Recent content by vnikoofard

Suppose we have $$[Q^a,Q^b]=if^c_{ab}Q^c$$ where Q's are generators of a Lie algebra associated a SU(N) group. So Q's are traceless. Also we have $$[P^a,P^b]=0$$ where P's are generators of a Lie algebra associated to an Abelian group. We have the following relation between these generators...

Thanks very much for response. I wonder myself maybe appear a minus sign in the second line. Are you sure? Maybe I am confusing this situation with part by part integration!

Hi friends I am trying to drive constraints of a Lagrangian density by Dirac Hamiltonian method. But I encountered a problem with calculating one type of Poisson Bracket: {\varphi,\partial_x\pi}=? where \pi is conjugate momentum of \varphi. I do not know for this type Poisson Bracket I can...

Thanks Haushofer! :)

Thank you again for your help, my friend!

Dear Dickfore, I am really poor on Topology and such kinds of mathematics. Recently I decided to begin studying this topics. Can you please suggest me some good textbooks for self-study. I am thinking about 3rd edition of Frankle's book: "Geometry of Physics".

Would you please explain more about the integral? It seems interesting.

Thanks! Now I got it. When W_{\mu \nu} =0 means \partial_\mu V_{\nu} -\partial_{\nu}V_{\mu}=0. So for having this expression we must suppose that V_{\mu} =\partial_\mu\lambda where \lambda is a scalar. Because we can change order of derivations \partial_{\mu}, \partial_{\nu}. Is it correct?

Unfortunately I do not know. Is it related to Stokes's theorem?

Yes, I am sure. :( You can check it in the mentioned paper.

Hi Friends I am reading the following paper http://arxiv.org/abs/hep-th/9705122 In the page 4 he says that \tilde{W}_{\mu\nu}=0\Rightarrow V_{\mu}=\partial_{\mu}\lambda Where \tilde{W}^{\mu\nu}\equiv\frac{1}{2}\epsilon^{\mu \nu\rho\sigma}W_{\rho\sigma} and...