According to Griffiths's book on electrodynamics, including magnetic charge the Maxwell equations become

$$ \begin{align*} \nabla \cdot \vec{E} &= \frac{\rho_e}{\epsilon_0} &&& \nabla \times \vec{E} + \partial_t \vec{B} &= -\mu_0 \vec{j}_m \\ \nabla \cdot \vec{B} &= \mu_0 \rho_m &&& \nabla \times \vec{B} - \partial_t \vec{E} &= \mu_0 \vec{j}_e \end{align*} $$

where $\rho_e$ and $\rho_m$ are the electric and magnetic charge densities, and $\vec{j}_e$ and $\vec{j}_m$ are the electric and magnetic current densities. I've taken $\mu_0 \epsilon_0 = c = 1$ for simplicity.

In problem 7.60 he asks to show the invariance of these equations under a duality transformation

$$ \begin{pmatrix} E' \\ B' \\ \end{pmatrix} = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{pmatrix} \begin{pmatrix} E \\ B \end{pmatrix} $$

with the same matrix applied to a "row vector" of the charge and current densities.

This looks like a $SO(2)$ symmetry. However, when I presented this fact in an exam, the professor said this was not the entire symmetry group showing electromagnetic duality. Sadly, he didn't know what the full group was either.

My idea now was to define a complex vector field $\vec{F} = \vec{E} + i \vec{B}$ with corresponding complex sources $\rho = \rho_e + i \rho_m$ and $\vec{j} = \vec{j}_e + i \vec{j}_m $. Then Maxwell's equations turn into

$$ \begin{align*} \nabla \cdot \vec{F} &= \rho \\ \nabla \times \vec{F} -i \partial_t \vec{F} &= i \mu_0 \vec{j} \end{align*} $$

The duality symmetry I can spy in this is multiplying $\vec{F}$, $\rho$ and $\vec{j}$ with the same complex number, which is equivalent to the earlier $SO(2)$ and a rescaling.

Is this the full group? If not, what am I still missing?

  • $\begingroup$ That second equation is trivial, so I doubt we gain anything from it. I guess separating time and position derivatives would be more beneficial (I'm trying but you should too). $\endgroup$
    – acarturk
    Commented Sep 3, 2019 at 14:18

1 Answer 1


Yes, OP is right. The EM duality symmetry group of Maxwell equations is $SO(2)\times \mathbb{R}_+$, where $SO(2)$ is the EM rotation group and $\mathbb{R}_+$ is dilations. For generalizations, see Ref. 1.


  1. P. Aschieri, S. Ferrara & B. Zumino, 3 Lectures on EM Duality, 2008.
  • $\begingroup$ So, it simply reflects the fact that the theory can be written as a $U(1)$ gauge theory? Or is this a true symmetry and not redundancy? I suppose this corresponds to the global $U(1)$ symmetry and is thus a true symmetry, unlike the gauged $U(1)$. Correct me if I am wrong. Thanks. $\endgroup$
    – user87745
    Commented May 14, 2020 at 19:56

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