# B Does a moving bar magnet create charges?

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1. Feb 12, 2017

### ealbers

So I understand that a charged particle moving relative to me has a magnetic field, and does not have one if its stationary relative to me...
So is it true that a magnet moving relative to me has a charge?
Say I took a bar magnet and threw it away from me way out in space, would it acquire a charge? If so, would it get 2 charges one - and one + on each end of the magnet??
Thanks!

2. Feb 12, 2017

### rootone

Magnetic fields are not relative to you.
A bar magnet has a magnetic field which exists whether or not you throw it.

3. Feb 12, 2017

### Staff: Mentor

There has to be charge present in the magnet for it to be a magnet; any EM field has to have a source somewhere. If the magnet is electrically neutral when it is at rest relative to you, that's because it has equal amounts of positive and negative charge and the static electric fields of those charges cancel each other out, leaving only the magnetic fields.

If the magnet is moving relative to you, all of its charges will be moving at the same speed relative to you, so it will still appear electrically neutral to you.

4. Feb 12, 2017

### Staff: Mentor

Well, it is overall electrically neutral, but it does become electrically polarized. What is magnetization in one frame is polarization in another frame.

5. Feb 12, 2017

### Staff: Mentor

Yes agreed. To put it another way, the charge density in the frame in which the magnet is moving is still zero, but the electric field and the current density are nonzero.

6. Feb 13, 2017

### Battlemage!

I created a contraption for intro physics that spun the most powerful hobby bar magnets you can buy and using an oscilloscope my group measured a very obvious alternating current.

7. Feb 13, 2017

### Staff: Mentor

Yes, there should be a nonzero current density (see my previous post). But should there be a nonzero charge density? That would seem to contradict Maxwell's Equations (because the electric field in the frame in which the magnet is moving should be divergence-free--at least, that's what transforming the fields from the magnet's rest frame seems to show).

8. Feb 13, 2017

### ealbers

Just trying to understand how a stationary electron next to a bar magnet will stay stationary if I start moving towards both of them...a magnetic field appears around the electron from my relative motion, a charge must appear to cancel out the magnetic field??
Or picture 2 identical bar magnets and a electron and a proton the electron next to magnet 1 and the proton next to magnet 2, then start moving towards them, the electron and proton get opposite magnetic fields, but the magnets have the same NS orientation, so something has to give, I assume somehow a charge on the magnet cancels out any effect of the magnetic field I observe as my motion is now not in the same frame as the magnet/particles.

9. Feb 13, 2017

### Staff: Mentor

No, an electric field appears whose action on the electron cancels out the action of the transformed magnetic field. But the electric field is divergence-free (it is due to current density, not charge density).

10. Feb 13, 2017

### ealbers

Is that the same for the proton? Does its field also transform to cancel a identical magnet next to it from the same motion?
Say both particles are in the S end of their respective bar magnets fields...
Now the electron gets a magnetic field and the proton also gets a magnetic field around it, opposite from the electron...
How do the electron and protons electric fields stop the magnetic field of the bar magnet from interacting with the now existing magnetic field around each particle?
It seems weird

11. Feb 13, 2017

### Battlemage!

We didn't even consider that. We used spinning magnets and loops of wire, and measured the current in the wire. I suppose the current density would be our measured current divided by the total volume of the wire, right?

As for charge density, it should be zero as you say, but respect to an induction problem like a spinning magnet inside a coil of wire, what would the physics of that look like? (Which presumably show a charge density of zero)

12. Feb 13, 2017

### Staff: Mentor

It's current divided by the cross sectional area of the wire. (And if you're using conventional units you have to throw in factors of $c$ here and there. In relativistic units where $c = 1$ it's actually easier conceptually: current is charge per unit time, and current density must have the same units as charge density so it's current per unit area, or charge per unit time per unit area, which is the same as charge per unit volume.)

13. Feb 14, 2017

### vanhees71

The way to answer this question is to write the quantities in covariant form. I use Heaviside-Lorentz units as usual, which makes things most convenient. Electromagnetically the matter is characterized by charge and current densities as well as electric and magnetic polarization densities. The charge and current densities are combined to a four-vector,
$$j^{\mu}=\begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix}.$$
The polarizations are a bit more cumbersome to explain. First one must remember that the Maxwell equations in matter are written in a form which splits the fields in parts created by the free sources and polarization fields resulting from intrinsic charge, current, and magnetization distributions. The total electromagnetic field $(\vec{E},\vec{B})$ and the ones created by the free charges and currents $(\vec{D},\vec{H})$. These (and only these combinations) build antisymmetric tensors
$$(F_{\mu \nu}) \equiv (\vec{E},\vec{B}) = \begin{pmatrix} 0 & -E_x & -E_y & - E_x \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}.$$
Analogously one has
$$(D_{\mu \nu})=(\vec{D},\vec{H}).$$
Now for a historical reason and a misunderstanding about which fields combine naturally together the definition of the electric and magnetic polarization are different in sign, i.e., one defines
$$\vec{E}=\vec{D}+\vec{P}, \quad \vec{B}=\vec{H} - \vec{M},$$
and that's why the polarization tensor combines to
$$P_{\mu \nu} =(\vec{P},-\vec{M}).$$
Despite this confustion with the signs, now it's easy to understand what's going on concerning how different observers see the electromagnetic field of a permanent magnet.

In its rest frame it's uncharged, and there are no currents, i.e., in the rest frame you have
$$\rho=0, \quad \vec{j}=0 \; \Rightarrow \; j^{\mu}=0.$$
Under Lorentz boosts this transforms like a Minkowski vector, and thus
$$j^{\prime \mu}={\Lambda^{\mu}}_{\nu} j^{\nu}=0.$$
Also for the observer relative to which the magnet is moving with constant velocity, there are neither charge nor current densities.

For the observer at rest, we can also assume that $\vec{P}=0$ and $\vec{M} \neq 0$ (i.e., it's electrically unpolarized, but it has a magnetization of course). Now the Polarization tensor transforms as any 2nd-rank tensor,
$$P^{\prime \mu \nu}={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} P^{\rho \sigma},$$
and this gives you entries also for the electric polarization, i.e., in the reference frame, where the magnet is moving, there's both an electric polarization and magnetization, and that's why in this frame there's also a electric field, while in the frame where the magnet is at rest there's only a magnetic field. In other words, a body which has only a magnetization when viewed in its rest frame becomes also electrically polarized.

Last edited: Feb 16, 2017
14. Feb 15, 2017

### Mister T

But his original question was whether or not the magnet acquires a charge. I take it the answer is that it doesn't. That is, there's an electric field, but there is no electric charge that can be claimed as the source of that electric field.

15. Feb 16, 2017

### vanhees71

There's electric polarization in the frame where the magnet moves, but no charges. This must be so since in the restframe there were neither charge nor current distributions.

16. Feb 16, 2017

### Mister T

When I first saw this I noticed the lack of symmetry in your question. I didn't say anything then because I didn't want to create a confusion.

I think the symmetry of the situation you describe prompts us to ask if there's an electric field, not if there's an electric charge. You need electric charge to create an electric field, but only in the electrostatic case, not in general.

Vanhees tells us this ...