# Does the exact string theory $S$-matrix describe all physics there is?

Suppose someone manages to evaluate the string theory $S$-matrix to all orders for any and all vertex operator insertions including non-perturbative contributions from world-sheet instantons and re-sum the whole series to obtain the exact non-perturbative string theory $S$-matrix for any combination of in- and out-states. Suppose further that the analytic result is compact, tractable, and easily amenable to numerical evaluations (say, some special function). Would such a result tell us "what string theory is"? Would it be enough in principle to answer all sensible questions about physics described by string theory? If not, what else is there we should care about?

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Are you excluding cosmological questions? One must be clear that there are many different string S-matrices, which are linked by non-S-matrix operations, which involve turning on moduli and such, and infinite number of particles. So you shouldn't say "the" string S-matrix, but "the string S-matrix for a flat-version of our vacuum". – Ron Maimon Nov 19 '12 at 4:51
Of course not, I asked about "all sensible questions about physics" which certainly includes cosmology, doesn't it? If you know the S-matrix "for any and all vertex operator insertions", as I supposed, that should allow for arbitrary moduli and geometries, no? If not, please explain why not. – Udo Kamilla Nov 19 '12 at 4:55
It's not enough because the S-matrix is for a finite number of particles--- it doesn't even describe what happens when you move a charged particle from one momentum to another, this involves infinite number of soft photons, let alone change moduli over a region where the cosmology changes. – Ron Maimon Nov 19 '12 at 5:09
Perhaps you have a point, that's what I'm trying to find out. But you are not giving a coherent argument. Changing the momentum of just one particle is not a physically allowed process. At least you'd have to construct a physically possible process described in terms of actual observables (--> presumably S-matrix elements?!) and use that to show that there is something more to be learned... – Udo Kamilla Nov 19 '12 at 5:58

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You can't describe electron electron scattering, as this is infrared divergent (you find the same infrared divergences in string theory--- you fix them by adding a soft classical changing background--- this is the dirty secret). You can't describe T-duality, as this is condensation of infinite number of zero mass particles. But you are absolutely right in your insistance that the S-matrix is complete for a given background, so I don't want to give fuel to string critics by saying "there's more than S-matrix", because it is more correct to say there isn't. – Ron Maimon Nov 20 '12 at 15:44
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## 2 Answers

Well, for starters, the scattering matrix picture of interactions does not include the dynamics of spacetime, it is instead assumed as a background space where everything happens

Even string theory is just classical general relativity in a more fair description such that it can be quantized in a way that gives finite results for measurable quantities: it assumes that string modes contribute to $T_{\mu \nu}$ and as such, produce a curvature. The curvature of coherent excitations of a closed string has been proved to be equivalent to a small perturbation of the metric (see this question for details) and this gives string theorists confidence that such excitations describe gravitons.

But the picture of space-time is still classical, and a proper nonperturbative formulation of quantum spacetime is a revolution that still has not happened. Until that happens, no scattering matrix description can hope to be complete

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That's obviously not true as there are matrix operators for the graviton which perturb the metric and hence the spacetime. That's the whole point why people got excited about strings, they include quantum gravitons in a dynamic theory. In principle it should be possible to obtain any dynamic spacetime background by appropriate vertex insertions, no? If this is not the case then I would be interested in hearing a rational argument why not. – Udo Kamilla Nov 19 '12 at 5:04

No, first because string theory is based in a number of approximations/assumptions and second because not every physical question can be answered assuming that processes take an infinite time and involve objects separated infinitely as is assumed in the S-matrix approach.

The S-matrix approach is excellent for particle physics, which deals with few particles (usually two or three) in a large mostly empty volume and only considers initial and final states of free particles. The S-matrix approach fails when you start to study many-body motion in condensed phases. This is the reason why chemists have developed other theories beyond the S-matrix formalism for the study of chemical reactions, for instance.

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String theory is obviously not based on any approximation whatsoever. I did not ask what an S-matrix (in QFT) is usually used for. My question is conceptual and concerns string theory. – Udo Kamilla Nov 20 '12 at 2:10

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Evidently string theory is based in a number of approximations and this is why the string S-matrix fails to explain even the most elementary chemical reactions in condensed phases. Moreover, even in its supposed strong point (as 'candidate' for a quantum gravity theory) the string theory approach is based in a set of gross assumptions and this is why generalizations to string theory are under active research. – juanrga Nov 20 '12 at 15:55
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