2
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

  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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Commented 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. $\endgroup$ Commented 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 $\partial_{\mu}A^{\nu}$ $\endgroup$ Commented May 17, 2022 at 21:25

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.