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Fragmentation functions

Assume a $\mathrm{q}\overline{\mathrm{q}}$ jet system, in its c.m. frame, with the quark moving out in the $+z$ direction and the antiquark in the $-z$ one. We have discussed how it is possible to start the flavour iteration from the $\mathrm{q}$ end, i.e. pick a $\mathrm{q}_1 \overline{\mathrm{q}}_1$ pair, form a hadron $\mathrm{q}\overline{\mathrm{q}}_1$, etc. It has also been noted that the tunnelling mechanism is assumed to give a transverse momentum $p_{\perp}$ for each new $\mathrm{q}_i \overline{\mathrm{q}}_i$ pair created, with the $p_{\perp}$ locally compensated between the $\mathrm{q}_i$ and the $\overline{\mathrm{q}}_i$ member of the pair, and with a Gaussian distribution in $p_x$ and $p_y$ separately. In the program, this is regulated by one parameter, which gives the root-mean-square $p_{\perp}$ of a quark. Hadron transverse momenta are obtained as the sum of $p_{\perp}$'s of the constituent $\mathrm{q}_i$ and $\overline{\mathrm{q}}_{i+1}$, where a diquark is considered just as a single quark.

What remains to be determined is the energy and longitudinal momentum of the hadron. In fact, only one variable can be selected independently, since the momentum of the hadron is constrained by the already determined hadron transverse mass $m_{\perp}$,

\begin{displaymath}
(E+p_z)(E-p_z) = E^2 - p_z^2 = m_{\perp}^2 = m^2 + p_x^2 + p_y^2 ~.
\end{displaymath} (238)

In an iteration from the quark end, one is led (by the desire for longitudinal boost invariance and other considerations) to select the $z$ variable as the fraction of $E + p_z$ taken by the hadron, out of the available $E + p_z$. As hadrons are split off, the $E + p_z$ (and $E - p_z$) left for subsequent steps is reduced accordingly:
$\displaystyle (E+p_z)_{\mathrm{new}}$ $\textstyle =$ $\displaystyle (1-z) (E+p_z)_{\mathrm{old}} ~,$  
$\displaystyle (E-p_z)_{\mathrm{new}}$ $\textstyle =$ $\displaystyle (E-p_z)_{\mathrm{old}} -
\frac{m_{\perp}^2}{z (E+p_z)_{\mathrm{old}}} ~.$ (239)

The fragmentation function $f(z)$, which expresses the probability that a given $z$ is picked, could in principle be arbitrary -- indeed, several such choices can be used inside the program, see below.

If one, in addition, requires that the fragmentation process as a whole should look the same, irrespectively of whether the iterative procedure is performed from the $\mathrm{q}$ end or the $\overline{\mathrm{q}}$ one, `left-right symmetry', the choice is essentially unique [And83a]: the `Lund symmetric fragmentation function',

\begin{displaymath}
f(z) \propto \frac{1}{z} z^{a_{\alpha}} \left( \frac{1-z}{z}...
...^{a_{\beta}} \exp \left( - \frac{bm_{\perp}^2}{z}
\right) ~.
\end{displaymath} (240)

There is one separate parameter $a$ for each flavour, with the index $\alpha$ corresponding to the `old' flavour in the iteration process, and $\beta$ to the `new' flavour. It is customary to put all $a_{\alpha,\beta}$ the same, and thus arrive at the simplified expression
\begin{displaymath}
f(z) \propto z^{-1} (1-z)^a \exp (-bm_{\perp}^2/z) ~.
\end{displaymath} (241)

In the program, only two separate $a$ values can be given, that for quark pair production and that for diquark one. In addition, there is the $b$ parameter, which is universal.

The explicit mass dependence in $f(z)$ implies a harder fragmentation function for heavier hadrons. The asymptotic behaviour of the mean $z$ value for heavy hadrons is

\begin{displaymath}
\langle z \rangle \approx 1 - \frac{1+a}{bm_{\perp}^2} ~.
\end{displaymath} (242)

Unfortunately it seems this predicts a somewhat harder spectrum for $\mathrm{B}$ mesons than observed in data. However, Bowler [Bow81] has shown, within the framework of the Artru-Mennessier model [Art74], that a massive endpoint quark with mass $m_Q$ leads to a modification of the symmetric fragmentation function, due to the fact that the string area swept out is reduced for massive endpoint quarks, compared with massless ditto. The Artru-Mennessier model in principle only applies for clusters with a continuous mass spectrum, and does not allow an $a$ term (i.e. $a \equiv 0$); however, it has been shown [Mor89] that, for a discrete mass spectrum, one may still retain an effective $a$ term. In the program an approximate form with an $a$ term has therefore been used:
\begin{displaymath}
f(z) \propto \frac{1}{z^{1 + r_Q b m_Q^2}}
z^{a_{\alpha}} \...
...)^{a_{\beta}} \exp \left( - \frac{b m_{\perp}^2}{z} \right) ~.
\end{displaymath} (243)

In principle the prediction is that $r_Q \equiv 1$, but so as to be able to extrapolate smoothly between this form and the original Lund symmetric one, it is possible to pick $r_Q$ separately for $\c $ and $\b $ hadrons.

For future reference we note that the derivation of $f(z)$ as a by-product also gives the probability distribution in proper time $\tau$ of $\mathrm{q}_i \overline{\mathrm{q}}_i$ breakup vertices. In terms of $\Gamma = (\kappa \tau)^2$, this distribution is

\begin{displaymath}
{\cal P}(\Gamma) \, \d\Gamma \propto \Gamma^a \, \exp(-b \Gamma) \,
\d\Gamma ~,
\end{displaymath} (244)

with the same $a$ and $b$ as above. The exponential decay allows an interpretation in terms of an area law for the colour flux [And98].

Many different other fragmentation functions have been proposed, and a few are available as options in the program.

$\bullet$
The Field-Feynman parameterization [Fie78],
\begin{displaymath}
f(z) = 1 - a + 3a(1-z)^2 ~,
\end{displaymath} (245)

with default value $a = 0.77$, is intended to be used only for ordinary hadrons made out of $\u $, $\d $ and $\mathrm{s}$ quarks.
$\bullet$
Since there are indications that the shape above is too strongly peaked at $z = 0$, instead a shape like
\begin{displaymath}
f(z) = (1+c) (1-z)^c
\end{displaymath} (246)

may be used.
$\bullet$
Charm and bottom data clearly indicate the need for a harder fragmentation function for heavy flavours. The best known of these is the Peterson/SLAC formula [Pet83]
\begin{displaymath}
f(z) \propto \frac{1}{ z \left( 1 -
\frac{\displaystyle 1...
...displaystyle \epsilon_Q}{\displaystyle 1-z}
\right)^2 } ~,
\end{displaymath} (247)

where $\epsilon_Q$ is a free parameter, expected to scale between flavours like $\epsilon_Q \propto 1/m_Q^2$.
$\bullet$
As a crude alternative, that is also peaked at $z=1$, one may use
\begin{displaymath}
f(z) = (1+c) z^c ~.
\end{displaymath} (248)

$\bullet$
In [Edé97], it is argued that the quarks responsible for the colour fluctuations in stepwise diquark production cannot move along the light-cones. Instead there is an area of possible starting points for the colour fluctuation, which is essentially given by the proper time of the vertex squared. By summing over all possible starting points, one obtains the total weight for the colour fluctuation. The result is a relative suppression of diquark vertices at early times, which is found to be of the form $1-\exp(-\rho \Gamma)$, where $\Gamma\equiv\kappa^2\tau^2$ and $\rho \approx 0.7{\mathrm{GeV}}^{-2}$. This result, and especially the value of $\rho$, is independent of the fragmentation function, $f(z)$, used to reach a specific $\Gamma$-value. However, if using a $f(z)$ which implies a small average value $\langle\Gamma\rangle$, the program implementation is such that a large fraction of the $\mathrm{q}\rightarrow B +\mathrm{q}\mathrm{q}$ attempts will be rejected. This dilutes the interpretation of the input $P(\mathrm{q}\mathrm{q})/P(\mathrm{q})$ parameter, which needs to be significantly enhanced to compensate for the rejections.
A property of the Lund Symmetric Fragmentation Function is that the first vertices produced near the string ends have a lower $\langle\Gamma\rangle$ than central vertices. Thus an effect of the low-$\Gamma$ suppression is a relative reduction of the leading baryons. The effect is smaller if the baryon is very heavy, as the large mass implies that the first vertex almost reaches the central region. Thus the leading-baryon suppression effect is reduced for $\c $- and $\b $ jets.


next up previous contents
Next: Joining the jets Up: String Fragmentation Previous: String Fragmentation   Contents
Stephen_Mrenna 2012-10-24