Are magnets charged with rotation?

[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.

 

[SleepDeprived]

I wonder whether electromagnet is charged with rotation.

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

 

[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.

 

[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.

Do you mean if you rotate a wire, will it create a magnetic field?

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.

 

[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.
...

 

[SleepDeprived]

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.
...

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.

 

[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

 

[SleepDeprived]

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.

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

 

[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.

 

[jartsa]

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.

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)

 

[SleepDeprived]

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.

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.

is there a different magnetic field and a new electric field?

I am not quite sure what this means.

 

[jartsa]

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.

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.

 

[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.

I am not sure at all about this speculation. Your teaching is quite appreciated.

 

[jartsa]

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".

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

So rotation of electrically neutral magnet generate charge on it.

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

 

[sweet springs]

Thanks jartsa.

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

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.

 

[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.
Your help is appreciated.

 

[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.

 

[jartsa]

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.
Your help is appreciated.

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 v_m. Electron2 thinks the whole magnet is moving to the right at speed v_m.

Electron1 thinks electron2 is moving to the left at speed v_m + v_e. Electron2 thinks electron1 is moving to the right at speed v_m + v_e. (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?v_m means magnet's speed relative to electron
v_e means electron's speed relative to magnet

 

[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 v_m. Electron2 thinks the whole magnet is moving to the right at speed v_m.

Electron1 thinks electron2 is moving to the left at speed v_m + v_e. Electron2 thinks electron1 is moving to the right at speed v_m + v_e. (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?v_m means magnet's speed relative to electron
v_e means electron's speed relative to magnet

... 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 v_m and v_e zero.

Now every electron sees every proton having the same speed v_p. If an electron sees protons being densely packed because of the v_p, 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.

 

[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

I hope these will help you to catch my point. Regards.

 

[sweet springs]

For the case of Fig.1 now I come to consider that $rho'=0$.

In the original inertial frame of reference, rest frame of free electrons is not an inertial frame but rotation frame. Non Euclidean geometry tells circumference of the ring is $2\pi\gamma r$ where r is radius of the ring. In order to mention the rest frame of reference is inertiaｌ or rotation, I introduce suffix .in and .rot. Number density is
n_{0.rot}=\frac{\mathfrak{N}_e}{2\pi\gamma rS} where $\mathfrak{N}_e$ is number of free electrons.
In the inertial frame of reference the free electrons move at sped v so as same as Feynman tells for straight wires, Lorentz transformation put
n_{v,in}=\frac{\mathfrak{N}_e}{2\pi rS}
Thus non Euclid geometry and Lorentz transformation cancel.

N_{0.in}=\frac{\mathfrak{N}_p}{2\pi rS} where $\mathfrak{N}_p$ is number of positive ions, Thus
n_{v.in}=N_{0.in}, \mathfrak{N}_p=\mathfrak{N}_p, \rho=e\{N_{0.in}-n_{v.in}\}=0

Then we rotate the ring so that free electrons are still as the right hand side of Fig.1.
Free electrons are still. Positive ions rotate. The numbers of them are same. It is just changing the roles of still and rotating between free electrons and positive ions. So the result is same.

\rho'=0,n'_{0.in}=N'_{-v.in},\mathfrak{N}_e=\mathfrak{N}_p

Even if we rotate neutral ring in which current run, that represents magnet in general, or change current in any way, there appear no charge and thus no electric field.

 

[sweet springs]

As for the case Fig.2, number density of free electrons is

n_{0.rot}=\frac{\mathfrak{N}_e}{2\pi\gamma rS}

Number density of positive ions are by cancel of non Euclidean geometry and Lorentz transformation
N_{-v.rot}=\gamma N_{0.rot}=\gamma \frac{\mathfrak{N}}{2\pi\gamma rS},\ \mathfrak{N}=\mathfrak{N}_p=\mathfrak{N}_e
Thus charge density is
\rho"=e\{N_{-v.rot}-n_{0.rot}\}=(1-\frac{1}{\gamma})\frac{\mathfrak{N}}{2\pi rS}>0,

Neutral ring wire with current going through, that represents magnet in general, is observed positive charged in the rotation frame of reference in which current is still. Positive charge generate electric field. Electric field and magnetic field of magnet generate Poynting vector. Poynting vector show that electromagnetic field have angular momentum. Electromagnetic field of zero angular momentum in inertial frame of reference is observed to have non zero angular momentum in rotation frame of reference. It is obvious.

 

[sweet springs]

I am almost sure of homogeneous positive charge of ring wire in rotation frame of reference. but my #15 and #16 still puzzles me. Best.

 

[vanhees71]

I've not followed the details of the discussion, but you should note that a rotating reference frame (i.e., a frame rotating against an inertial and thus against any inertial frame) is not a inertial frame, and thus you have to apply the machinery of general covariance (you are very close to General Relativity doing that), i.e., you have to introduce the non-trivial metric and covariant derivatives, etc. etc. For a starting point, see the folowing problem of our current GR lecture in Frankfurt:

http://th.physik.uni-frankfurt.de/~hees/art-ws16/sheet02.pdf

and solutions

http://th.physik.uni-frankfurt.de/~hees/art-ws16/sheet02.pdf