# How to raise and lower indices of field equations in curvilinear coordinates?

I am confused about a basic question. Consider the field equation of a gauge theory written in terms of curvilinear coordinates $$\frac{1}{\sqrt{-g}}D_\mu(\sqrt{-g}F^{\mu\nu})=j^\nu.$$ My questions are:

1. Do the partial derivatives in the gauge covariant derivative and the field strength remain partial, that is $D_\mu=\partial_\mu+ieA_\mu$ and $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$, or they became covariant, $D_\mu=\nabla_\mu+ieA_\mu$ and $F^{\mu\nu}=\nabla^\mu A^\nu-\nabla^\nu A^\mu$ ?
2. I would like to lower all indices. Why should I write $$\frac{1}{\sqrt{-g}}D_\mu(\sqrt{-g}g^{\mu\sigma}g^{\nu\rho}F_{\sigma\rho})$$ instead of $$\frac{1}{\sqrt{-g}}g^{\mu\sigma}g^{\nu\rho}D_\mu(\sqrt{-g}F_{\sigma\rho}),$$ when the metric is position dependent? If the partial derivatives become covariant, then the metric can go in and out freely but I think that the derivatives are partial.
• With the explicit factor of $\sqrt{-g}$ in there, both are correct since $\nabla_\mu \sqrt{-g} = 0$. – Prahar Dec 15 '17 at 12:18

The general recipe for getting a curved space Lagrangian is to replace partial derivatives with covariant ones $\partial_\mu \rightarrow \nabla_\mu$. However, in the case of the electromagnetic field strength tensor it doesnt matter because it is antisymmetric. $\nabla_\mu A_\nu = \partial_\mu A_\nu -\Gamma_{\mu\nu}^\lambda A_\lambda$. The second term is symmetric in the interchange of $\mu$ and $\nu$ and so doesn't contribute to $F_{\mu\nu}$.

The action in curved space is

\begin{equation} S = \int\mathrm d^4x\sqrt{-g}\mathcal L \end{equation}

Just like in flat space we use Stokes' theorem to derive the equations of motion. You can do this in two ways. You can use the curved space version in which case

\begin{equation} \int\mathrm d^4x\sqrt{-g}\nabla_\mu(...) \end{equation} gives a boundary term which we throw away, leading to the equation of motion of type

\begin{equation} \nabla_\mu F^{\mu\nu} + ... \end{equation} Alternatively you can treat $\sqrt{-g}$ as some function and use the usual, flat space version of Stokes' theorem where the boundary term is

\begin{equation} \int\mathrm d^4x\partial_\mu(...). \end{equation} Then the equation of motion will be of the form

\begin{equation} \frac{1}{\sqrt{-g}}\partial_\mu \left(\sqrt{-g}F^{\mu\nu}\right) + ... \end{equation} Both of these will, of course, give the same equation of motion.

• Very elucidating, thanks. How about the second question? It does matter whether the metric is in or out the derivatives since it is position dependent in general. – Diracology Dec 15 '17 at 12:45
• If you are using covariant derivatives then you can freely commute the derivative and the metric. If you are using partial derivatives then you cannot and must instead take derivatives of the metric as well. – Philo Dec 15 '17 at 12:55
• Why my second equation (metric inside the derivative) is correct while the third is wrong? Both using partial derivatives. Is it because $\partial_\mu(\sqrt{-g}F^{\mu\nu})$ does not transform as a four vector when using general curvilinear coordinates? – Diracology Dec 15 '17 at 14:28
• Yes, that's one way of looking at it. On a purely mathematical level it is because $\partial_\mu g_{\alpha\beta} \neq 0$ so you cannot commute the metric and the partial derivatives. In contrast, $\nabla_\mu g_{\alpha\beta} = 0$ so you can commute the metric past the derivative. – Philo Dec 15 '17 at 15:33