Question about electric flux in curved spacetime

Question

I just started physics II (electricity and magnetism) and in learning about Gauss's Law, I came across this definition on Wikipedia: $$\Phi_E=c\oint_{S}F^{\kappa0}\sqrt{-g}\ dS_{\kappa}$$ Where $c$ is the speed of light, $F^{\mu\nu}$ is the electromagnetic field tensor $$F^{\mu\nu}= \begin{bmatrix} 0 & -E_{x}/c & -E_{y}/c & -E_{z}/c \\ E_{x}/c & 0 & -B_{z} & B_{y} \\ E_{y}/c & B_{z} & 0 & -B_{x} \\ E_{z}/c & -B_{y} & B_{x} & 0 \end{bmatrix}$$ and $g$ is the determinant of the metric tensor $g_{ij}$. Below the statement of the formula the article explains the $dS_{\kappa}$ term, which I am unfamiliar with. I am assuming this is a differential form that is also a tensor, but I'm not sure. $$dS_{\kappa}=dS^{ij}=dx^{i}dx^{j}$$ From what I understand about tensors, $F^{\kappa0}$ is just the first column of $F^{\mu\nu}$ since the zero in the second index holds the column constant. $$F^{\kappa0}=\begin{bmatrix} 0 \\ E_{x}/c \\ E_{y}/c \\ E_{z}/c \end{bmatrix}$$ Going back to the initial integral, it seems like the shared index kappa implies using Einstein notation for $F^{\kappa0}$ and $dS_{\kappa}$. $$F^{\kappa0}dS_{\kappa}=\sum_{m=0}^{3}{F^{m0}dS_{m}} =F^{00}dS_{0}+F^{10}dS_{1}+F^{20}dS_{2}+F^{30}dS_{3}$$

Which would mean $$cF^{\kappa0}dS_{\kappa}=(0)dS_{0}+E_{x}dS_{1}+E_{y}dS_{2}+E_{z}dS_{3}$$ Someone on math stack exchange clarified the notation because I thought the $E_{x},E_{y},E_{z}$ terms denoted partial derivatives, but apparently these are just the components of $\vec{E}$.

If that is the case and $F^{\kappa0}=(E^{0},\vec{E})$, $dS_{\kappa} = (dS_{0}, d\vec{S})$ the above looks like $\vec{E}$ dotted with $d\vec{S}$

i.e. $$cF^{\kappa0}dS_{\kappa}=\vec{E}\cdot d\vec{S}$$ Which would mean that the initial definition of $\Phi_{E}$ can be re-written as $$c\oint_{S}\sqrt{-g}F^{\kappa0}dS_{\kappa}=\oint_{S}\sqrt{-g}\vec{E}\cdot d\vec{S}$$

I have almost no experience with tensors so I was wondering if this is what is implied by this definition of Gauss's law, but this definition is just more general. Maybe this is obvious but I'm not sure. I'm also curious about a few things, namely:

Why is $g$ negative under the root?

Is $\sqrt{-g}$ functioning in a similar way to the determinant of the Jacobian, used for integrating in a change of coordinates?

What exactly is the differential $dS_{\kappa}$?

I'm also curious what I can look into to learn more about this as I find it pretty interesting. Thanks for any responses.

Answer 1

I'll presume you know what a metric is, so I'll just build on that. Imagine you have a manifold $(M,g)$ of dimension $m$. Let's say we are working on a coordinate system $X^0,X^1,\cdots ,X^{m-1}$ on $M$. We can define a volume element on $M$ as $$dV_M = \sqrt{|g|} \epsilon_{ab\cdots l}dX^adX^b\cdots dX^l$$where $|g|$ is the absolute value of determinant of metric $g$. $\epsilon_{ab\cdots l}$ is the Levi-Civita symbols (see https://en.wikipedia.org/wiki/Levi-Civita_symbol#). However, I'll mention that $g$ in your flux expression is not the Minkowski metric here , unlike what @Ghoster has mentioned in the comment, rather it's an induced metric wrt a stationary observer in a (+---) metric convention. The reason for this is that the hypersurface $S$ over which we are estimating the flux is 2 dimensional therefore the area element whould have 4-2=2 indices rather than 1. Now, what is an induced metric? In this case we can think of hyper-surface $N\subset M$ . $N$ could represent the plane of simultaneity of the stationary observer. Let's say ({$x^a$}) with $a=0,1,\cdots m-2$ represents the coordinates on $N$. On this hypersurface , you can write $dX^a=\frac{\partial X^a}{\partial x^p}dx^p$ , and then the line element on $N$ would look like $g_{ab}dX^adX^b|_N = h_{ab}dx^adx^b$ . This $h$ is essnetially the $g$ in your electric flux expression. You can workout the expression of $h$ starting from a Minkowski line element and see that $det(h)<0$, and therefore, you need a minus sign inside the square-root to make up for that. Indeed a stationary observer will have a velocity field $v^a=(1,0,0,0)$ and so you can choose $x^a=X^a$ ($a\neq 0$) which will ensure $h_{ab} = diag(-1,-1,-1)$ as one should expect. In general, for a curved space-time, you can choose $v^a$ to be any unit time-like vector and then choose $x^a$ orthogonal to $v$.

Now, once we are in hypersurface $N$, we can write the volume element as $$dV_N = \sqrt{|h|}\epsilon_{ab\cdots r}dx^a\cdots dx^r$$ Let's take the choice $x^a=X^a$ ($a\neq0$) as considered above, these are my spatial coordinates with basis , say, ${e^a}_k = \delta^a_k$ ($k$-th component of $a$-th basis), then the quantity $$dS_k = {e_k}^a\epsilon_{ab\cdots r}dx^b\cdots dx^r$$ Then you will see that ${e^k}_adS_k=0$ due to anti-symmetry property of $\epsilon$. The density $\sqrt{|h|}dS^k$ represents the area element on the 2-dimensional surface whose unit normal vector is ${e^k}_a$.

And indeed this $\sqrt{|g|}$ acts as a Jacobian since a coordinate transformation from $X^0,\cdots ,X^{m-1} \to Y^0,\cdots ,Y^{m-1}$ will keep $dV_M$ invariant $$\sqrt{|g(X)|}\epsilon_{ab\cdots l}dX^a\cdots dX^l=\sqrt{|g(Y)|}\epsilon_{pq\cdots r}dY^p\cdots dY^r$$.