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

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


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

and obtain the same solution?


1 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.

  • $\begingroup$ 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? $\endgroup$ May 17, 2022 at 18:21
  • 1
    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    May 17, 2022 at 19:07
  • $\begingroup$ why would we want it as a two form, we could equally solve the Euler Lagrange equations for the rank (1,1) tensor $\partial_{\mu}A^{\nu}$ $\endgroup$ May 17, 2022 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.