This post directly follows this question and is very similar in nature:

Consider the following Lagrangian: $$\mathcal{L}=\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\partial_\mu\phi^\ast\partial^\mu\phi$$ $$+ieA^\mu\left(\phi\partial_\mu\phi^\ast - \phi^\ast\partial_\mu\phi\right) +e^2A_\mu A^\mu\phi^\ast\phi-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2\tag{1}$$ [...]

The equation of motion obtained from the Lagrangian, $(1)$, for $A_\color{red}{\mu}$ is more interesting, $$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=ie\left(\phi\partial_{\color{red}{\nu}}\phi^\ast-\phi^\ast\partial_{\color{red}{\nu}}\phi\right)$$ $$+2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{2}$$ This can be written as $$\partial_\mu F^{\mu\nu}=ie\left[\phi\left(D^\nu\phi\right)^\ast-\phi^\ast D^\nu\phi\right]\tag{3}$$

Where $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$ and $D^\nu\phi\equiv \partial^\nu\phi + ieA^\nu\phi$ - similar to the previous question.

Before even doing any calculations I believe there are some typos present on the indices I marked in red. In particular, I think the $A_\mu$ should be $A^\nu$ as there is no equation of motion for $A_\mu$: equation $(2)$ does not even contain $A_\mu$. I think the corrected version of $(2)$ should read $$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)$$ $$+2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{A}$$

A quick check is needed to make sure equation $(\mathrm{A})$ is actually correct, to verify this I will use it to prove eqn. $(3)$: $$\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=\partial_\mu F^{\mu\nu}$$ $$=ie\phi\partial^\nu\phi^\ast-ie\phi^\ast\partial^\nu\phi+e^2A^\nu\phi^\ast\phi+e^2A^\nu\phi^\ast\phi$$ $$ie\phi\left(\partial^\nu\phi^\ast-ieA^\nu\phi^\ast\right)-ie\phi^\ast\left(\partial^\nu\phi+ieA^\nu\phi\right)$$ $$=ie\left[\phi\left(D^\nu\phi\right)^\ast-\phi^\ast D^\nu\phi\right]$$ which is eqn. $(3)$, so equation $(\mathrm{A})$ appears correct.

The objective is now to prove equation $(\mathrm {A})$:

To make the proof easier I first rewrite the Lagrangian, $(1)$ as $$\mathcal{L}=-\frac14\partial_\mu A_\nu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\frac14\partial_\nu A_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)$$ $$+ieA_\nu\left(\phi\partial^\nu\phi^\ast - \phi^\ast\partial^\nu\phi\right) +e^2A_\nu A^\nu\phi^\ast\phi$$ $$-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2+\partial_\mu\phi^\ast\partial^\mu\phi$$ where in the second line of eqn. $(1)$ I lowered and raised the respective indices on the terms involving $A^\mu \partial_\mu\phi$ and let $\mu\to \nu$ in these terms and the next term involving $e^2$. So $$\frac{\partial\mathcal{L}}{\partial A_\nu}=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)+e^2A^\nu\phi^\ast\phi$$ and $$\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}=-\frac14\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)$$ Putting these together gives $$\frac{\partial\mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu A_\nu}$$ $$=ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right)+e^2A^\nu\phi^\ast\phi+\frac14\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{!}$$ But this is not the same as equation $(\mathrm{A})$: $$ie\left(\phi\partial^\nu\phi^\ast-\phi^\ast\partial^\nu\phi\right) +2e^2A^\nu\phi^\ast\phi-\partial_\mu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=0\tag{A}$$ differing by a factor of $2$ and $1/4$ in the second and third terms, respectively. I'm sure equation $(\mathrm{A})$ is the correct one as I derived eqn. $(3)$ from it. It is likely I am making mistakes in $(\mathrm{!})$, but I just cannot seem to find them.

May I please have some help in proving eqn. $(\mathrm{A})$ or identifying errors in $(\mathrm{!})$?

  • $\begingroup$ That post contains a lot of questions. "equation (2) does not even contain $A_\mu$" Imo it does. Raise the index of $A$ and contract with the metric and $A_\mu$ enters the stage. Similar things were discussed here. Please think about this and reduce this post to a parsimonious size. $\endgroup$
    – Kurt G.
    Feb 17 at 8:41
  • 1
    $\begingroup$ @KurtG. "That post contains a lot of questions." No, there is only one question here, and that question is how to prove eqn. $(\mathrm{A})$. I even put it in boldface for clarity. What other questions do you see in this post? "[...] reduce this post to a parsimonious size". If I do that I cannot show my working and hence convey my confusion/lack of understanding. Do you think I enjoy writing long posts? $\endgroup$ Feb 17 at 14:28
  • $\begingroup$ I have already highlighted a misconception you had. You should address that. I usually stop reading long posts when there is some disagreement about the basics. $\endgroup$
    – Kurt G.
    Feb 17 at 18:19
  • $\begingroup$ @KurtG. Thanks, I checked that link and am now very confused, in particular I don't know how to deal with derivatives of four-vectors, for instance, in the question of the link you posted I thought that $\frac{\partial (x_{\mu} x^{\mu})}{\partial x^{\mu}}=x_\mu$ and I don't know why the OP wrote $2x_\mu$. You call this "the basics" yet I have not come across it. Moreover, in your first comment in that link you write that $\frac{\eta_{\nu\rho}\partial (x^{\rho} x^{\nu})}{\partial x^{\mu}}$ and caution the OP to take care when doing the derivative, I would not know how to take the compute .. $\endgroup$ Feb 18 at 6:30
  • $\begingroup$ ... the derivative when the Minkowski metric is involved. For the case of my question this could explain the factor of $2$ in front of the $e^2A^\nu\phi^\ast\phi$ term if writing $A^\nu=\eta^{\sigma\nu}A_\sigma$. But that still doesn't explain the factor of $1/4$ in front of the field strength tensor, $F^{\mu\nu}$. While I acknowledge that hints/tips are more useful to someone when learning, but in this case could you please be at bit less cryptic? I mean, you've pointed out where I am going wrong but haven't yet said what is the correct thing to do (at least to start) to prove $(\mathrm{A})$. $\endgroup$ Feb 18 at 6:40

1 Answer 1


The key things that lead to the correct factors in (A) are

  • the basic expression $A_\mu A^\mu=\eta^{\mu\rho}A_\mu A_\rho=-A_0^2+A_1^2+A_2^2+A_3^2$ in (1) from OP that you have to differentiate w.r.t. each of $A_0,A_1,A_2,A_3\,.$ This differentiation obviously yields (basic calculus) $-2A_0,+2A_1,+2A_2,+2A_3$ respectively. This can be written as $$\tag{1} \frac{\partial}{\partial A_\nu}(A_\mu A^\mu)=2\eta^{\nu\rho}A_\rho=2A^\nu $$ (c.f. (A)).

  • To the term $F_{\alpha\beta}F^{\alpha\beta}=(\partial_\alpha A_\beta-\partial_\beta A_\alpha)(\partial^\alpha A^\beta-\partial^\beta A^\alpha)$ we are now going to apply the same basic rules: \begin{align}\tag{2} &(\partial_\alpha A_\beta-\partial_\beta A_\alpha)(\partial^\alpha A^\beta-\partial^\beta A^\alpha)= (\partial_\alpha A_\beta-\partial_\beta A_\alpha)\,\underbrace{\eta^{\alpha\rho}\eta^{\beta\sigma} (\partial_\rho A_\sigma-\partial_\sigma A_\rho)} \end{align} From $\eta^{\mu\nu}=\operatorname{diag}(-1,1,1,1)\,,$ the underbraced term is numerically equal to \begin{align} \cases{0&$\alpha=\beta\,,$\\[2mm] -(\partial_\color{red}{0} A_\beta-\partial_\beta A_\color{red}{0})\,,&$\beta=1,2,3\,,$\\[2mm] -(\partial_\alpha A_\color{red}{0}-\partial_\color{red}{0} A_\alpha)\,,&$\alpha=1,2,3\,,$\\[2mm] (\partial_\alpha A_\beta-\partial_\beta A_\alpha)\,,&$\alpha,\beta=1,2,3,\alpha\not=\beta\,.$\\ } \end{align} This leads to \begin{align} \\[2mm] (2)&=-F_{01}^2-F_{02}^2-F_{03}^2+F_{12}^2+F_{13}^2+F_{23}^2 -F_{10}^2-F_{20}^2-F_{30}^2+F_{21}^2+F_{31}^2+F_{32}^2\\[2mm] &=-2F_{01}^2-2F_{02}^2-2F_{03}^2+2F_{12}^2+2F_{13}^2+2F_{23}^2\,. \end{align} The last line uses again that $F_{\alpha\beta}$ is anti symmetric which makes $F_{\alpha\beta}^2$ symmetric. Differentiating this w.r.t. $\partial_\mu A_\nu$ for a fixed index pair $\mu,\nu$ gives $$ \frac{\partial F_{\alpha\beta}F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)}=\cases{-4\,F_{\color{red}{0}\nu}=-4\,(\partial_\color{red}{0} A_\nu-\partial_\nu A_\color{red}{0})\,,&$\nu=1,2,3\,,$\\[2mm] -4\,F_{\mu\color{red}{0}}=-4\,(\partial_\mu A_\color{red}{0}-\partial_\color{red}{0} A_\mu)\,,&$\mu=1,2,3\,,$\\[2mm] 4\,F_{\mu\nu}=4\,(\partial_\mu A_\nu-\partial_\nu A_\mu)\,,&$\mu,\nu=1,2,3\,.$} $$ This can be written as $$ \frac{\partial F_{\alpha\beta}F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} =4\,\eta^{\mu\rho}\eta^{\nu\sigma}\,F_{\rho\sigma}=4\,F^{\mu\nu}=4\,(\partial^\mu A^\nu-\partial^\nu A^\mu)\,. $$ This confirms again what we see in (A): the $\frac14$ cancels and a minus sign comes from $-\partial_\mu\frac{\mathcal{L}}{\partial(\partial_\mu A_\nu)}\,.$

  • $\begingroup$ and similarly for the $\mu=0$ case in $\frac{\partial F_{\alpha\beta}F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)}$ expression. Otherwise your answer makes perfect sense. Sorry haven't had time to look at your answer until today. Thank you. $\endgroup$ Mar 1 at 7:35
  • $\begingroup$ Thanks for the editing, now it is clear. Just one more question for you if I may, in my (now deleted answer) I wrote $\frac{\partial (\eta_{\color{red}\nu\rho}\,x^{\rho}\,x^{\color{red}\nu})}{\partial x^{\color{red}\mu}}=-\frac{\partial \left(x^0\right)^2}{\partial x^0}+\frac{\partial \left(x^1\right)^2}{\partial x^1}+\frac{\partial \left(x^2\right)^2}{\partial x^2}+\frac{\partial\left(x^3\right)^2}{\partial x^3}=2\left(-x^0+x^1+x^2+x^3\right)=2\eta_{\sigma\mu}x^\sigma=2x_\mu$ and you wrote a comment below saying.... $\endgroup$ Mar 2 at 4:09
  • $\begingroup$ .... "The expressions $-\frac{\partial \left(x^0\right)^2}{\partial x^0}+\frac{\partial \left(x^1\right)^2}{\partial x^1}+\frac{\partial \left(x^2\right)^2}{\partial x^2}+\frac{\partial\left(x^3\right)^2}{\partial x^3}$ and $2(-x^0+x^1+x^2+x^3)$ are garbage" Do you mean 'garbage' as in unnecessary steps that are not needed or garbage as in mathematically incorrect? I'm sure what I wrote was perfectly fine mathematically, but now you have me questioning this. $\endgroup$ Mar 2 at 4:10
  • $\begingroup$ @SiriusBlack let me know if you cannot see that deleted answer so that I can better refer to it. Hints: that derivative is not a sum over $\mu\,$ and your end result $2x_\mu$ needs a minus sign when $\mu=0.$ $\endgroup$
    – Kurt G.
    Mar 2 at 4:48
  • $\begingroup$ Thanks, I can see the deleted answer, but I don't think you can. I have therefore copied the contents of that answer as a new answer (it's really a comment, but too long). It should never have been deleted in the first place. $\endgroup$ Mar 2 at 5:03

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