Why do Maxwell's equations contain each of a scalar, vector, pseudovector and pseudoscalar equation?

Question

Maxwell's equations, in differential form, are

$$\left\{\begin{align} \vec\nabla\cdot\vec{E}&=~\rho/\epsilon_0,\\ \vec\nabla\times\vec B~&=~\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t},\\ \vec\nabla\times\vec E~&=-\frac{\partial\vec B}{\partial t},\\ \vec\nabla\cdot\vec{B}~&=~0, \end{align}\right.$$

which are, respectively, scalar, vector, pseudovector and pseudoscalar equations. Is this purely a coincidence, or is there a deeper reason for having one of each type of equation?

If I'm not mistaken, these objects correspond to (or, at least, a correspondence can be made with) the ranks of differential forms on a 3-dimensional manifold, so I guess there might be some connection with the formulation of Maxwell's equations in terms of differential forms. If this is this case, is there an underlying physical reason that our expression of the equations turns out with one of each rank of equation, or is it a purely mathematical thing?

As a final note, I might also be totally wrong about the rank of each equation. I'm going on the contents of the right-hand side. (e.g. a "magnetic charge density" would be pseudoscalar.)

Answer 1

In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.

But this is rather illusory. In relativity, the equations look quite different:

$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$

where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).

The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.

What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit more. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.

Answer 2

In the multivector formulation of electromagnetism as developed by David Hestenes, there's only one equation: $\Box F = J$.

F and J are space-time objects. F is a "bivector", a 2nd order antisymmetric entity, pretty much like the usual $F_{\mu\nu}$ in relativity. Looking at it's space and time parts separately, it has a vector part, $\vec E$ (really a time-space bivector) and an axial vector part $\vec B$ (really a space-space bivector). $J$ is a four-vector, which of course in the space + time view is a scalar charge density plus a vector current density.

In multivector algebra, when we multiply two vectors the result is the sum of the inner (dot) and outer (cross) products, creating an entity with a scalar part and a bivector part. You get two equations in three-space + time, for the price of one multivector equation in spacetime. The $\Box$ differential operator applies like that, making a divergence and a cross product. The two parts of $F$ and the two ways of applying $\Box$ result in four equations, the familiar Maxwell ones.

This might seem like gibberish to anyone unfamiliar with it. It's a form of Clifford Algebra. Book: Space-Time Algebra by David Hestenes, Gordon and Breach 1966. He has numerous papers in scholarly journals which I don't have the time to look up just now.

Answer 3

There is indeed a reason for this. First, let's translate the Maxwell equations to the language of differential forms on Euclidean 3-space (not Minkowski space-time): $$ {\rm d}B = 0 \qquad {\rm d}E + \frac{\partial B}{\partial t} = 0 \\ \star{\rm d}\star E = \rho \qquad \star{\rm d}\star B - \frac{\partial E}{\partial t} = J $$ in terms of electric 1-form $E$ and magnetic 2-form $B$.

The benefit is that the exterior algebra is graded, and we can combine these equations into a single one without the terms of the four equations interfering.

We're free to choose prefactors (in particular signs) as we see fit. The interesting combinations I could come up with are $$ (\star{\rm d}\star - {\rm d} - \frac{\partial}{\partial t})(E + B) = \rho + J $$ and $$ (\star{\rm d}\star - {\rm d} + \frac{\partial}{\partial t})(E - B) = \rho - J $$ which more or less correspond to the combined Maxwell's equations in bivector and 2-form representation. The former could be written as $$ {\rm D}F = J $$ in terms of geometric algebra and the latter as $$ (\delta + d)F = J $$ in terms of differential forms and codifferential $\delta$, both on Minkowski space-time and with definitions of $F$ and $J$ appropriate for the chosen formalism.

So basically, we're faking relativistic relations on top of Euclidean geometry. For example, the invariant length of the '4-vector' $\rho + J$ can be expressed in the exterior algebra as $$ (\rho + J)\wedge\star(\rho - J) = \star(\rho^2 - \langle J,J\rangle) $$

Answer 4

This is not entirely a coincidence. In covariant (relativistic space-time) formulation, the scalar and vector equations (the equations with a source term) combine into one space-time tensor equation, and so do the pseudovector and pseudoscalar equations (the source-free equations).