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.

A simple langragian that gives the simplest interaction is $\mathcal{L}=(\partial\phi)^2+(m\phi)^2$ where $m$ is some constant. Does anyone know of theory in four dimensions which is physically interesting and couples three vector fields over a 4-dim'l manifold.

share|improve this question
    
what about $\mathcal{L}_{\text{I}} = \kappa \, \partial^{\alpha}A^{\beta}\partial_{\alpha}A^{\gamma}\partial_{\beta}A_{\gamma}$‌​? That seems like a lorentz invariant with three vector potentials. –  kηives Jan 24 '13 at 4:45
    
thanks for the comment. do you know if such a term appears in the lagrangian for some physically relevant theory? that's the heart of my question. like you, i can write different terms that would be acceptable but i am interested in a theory which is already of some importance. –  yca Jan 24 '13 at 5:10
    
Sorry, I don't . . . but if you can make one that is Lorentz invariant, and, under a group of your choice, gauge invariant and self consistent, it's a good way to learn and have fun. –  kηives Jan 24 '13 at 5:14
1  
The hugely important example is Yang-Mills theory (en.wikipedia.org/wiki/Yang%E2%80%93Mills_theory or any passable QFT textbook). Nonabelian Yang-Mills theory has three and four point vertices for the gauge bosons. If you only want a three point vertex you'll have to go to Chern-Simons theory in three dimensions (arxiv.org/pdf/hep-th/9902115v1.pdf). I'm afraid I don't know of any well defined 4D theory which has 3-point vector couplings but not 4-point ones. –  Michael Brown Jan 24 '13 at 8:12
    
@MichaelBrown: I suppose that by well-defined, you mean UV complete? Otherwise, practically any Lagrangian will do. –  Vibert Jan 24 '13 at 9:50
show 1 more comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.