We are given the following Lagrangian density: $$\mathcal{L}=F_{\mu \nu} A^{\mu} \mathcal{J}^{\nu}$$ where $F_{\mu \nu}$ is the electromagnetic field tensor, $ A^{\mu}$ the 4-vector of the vector potential and $\mathcal{J}^{\nu}$ is the 4-vector of current density.

By making use of the Euler-Lagrange equations, determine the differential equations describing the systems' evolution over time.

Euler-Lagrange equation: $$\frac{\partial \mathcal{L}}{\partial A_\alpha} - \partial_{\rho} \left( \frac{\partial \mathcal{L}}{\partial(\partial_{\rho}A_{\alpha})}\right)=0.$$


(1) Finding $\frac{\partial \mathcal{L}}{\partial A_\alpha}$ is, I believe, straightforward since there is only one explicit dependence on $A$:

$$\frac{\partial \mathcal{L}}{\partial A_\alpha}=F_{\alpha \nu} \mathcal{J}^{\nu}$$ where $\mu = \alpha.$

(2) Then, let's write the EM tensor as $$F_{\mu \nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$

We know that $$\frac{\partial(\partial_{\mu}A_{\nu})}{\partial(\partial_{\rho}A_{\alpha})}=\delta_{\mu}^{\rho}\delta_{\nu}^{\alpha},$$ since the indices must coincide for the derivative to exist.

What does the location of the indices mean on the Kronecker delta? I have seen both up/down and up/up.

Then, evaluating the derivative, I get:

$$(\delta_{\mu}^{\rho}\delta_{\nu}^{\alpha}-\delta_{\nu}^{\rho}\delta_{\mu}^{\alpha}) A^{\mu} \mathcal{J}^{\nu}.$$

From this point on, I'm not really sure how to continue. Can I simply evaluate the expression for when indices are equal and not equal to each other? I can't see how this would yield a general equation of motion. Else, can I plug in the metric tensor via the Kronecker deltas?

  • $\begingroup$ The general rule is that a derivative with respect to an up index is a down index and vice versa. Some people are sloppy with index placement though, since it doesn't matter (you can always fix it at the end). In order to continue, what's wrong with just taking $\partial_\rho$? Note that $\partial_\rho \delta^\rho_\mu = \partial_\mu$, etc. $\endgroup$
    – knzhou
    Jan 9, 2018 at 12:17
  • $\begingroup$ Do you mean, like Per suggested in his answer below, in part (1) of my attempt it should be F$^{\alpha}$ $_{\nu}$? Thanks for the suggestion on taking $\delta_{\rho}$. Please see my answer below for what I end up with. $\endgroup$
    – Samalama
    Jan 9, 2018 at 17:04

2 Answers 2


You use Kronecker delta as a Kronecker delta, it is one when indices are equal otherwise zero.

You might have made a mistake, note that the Lagrangian may be written as ${\cal L} = {F^{\mu}}_{\nu}A_{\mu}J^{\nu}$, which helps before evaluating the derivative $\frac{\partial \cal L}{\partial A_{\alpha}}$

  • $\begingroup$ Thanks for responding! You rewrote the Lagrangian using by lowering a single index using the metric tensor, correct? When I do this explicitly, I end up with a different variable than $\mu$ but I guess since this is just a dummy, we can change it back to $\mu$ at the end? I'm not yet very confident with operating on different indices, so I'm not really sure how else to continue or how re-writing the Lagrangian as you suggested might change things. $\endgroup$
    – Samalama
    Jan 9, 2018 at 16:50
  • $\begingroup$ @Samalama Don't neglect the position of the index, up or down. When you take the derivative with respect to $A_{\alpha}$, you want $\cal L$ to be written the way I showed you, with the index of A in the same position as $A_{\alpha}$. Then the result will have the index $\alpha$ up, which it should have. Note that when you have two equal indices, one is always upper and the other lower, e.g. $\partial_{\mu}A^{\mu}$. $\endgroup$
    – Per Arve
    Jan 10, 2018 at 17:16

Following advice on indices I continue from where I stopped above and take the derivative $\partial_{\rho}$. As suggested by knzhou, $\partial_{\rho} \delta^{\rho}_{\mu}=\partial_{\mu}$. Looking at the first Kronecker delta product in my bracket, I need to specify $\rho=\mu$ and $\alpha = \nu$ for this term to exist. By doing this, I automatically lose the second Kronecker delta product in the bracket, such that

$$ \partial_{\rho} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}A_{\alpha})} \right)=\partial_{\mu}A_{\mu}\mathcal{J}^{\nu}$$

Finally, I get

$$ \mathcal{J}^{\nu}(\partial^{\alpha}A_{\nu}-\partial^{\nu}A_{\alpha}-\partial_{\mu}A_{\mu})=0$$

$$ {F^{\alpha}}_{\nu}=\partial_{\mu}A_{\mu}$$

If this is correct (wouldn't be surprised if it's wrong), I feel like this is no final answer? Can it be simplified or neatened up somehow?

  • $\begingroup$ This shouldn't be posted as an "answer". Please edit it into your question, then delete this answer. $\endgroup$
    – J.G.
    Feb 15, 2022 at 10:44

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