Joining the jets

The $f(z)$ formula above is only valid, for the breakup of a jet system into a hadron plus a remainder-system, when the remainder mass is large. If the fragmentation algorithm were to be used all the way from the $\mathrm{q}$ end to the $\overline{\mathrm{q}}$ one, the mass of the last hadron to be formed at the $\overline{\mathrm{q}}$ end would be completely constrained by energy and momentum conservation, and could not be on its mass shell. In theory it is known how to take such effects into account [Edé00], but the resulting formulae are wholly unsuitable for Monte Carlo implementation.

The practical solution to this problem is to carry out the fragmentation both from the $\mathrm{q}$ and the $\overline{\mathrm{q}}$ end, such that for each new step in the fragmentation process, a random choice is made as to from what side the step is to be taken. If the step is on the $\mathrm{q}$ side, then $z$ is interpreted as fraction of the remaining $E + p_z$ of the system, while $z$ is interpreted as $E - p_z$ fraction for a step from the $\overline{\mathrm{q}}$ end. At some point, when the remaining mass of the system has dropped below a given value, it is decided that the next breakup will produce two final hadrons, rather than a hadron and a remainder-system. Since the momenta of two hadrons are to be selected, rather than that of one only, there are enough degrees of freedom to have both total energy and total momentum completely conserved.

The mass at which the normal fragmentation process is stopped and the final two hadrons formed is not actually a free parameter of the model: it is given by the requirement that the string everywhere looks the same, i.e. that the rapidity spacing of the final two hadrons, internally and with respect to surrounding hadrons, is the same as elsewhere in the fragmentation process. The stopping mass, for a given setup of fragmentation parameters, has therefore been determined in separate runs. If the fragmentation parameters are changed, some retuning should be done but, in practice, reasonable changes can be made without any special arrangements.

Consider a fragmentation process which has already split off a number of hadrons from the $\mathrm{q}$ and $\overline{\mathrm{q}}$ sides, leaving behind a a $\mathrm{q}_i \overline{\mathrm{q}}_j$ remainder system. When this system breaks by the production of a $\mathrm{q}_n \overline{\mathrm{q}}_n$ pair, it is decided to make this pair the final one, and produce the last two hadrons $\mathrm{q}_i\overline{\mathrm{q}}_n$ and $\mathrm{q}_n\overline{\mathrm{q}}_j$, if

$$( (E+p_z)(E-p_z) )_{\mathrm{remaining}} = W_{\mathrm{rem}}^2 < W_{\mathrm{min}}^2 ~.$$ (236)

The $W_{\mathrm{min}}$ is calculated according to

$$W_{\mathrm{min}} = ( W_{\mathrm{min}0} + m_{\mathrm{q}i} + m_{\mathrm{q}j} + k \, m_{\mathrm{q}n} ) \, (1 \pm \delta) ~.$$ (237)

Here $W_{\mathrm{min}0}$ is the main free parameter, typically around 1 GeV, determined to give a flat rapidity plateau (separately for each particle species), while the default $k=2$ corresponds to the mass of the final pair being taken fully into account. Smaller values may also be considered, depending on what criteria are used to define the `best' joining of the $\mathrm{q}$ and the $\overline{\mathrm{q}}$ chain. The factor $1 \pm \delta$, by default evenly distributed between 0.8 and 1.2, signifies a smearing of the $W_{\mathrm{min}}$ value, to avoid an abrupt and unphysical cut-off in the invariant mass distribution of the final two hadrons. Still, this distribution will be somewhat different from that of any two adjacent hadrons elsewhere. Due to the cut there will be no tail up to very high masses; there are also fewer events close to the lower limit, where the two hadrons are formed at rest with respect to each other.

This procedure does not work all that well for heavy flavours, since it does not fully take into account the harder fragmentation function encountered. Therefore, in addition to the check above, one further test is performed for charm and heavier flavours, as follows. If the check above allows more particle production, a heavy hadron $\mathrm{q}_i\overline{\mathrm{q}}_n$ is formed, leaving a remainder $\mathrm{q}_n\overline{\mathrm{q}}_j$. The range of allowed $z$ values, i.e. the fraction of remaining $E + p_z$ that may be taken by the $\mathrm{q}_i\overline{\mathrm{q}}_n$ hadron, is constrained away from 0 and 1 by the $\mathrm{q}_i\overline{\mathrm{q}}_n$ mass and minimal mass of the $\mathrm{q}_n\overline{\mathrm{q}}_j$ system. The limits of the physical $z$ range is obtained when the $\mathrm{q}_n\overline{\mathrm{q}}_j$ system only consists of one single particle, which then has a well-determined transverse mass $m_{\perp}^{(0)}$. From the $z$ value obtained with the infinite-energy fragmentation function formulae, a rescaled $z'$ value between these limits is given by

$$z' = \frac{1}{2} \left\{ 1 + \frac{m_{\perp i n}^2}{W_{\math... ...erp n j}^{(0)2}}{W_{\mathrm{rem}}^2} } \, (2 z -1) \right\} ~.$$ (238)

From the $z'$ value, the actual transverse mass $m_{\perp n j} \geq m_{\perp n j}^{(0)}$ of the $\mathrm{q}_n\overline{\mathrm{q}}_j$ system may be calculated. For more than one particle to be produced out of this system, the requirement

$$m_{\perp n j}^2 = (1-z') \, \left( W_{\mathrm{rem}}^2 - \fra... ...}{z'} \right) > (m_{q j} + W_{\mathrm{min}0})^2 + p_{\perp}^2$$ (239)

has to be fulfilled. If not, the $\mathrm{q}_n\overline{\mathrm{q}}_j$ system is assumed to collapse to one single particle.

The consequence of the procedure above is that, the more the infinite energy fragmentation function $f(z)$ is peaked close to $z=1$, the more likely it is that only two particles are produced. The procedure above has been constructed so that the two-particle fraction can be calculated directly from the shape of $f(z)$ and the (approximate) mass spectrum, but it is not unique. For the symmetric Lund fragmentation function, a number of alternatives tried all give essentially the same result, whereas other fragmentation functions may be more sensitive to details.

Assume now that two final hadrons have been picked. If the transverse mass of the remainder-system is smaller than the sum of transverse masses of the final two hadrons, the whole fragmentation chain is rejected, and started over from the $\mathrm{q}$ and $\overline{\mathrm{q}}$ endpoints. This does not introduce any significant bias, since the decision to reject a fragmentation chain only depends on what happens in the very last step, specifically that the next-to-last step took away too much energy, and not on what happened in the steps before that.

If, on the other hand, the remainder-mass is large enough, there are two kinematically allowed solutions for the final two hadrons: the two mirror images in the rest frame of the remainder-system. Also the choice between these two solutions is given by the consistency requirements, and can be derived from studies of infinite energy jets. The probability for the reverse ordering, i.e. where the rapidity and the flavour orderings disagree, is given by the area law as

$${\cal P}_{\mathrm{reverse}} = \frac{1}{1 + e^{b\Delta}} ~~~... ...2 + m_{\perp n j}^2)^2 - 4 m_{\perp i n}^2 m_{\perp n j}^2} ~.$$ (240)

For the Lund symmetric fragmentation function, $b$ is the familiar parameter, whereas for other functions the $b$ value becomes an effective number to be fitted to the behaviour when not in the joining region.

When baryon production is included, some particular problems arise (also see section ). First consider $B\overline{B}$ situations. In the naïve iterative scheme, away from the middle of the event, one already has a quark and is to chose a matching diquark flavour or the other way around. In either case the choice of the new flavour can be done taking into account the number of SU(6) states available for the quark-diquark combination. For a case where the final $\mathrm{q}_n \overline{\mathrm{q}}_n$ breakup is an antidiquark-diquark one, the weights for forming $\mathrm{q}_i\overline{\mathrm{q}}_n$ and $\mathrm{q}_n \overline{\mathrm{q}}_i$ enter at the same time, however. We do not know how to handle this problem; what is done is to use weights as usual for the $\mathrm{q}_i\overline{\mathrm{q}}_n$ baryon to select $\mathrm{q}_n$, but then consider $\mathrm{q}_n \overline{\mathrm{q}}_i$ as given (or the other way around with equal probability). If $\mathrm{q}_n \overline{\mathrm{q}}_i$ turns out to be an antidiquark-diquark combination, the whole fragmentation chain is rejected, since we do not know how to form corresponding hadrons. A similar problem arises, and is solved in the same spirit, for a $BM\overline{B}$ configuration in which the $B$ (or $\overline{B}$) was chosen as third-last particle. When only two particles remain to be generated, it is obviously too late to consider having a $BM\overline{B}$ configuration. This is as it should, however, as can be found by looking at all possible ways a hadron of given rank can be a baryon.

While some practical compromises have to be accepted in the joining procedure, the fact that the joining takes place in different parts of the string in different events means that, in the end, essentially no visible effects remain.