Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In QFT based particle theory, SU(N)-colored particles are not really present as asymptotic states, then raising some problems to build a S-matrix or other more axiomatic approaches to the theory.

Given that I have the imaginery of a oriented string as one terminated in a particle/antiparticle pair, and an unoriented string as one terminated in a pair of particles or antiparticles, I would expect that only oriented string theories have a S-matrix formulation. Is this true? If not, which is the fail in my visualisation?

share|improve this question
The right way to visualize perturbative string theory is as defined by Riemann surfaces (possibly with boundaries) with marked points corresponding to the asymptotic states. (A conformal rescaling of the worldsheet is used to map the infinite string history onto the finite Riemann surface.) An unoriented string could be conceived as an equal superposition of oppositely directed oriented strings... Concerning the relation with gauge theory: an individual "quark" would be a string between a color brane and a flavor brane; a meson, a string between two flavor branes... –  Mitchell Porter Dec 4 '12 at 11:50
... a baryon, a collection of N strings terminating on a compact brane. A non-gauge-invariant object like a diquark might be part of a baryon, e.g. you would think of a brane with 3 strings attached, and two of the strings very short, so that the third, long string can then be conceived as connecting a quark and a "diquark object". But all this would be just one way that gauge theory embeds in string theory. A truly comprehensive account might list half a dozen ways to get one from the other. –  Mitchell Porter Dec 4 '12 at 11:54
btw, a nice thing of brane stacks is that their mass spectrum, for not interesecting, has a Koide-like flavour. –  arivero Dec 5 '12 at 11:57
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.