I need to verify the 4-current conservation from the Euler-Lagrange equations for $A^\mu$. The Lagrangian defining electromagnetic field is: $$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-j^{\mu}A_\nu$$ I derived E-L EOM as: $$\partial_\mu F^{\mu\nu}= j^\nu.$$ Then I did the following:

$$\partial_\nu j^\nu= \partial_\nu \partial_\mu F^{\mu\nu}= -\partial_\mu \partial_\nu F^{\nu\mu}= -\partial_\mu F^\mu=0$$

This verifies the conservation law for current. But can I contract the electromagnetic tensor in that way?

  • 2
    $\begingroup$ What is $F^{\mu}$? $\endgroup$
    – Qmechanic
    May 13, 2018 at 16:00
  • $\begingroup$ I just got it after contracting F$^{\nu\mu}$ with $\partial_\nu$. I know it has no meaning, but is that step allowed? $\endgroup$
    – Zen
    May 13, 2018 at 16:06

2 Answers 2


First of all, you defined

$$F^\mu = \partial_\nu F^{\nu\mu},$$

you can do it, but from the equation you found $\partial_\mu F^{\mu\nu} = j^\nu$ it should be obvious that

$$F^\mu = j^\mu.$$

Thus, you wanted to show $\partial_\nu j^\nu = 0$ and you used that $\partial_\mu F^\mu = 0$. In other words, you have shown that $\partial_\nu j^\nu =0$ by assuming it. That's clearly wrong.

If you had notice that $F^\mu = j^\mu$ and said: well, I've found that $\partial_\mu j^\mu = -\partial_\mu j^\mu$ and hence it must be zero, then it would be fine.

Another way, which ends up the same thing, would be as follows. You noticed that

$$\partial_\mu F^{\mu\nu}=j^\nu.$$

Hence, you have

$$\partial_\nu j^\nu = \partial_\nu \partial_\mu F^{\mu\nu}.$$

Now notice that $F^{\mu\nu}$ is skew-symmetric, while $\partial_\nu\partial_\mu$ is symmetric. Thus you are contracting something symmetric with something skew-symmetric. That gives you zero and hence

$$\partial_\nu j^\nu = 0.$$

As for the last piece of argument, let $a_{\mu\nu}$ symmetric and $F^{\mu\nu}$ skew-symmetric. Now consider $a_{\mu\nu}F^{\mu\nu}$. Since $F^{\mu\nu}$ is skew symmetric you have


and since $a_{\mu\nu}$ is symmetric you have


These are dummy indices however, summed over. So on the LHS, rename $\mu\leftrightarrow \nu$ (this is just a ralabeling). With it


and thus the number $a_{\mu\nu}F^{\mu\nu}$ is equal to minus itself, so it can only be zero.


Tensors can indeed be contracted such that a rank-(2,2) tensor $T^{ab}_{\phantom{ab} bc}=W^{a}_{\phantom{a} c}$, where your rank-(2,2) tensor is $\partial_{\nu}\partial_{\mu}F^{\mu\nu}$. However this expression comes from the definition of the 4-current $j^{\nu}$, and by contracting the tensor in that way you are simply going back to the initial problem you need to prove, without adding any new logical steps to actually prove the conservation itself.

A straightforward way to easily achieve the result you want is instead simply considering explicitly the expression for $F^{\mu\nu}$ $$F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ To then prove that $\partial_{\nu}j^{\nu}=0$ (and remembering the commutativity of partial derivatives) \begin{align} \partial_{\nu}\partial_{\mu}F^{\mu\nu}&=\partial_{\nu}\partial_{\mu}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})\\ &=\partial_{\nu}\partial_{\mu}\partial^{\mu}A^{\nu}-\partial_{\mu}\partial_{\nu}\partial^{\nu}A^{\mu}\\ &=\partial_{\nu} \Box A^{\nu}-\partial_{\mu} \Box A^{\mu} \end{align} where $\Box \equiv \partial_{\alpha}\partial^{\alpha}=\partial^{\alpha}\partial_{\alpha}$ is the d'Alembert operator. Since the $\mu$ and $\nu$ are only dummy indices, $\partial_{\nu} \Box A^{\nu}=\partial_{\mu} \Box A^{\mu}$ therefore their subtraction yields zero, and the current is conserved.


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