Finding equation of motion of Lagrangian density: What does the location of the indices mean?

Question

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

Attempt:

(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?

Per Arve · Answer

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  $frac{partial cal L}{partial A_{alpha}}$

Samalama · Answer

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?

Finding equation of motion of Lagrangian density: What does the location of the indices mean?

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