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 string theory, the first excited level of the bosonic string can be decomposed into irreducible representations of the transverse rotation group, $SO(D-2)$. We then claim that the symmetric traceless part (i.e. the 35 rep) is the spin-2 graviton - but isn't the label "spin-2" intrinsically 3+1 dimensional? I.e. it labels the representation under the little group $SU(2)$?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

It is traditional to label massless (and some massive) states in higher dimension by their 3-d "spin" counterparts, even thought he label is completely inaccurate, as you say. All antisymmetric forms are "spin-1", the symmetric two-index object is "spin-2", a fundamental spinor is "spin 1/2" and a vector of spinors is "spin 3/2". These labels refer to the maximum helicity of the associated massless particle, although the number of components is completely different than in 4d. For learning the higher dimensional rotation group, there is an article by Scherk from the 1970s.

share|improve this answer

I don't know anything about string theory, but the graviton has been described as a spin-2 particle well before string theory. In his "Lectures on Gravitation" Feynman explains why the graviton must be integer spin, then explains why it can't be 0 or 1, then proceeds to attempt to build the quantum field theory of a spin-2 graviton, simply because it's the next simplest thing that could work. (It doesn't, Feynman abandoned this line of research.)

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.