(for a more muddled version, see physics.stackexchange: http://physics.stackexchange.com/questions/14020/whats-with-mandelstams-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.

This post has been migrated from (A51.SE)