Electromagnetic duality and vector/pseudovector transformation properties

Question

One consequence of electromagnetic duality (see e.g., https://doi.org/10.1038/s41467-023-36420-4) is that if we have a system described by permittivity and permeability profile $(\varepsilon, \mu)(\mathbf{r})$ that has an eigensolution with associated electric and magnetic fields $(\mathbf{E}, \mathbf{H})(\mathbf{r})$, then the "dual" problem $$ (\varepsilon', \mu') = (\mu, \varepsilon) $$ will have an eigensolution ($Z_0 = \sqrt{\mu_0/\varepsilon_0}$) $$ (\mathbf{E}', \mathbf{H}') = (Z_0\mathbf{H}, -Z_0^{-1}\mathbf{E}). $$

While I can reconcile this with Maxwell's equations, I'm having trouble reconciling it with the respectively vectorial and pseudovectorial transformation properties of $\mathbf{E}$ and $\mathbf{H}$ under spatial transformations.

In particular: suppose there is an affine transformation (i.e., proper or improper rotation) $R$ with matrix form $\mathbf{R}$ acting on the fields. We know that it acts on $\mathbf{E}(\mathbf{r})$ and $\mathbf{H}(\mathbf{r})$ in different ways since one is a vector and the other a pseudovector: $$ R \mathbf{E}(\mathbf{r}) = (\mathbf{R}\mathbf{E})(\mathbf{R}^{-1}\mathbf{r}),\\ R \mathbf{H}(\mathbf{r}) = \det(\mathbf{R})(\mathbf{R}\mathbf{H})(\mathbf{R}^{-1}\mathbf{r}). $$

Now, if I try to apply these rules to the dual solutions $(\mathbf{E}', \mathbf{H}')$ I become confused. Because, on one hand (first applying the transformation rule, then using duality): $$ R \mathbf{E}'(\mathbf{r}) = (\mathbf{R}\mathbf{E}')(\mathbf{R}^{-1}\mathbf{r}) = Z_0(\mathbf{R}\mathbf{H})(\mathbf{R}^{-1}\mathbf{r}) $$ but on the other hand (first using duality, then applying the transformation rule): $$ R \mathbf{E}'(\mathbf{r}) = Z_0 R\mathbf{H}(\mathbf{r}) = Z_0\det(\mathbf{R})(\mathbf{R}\mathbf{H})(\mathbf{R}^{-1}\mathbf{r}). $$ So... which is it?

More broadly, I'm trying to understand how in one sense duality is placing the electric and magnetic fields on the same footing - and in another sense, they are endowed with different transformation rules. I.e., why is one a vector and the other a pseudovector? Similarly, how to think about the Lorentz force equation under duality transformations?

Answer 1

Both are solutions. If $(\mathbf{E},\mathbf{H})$ is a solution, then so is $(-\mathbf{E},-\mathbf{H})$. Thus, $\mathbf{E}'=-Z_0\mathbf{H}$, $\mathbf{H}'=Z_0^{-1}\mathbf{E}$ is also a solution.

On your broader question, the way to understand it is to recognise that the electromagnetic field is a bivector (a plane element with orientation an magnitude, in the same way a vector is a line element), with six basis components, one for each pair of spacetime coordinates.

I'm going to use the (Clifford) spacetime algebra approach described in "Spacetime Algebra as a Powerful Tool for Electromagnetism" by Dressel, Bliokh, and Nori. This gives a detailed description of the duality symmetry, which turns out to be a continuous $U(1)$ symmetry that includes both the above solutions. Spacetime algebra is (perhaps unfortunately) not the mainstream way of doing it, but I find it a lot clearer in terms of developing a geometric intuition for what it all means.

This answer is only a brief outline sketch - for the details and a much better explanation, see Dressel, Bliokh, and Nori.

We have a basis for the space of bivectors: $\gamma_{xt}$, $\gamma_{yt}$, $\gamma_{zt}$, $\gamma_{yz}$, $\gamma_{xz}$, $\gamma_{xy}$. Because Gibbs' vector algebra can't handle bivectors, we can split this into temporal and spatial components by multiplying the whole thing by the constant basis vector along the time axis $\gamma_t$, which is equivalent to picking a reference frame to observe the field from. Then the temporal components $\gamma_{xt}$, $\gamma_{yt}$, $\gamma_{zt}$ turn into a vector with basis $\gamma_{x}$, $\gamma_{y}$, $\gamma_{z}$, which turns out to be the electric field, and the spatial components $\gamma_{yz}$, $\gamma_{xz}$, $\gamma_{xy}$ turn into the trivector basis $\gamma_{yzt}$, $\gamma_{xzt}$, $\gamma_{xyt}$. Finally, we dualise the trivector into its orthogonal complement (which can be done in this algebra by multiplying by the pseudoscalar $I=\gamma_{xyzt}$), getting a pseudovector $\gamma_{x}$, $\gamma_{y}$, $\gamma_{z}$, and this is the magnetic field.

What happens if we reflect our bivector basis in the $x$ axis? The terms with an $x$ component flip sign, so we have $-\gamma_{xt}$, $\gamma_{yt}$, $\gamma_{zt}$, $\gamma_{yz}$, $-\gamma_{xz}$, $-\gamma_{xy}$. That is, in the temporal part only the $x$ axis of the electric field flips, but in the spatial part the $y$ and $z$ components of the magnetic field flip. This is why the electric field transforms like a vector, and the magnetic field like a pseudovector, even though they are components of the same structure, and on an equal footing.

Multiplying by $e^{\theta I}$ for real numbers $\theta$ gives another solution. Setting $\theta=-\pi/2$ gives the familiar duality transformation you started with. Using $\theta=\pi/2$ gives the other solution. Essentially, the reason for the flip is that the pseudoscalar $I$ used to generate the orthogonal complement is transformed by a reflection to $-I$. The confusion arises entirely because of our efforts to represent a bivector structure in Gibbs' vector algebra.