On the Regge slopes intramultiplet relation

Abstract

We show that only additivity of inverse Regge slopes is consistent with both the formal chiral limit m(n) → 0 and the heavy quark limit M (Q) ≫ M (n), where n = u, d, and m, M are current and constituent quark masses, respectively.

Some mass relations for mesons and baryons in Regge phenomenology

“…The paper by Burakovsky and Goldman [42] showed that only the additivity of inverse Regge slopes is consistent with the formal chiral and heavy quark limits for both mesons and baryons, and that the factorization of Regge slopes, although consistent with the formal chiral limit, fails in the heavy quark limit. Besides, in…”

“…Also, it saturates the inequality for Regge intercepts [52] which follows from the Schwarz inequality and the unitarity relation. The above two relations are usually generalized to the baryon case [23,42,51], in which one has…”

“…[45], the slopes and intercepts of the Regge trajectories are fundamental constants of hadron dynamics, perhaps in general more important than the masses of particular states. Thus, the determination of Regge slopes and intercepts of hadrons is of great importance since this provides opportunities for a better understanding of the dynamics of strong interactions [42].…”

“…[21,42], Regge intercepts and slopes are useful for many spectral and nonspectral purposes, for example, in the recombination [43] and fragmentation [44] models. There-fore, as pointed out in Ref.…”

Regarding the Charmed‐Strange Member of the 23S1 Meson State

“…The parameters ${\alpha }_{i\stackrel{-}{i^{\prime }}}^{\prime }$ and ${\alpha }_{i\stackrel{-}{i^{\prime }}}false(\mathrm{normal0}false)$ are, respectively, the slope and intercept of the trajectory. The intercepts and slopes can be described by [15, 29] $\begin{array}{c}{\alpha }_{i\stackrel{-}{i}}\left(\mathrm{normal0}\right)+{\alpha }_{j\stackrel{-}{j}}\left(\mathrm{normal0}\right)=\mathrm{normal2}{\alpha }_{i\stackrel{-}{j}}\left(\mathrm{normal0}\right),\end{array}$ $\begin{array}{c}\frac{\mathrm{normal1}}{{\alpha }_{i\stackrel{-}{i^{\prime }}}^{\prime }}+\frac{\mathrm{normal1}}{{\alpha }_{j\stackrel{-}{j^{\prime }}}^{\prime }}=\frac{\mathrm{normal2}}{{\alpha }_{i\stackrel{-}{j^{\prime }}}^{\prime }}.\end{array}$ …”