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?

Answer 1

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

Answer 2

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?