# What's with Mandelstam's argument that only linear regge trajectories are stable?

While thinking about how to answer a "describe string theory" question, I remembered an old argument of Stanley Mandelstam's that linear Regge trajectories implies stability. I never fully understood the exact argument, or its limitations. (The article is "Dual-Resonance Models" from 1974, sciencedirect.com/science/article/pii/0370157374900349, thanks for digging it up)

Here is the argument as I remember it: consider the Regge trajectory function $\alpha(s)$, and expand it in a dispersion relation with two subtractions:

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

Where a,b are constants. Mandelstam says that the imaginary part of $\alpha(s)$ is a measure of some sort of instability or decay of the string states, 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 with no subtractions, with one subtraction, with two subtractions, etc. Can you conclude that exactly constant, exactly linear, exactly quadratic Regge trajectories are also stable? Maybe exactly constant trajectories at special values can be interpreted as free stable point particles of a given spin and mass, but what are quadratic trajectories exactly? Does he mean to argue that nearly linear trajectories are necessarily stable when they are exactly linear? What is the precise conclusion?
• The Regge trajectory function appears in an exponent in the scattering amplitude, so you have to take a log to extract it. Why you don't get a cut contribution from taking the log of negative values which has nothing to do with physics, just from the Regge scattering ansatz?
• Actual scattering amplitudes are combinations of different trajectory contributions, so why does each individual trajectory have to be separately analytic, with singularities that are determined by physics? Is this a heuristic assumption?
• Ok, even if you don't have cuts from the log, the expansion suggests that an imaginary part of the trajectory function, like the imaginary part of a two-point function, has some sort of physical interpretation that allows you to interpret it immediately as some sort of decay rate. What is that physical interpretation? Does it work far away from linear trajectories?
• How does this argument translate to modern string theory? I have never seen this kind of argument anywhere else.

Mandelstam is still around, but I don't know him, and I think that someone else who is better versed in Regge theory than I am will know the answer immediately, because Mandelstam states it in two lines, without explanation, so it must be obvious.

Later Comment: Having received a short cryptic answer from Mandelstam regarding this, I started thinking about it again, and I gave a partial answer below. But I am not really satisfied with it yet. But one of my complaints, though, that separating out the individual trajectory and expanding it by analytic dispersion relations might fail, is not really serious. The trajectory exchange amplitude is the sum of the resonances on the trajectory, so it is always separately analytic. Why it's log should also be analytic to justify the dispersion expansion is probably equally easy to see, but I don't see it

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I wish I knew enough to answer this... as it stands I'll be interested to see what anyone else can come up with. – David Z Aug 27 '11 at 22:12
I asked Mandelstam by email just now, maybe he'll respond. I already know that this is true in string perturbation theory, but I want to understand how he knew this before having a theory. – Ron Maimon Aug 27 '11 at 22:16
Enlosing your main equation in \$\$...\$\$ to give it a line by itself would make it more readable – Larry Harson Aug 28 '11 at 13:45
I got this answer from Mandelstam: "The reason is tht the decay time of a resonance is inversely proportional to the imaginary part of the pole position. With linear trajectories the imaginary part is zero, so the decay time is infinite, i. e., the resonance is stable." Well, I knew that already, and it doesn't help at all. – Ron Maimon Aug 28 '11 at 14:40
@Ron: maybe you could point him to this site. It'd be awesome, if someone like him would leave an answer here. – luksen Aug 29 '11 at 15:59