How do I verify the invariance on Yang-Mills' Lagrangian:

$$L = -\frac{1}{4} \sum_{a} \left(\partial_\mu A_\nu^a - \partial_\nu A_\mu^a + gf^{abc}A_\mu^bA_\nu^c \right)^2$$

under the transformation:

$$A_\mu^a (x) \rightarrow A_\mu^a (x) + \frac{1}{g} \partial_\mu \alpha^a(x) -f^{abc}\alpha^b(x)A_\mu^c(x)$$

The first term on the gauge transformation obviously recovers the old lagrangian, but the second and third terms are getting very problematic. Can someone help me with those? I'm note sure about my argument on terms $\partial_\nu \partial_\mu \alpha^a$ and $\partial_\mu \partial_\nu \alpha^a$, and the third one just get worse. I'm pretty sure that I must use the fact that the structure constant is totally antisymmetric, but it's not getting better.

  • $\begingroup$ Not totally sure what the issue is, but $\alpha$ is infinitesimal, so $\alpha^2$ terms should be dropped $\endgroup$ May 21 '19 at 18:55
  • 1
    $\begingroup$ Look here physics.stackexchange.com/q/456875 $\endgroup$
    – Vicky
    May 21 '19 at 19:00
  • $\begingroup$ The problem is entirely mathematical. The $\alpha^2$ terms will sure drop, but there are many other terms from the gauge transformation, mainly at $A_\mu^b A_\nu^c$, that i couldn't make it cancel with each other to show the lagrangian invariance. $\endgroup$
    – otto
    May 21 '19 at 19:26
  • $\begingroup$ @otto First, if you want to write an answer to a comment use '@nickname'. Otherwise the person you are answering will see nothing. Second, in the link I provided your question is solved in an 'exact' way, I mean without Taylor expansions. Make a full reading of it $\endgroup$
    – Vicky
    May 24 '19 at 11:17

As already mentioned in the comment, the exact version was proven in the link. However, if you insisted to show it infinitesimally in your notation, here is how.

First, consider the variation (warning: sloppy of notation use in the following but it is clear in the context I guess)

\begin{equation} \label{delL} \delta L = -\frac{1}{2} \sum_a \left( \partial_\mu A_\nu^a - \partial_\nu A^a_\mu + gf^{abc}A^b_\mu A^c_\nu \right) \left( \partial_\mu \delta A^a_\nu + gf^{abc}A^b_\mu \delta A^c_\nu - (\mu \leftrightarrow \nu) \right), \end{equation} where \begin{equation} \label{delA} \delta A^a_\mu = \frac{1}{g} \partial_\mu A^a - f^{abc} \alpha^b A^c_\mu. \end{equation} Now, if you substitute \tag{delA} into \eqref{delL} and patiently listed every terms, you will encounter the terms you noticed already: \begin{equation} \delta L = -\frac{1}{2}\sum_a F_{\mu\nu}^a \left[ \partial_\mu \left( \frac{1}{g}\partial_\nu \alpha^a - f^{abc}\alpha^b A^c_\nu \right) + f^{abc}A^b_\mu \left( \frac{1}{g}\partial_\nu\alpha^c - f^{cde}\alpha^d A^e_\nu \right) - (\mu\leftrightarrow\nu) \right] \end{equation} I will not show in detail every step, but describe the computation. Expanding the first big round bracket inside the big square bracket, you will have the derivative hitting on the (infinitesimal) gauge parameter, $ \frac{1}{g}\partial_\mu \partial_\nu \alpha^a - (\mu\leftrightarrow\nu). $ The above two terms (notice the term with $\mu,\nu$ exchanged actually cancel each other, because \begin{equation} \partial_\mu \partial_\nu \alpha^a = \partial_\nu \partial_\mu \alpha^a, \end{equation} i.e. partial derivatives commute. There are also terms, $ -f^{abc} (\partial_\mu \alpha^b) A^c_\nu + f^{abc}A^b_\mu \partial_\nu \alpha^c - (\mu\leftrightarrow\nu), $ where the above first term comes from the first round bracket and the above second term from the second round bracket inside the big square bracket. One can easily show that they cancel each other after renaming the indices and considering the anti-symmetric properties of the spacetime indices $\mu,\nu$ and the internal indices $a,b,c$.

Finally, the most tricky part is the remaining terms, \begin{equation} -f^{abc} \alpha^b \partial_\mu A^c_\nu - f^{abc}f^{cde} \alpha^d A^b_\mu A^e_\nu - (\mu \leftrightarrow \nu). \end{equation} The second term above together with their $- (\mu \leftrightarrow \nu)$ counter part can be rewritten using Jacobi identity and then combine with the rest term to get \begin{equation} f^{abc}\alpha^b F_{\mu\nu}^c \end{equation} Substitute this back to the original variation equation \eqref{delL}, you will get \begin{equation} f^{abc} \alpha^b F_{\mu\nu}^c F^{\mu\nu\, a}. \end{equation} You may want to verify that the last pieces in this form vanish (relatively) manifest.

  • $\begingroup$ Anyone has a clue how Latex equation reference is done here? I tried \eqref{} and \label{} but it does not work. $\endgroup$
    – chichi
    Aug 13 '19 at 0:50
  • $\begingroup$ Hmm...both \ eqref and \ tag do not work, I gave up. $\endgroup$
    – chichi
    Aug 13 '19 at 3:08

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.