Faraday was mathematically untrained and conceptualized fields in terms of lines of force. Maxwell reformulated this into what we would today describe as vector fields. My question is about formalizing the connection between these two pictures.
On a smooth manifold, there is something that Lee calls the fundamental theorem on flows. Basically it says that given a smooth vector field, we can define a smooth flow (but not always globally, because we may hit boundaries, etc.). It seems more trivial to show that given a flow, you can define a vector field by differentiating the flow with respect to the parameter.
This could sort of be considered a formalization of the notion of going from Maxwell's picture to Faraday's and vice versa. However, in the Faraday picture, we don't really get all the info about the flow, but only about the orbits, i.e., the images of the integral curves, omitting the information about the parametrization.
Is there a more satisfactory way of formalizing this connection between the Faraday and Maxwell pictures? Does it only work if the field is divergence-free? Can it be divergence-free everywhere except at certain holes, like point charges?
Example 1: Intuitively, I can imagine, say, having a charge +37 and a charge -11 at two points in space. I could pick n=100 unit vectors pointing outward from the +37 charge and distributed approximately uniformly (or just randomly) on the unit sphere. Then I could integrate all of the relevant orbits on a computer. I would find that about 11/37 of these terminated on the -11 charge, while the rest went off to infinity. Is there any reason to think that this process converges in some sense to a correct result when n approaches infinity, and that we can also reverse the process, and recover the original field information, with an error that approaches zero in the same limit?
Example 2: Here is an example of why it seems to me like we need some kind of condition on the divergence. Consider the fields $\textbf{F}=\textbf{r}/r^3$ (Newtonian gravity) and $\textbf{G}=\textbf{r}e^{-r}/r^3$ (a Yukawa field). $\textbf{F}$ is divergence-free except at the origin (which I guess would show up as a hole in the manifold), but $\textbf{G}$ isn't. The orbits of the two fields are identical, so we can't recover the fields from the Faraday picture.
Edit: Flows may not be the right way to formalize Faraday's intuition, in which case I'd be interested in other proposals for how to do so. It seems as though we actually get more information from discretization, as in example 1, than from the images of all the flow lines. For one thing, the images of the flow lines are invariant under a transformation $\textbf{F}\rightarrow c \textbf{F}$. If you zoom in on the discretized picture, you can infer the magnitude of the field from the density of the field lines, but it seems like you lose that ability if you pass directly to the limit of a continuum of field lines.
What I have written below is speculative, proving that it works would take some serious efforts and might make a very nice PhD thesis. Nevertheless, at least it restates Faraday's physical intuition in the form of a provable/refutable mathematical conjecture.
Let $X$ denote the space of all smooth nowhere vanishing vector fields on some domain $D$ the 3-dimensional space $R^3$. (A caveat: I am simplifying quite a bit, as one needs to allow vector fields with zeroes as well as with singularities in order to get a physically meaningful statement. If you like, you can assume $D$ is the complement to a finite subset in $R^3$ which contains all possible zeroes and blow-up points of the vector fields. However, this is still cheating.) Let $Y$ denote the space of smooth 1-dimensional oriented foliations on $\Omega$. Each foliation is simply a family of smooth pairwise disjoint consistently oriented curves in $D$, which are trajectories of vector fields on $D$. There is a map $$ f: X\to Y $$ which sends each vector field to its family of trajectories. One can (and I will!) equip $X$ and $Y$ with structures of smooth infinite dimensional manifolds; the tangent spaces to both manifolds are isomorphic to spaces of vector fields (possibly vanishing somewhere) on $D$. One can verify that the map $f$ is smooth (has derivatives of all orders). The trouble is that $f$ is not one-to-one: The set of trajectories $y=f(x), x\in X$, does not uniquely determine the vector field $x$: The field $x$ is only unique up to reparameterizations of its trajectories.
Conjecture: There exists an open and dense subset $Y'\subset Y$ such that the restriction of $f$ to $f^{-1}(Y')$ is a fiber bundle whose fibers $F_y$ are diffeomorphic to spaces of smooth maps $$ \psi: D\times {\mathbb R}\to {\mathbb R} $$ such that for each $p\in D$ the map $\psi: \{p\}\times {\mathbb R}\to {\mathbb R}$ is monotonically increasing.
You might not know what a fiber bundle is, in which case just think of the projection (to the second factor) map $$ U\times V\to V, $$ where $U$ and $V$ are some infinite dimensional vector spaces.
Now, suppose that $M\subset X$ is some finite dimensional smooth submanifold. You can think of $M$ as some family of vector fields $x_a$ on $D$ parameterized by the parameter $a$ which belongs to some domain in ${\mathbb R}^n$. Physically (I am cheating here, but not too much!) you can think of each $x_a$ as the gravitational fields determined by $k$ massive point-objects located in $D$. Then the number $n$ of parameters is $(3+1)k$: Each 3 spacial parameters for each point-object plus one parameter for its mass.
One can then prove (subject to the Conjecture above) that for "generic" choice of the submanifold $M$ (think of choosing points and masses in some generic fashion), the restriction of $f$ to $M$ is one-to-one. In other words, if we restrict to the vector fields $x$ which are from the family $M$ then the set of trajectories of $x$ uniquely determines the vector field. (If $x, x'$ are two vector fields in $M$ which have the same set of trajectories the $x=x'$.)
In order to get why this might be true, think of $X$ as $R^4$, $Y'$ as a hyperplane in $R^4$ and $f$ as the orthogonal projection $X\to Y$. Then for a generic simple curve $M\subset R^4$ the projection $f(M)$ is a simple curve in $R^3$. (This of course will not work if I consider curves in $R^3$ and project them to a hyperplane, but remember that the actual $X$ and $Y'$ are both infinite-dimensional, while $M$ is finite dimensional.)