# Magnetic field “lines” in $D \ne 4$ spacetimes

The electromagnetic field is represented by an antisymetric tensor $$F_{ab} = -\, F_{ba}$$ (the faraday). In $$D = 4$$ spacetimes, it has 6 independent componenents: 3 describing the electric field: $$F_{0 i}$$, and 3 describing the magnetic field: $$F_{ij}$$ (where $$i, j = 1, 2, 3$$). Both fields can be described as vectors in the 3D space section.

In $$D = 3$$ spacetimes, the faraday has 3 independent components: 2 describing the electric field, and only 1 describing the magnetic field, which is then a scalar field in the 2D space section, instead of a vector.

In $$D = 5$$ spacetimes, the electric field is again a vector in 4D space, with 4 components, and the magnetic field has 6 components (so it's not a vector in 4D space).

In general, for spacetimes of $$D$$ dimensions, the electric field $$F_{0i}$$ (where $$i = 1, 2, \dots, D-1$$) is always a vector in the $$D - 1$$ space section while the magnetic field has $$\frac{1}{2} (D -1)(D-2)$$ components.

$$D = 4$$ spacetimes are the only ones for which both the electric field and the magnetic field can be described as vectors, and magnetic field lines can thus be defined.

So what should be the most general graphical representation of magnetic fields for any $$D$$, if "lines" in the space section cannot be drawn from the field components themselves?

If the magnetic field could be described as surfaces in space, how do you define those surfaces, if you have the field components $$B_{ij} \equiv F_{ij}(x)$$ and cartesian coordinates $$x^i$$ (where $$i, j = 1, 2, 3, \dots, D - 1$$)? Suppose for simplicity that the field is static (no dependance on $$t$$ and no electric field). The faraday tensor is a 2-form: $$$$\tag{1} \mathbf{F} = F_{ij}(x) \, \mathbf{d}x^i \wedge \mathbf{d}x^j.$$$$ So how can we find the magnetic surfaces associated to this 2-form? What is the differential equation equivalent to the field lines equation in 3D space: $$$$\tag{2} \frac{d\mathbf{r}}{d s} = \lambda(s) \, \mathbf{B}(\mathbf{r}(s)).$$$$ If the variables $$u$$ and $$v$$ are an arbitrary parametrisation on a given surface, would it be $$$$\tag{3} \frac{\partial x^i}{\partial u} \, \frac{\partial x^j}{\partial v} - \frac{\partial x^j}{\partial u} \, \frac{\partial x^i}{\partial v} = \lambda(u, v) \, B_{ij}(x(u, v)) \quad ?$$$$ where $$\lambda(u, v)$$ is an arbitrary scalar function. I never saw equation (3) anywhere before.

• The magnetic field is a 2-form in any dimension, so it should be drawn in terms of surfaces, not lines. – knzhou Oct 15 at 3:34
• @SV, knzhou is right. If there's no electric field, then $\mathbf{F} = F_{ij} \, \mathbf{d}x^i \wedge \mathbf{d}x^j$ is a 2-form. This express a magnetic field in some reference frame using coordinates $x^i$. But I wonder how we could use it to define the surface describing the local magnetic field. – Cham Oct 17 at 0:40
• Not an answer, just something to keep in mind: consider a 4-d space with coordinates $w,x,y,z$ and consider a magnetic field proportional to the 2-form $$dw\wedge dx + dy\wedge dz.$$ This 2-form cannot be written as $dX\wedge dY$ for any functions $X,Y$, so this magnetic field cannot be represented by any single surface element at any given point. It must instead be written as a superposition of two surface elements per point. Also, the superposition is not unique, because the same 2-form is proportional to $$(dw+dy)\wedge(dx+dz)+(dw-dy)\wedge(dx-dz).$$ – Chiral Anomaly Oct 17 at 13:22
• It might be relevant to note that even in the $4D$ spacetimes, magnetic field is not a vector. It's a pseudo-vector. That's related to why the electric field continues to have a vector representation but magnetic field does not at the dimension changes. – Brick Oct 17 at 14:59
• @ChiralAnomaly, it may help visualize your field superposition by studying the motion of a charged particle in such a field: $$\frac{du^a}{d\tau} = \frac{q}{m} \, F^a_{\;\, b} \, u^b.$$ Apparently, the particle would rotate independantly and simultaneously in both planes. This is 4D space weirdness! – Cham Oct 19 at 3:17

## 1 Answer

This isn't an answer, just an extended comment documenting a failed attempt, and why it fails in higher-dimensional space even though it works fine in 2- and 3-dimensional space. To keep this brief, I'll assume that the calculus of differential forms is familiar.

## Canonical form for the field

Let $$\mathbf{x}$$ denote a point in flat $$N$$-dimensional space, in the usual Cartesian coordinate system with coordinates $$x^j$$. For any $$N$$, the magnetic field is a two-form: $$B(\mathbf{x}) = \sum_{j,k} B_{jk}(\mathbf{x})\, dx^j\wedge dx^k. \tag{1}$$ The coefficients $$B_{jk}(\mathbf{x})$$ are ordinary functions. The one-forms $$dx^j$$ anticommute with each other, so WLOG we can take $$B_{jk}$$ to be antisymmetric. Then at each point $$\mathbf{x}$$, we can choose the coordinate system so that $$B_{jk}$$ is block-diagonal, with antisymmetric blocks of size $$2\times 2$$, and with a zero in the leftover $$1\times 1$$ block when $$N$$ is odd. (This is mentioned in https://math.stackexchange.com/questions/142241. Also see theorem 6.15 on pages 203 in Shafarevich and Remizov's Linear Algebra and Geometry.)

In other words, at any given point $$\mathbf{x}$$, we can choose the coordinate system so that $$B$$ has the canonical form \begin{align} B(\mathbf{x}) &= B_{12}(\mathbf{x})\, dx^1\wedge dx^2 \\ &+ B_{34}(\mathbf{x})\, dx^3\wedge dx^4 \\ &+ \cdots, \tag{2} \end{align} continuing the sum until we run out of indices. If we define $$n$$ by either $$N=2n$$ or $$N=2n+1$$, depending on whether $$N$$ is even or odd, then the sum has up to $$n$$ terms.

## Geometric representation in the typical case

If the coefficients of these $$n$$ terms all have distinct magnitudes (this is the typical case), then the decomposition (2) is unique up to permutations of the terms. This suggests that we can "visualize" the magnetic field in terms of $$n$$ tiny disks per point, one per term in (2), with some monotonic relationship between the disk-radius and the magnitude of the coefficient, such that the disk-radius goes to zero as the magnitude goes to zero. The orientations of each disk is determined by the basis form $$dx^j\wedge dx^k$$ in the corresponding term in (2). (Remember that a disk has a signed orientation, where the sign can be depicted as a "direction of travel" around the boundary of the disk. Flipping the disk reverses the sign, which corresponds to the identity $$dx^j\wedge dx^k=-dx^k\wedge dx^j$$.)

When $$N=3$$, the sum (2) has only one term ($$n=1$$). The traditional magnetic field "lines" thread through the centers of the disks with a consistent orientation, and the radii of the disks encodes the magnitude of the field in the same way that the density of "lines" does in the traditional picture. However, the traditional "lines" picture relies on having a unique line orthogonal to any given disk, which is only true in 3-d space. This is why we need to use disks (or other surface elements) instead of lines for general $$N$$.

As long as the coefficients of these $$n$$ terms all have distinct magnitudes (which is trivially true when $$N=3$$), this picture is uniquely determine by the magnetic field $$B$$, and conversely. This assumes that we have chosen a monotonic relationship between the disk-radius and the coefficient-magnitude. Maybe this monotonic relationship can be chosen to make the picture "natural" in some sense, but I won't try to address that, because even a "natural" choice does exist, the picture has a bigger problem...

## Geometric representation in the degenerate case

The problem occurs when the coefficients of the $$n$$ terms in (2) are not all distinct. This is the issue that I originally raised in a comment. To see the problem, consider the case $$B\propto dx^1\wedge dx^2 + dx^3\wedge dx^4. \tag{3}$$ With $$c := \cos\theta$$ and $$s:=\sin\theta$$ for any angle $$\theta$$, the identity \begin{align} dx^1\wedge dx^2 & + dx^3\wedge dx^4 \\ &= (c\,dx^1 - s\,dx^3) \wedge (c\,dx^2 - s\,dx^4) \\ &+ (s\,dx^1 + c\,dx^3) \wedge (s\,dx^2 + c\,dx^4) \tag{4} \end{align} holds. This shows that such a $$B$$ does not uniquely determine the orientations of the disks in the picture proposed above. This non-uniqueness makes the picture unnatural. This problem occurs only when $$N\geq 4$$ ($$n\geq 2$$).

I don't know of a natural way to fix this. I thought about using something like a (topological) Cartesian product of disks, but that doesn't seem to help (and it loses important information about the orientations). I also thought about using the motion of a standardized set of charged particles in an infinitesimal neighborhood of each point, but I didn't think of a way to make that work, either. A quick search on Math SE turned up a few posts about visualizing $$p$$-forms, and none of them offered anything better.

• I'm surprised by all this, I was expecting some unique surface to represent any magnetic field in any spacetime of dimension $D > 4$ (or $N > 3$). This subject appeared to me as pretty standard and should have been well studied before. It appears to not be the case, which suprrises me alot. Thanks a lot for your effort. After the bounty ends, I'll give you the points. – Cham Oct 26 at 18:55