Currently at the Wikipedia page on Yang-Mills theory, you see that [a screenshot],

enter image description here

Isn't that an obvious mistake? Based on this normal convention: $$ F^2 \equiv F \wedge \star F $$ Shouldn't it be:

$$ -\frac{1}{2}\operatorname{Tr}(F^2)\equiv -\frac{1}{2}\operatorname{Tr}(F \wedge \star F)= -\frac{1}{4}\operatorname{Tr}(F_{\mu \nu} F^{\mu \nu})d^4x =- \frac{1}{4} \operatorname{Tr}[T^a T^b] F^{a\mu \nu} F_{\mu \nu}^b d^4x $$ $$ = - \frac{1}{4} \frac{1}{2} \delta^{ab} F^{a\mu \nu} F_{\mu \nu}^b d^4x =- \frac{1}{8} \sum_a F^{a\mu \nu} F_{\mu \nu}^a d^4x =- \frac{1}{8} F^{a\mu \nu} F_{\mu \nu}^a d^4x, $$

where Einstein summation notation is assumed in the end. Based on the SU(N) Lie algebra, $\operatorname{Tr}[T^a T^b] =\frac{1}{2} \delta^{ab} $. The $a,b$ are the indices for fundamental representation of SU(N). There is an overall factor mismatch between left and right hand side.

$$ \operatorname{Tr}(F^2)=\operatorname{Tr}(F \wedge \star F)= \frac{1}{2}\operatorname{Tr}(F_{\mu \nu} F^{\mu \nu})d^4x = \frac{1}{4} F^{a\mu \nu} F_{\mu \nu}^ad^4x, $$ Alternatively, we can write $$ \operatorname{Tr}(F_{\mu \nu} F^{\mu \nu})d^4x = \frac{1}{2} F^{a\mu \nu} F_{\mu \nu}^ad^4x $$

For Maxwell theory, we have $$ \mathcal{L}_\mathrm{Maxwell}d^4x = -\frac{1}{2}\operatorname{Tr}(F^2)\equiv -\frac{1}{2}\operatorname{Tr}(F \wedge \star F)= -\frac{1}{4}\operatorname{Tr}(F_{\mu \nu} F^{\mu \nu})d^4x=- \frac{1}{4} \operatorname{Tr}[T^a T^b] F^{a\mu \nu} F_{\mu \nu}^b d^4x $$ $$ = -\frac{1}{2}(F^2)= -\frac{1}{2}(F \wedge \star F)= - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}d^4x, $$

Some correction on Wiki page is required if my above is correct.

  • 4
    $\begingroup$ Wikipedia is correct. Note that $\mathrm{tr}(F^2)\equiv \mathrm{tr}(F_{\mu\nu}F^{\mu\nu})$ instead of, as you wrote, $\mathrm{tr}(F^2)\overset{\mathrm{no}}=\color{red}{\frac12} \mathrm{tr}(F_{\mu\nu}F^{\mu\nu})$. $\endgroup$ – AccidentalFourierTransform Jun 21 '18 at 22:33
  • $\begingroup$ Thanks, then I guess it is the convention issue. Usually at the context I am familiar, we use $$ F^2 \equiv F \wedge \star F $$ $\endgroup$ – wonderich Jun 21 '18 at 22:38
  • $\begingroup$ @wonderich It is pretty common to use $T^2$ for the "standard" contraction of any tensor with itself (i.e. pairing first with first, second with second, etc, indices). $\endgroup$ – Danu Jun 21 '18 at 22:47
  • $\begingroup$ @AccidentalFourierTransform I like your equals sign with the "no" above it; how do you do that? $\endgroup$ – probably_someone Jun 22 '18 at 3:26
  • $\begingroup$ @probably_someone right click > show math as > TeX commands :-) $\endgroup$ – AccidentalFourierTransform Jun 22 '18 at 3:28

Your Answer

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

Browse other questions tagged or ask your own question.