the string four momentum and |
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| 14 | $\mu = 0,1,2,3$. |
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| 15 | |
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| 16 | The fragmentation of a string follows an iterative scheme: |
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| 17 | \begin{equation} |
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| 18 | string \Rightarrow hadron + new \ string, |
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| 19 | \end{equation} |
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| 20 | \textit{i. e.} a quark-antiquark (or diquark-antidiquark) pair is |
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| 21 | created and placed between leading quark-antiquark (or diquark-quark or |
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| 22 | diquark-antidiquark or antiquark-antidiquark) pair. |
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| 23 | |
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| 24 | The values of the strangeness suppression and diquark suppression |
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| 25 | factors are |
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| 26 | \begin{equation} |
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| 27 | \label{LSD2} u:d:s:qq = 1:1:0.35:0.1. |
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| 28 | \end{equation} |
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| 29 | |
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| 30 | A hadron is formed randomly on one of the end-points of the string. The |
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| 31 | quark content of the hadrons determines its species and charge. In the |
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| 32 | chosen fragmentation scheme we can produce not only the groundstates of |
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| 33 | baryons and mesons, but also their lowest excited states. If for baryons the |
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| 34 | quark-content does not determine whether the state belongs to the lowest |
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| 35 | octet or to the lowest decuplet, then octet or decuplet are choosen with |
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| 36 | equal probabilities. |
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| 37 | In the case |
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| 38 | of mesons the multiplet must also be determined before a type of hadron |
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| 39 | can be assigned. The probability of choosing a certain multiplet depends |
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| 40 | on the spin of the multiplet. |
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| 41 | |
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| 42 | The zero transverse |
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| 43 | momentum of created quark-antiquark (or diquark-antidiquark) pair is |
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| 44 | defined by the sum of an equal and opposite directed transverse momenta |
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| 45 | of quark and antiquark. |
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| 46 | |
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| 47 | The transverse momentum of created quark is randomly sampled according |
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| 48 | to probability ($\ref{LSE4}$) |
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| 49 | with the parameter $a = 0.25$ GeV$^{-2}$. Then a hadron transverse momentum |
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| 50 | ${\bf p_t}$ is determined by the sum of the transverse momenta of its |
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| 51 | constituents. |
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| 52 | |
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| 53 | The fragmentation function $f^h(z,p_t)$ represents the probability |
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| 54 | distribution for hadrons with the transverse momenta $\bf{p_t}$ to |
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| 55 | aquire the light cone momentum fraction |
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| 56 | $z=z^{\pm}=(E^h \pm p^h_z/(E^q \pm p^q_z)$, where $E^h$ and $E^q$ are |
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| 57 | the hadron and fragmented quark energies, respectively |
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| 58 | and $p^h_z$ and $p^q_z$ are hadron and fragmented quark longitudinal momenta, |
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| 59 | respectively, |
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| 60 | and $z^{\pm}_{min} \leq z^{\pm} \leq z^{\pm}_{max}$, from the |
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| 61 | fragmenting string. The values of $z^{\pm}_{min,max}$ are determined by |
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| 62 | hadron $m_h$ and constituent transverse masses and the available string |
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| 63 | mass. One of the most common fragmentation function is used in |
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| 64 | the LUND model \cite{LUND83}: |
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| 65 | \begin{equation} |
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| 66 | \label{LSD5} f^h (z, p_t) \sim \frac{1}{z}(1-z)^a |
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| 67 | \exp{[-\frac{b(m_h^2+p_t^2)}{z}]}. |
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| 68 | \end{equation} |
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| 69 | One can use this fragmentation function for the decay of the excited |
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| 70 | string. |
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| 71 | |
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| 72 | One can use also the fragmentation functions are derived in \cite{Kai87}: |
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| 73 | \begin{equation} |
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| 74 | \label{LSD6} f^{h}_q (z, p_t)=[1+\alpha^h_q(<p_t>)] (1-z)^{\alpha^h_q(<p_t>)}. |
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| 75 | \end{equation} |
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| 76 | The advantage of these functions as compared to the LUND fragmentation |
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| 77 | function is that they have correct three--reggeon behaviour at |
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| 78 | $z\rightarrow 1$ \cite{Kai87}. |
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| 79 | |
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| 80 | \subsection{The hadron formation time and coordinate.} |
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| 81 | |
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| 82 | \hspace{1.0em}To calculate produced hadron formation times and longitudinal |
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| 83 | coordinates we consider the $(1+1)$-string with mass $M_S$ and string |
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| 84 | tension $\kappa$, which decays into hadrons at string rest frame. The |
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| 85 | $i$-th produced hadron has energy $E_i$ and its longitudinal momentum |
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| 86 | $p_{zi}$, respectively. Introducing light cone variables $p^{\pm}_{i}= |
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| 87 | E_{i} \pm p_{iz}$ and numbering string breaking points consecutively |
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| 88 | from right to left we obtain $p^{+}_{0} = M_{S}$, $p_{i}^{+}=\kappa |
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| 89 | (z^{+}_{i-1}-z_i^{+})$ and $p_{i}^{-} = \kappa x^{-}_i$. |
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| 90 | |
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| 91 | We can identify the hadron formation point coordinate and time as the |
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| 92 | point in space-time, where the quark lines of the quark-antiquark pair |
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| 93 | forming the hadron meet for the first time (the so-called 'yo-yo' |
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| 94 | formation point \cite{LUND83}): |
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| 95 | \begin{equation} |
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| 96 | \label{LSD7}t_i = \frac{1}{2\kappa}[M_S - 2 \sum_{j=1}^{i-1}p_{zj} + E_i - |
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| 97 | p_{zi}] |
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| 98 | \end{equation} |
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| 99 | and coordinate |
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| 100 | \begin{equation} |
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| 101 | \label{LSD8} |
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| 102 | z_i = \frac{1}{2\kappa}[M_S - 2 \sum_{j=1}^{i-1}E_{j} + p_{zi}- E_i]. |
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| 103 | \end{equation} |
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