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

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?

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.
• 1. Interesting, so if we treat the metric, $g_{\mu\nu}$ as a constant between frames, it doesn't matter which one we use? 2. Is there any reason why we conventionally use $\partial_{\mu}A_{\mu}$, other than the fact that it makes the problem much easier to solve? May 17, 2022 at 18:21
• 1. Yes, except that we don't 'treat' the metric, which may wrongly suggest that we have options. For fixed coordinate system, the metric components are what they are. 2. Yes, to make $F$ a closed 2-form. May 17, 2022 at 19:07
• why would we want it as a two form, we could equally solve the Euler Lagrange equations for the rank (1,1) tensor $\partial_{\mu}A^{\nu}$ May 17, 2022 at 21:25