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  Gauge Invariance of Yang Mills Lagrangian

+ 3 like - 0 dislike

I am trying to show the invariance of the following Yang Mills Lagrangian: $$L= -\frac{1}{4} F^a_{\mu \nu} F_a^{\mu\nu} + J_a^\mu A_\mu^a$$ under the following gauge transformation ($\theta$ being a rotation in color space and $g$ related to the structure constant): $$L \rightarrow -\frac{1}{4} \left( F^a_{\mu\nu} +\epsilon ^a_{jk} \theta^j F^k_{\mu \nu}\right)\left(F_a^{\mu\nu} + e_a^{jk}\theta_j F_k^{\mu\nu}\right) + \left( J_a^\mu + \epsilon_a^{jk} \theta_j J_k^\mu \right) \left( A_\mu^a + \epsilon^a_{jk}\theta^jA_\mu^k -\frac{1}{g} \partial^\mu \theta^a \right),$$

where each term is now transformed accordingly.

I was able to simplify the above to and obtain: $$L \rightarrow -\frac{1}{4} \left( F^a_{\mu\nu}F_a^{\mu\nu} + \epsilon^a_{jk} \theta^j F^k_{\mu\nu} F_a^{\mu\nu} \epsilon_a^{j' k'} \theta_{j'} F_{k'}^{\mu\nu}\right) + J_a^\mu A_\mu^a - J_a^\mu \frac{1}{g} \partial^\mu \theta^a + \epsilon_a^{jk} \theta_j J_k^\mu \epsilon^a_{j'k'} \theta^{j'}A_\mu^{k'} - \epsilon_a^{jk} \theta_{j} J_k^{\mu}\frac{1}{g} \partial^\mu \theta^a.$$

How could I possibly reduce it to a form similar to the original, untransformed Lagrangian? There are about 4 terms I can't get rid of, though it has been suggested to me that I use the equation of motion of YM, which I have handy but can't seem to use them appropriately. Any help would be greatly appreciated. Also note that I may end up with a boundary term which would vanish when varying the action, thus possibly giving me say 3 terms instead of the original 2 (which is fine, though I can't identify them yet).

This post imported from StackExchange Physics at 2014-07-01 10:33 (UCT), posted by SE-user user44212
asked Jun 30, 2014 in Theoretical Physics by user44212 (15 points) [ no revision ]
where the do the 3 Field strengths in the first term come from? Given that you are doing a linear transformation on $F^2$ it can't be right.

This post imported from StackExchange Physics at 2014-07-01 10:33 (UCT), posted by SE-user TwoBs

1 Answer

+ 2 like - 0 dislike
From the $F^2$ part, you have 4 terms. 2 cancel if you use that the structure constants are completely antisymmetric, the last is second order and can be discarded, the original $F^2$ term is left. From the interaction part there are six terms. One is second order and two cancel directly from antisymmetry again. Another will also cancel from antisymmetry if you do an integration by parts, leaving only the original interaction term.This post imported from StackExchange Physics at 2014-07-01 10:33 (UCT), posted by SE-user Robin Ekman
answered Jun 30, 2014 by Robin Ekman (215 points) [ no revision ]

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