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We frequently see: A certain action and then we are asked to solve for Bianchi identity and Maxwell equation.

I have often solved for them but I never knew what is the difference between the two? In more specific words, I know that Maxwell equation is derived from the action but what does a Bianchi identity mean?

EDIT: Now that in the comments we faced a new issue, I am editing this with a example.

Example: Lagragian is $$L=-\frac{1}{4}ImZF_{\mu\nu}F{\mu\nu}-\frac{1}{8}ReZ\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$

So, the book (Van Proeyen's SUGRA book) says tat the equation of motion of this theory will be $$\partial_{\mu}[(ImZ)F^{\mu\nu}+i(ReZ)\tilde{F}^{\mu\nu}]=0$$ How come this is the same as te usual Maxwell equation $d\star F=0$ (I believe this is a dual teory so maybe the equation Evan presented in the comments will be instead dF=0 but in all cases I could cut the story short and ask how did Van Proeyen reach this $$\partial_{\mu}[(ImZ)F^{\mu\nu}+i(ReZ)\tilde{F}^{\mu\nu}]=0?$$

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The difference is perhaps illuminated best in the language of differential forms. Recall that the vector potential $A$ can be considered a 1-form field $A$, which we can act on with the exterior derivative operator $d$ to get the electromagnetic field strength tensor \begin{equation} F=dA. \end{equation} Since $F$ is an exact form we know that it is closed. So \begin{equation} dF = d^2A=0. \end{equation} This is the Bianchi Identity $\nabla_{[\alpha}F_{\beta\gamma]}=0$, expressed in the language of differential forms. Now, another operation that we can apply to a differential form is the Hodge dual operator *, which maps $k$-forms to $(n-k)$-forms in $n$-dimensional space. So, in 4 spacetime dimensions the Hodge dual of the 2-form $F$ will be another 2-form, $*F$. Now, $*F$ is in general not an exact form and so it's exterior derivative need not vanish. Instead, we have \begin{equation} d*F=\mu_0 J, \end{equation} where $J$ is the current 3-form. This is Maxwell's equations $\partial_{\alpha}F^{\alpha\beta}=\mu_o J^{\beta}$ expressed in the language of differential forms.

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  • $\begingroup$ Yes, thank you for mentioning that. But if for example we have, $S=\int{F_{\mu\nu}\tilde{F}_{\mu\nu}}$ and they ask to find the field equations. Then one will have to solve for this: $\frac{\delta S}{\delta F^{\rho\sigma}}=0$ which does not agree with $d\star F= 0$ (if we assume we have no sources). In other words, isn't maxwell equation supposed to obtained from Euler-Lagrange equation? $\endgroup$ Sep 18, 2015 at 21:18
  • $\begingroup$ @Beyond-formulas It is obtainable from the Euler-Lagrange equations. The trick is that $F$ is not the field that is varied, you must look at $A$ in $F=\mathrm{d}A$. $F$ is the field strength, the actual "field" as far as Euler-Lagrange is concerned is the potential $A$. $\endgroup$
    – Ryan Unger
    Sep 18, 2015 at 22:03
  • $\begingroup$ Is it always the case? I see people varying the action with respect to F too. @0celo7 $\endgroup$ Sep 18, 2015 at 22:25
  • $\begingroup$ In one post I read @0celo7, it says that if $A$ is not preset explicitly in the action then it is safe to derive with respect to F. Otherwise, one should derive with respect to A. $\endgroup$ Sep 18, 2015 at 23:14
  • $\begingroup$ @Beyond-formulas That is correct. Even when $A$ appears explicitly you can rewrite in terms of only $F$ but then your expressions will not be manifestly local. That's why it's easier to do everything in terms of $A$. $\endgroup$
    – Evan Rule
    Sep 18, 2015 at 23:25

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