Given an electric potential in curved spacetime, how to calculate the total electric charge?

Question

In general $d$ dimensional setup ($d-1$ spatial coordinates and $1$ temporal), with a nontrivial metric $g_{\mu\nu}(t,\vec{r})$, having for simplicity a static electric potential $\phi(\vec{r})$. How to calculate the electric charge that is the source of this potential?

I know that when considering classical physics, the charge $Q$ is given by the spatial integration on the laplacian of the potential over all the space $V$:

$$Q=-\int_V \nabla^2\phi(\vec{r})d\vec{r}$$

Three options make sense to me when moving to curved spacetime:

1. There is no change at all and we can calculate it as the classical integration above
2. All the components of the metric should be taken into account according to $$Q = -\int_V d^{d-1}r \sqrt{-g}g^{\mu\nu}\nabla_\mu\partial_{\nu}\phi(\vec{r})$$ Here $\mu,\nu\in\{\vec{r},t\}$ Where $\nabla_\mu$ is the covariant derivative which operate on a vector field $\partial_{\nu}\phi(\vec{r})$ as follow: $$\nabla_\mu\partial_{\nu}\phi(\vec{r})=\partial_\mu\partial_{\nu}\phi(\vec{r})-\Gamma^{\alpha}_{\mu\nu}\partial_{\alpha}\phi(\vec{r}) $$ and $g=\det{g_{\mu\nu}}$
3. Only the spatial part of the metric is joining the spatial integration: $$Q = - \int_V d^{d-1}r \sqrt{-\tilde{g}}\tilde{g}^{\mu\nu}\nabla_\mu\partial_{\nu}\phi(\vec{r})$$ Here $\mu,\nu\in\{\vec{r}\}$ Where $\tilde{g}$ represents the spatial part of the metric, such that, the full metric is given by: $$g_{\mu\nu}=\begin{pmatrix} g_{tt} & g_{t\vec{r}} \\ g_{\vec{r}t} & \tilde{g}_{\vec{r}\vec{r}} \\ \end{pmatrix}$$

From (1)-(3) what is the right way to calculate the total charge?

Answer 1

The invariant equation for the induction field (tangent 2-form at the surface of a conducting sphere) is

$$D = D_{ik}(t,x) dx^i\wedge dx^k : \quad D = \frac {q}{4 \pi r^2} \sin \theta d\phi\wedge d \theta$$

It counts the charge inside any sphere by

$$\int_{r*S^2} D = \int_{\phi,\theta \in S^2} \frac {q}{4 \pi } \sin \theta d\phi\wedge d \theta = q $$

D has a Hodge dual, metric product with the volume 4-form, independent of $r$ with zero exterior derivative.

$$E=*D = \left< dct\wedge dr \wedge r \sin \theta d\phi \wedge r d\theta ,D\right> = \frac {q}{4 \pi} dct \wedge dr $$

with $d E = 0$ in vacuum.

Exctly these set of equations can be transferred to any pseudo-Riemann 4-manifold.

If the manifold is spacelike open, the integral over any n-1-sphere of a Cauchy (spacelike) n-space (at constant time) yields the charge inside.

If the manifold has no spacelike boundaries and is simply connected, the total charge must be zero.

Confer Misner/Thorne/Wheeler, Gravitation chpt 2: Forms and electromagnetism.

The classical expositions by mostauthors, using Riemannian manifolds with vector fields, are as confusing, as are the classical tables of vector field differential algebra for the student.