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 $$\partial_{\mu}[(ImZ)F^{\mu\nu}+i(ReZ)\tilde{F}^{\mu\nu}]=0?$$


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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