Recent content by creepypasta13

The corrections in the typos in the OP are: B_{\mu}(x) = \Sigma_{a}B^{a}_{\mu}(x)\frac{\lambda^{a}}{2} C_{\mu}(x) = \Sigma_{a}C^{a}_{\mu}(x)\frac{\lambda^{a}}{2}

Homework Statement Consider a non-abelian gauge theory of SU(N) × SU(N) gauge fields coupled to N^{2} complex scalars in the (N,N^{_}) multiplet of the gauge group. In N × N matrix notations, the vector fields form two independent traceless hermitian matrices Bμ(x) =\Sigma_{a}...

I forgot to say that I recalculated \frac{\partial L}{\partial (\partial_{\mu}A_{\nu})} = -(\partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu})) + (m/2)\epsilon_{\lambda}^{\mu\nu}A^{\lambda}

so you're saying that to fix the error, I should set F_{\lambda}F^{\lambda} = (1/4)\epsilon_{\lambda\mu\nu}F^{\mu\nu}\epsilon^{\lambda\alpha\beta}F_{\alpha\beta}? If so, that still would not get rid of the m term in the equation I got for \frac{\partial L}{\partial (\partial_{\mu}A_{\nu})}...

The prof just told me the indices not being ALL up or downstairs is not a typo. I still would like to know if I'm calculating the Euler lagrange equations correctly or not.

is the epsilon that does not have ALL of its indices either upstairs or downstairs a typo? If not, then the usual product of epsilons is of no use since they will now follow the relation the prof gave us in 2+1 dimensions

I'm having a lot of problems with tensors. Here is what the professor in class told us in the lecture notes In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor F^{\mu\nu} = -F^{\nu\mu} is equivalent to an axial Lorentz vector, F^{\mu\nu} =...

The full Lagrangian I have is -(1/2)[ \partial^{\lambda}A^{\mu}\partial_{\lambda}A_{\mu} - \partial^{\mu}A^{\lambda}\partial_{\mu}A_{\lambda} ] + (m/2)(\epsilon_{\lambda\mu\nu}\partial^{}\muA^{\nu})A^{\lambda} so it seems that I have the same kinetic term that you have After I got the...

It seems that I got my 3rd term instead of the 3rd term you got. I double-checked my work, and I don't see how you got the 3rd term that is different from mine But my \frac{\partial L}{\partial A^{\lambda}} = (m/2)F_{\lambda} and \partial_{\mu}( \frac{\partial L}{\partial...

From the work I did above or from computing the E-L equations from writing the Lagrangian in (6) wrt A? After getting the new E-L equations, I obtained (m/2)F^{\lambda} + (1/2)\epsilon^{\lambda\mu\nu}\partial_{\mu}F_{\lambda} -...

I tried substituting the equation I got just after the E-L eq into the last equation, but then that gives ε_{\nu}^{αβ}\partial^{2}F^{λ} - ((m^2)/16)F^{\muβ} + ((m^2)/16)F^{\mu\nu} + ((m^2)/4)ε_{\nu}^{αβ}[/itex]F^{λ} = 0 Can I get the middle two terms to cancel by setting F^{\muβ} =...

No, we've never covered the Proca equation. As for the epsilon with one index down and 2 up, that's just the Hint the prof gave us. So if that's wrong, then he made a typo. Same with the Lagrangian (6), as that is what the prof gave us

Thanks for the reply. I asked this question because I was having trouble with this problem:In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor F^{\mu\nu} = -F^{\nu\mu} is equivalent to an axial Lorentz vector, F^{\mu\nu} = e^{\mu\nu\lambda}F_{\lambda}. We have...

Homework Statement Let's say I have (g^{\nu\alpha}g^{\mu\beta} - g^{\nu\beta}g^{\mu\alpha})F_{\nu} The Attempt at a Solution Would this just equal g^{\mu\beta}F_{\alpha} - g^{\mu\alpha}F_{\beta} = \delta^{\mu}_{\alpha}F_{\alpha} - \delta^{\mu}_{\beta}F_{\beta} = 0?

Homework Statement So I'm having some difficulty with my QFT assignment. I have to solve the following problem. In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor F^{\mu\nu} = -F^{\nu\mu} is equivalent to an axial Lorentz vector, F^{\mu\nu} =...