I understand that trace is needed in order to preserve gauge invariance of the lagrangian equation by using the cycling property. But I fail to see why the following equation holds true: $$-\frac{1}{2}Tr(F^{\mu\nu}F_{\mu\nu})=-\frac{1}{4}F^{a\mu\nu}F_{a\mu\nu}.$$ I also fail to understand how is the gauge symmetry preserved in the right term of this equation, I know that the $a$ index means that we are considering only the components and not the whole field but then I'm not even clear on how should I apply the field transformation such as an $SU(2)$ matrix to such components.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
This is simply because $$\text{tr } F^{\mu\nu} F_{\mu\nu} = \text{tr } F^{\mu\nu a} F_{\mu\nu}^b T_a T_b = F^{\mu\nu a} F_{\mu\nu}^b\, \text{tr } T_a T_b = \frac12 F^{\mu\nu a} F_{\mu\nu}^b \, \delta_{ab} = \frac12 F^{\mu\nu a} F_{\mu\nu}^a$$ where the normalization $\text{tr}(T_aT_b) = \delta_{ab}/2$ is conventional.
-
-
$\begingroup$ $T_a$ is one of the generator of the group (for example in SU(2) $T_a=\frac{\sigma_a}{2}$). Thank you for your answer. In such a term then the gauge transformation simply gets "eaten up" by the trace and then is automatically gauge invariant? $\endgroup$– Ringo_00Nov 28, 2018 at 17:19
-
1$\begingroup$ @Ringo_00 I suppose you could put it that way, yeah! $\endgroup$– knzhouNov 28, 2018 at 17:36