#### Magnetic Charge
emd.Me.mc

In free space, where \(\rho\) and \({\bf J}\) vanish:

\begin{align} (i) &{\boldsymbol \nabla} \cdot {\bf E} = 0, &(iii) {\boldsymbol \nabla} \times {\bf E} + \frac{\partial {\bf B}}{\partial t} = 0, \\ (ii) &{\boldsymbol \nabla} \cdot {\bf B} = 0, &(iv) {\boldsymbol \nabla} \times {\bf B} - \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} = 0, \end{align}

Symmetry: replace \({\bf E}\) by \({\bf B}\) and \({\bf B}\) by \(-\mu_0 \varepsilon_0{\bf E}\) in the first pair. They turn into the second pair. This symmetry is spoiled by \(\rho\) and \({\bf J}\). What if we had a truly symmetric situation, i.e.

\begin{align} (i) &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho_e}{\varepsilon_0}, &(iii) {\boldsymbol \nabla} \times {\bf E} = -\mu_0 {\bf J}_m -\frac{\partial {\bf B}}{\partial t}, \\ (ii) &{\boldsymbol \nabla} \cdot {\bf B} = \mu_0 \rho_m, &(iv) {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}_e + \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t}, \label{Gr(7.43)} \end{align}

where \(\rho_m\) would represent the density of magnetic charge and \({\bf J}_m\) would be the current of magnetic charge. Both charges would be conserved: \[ {\boldsymbol \nabla} \cdot {\bf J}_m = -\frac{\partial \rho_m}{\partial t}, \hspace{1cm} {\boldsymbol \nabla} \cdot {\bf J}_e = -\frac{\partial \rho_e}{\partial t}. \label{Gr(7.44)} \] Maxwell's equations beg for magnetic charges. But we've never found any!

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42