# Is the Lorentz transform of a bound current - a bound current?

$$\mathbf{J}_\mathrm{M} = \nabla \times \mathbf{M}$$.

The charge density of a moving magnet that is electrically neutral in its own rest frame satisfies the condition

$$\rho' = 0$$.

The non-relativistic approximation the charge density $$\rho$$ observed in our inertial (lab) frame is

$$\rho \approx \left( \rho' +\frac{1}{c^2}\mathbf{J}\cdot \mathbf{v} \right)$$

where the magnet is observed to have velocity $$\mathbf{v}$$. Therefore

$$\rho \approx \left( \frac{1}{c^2}\mathbf{J}_\mathrm{M}\cdot \mathbf{v} \right)$$.

The current density is the product of the charge density and its velocity. The current density associated with charge density on the magnet observed to be moving at velocity $$\mathbf{v}$$ in our inertial (lab) frame is the polarization current

$$\mathbf{J}_\mathrm{P} \approx \left( \frac{1}{c^2}\mathbf{J}_\mathrm{M}\cdot \mathbf{v} \right) \mathbf{v}$$.

The divergence of this gives us

$$\nabla \cdot \mathbf{J}_\mathrm{P} \approx \nabla \cdot\left( \left( \frac{1}{c^2}\mathbf{J}_\mathrm{M}\cdot \mathbf{v} \right) \mathbf{v} \right)$$.

The divergence of the current density gives us a time-varying charge density as implied by charge conservation

$$\nabla \cdot \mathbf{J} = - {\partial \rho \over \partial t}$$.

Per the differential form of Gauss' law, the charge density is related to the divergence of the electric field

$$\rho = \varepsilon_0 \nabla \cdot \mathbf{E}$$.

Therefore, it follows that a time-varying charge density gives rise to a displacement current with some divergence

$$\frac{\partial \rho}{\partial t} = \varepsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$.

According to the Ampère-Maxwell equation

$$\mathbf{\nabla}\times \mathbf{B} = \mu_0\mathbf{J}+\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$

where

$$\mathbf{J} = \mathbf{J}_\mathrm{f}+\mathbf{J}_\mathrm{P} +\mathbf{J}_\mathrm{M}$$.

In the case of a moving magnet, we have

$$\mathbf{J} = \mathbf{J}_\mathrm{P} + \mathbf{J}_\mathrm{M}$$

and thus

$$\mathbf{\nabla}\times \mathbf{B} = \mu_0\mathbf{J}_\mathrm{P}+\mu_0\mathbf{J}_\mathrm{M}+\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$.

which has a divergence of zero. Therefore

$$\nabla \cdot \mathbf{J}_\mathrm{P} + \nabla \cdot \mathbf{J}_\mathrm{M} = -\varepsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$.

Since the magnetization current is the curl of the magnetization, it has no divergence. Therefore

$$\nabla \cdot \mathbf{J}_\mathrm{P} = -\varepsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$,

thus, the divergence of the displacement current is

$$\varepsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} \approx -\nabla \cdot\left( \left( \frac{1}{c^2}\mathbf{J}_\mathrm{M}\cdot \mathbf{v} \right) \mathbf{v} \right)$$.

Therefore this displacement current is required to ensure that the total current flows in a closed path.

However, this displacement current may extend beyond the confines of our magnet. Is this really a bound current anymore?

If real, it would imply that this displacement current could couple to some unshielded charges in some circuit that we set up, "feeding" displacement current into and out of our circuit. This would be capacitive coupling.

In essence, it would imply that we could "connect" the bound (polarization) currents of moving magnets into our circuit, so as long as displacement currents "diverge" and "converge" at both our magnet and our circuit to ensure the continuity of current.

According to certain physicists commenting on the subject of the Aharonov-Casher effect, the force exerted onto the electric dipole (of a moving magnetic dipole) by the electric field of a line charge results in a rate of change of the "internal mechanical momentum" of the dipole but does not influence its motion and therefore does not cause it to accelerate. The acceleration of the line charge due to the electric field of the moving magnetic dipole did not appear to be considered.

But what if the unshielded charges on our circuit were not fixed? Is it actually allowable to connect the bound currents of our moving magnet to unshielded, mobile charges on our circuit via the electric displacement current from our moving magnet, or is there some underlying physics which forbids even a small amount of this from happening? Isn't it obvious or not that you're not supposed to be able to get energy from magnets (a corollary being that we should not be able to connect its bound currents to some circuit we construct) and that to do so would violate quantum mechanics? How does one reconcile these facts?

• Is it possible to briefly formulate a question? – Alex Trounev Mar 17 at 22:13
• Let's see if I can reduce all the questions in the above post to a single question which motivates them all. Here's the list of the questions I have in the post: * Is the Lorentz transform of a bound current - a bound current? * However, this displacement current may extend beyond the confines of our magnet. Is this really a bound current anymore? * [All the text in the last paragraph of the post] In one question: Are bound currents on magnets fundamental in that they cannot supply current to elements outside the bounds of the atoms in which they reside? – Kevin Marinas Mar 17 at 22:39