maxwell equations - How seriously should I take the notion of "magnetic current density"

Increasingly I've noticed that people are using a curious quantity $\vec M$ to denote something called magnetic current density in the formulation of the maxwell's equations

where instead of $\nabla \times \vec E = - \partial_t\vec B$, you would have $\nabla \times \vec E = - \partial_t\vec B - \vec M$

(i.e. http://my.ece.ucsb.edu/York/Bobsclass/201C/Handouts/Chap1.pdf)

Under what additional assumptions can $\vec M$ be made zero so that the conventional maxwell's equation is consistent with the extended maxwell's equation?

Thank you

Answer

If we let $\mu_0=1$, $\epsilon_0 =1$ (adopting a system of units where $c=1$), then Maxwell's equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ via a rotation (see below). $$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t}$$

If "source" terms $\rho$ and ${\bf J}$, the electric charge and current density, are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations could be written using a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$ (what you refer to as ${\bf M}$, though I prefer to reserve that for magnetisation), then we write $$ \nabla \cdot {\bf D} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf H} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$

With these definitions, Maxwell's equations acquire symmetry to duality transformations. If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form $$ \left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$ where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.

Now you could argue about what we define as electric and magnetic charges, but that comes down to semantics. What is clear though is that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations above), all particles appear to have the same ratio, so we choose to fix it so that one of the charge types is always zero - i.e. no magnetic monopoles and no magnetic current density.