Relativistic Euler-Lagrange equations for a four-vector (or one-form) field

Question

I think the best way to ask my question is by considering the maxwell-Lagrangian,

$$\mathcal{L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}=-\frac{1}{2}(\partial^{\mu}A^{\nu}\partial_{\mu}A_{\nu}-\partial^{\mu}A^{\nu}\partial_{\nu}A_{\mu}).$$

For scalar fields, $\phi_{i}$, we have Euler-Lagrange equations

$$\frac{\partial\mathcal{L}}{\partial \phi_{i}} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{i})}\right)=0.$$

I'm not sure i fully understand what we have for a four vector field. I initially learnt that it would be this

$$\frac{\partial\mathcal{L}}{\partial A^{\mu}} - \partial_{\nu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A^{\mu})}\right)=0.$$

(obviously you would need to change the indices in $\mathcal{L}$ to something other than $\mu$ and $\nu$ before calculating.)

My question is, does it matter whether we treat $A^{\mu}$ as a vector or can we apply the Euler Lagrange equations to $A_{\mu}$. I.e, is this expression

$$\frac{\partial\mathcal{L}}{\partial A_{\mu}} - \partial^{\nu}\left( \frac{\partial \mathcal{L}}{\partial(\partial^{\nu}A_{\mu})}\right)=0~?$$

Equally, does it matter whether or the differential $\partial$ and the vector field $A$ are of opposite form, (i.e, if one is one form should the other necessarily be 4-vector). Or can we equally solve

$$\frac{\partial\mathcal{L}}{\partial A_{\mu}} - \partial_{\nu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})}\right)=0$$

or

$$\frac{\partial\mathcal{L}}{\partial A^{\mu}} - \partial^{\nu}\left( \frac{\partial \mathcal{L}}{\partial(\partial^{\nu}A^{\mu})}\right)=0$$

and obtain the same solution?

Answer 1

1. It may matter if the metric components $g_{\nu\lambda}=g_{\nu\lambda}(x)$ [that we use to raise and lower indices with] depend on the spacetime coordinate $x^{\mu}$. This happens e.g. in GR.

2. In this case, a spacetime derivative $\partial_{\mu}:=\frac{\partial}{\partial x^{\mu}}$ and the gauge potential $A_{\mu}$ should have a lower/sub-index by standard convention.

3. In particular, $\partial^{\mu}:=g^{\mu\nu}\partial_{\nu}$ and $A^{\mu}:=g^{\mu\nu}A_{\nu}$ with an upper/super-index are composite objects by standard convention.

4. Apropos EL equations and differentiation, see also this & this related Phys.SE posts.