# 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 $\nabla \times \vec E = - \partial_t\vec B - \vec M$

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

• Like magnetic charge density, it's always zero, and is added just to make symmetry apparent. – Ruslan Nov 23 '14 at 9:58

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.

• Hi thank you! can you explicitly show how you can use that rotational matrix on the maxwell's equation? I've never seen that done before – Carlos - the Mongoose - Danger Nov 23 '14 at 18:23
• @IllegalImmigrant Just try it with $\phi = \pi/2$. You will see that $\rho \rightarrow -\rho_m$ whilst ${\bf D} \rightarrow {-\bf B}$ and thus $\nabla \cdot {\bf D} = \rho$ transforms to $\nabla \cdot {\bf B} = \rho_m$ and so on. – Rob Jeffries Nov 23 '14 at 22:31

If it is eventually found that magnetic monopoles exist, then M would not be always zero and Maxell's equations would become fully symmetrycal on E and B.

What York is doing is not "physics" but an ancient antenna engineering trick whereby one first solves for the field of the linear electric Hertzian dipole and then to solve for the radiated field of a closed loop current one considers the latter as a formally equivalent "magnetic dipole" source resulting in the interchange of E and B everywhere. The same idea can be used to find the scattered/diffracted field of a slot from knowing what happens with its complement (Babinet).

• Hi can you provide a source to this? Seems to be very interesting! – Carlos - the Mongoose - Danger Nov 23 '14 at 18:25
• Any book on antenna engineering, say, by Orfanidis or Krauss or Schelkunoff. – hyportnex Nov 23 '14 at 20:05

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?

The standard electromagnetic theory assumes there are no magnetic charges and no corresponding current density. The author of the text you referenced included $\vec{M}$ there for unclear reasons. In all practical uses of Maxwell's equations, there is no magnetic current density present at all.

For regions where $\vec{M} = \vec{0}$, his equation becomes the same as the standard equation, but the EM field will be different in his theory than in the standard theory if there is at least one point of space where there is non-vanishing magnetic charge.