# I Are magnets charged with rotation?

1. Dec 25, 2016

### sweet springs

Hi I am confused about rotating magnet.
Feynman lectures on physics 13–6 The relativity of magnetic and electric fields, http://www.feynmanlectures.caltech.edu/II_13.html , deals with two inertial frames. Wire is charged in some inertial frame. Can we apply it to tangential parts of curved wire rings as part of an electromagnet?
I wonder whether electromagnet is charged with rotation.

2. Dec 25, 2016

### SleepDeprived

I am not quite sure what you meant by this? Do you mean if you rotate a wire, will it create a magnetic field?

3. Dec 26, 2016

### brycenrg

Good question, there are magnets that we are used to, these are called permanent magnets. The actual type of material gives them this property.

There are also electromagnets which are more common, these produce the magnetic effect by induced electric current. If you have a electric stove, every time you turn it on it produced a magnetic field. You can test it by moving a magnet around it, be careful.

4. Dec 26, 2016

### sweet springs

Thanks. I try to explain my point. Feynman formula (13.24) and (13.25) describe transformation of charge density due to length contraction. When we rotate electromagnet along the ring current, though I am not sure about length contraction in rotation, charge density keeps constant or is transformed in the same manner as Feynman.
I mean if we rotate a neutral ring with electric current, will it create an electric field coming from generated charge on the ring wire.

5. Dec 26, 2016

### vanhees71

To summarize Feynman's explanations in a compact form (which helps to avoid confusion and was discovered by Minkowski in 1908) you just note that charge and current densities together build a Minkowski four-vector field
$$(j^{\mu})=\begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix},$$
which transforms as such under Lorentz transformations. Feynman carefully has to derive this from the assumption that electric charge is a scalar and the kinematics of special relativity, since he has chosen a conventional approach to teach electrodynamics (rightfully so for a freshman course in physics, as which the Feynman Lectures have been invented to begin with, but they turn out to be rather advanced and thus rather make up marvelous theory books for advanced undergraduates to beginning graduates). Of course, a really modern approach, which starts from relativity and formulates electrodynamics as a relativistic field theory, which it indeed is since Maxwell formulated it, is much simpler conceptually (see Landau, Lifshitz vol. II).

In the same way the scalar and the vector potential together (in Lorenz gauge!) build a four vector field,
$$(A^{\mu})=\begin{pmatrix} \Phi \\ \vec{A} \end{pmatrix},$$
and thus the (gauge invariant!) field-strength tensor (or to honour Faraday also known as Faraday tensor)
$$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
an antisymmetric 2nd-rank four tensor. It's components consist of the electric and magnetic field components, because
$$F_{0 j}=\frac{1}{c} \partial A_{j} - \partial_j \Phi=-\frac{1}{c} \partial_t A^{j} -\partial_j \Phi=E^j$$
and
$$F_{jk} = \partial_j A_k - \partial_k A_j = - (\partial_j A^k - \partial_k A^j)=-\epsilon^{jkl} B^l.$$
The latin indices run over $\{1,2,3 \}$.

In macroscopic electrodynamics there's a corresponding antisymmetric tensor combining $\vec{D}$ and $\vec{H}$, and the constitutive equations can be formulated in manifest covariant way too.

The so very easily to derive transformations of the field components and the charge-current vector shows you that, as special applications,

(a) a current conducting wire, which is electrically neutral in its rest frame becomes charged from the point of view of an observer moving relative to it.
(b) a pure electrostatic field also has magnetic components from the point of view of an observer moving relative to the corresponding charge distribution, which is entirely at rest in the frame, where you only have electric field components. Of course, from the moving observer's point of view there's also a convection-current density $\rho \vec{v}$.
(c) a moving uncharged permanent magnet, which has only magnetostatic field components in its rest frame, also has electric field components from the point of view of an observer moving with respect to the permanent magnet.
...

6. Dec 26, 2016

### SleepDeprived

Thanks for the explanation. I have a scenario in which it's not clear whether the force is a pure magnetic force or pure electric force I hope maybe you could explain.

For example, I have a current carry wire in the y-direction going from bottom to top. A charge Q1 is moving also in the y-direction with a velocity V1. Now for the reference frame of the lab (the wire), the force on the charge would be pure magnetic force. On the reference frame of the moving charge Q1, the force would be pure electric due to relativistic effect, but the total Force would be the same in both reference frame. And so far it is what we understand and it is what you said above.
Let's introduce a third participant, called N1, who is also traveling in the same y-direction but only at a velocity of 7/10 of charge Q1 or .7V1. To simplify the situation, this third participant, N1 is charge neutral so it would not be affected by either electric or magnetic force, BUT obviously N1 will measure that Q1 is attracted to the wire with the same force because all three references have to agree on the same amount of force. The complication is with respect to N1, the velocity of Q1 is only .3V1, not the full V1, so if N1 only uses the Lorentz function, Icurrent x .3V1 is the the same as Icurrent x V1. And if N1 only uses the force based only on relativistic effect, .3V1 won't give the full force either. How would we resolve this situation?
Thanks.

7. Dec 26, 2016

### vanhees71

Again, you can also write the equation of motion for the electron (neglecting radiation damping) in covariant form,
$$\frac{\mathrm{d}p ^{\mu}}{\mathrm{d} \tau}=\frac{q}{mc} F^{\mu \nu} p_{\nu}, \quad p_{\nu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
where $m$ is the invariant mass (scalar), $q=-e$ the charge of the electron, $p^{\mu}$ its four momentum, $(E/c,\vec{p})$, and $\tau$ the proper time.

For details see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

8. Dec 26, 2016

### SleepDeprived

Actually I am still not quite sure what you're asking. Your usage of the terms "electromagnet" and "ring current" might be a git general. I was wondering if you could be more specific

9. Dec 26, 2016

### arydberg

I think he is asking what is the result of a wire ring conducting a current rotating about it's axis through the center of the ring. Is there still the same magnetic field as there was when the ring was not rotating or is there a different magnetic field and a new electric field?

It is accepted that given a round permanet magnet in the shape of a disk with the north pole on top then rotating this magnet about it' s vertical axis causes no change in the magnetic field.

10. Dec 26, 2016

### jartsa

If N1 knows physics, he knows that any force on Q1 is smaller than what Q1 measures it to be, because:

Moving observer's force measurements are too large. For example a moving capacitor thinks the force between its plates is the Coulomb force, but the "real" force is smaller, it's Coulomb force minus magnetic force.

Let's consider a relativistic atom that become ionized in a weak electric field:

Observers co-moving with the atom think the force that pulled the electron out of the atom was large, but it was just the force of a weak electric field according to observers that co-move with the electric field.

(Or should I say that a magnetic force helped to ionize the atom ... well in this case the magnetic force is an internal force, and my point is that only a small external force is needed to break the atom, in the frame where the atom moves fast)

Last edited: Dec 26, 2016
11. Dec 26, 2016

### SleepDeprived

OK, I see now what he was asking. Let's look at a simpler example.
If we have a constant carrying current or a DC current, there is obviously a magnetic force proportional to the strength of the current, but the magnetic would be a constant therefore dB/dt = 0. So even if you place this DC current wire next to another wire, it would not induce any current on the other wire. Now if you physically jiggle or move about the DC current wire, now you would have a changing dB/dt since B is a vector so now you would have an induced current on the nearby wire.
Even though the current is the same, and B is the same even if the wire is moving about, but B is a vector, the near wire will experience changing B vector hence the induced current.
Based on Biot-Savart law, B is a function of Icurrent, mu, and R (R being the vector from the point of interest to the wire). When the wire is jiggled about, even I and mu are not changed, R is changed so that is where you get a change in dB/dt with respect to the nearby wire.
Now back to the current conducting ring, assuming the current is DC, the only thing difference in this case is now you have an accelerating component toward the center of the ring, so how would that affect the B field? My guess is it does not matter, because if you simply carry out the Biot-Savart equation, it doesn't really care about the accelerating component, so you would get the same result if you have the same I, mu, and R.

I am not quite sure what this means.

12. Dec 26, 2016

### jartsa

How about if we decompose a normal electromagnet to two electromagnets:

Electromagnet 1: In an electromagnet there is a ring shaped formation of electrons that rotates, and a charge moving relative to the center of that ring sees different length contractions at different parts of the ring, and therefore it sees different electron densities at different parts of the ring, and therefore it's attracted to some parts of the ring and repelled from some other parts of the ring.

Electromagnet 2: In an electromagnet there's a ring shaped positive crystal lattice that usually does not rotate. If we rotate the whole electromagnet to the same direction as the electron formation rotates, then a charge moving relative to the center of the electromagnet sees different length contractions at different parts of the positive crystal lattice, and therefore it sees different proton densities at different parts of the electromagnet.

When we rotate the whole electromagnet the positive crystal lattice becomes an electromagnet, and the electron formation becomes a stronger electromagnet.

The forces those two electromagnets exert on charges point to opposite directions.

13. Dec 26, 2016

### sweet springs

I further try to explain my point

Say free electrons of number density n_v in wire of cross section S go round with speed v in ring shaped wire. If we postulate that Feynman text applies in this rotation case, current is
$$i=-evS n_{v},\ e>0$$
$$n_{v}=\gamma n_{0}$$
Wire with current is electrically neutral
$$n_{0}<n_{v}=N_{0}$$
where N_0 is number density of positive ions of charge e that consist wire together with free electrons.

We move to reference frame of rotation where speed of electrons is zero and speed of positive ions is -v. Current I is
$$I=e(-v)S N_{-v}$$
$$N_{-v}=\gamma N_{0}=\gamma n_{v}$$
$$I=\gamma i$$
Charge desntity of wire is
$$e(N_{-v}-n_{0})=\gamma \beta^2 \ eN_{0} >0$$

We stay in the original inertial reference frame and rotate the wire ring to such an extent that free electrons are still. I postulate that the result is similar to the above case of "we rotate and wire is still". So rotation of electrically neutral magnet generate charge on it.

14. Dec 27, 2016

### jartsa

"We rotate" and "magnet rotates" are not the same things at all. Because rotation is absolute, not relative.

A neutral magnet must stay neutral, because charge is conserved.

15. Dec 27, 2016

### sweet springs

Thanks jartsa.
Following the way in Feynman's text, if we put an electron nearby the ring wire co-rotating with the ring current i, the electron is attracted to nearest side of wire and repelled from the farthest side by electric field. Thus in the rotating frame of reference in which the electron is still, the nearest side to the electron is charged plus and the farthest side is charged minus. In the rotating frame of reference such inhomogeneous charging appear.
In the inertial frame of reference in which the neutral ring is still and current i is running in it, after we rotate the ring there is no reason such an inhomegeneity take place. So I agree with you.

16. Dec 27, 2016

### sweet springs

However, on rotating frame of reference for the electron , say 1, in my previous post we put another electron 2 nearby the opposite side of the ring. Electron 2 as well as electron 1 get attractive force from its nearest part of ring thus charged plus and repulsive forth from the farthest side thus charged minus. Opposite signs to my previous post thought.
Consideration of forces acting on electron 1 and electron 2 give contradictory results about charging of the ring. Much puzzled.

I applied for rotating frame of reference which is a kind of non inertial frames
$$\mathbf{F}=e\{\mathbf{E}+\mathbf{v}\times\mathbf{B}\}$$
This might cause the trouble.

Last edited: Dec 27, 2016
17. Dec 27, 2016

### SleepDeprived

It seems like a lot of confusions arise from not agreeing on the terminology. I am still not sure what you're referring to when you're using the term "electromagnet". So I'll continue to illustrate based on what I know of the Maxwell equation and some basic circuitry.
Think of this situation by using a transformer. Obviously we all know what a transformer look like so it's a good starting point. The basic transformer has two winding coils - one primary and one secondary. Now, think of the winding coil as the "ring" that you've been referring to although the actual shape may be different it's the same electrically. So the way the transformer works is the primary coil has a AC current which induces a AC current on the secondary coil (or ring), but if the primary coil has on DC current, then it cannot induce any current on the secondary coil. Now obviously a real world transformer is stationary, if you want, you can have only DC current of the primary coil, but this time, you rotate the primary coil, then now you can induce a current on the secondary coil. So if you have a 60Hz AC current on the primary coil, then you'll have a 60Hz on the secondary coil. But if I have a DC current on the primary coil, but if somehow I can rotate the primary coil with a 60Hz cycle with my hand, then you have a 60Hz AC current on the secondary coil. It's the same thing.
Now, a little unexpected twist. All transformers have two winding coils in parallel to each other. But if you turn the secondary coil in such a way that it is fixed at 90 degree (or perpendicular) with respect to the primary coil, then there won't be any current induced on the secondary coil. Actually to be specific, there is a current induced this way but the current will be flowed in the coaxial direction of the wire, not along the length of the wire as a conventional transformer which what you normally want in a normal circuit. So in order to have an induced current, you need to have two coil in parallel and facing each other.
If you work out the relativistic effect on the contraction of length, it depends on how you rotate the ring, and how the rotate direction with respect to its neighbor.

Last edited: Dec 27, 2016
18. Dec 28, 2016

### jartsa

Well I refuse to consider rotating frames of references. But electron1 and electron2 moving to opposite directions at the opposite sides of the ring, see the situation like this:

Electron1 thinks the whole magnet is moving to the left at speed vm. Electron2 thinks the whole magnet is moving to the right at speed vm.

Electron1 thinks electron2 is moving to the left at speed vm + ve. Electron2 thinks electron1 is moving to the right at speed vm + ve. (I used Newtonian velocity addition for simplicity)

Electron1 thinks the opposite side of the ring is densely packed with fast moving electrons. And electron2 thinks the opposite side of the ring is densely packed with fast moving electrons. Is this a problem?

vm means magnet's speed relative to electron
ve means electron's speed relative to magnet

Last edited: Dec 28, 2016
19. Dec 28, 2016

### jartsa

... And now let us rotate the magnet so that the electrons stop rotating, I mean the electrons stop moving in the center of magnet frame. This makes both vm and ve zero.

Now every electron sees every proton having the same speed vp. If an electron sees protons being densely packed because of the vp, then the electron sees protons being equally densely packed everywhere. And if the electron sees protons being densely packed everywhere, it sees the radius of the magnet being reduced, and it sees electrons being densely packed as a result of that size reduction of the magnet.

20. Dec 28, 2016

### sweet springs

I draw two pictures to explain what I wonder.

There is a ring conductor in which free electron current is running and electrically neutral in total.

Fig.1　　Spin it. Free ekectrons stop. Positive ions rotate. $rho'=$?
http://fphys.4rm.jp/modules/xelfinder/index.php/view/158/スライド4.jpg

Fig.2 We move from inertial frame of reference to rotating frame of reference. There free ekectrons stop. Positive ions rotate. 　$rho"=$?
http://fphys.4rm.jp/modules/xelfinder/index.php/view/157/スライド5.jpg