0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is $T_a$ ? $\endgroup$ – Andrea Nov 28 '18 at 17:14
  • $\begingroup$ $T_a$ 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_00 Nov 28 '18 at 17:19
  • 1
    $\begingroup$ @Ringo_00 I suppose you could put it that way, yeah! $\endgroup$ – knzhou Nov 28 '18 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.