Raising and lowering indices

Question

I read multiple sources about raising and lowering indices and Einstein summary notation but am having problems doing it.

Here is an example. I am trying to derive Maxwell's inhomogeneous equations in the Minkowski vacuum \begin{equation} \partial_\mu F^{\mu\nu} = 0 \end{equation} using the alternative form in curved space: \begin{equation} \frac{1}{\sqrt{-g}}\partial_\mu\Bigg[(\sqrt{-g\,}g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}\Bigg] = 0 \end{equation} Substituting $g^{\phi\theta} = \eta^{\phi\theta} = \text{diag} (+,-,-,-)$: \begin{equation} \partial_\mu F_{\alpha\beta} = 0 \end{equation}

What is the proper way to convert this to $\partial_\mu F^{\mu\nu} = 0$ or $\partial_\alpha F^{\alpha\beta} = 0$ or $\partial_\lambda F^{\lambda\kappa} = 0$? I keep getting a mess.

edit:

I am at a level of trying to make sure I am doing the mechanics properly. I think in the back of my mind I knew I could contract indices to get your alternative form. Here is my messy approach: \begin{equation} \eta_{\mu\phi} \: \partial^\phi F_\alpha \, ^\rho \: \eta_{\beta\rho} = 0 \end{equation} \begin{equation} \eta_{\mu\phi} \: \eta^{\phi\sigma} \: \partial_\sigma F_\alpha \, ^\rho \: \eta_{\beta\rho} = 0 \end{equation} \begin{equation} \eta_{\mu\phi} \: \eta^{\phi\sigma} \: \partial_\sigma \: \eta_{\lambda\alpha} \: F^{\lambda\rho} \: \eta_{\beta\rho} = 0 \end{equation} Substituting $\eta^{\phi\theta} = \text{diag} (+,-,-,-)$: \begin{equation} \partial_\sigma F^{\lambda\rho} = 0 \end{equation} Then I would keep raising and lowering and substituting until I get the preferred form, if ever. I assume I am missing the proper approach.

Answer 1

You are overthinking this.

Multiplication by a $g$ tensor is what's usually meant by 'raising' or 'lowering' indices. Depending on how you structure your axioms, it is possible to take $g^{\mu\alpha}F_{\alpha\beta} = F^\alpha_\beta$ as a definition of what $F$ with an upper index means.

Generally, $F$ is defined as a covariant (bottom indices) tensor. This has various nice properties, chief among which being that it also makes $F$ a 2-form, which is a notion that is invariant under coordinate changes. The presence of a metric makes it possible to convert covariant objects to contravariant ones, so the top line could be equivalently written as

$$\frac{1}{\sqrt{-g}} \partial_\mu \left[\sqrt{-g} F^{\mu\nu} \right]$$ The only reason it is not written like that is to make the metric tensor dependence explicit, while leaving $F_{\mu\nu}$ in its 'natural' state.