Calculate the electric field of a moving infinite magnet, without boosting

Question

Consider a rectangular slab of permanently magnetized material. The slab's dimensions are $L_x$, $L_y$, and $L_z$, and the slab is uniformly magnetized in the $\hat{x}$-direction. The slab is not accelerating or spinning. Does the slab generate an electric field?

In a frame where the magnet is stationary, we know $\mathbf{E}$ is zero everywhere. In a frame where the magnet is moving, there are at least two ways to attack the problem:

1. Drop $d\mathbf{M}/d t$ into Maxwell's equations, solve for $\mathbf{E}$ and $\mathbf{B}$
2. Solve for $\mathbf{B}$ in the slab's rest frame, and use a relativistic boost to transform $\mathbf{B}$ in the slab's frame to $\mathbf{E}$ and $\mathbf{B}$ in the frame where the magnet is moving.

Both these methods give the result that $\mathbf{E}$ is nonzero in a frame where the magnet is moving.

Now let's consider a long, thin slab ($L_x, L_y \ll L_z$). In a frame where the slab is moving in the $\hat{z}$-direction, is there an electric field (external to the slab) near the 'center' of the magnet? Both the $d\mathbf{M}/d t$ argument and the Lorentz-boost argument seem unchanged. The magnetic field external to the slab does not vanish near the center of the slab, suggesting there is a nonzero electric field.

With the backstory laid out, here's my real question: In a frame where the slab is moving in the z-direction, is there still an electric field in the case where $L_z \rightarrow \infty$?

The Lorentz-boost argument seems unchanged, and suggests that there is. However, in the $L_z \rightarrow \infty$ case, $d\mathbf{M}/d t = 0$, suggesting no electric field. Can this case be calculated without Lorentz boosts? How do Maxwell's equations account for moving permanent magnets in the case where $d\mathbf{M}/d t = 0$?

EDIT: Followup questions:

What's a good reference for the electrodynamics of moving media?

What is the electric field generated by a spinning magnet?

Answer 1

First, one inevitably gets the same solutions if

1. he solves the problem in the slab's rest frame, and then Lorentz-transforms the result to the frame where the slab is moving;

2. or if one solves the problem directly in the frame where the slab is moving.

The reason is that Maxwell's equations are covariant under the Lorentz transformations. So if they're satisfied in one frame, they will be satisfied in any frame related by boosts, too. However, we must properly transform all the magnetizations and material relations etc. and add the corresponding moving sources which will be the main subtlety in the text below.

In your particular problem, one may say some generic statements about the magnetic (and electric) fields without much thinking. For example, if $\vec M$ is in the $x$-direction, it means that the electrons may be thought of to spin in the $yz$-plane. Take a surface of the slab parallel to the $yz$-plane - i.e. one face that belongs to a $x=x_0$ plane. It's pretty clear that there will inevitably be a component of the magnetic field $\vec B$ in the $x$-direction near the external side of the surface. If one boosts the $B_x$ magnetic field in the $z$-direction, there will inevitably be a nonzero electric field in the $y$-direction, $E_y$.

In the frame where the slab is moving, we seem to have no electric sources $\rho$ of the $\mbox{div}\,\vec D=\rho$ Gauss's law and no right-hand side of the Maxwell-Faraday equation, $\nabla\times \vec E = -\partial \vec B / \partial t$. So because there are no electric sources, you would think that the electric field should vanish. However, this is a flawed argument because the form of Maxwell's equations we're using here are only "Maxwell's equations for materials at rest".

In particular, the Gauss's law is optimized for $\vec D$ which we're imagining to be given by $\epsilon \vec E$, and is "purely electric". However, for a moving material, there should be an extra term of the type $\vec v\times \vec M$ included in $\vec D$. Because the latter has a nonzero $y$-component in the moving frame, there will be a nonzero $E_y$ in this frame, too.

The precise form of Maxwell's equations in a moving medium may be confusing and unfamiliar so I think it may be a good idea to try to transform the local physics to the rest frame of any material, whenever needed, and perhaps Lorentz-transform back. Whenever subtleties would occur, one would have to revisit the derivation of the "macroscopic Maxwell's equations" (for materials) and redo it with the possibility of moving materials.

Microscopic Maxwell's equations

Alternatively, you could always try to use the microscopic Maxwell's equations which include the Gauss's law in the form $\vec \nabla\cdot \vec E = \rho / \epsilon_0$. But in this form, $\rho$ includes not only free charges but also the "microscopic charges" related to the material.

Because the slab has nonzero values of $j_y$ and $j_z$ (currents inside the material) - recall that the electrons are kind of rotating in the $yz$-plane (to produce the magnetic $x$-field), it's also true that when we boost the system in the $z$ direction, the corresponding multiple of $j_z$ will produce a nonzero value of $\rho$ (microscopic charge density). This will be the source of the $E_y$ field discussed above. In particular, $j_z$ will be proportional to $[\delta(y-y_1)-\delta(y-y_2)]$ in the slab's frame which means that there will be $\rho \sim [\delta(y-y_1)-\delta(y-y_2)]$ in the frame where the slab is moving. It is this $\rho$ that will induce a nonzero value of $E_y$ right outside the material (in the frame where the slab is moving).

Answer 2

A very simple way of approaching this is to recognize that the electric and magnetic polarizations $(-\textbf{P},\textbf{M})$ transform in exactly the same way as the fields $(\textbf{E},\textbf{B})$ (Hnizdo 2011). In the lab frame, the magnetized slab is also electrically polarized, so clearly there is a nonvanishing electric field. One good way to see that $(-\textbf{P},\textbf{M})$ must mix is that otherwise we would get exactly the kind of paradox described in the question.

To see that $(-\textbf{P},\textbf{M})$ transforms exactly in the same way as $(\textbf{E},\textbf{B})$, it suffices to observe that when we split up the fields $(\textbf{E},\textbf{B})$ into their macrosopic part $(\textbf{D},\textbf{H})$ and their microscopic part $4\pi(-\textbf{P},\textbf{M})$, this macroscopic-microscopic split is frame-independent.

Hnizdo and McDonald, "Fields and Moments of a Moving Electric Dipole," 2011, http://www.physics.princeton.edu/~mcdonald/examples/movingdipole.pdf