(for a more muddled version, see physics.stackexchange: What's with Mandelstam's argument that only linear regge trajectories are stable?)

There is a 1974 argument of Mandelstam's that linear Regge trajectories implies stability, from "Dual-Resonance Models" from 1974, sciencedirect.com/science/article/pii/0370157374900349. Expand the Regge trajectory function $\alpha(s)$ in a dispersion relation with two subtractions:

$$ \alpha(s) = b + as + {1\over i\pi} \int_0^\infty {\mathrm{Im}(\alpha(s'))\over s-s'} ds'$$

The imaginary part of $\alpha(s)$ gives the decay of the string states, since where it hits an integer tells you where the poles are. So if the string resonances are exactly stable, then the imaginary part is zero, and the trajectory is linear.

This argument bugged me for these reasons:

  • It seems to work just as well with two subtractions, three subtractions, etc. Can you conclude that exactly quadratic or exactly cubic Regge trajectories are also stable? What is a quadratic or cubic trajectory?
  • The Regge trajectory function appears in the exponent, so you have to take a log to extract it. Why is it clear that it has a representation like the above, without a cut contribution at negative s?
  • In string theory, the trajectories are linear when they are long-lived, but the trajectory function doesn't look as fundamental today. Is there a more modern formulation of this, which would tell you which string limits are non-interacting just from a condition on the spectrum?

Mandelstam generously emailed me a short remark, saying essentially that the trajectory function imaginary part is a lifetime, and indeed this is obvious from the fact that it gives the position of the resonances, but I am still confused regarding the questions above.

Even a partial answer would be appreciated.

  • $\begingroup$ Ron, I was looking up strings and regge trajectories an this old question of yours came up.There have been some discussions of validation of string theory and I have been wondering whether you have unearthed any "predictions", checkable with the plethora of resonance data, that a string theory could give. $\endgroup$
    – anna v
    Commented Aug 28, 2014 at 12:53
  • $\begingroup$ @annav: This kind of string theory is now known as AdS/QCD, and nobody disagrees that it is predictive for hadron resonances, although the degree of quantitative agreement can be disputed. This stuff predicts things like hidden local symmetry (Hidden Regge theory in the 80s), topological baryons (Large N physics of the 80s), and Polchinski-Tan Pomerons (BFKL pomerons in string theory, early 2000s). There's a ton of work, this should give good handles, but AdS/QCD is most recent and most direct. This is not the same as predictions of fundamental string theory, the results are not controversial. $\endgroup$
    – Ron Maimon
    Commented Aug 29, 2014 at 3:18
  • $\begingroup$ thanks . you mean it is a second level string theory validation, validation by embedding more limited theories that themselves are validated by data. i.e. no direct line string theory which has not already been predicted by the minor ones. :( $\endgroup$
    – anna v
    Commented Aug 29, 2014 at 4:12

1 Answer 1


I don't have any sharp answer, but the argument seems sketchy. I think we know in the infinite-$N_c$ limit of QCD that we have exactly stable resonances, and some nearly-linear Regge trajectories in some region, but that they're not perfectly linear and they fail badly to be linear at negative s where BFKL describes Regge physics. http://arxiv.org/abs/hep-th/0603115 by Brower, Polchinski, Strassler, and Tan looks at this kind of thing in some detail, and might point to some older literature that has something to say.

  • $\begingroup$ The word "sketchy" is ambiguous, you probably mean "wrong". I know the Brower Polchinski Strassler Tan stuff, and their argument links up perturbative BFKL to nonperturbative pomeron. I believe that the turnaround location depends on N in a way consistent with Mandelstam, so that in the pure large N limit the pomeron trajectory is straight. I am not sure, however, but it is a good thing to verify. Thanks for the answer. $\endgroup$
    – Ron Maimon
    Commented Oct 11, 2011 at 19:50

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