# I Relativistic Explanation for the Unipolar Generator

1. Mar 11, 2013

### Jdo300

Hello All,

I was recently looking into explanations for the operation of Faraday's unipolar generator and came across the following article, titled "The Unipolar Generator, A Demonstration of Special Relativity" published by a professor at the University of Maryland:

http://www.physics.umd.edu/lecdem/outreach/QOTW/arch11/q218unipolar.pdf [Broken]

I'm an EE, though I am also very interested in physics. This article really intrigues me because I am wondering how special relativity is playing such a powerful role in the operation of a unipolar generator when the disk isn't spinning anywhere near relativistic velocities?

One question I have: Do relativistic effects manifest more strongly when dealing with microscopic things compared to the macroscopic things (as in charge distributions, etc)? Does the scale of an observed interaction play a role (microscopic vs. macroscopic reference frames)? Intuitively, I would expect that the disk would need to be rotating at some significant fraction of the speed of light to see measurable differences in charge distribution, but why is the manifestation so strong in this case?

Thanks,
Jason O

Last edited by a moderator: May 6, 2017
2. Mar 12, 2013

### Jano L.

Thanks for the reference, it is a nice experiment. The authors of the paper arrive at a reasonable numerical prediction, but their argumentation has some buts.

They claim that the phenomenon can be explained only with special relativity. They do not show why it is impossible to understand it without it, for example within the electromagnetic theory as used before Einstein. They only base their claim on the circumstance that when they use the relativistic transformation rule (first formula on the page 3) for electromagnetic field from one inertial frame to another, they are able to predict what voltmeter will show. This is not sufficient.

Their argumentation based on the transformation of fields is correct within good approximation, provided fields transform according to theory of relativity. It is true that the formula for transformation of the electric field can be applied, since the rotation is slow and the centrifugal force is negligible.

But it is not necessary to use these relativistic transformation rules and assume that $\mathbf E = 0$ in the frame of magnet. Instead, one can derive the same result by working in the lab frame all the time, thus entirely avoiding relativity. Here is the derivation.

Since the mobile charges do not move in radial direction, and the central force maintaining them in circular trajectories is negligible, we can expect that the macroscopic magnetic and electric force in the lab frame cancel each other:

$$\mathbf j'\times \mathbf B' + \rho' \mathbf E' = \mathbf 0,$$

Since in our situation $\mathbf j' = \rho' \mathbf v'$, we can conclude that

$$\mathbf E' = - \mathbf v' \times \mathbf B'.$$

Since the electric field in the lab frame is not zero, some charges have to be present. The magnet will contain spatial distribution of charge, positive charge near its axis, negative near its lateral surface, so the electric field $\mathbf E'$ at any point has radial direction and points from the axis to the lateral surface.

For this reason the axis will have higher electrostatic potential than the lateral area. If we make the assumption that the rotation of the magnet does not change its magnetic field, we can estimate $\mathbf B'$ by $\mathbf B$, which is 1 T and the potential difference is

$$\mathbf \int_{axis}^{lat.~surface} \mathbf E' \cdot d\mathbf r = \omega B \frac{R^2}{2}.$$

We thus arrive at the same conclusion. No transformation of forces was needed in this derivation; just the formula for magnetic force and the assumption that the rotation does not change the value of the magnetic field appreciably, which is quite natural and even physicist before the theory of relativity would most probably do.

(The formula #2 on p. 3, if it means $\mathbf v' \times \mathbf B$ is wrong by one minus, which however got canceled by another mistake of this kind in the integral.)

3. Mar 13, 2013

### jartsa

There are many relativistic effects happening in unipolar generator. The sum of these relativistic effects is that doubling the spinning rate doubles the voltage, although doubling the velocity more than doubles the relativistic effects of velocity.

One of the relativistic effects on a moving wire loop with electric current is that the velocity difference between the electrons and the wire decreases. Relativistic velocity addition works that way. This decrease of velocity difference decreases the length contraction difference.

Last edited by a moderator: May 6, 2017
4. Oct 20, 2016

### OriginalJames

I believe that Einstein demonstrated in one of his thought experiments that someone in one frame of reference might experience the motion of electrons, ie, an electric field while someone else in another frame of reference would experience a magnetic field for the same event, hence, the unification called the electromagnetic field which occurs at any velocity. See -- James

5. Oct 20, 2016

### vanhees71

6. Oct 28, 2016

### Jeronimus

Interesting video, but it seems to me that there is an error.

He says that even if there was a current flowing through the wire, the density of positive and negative charges would remain the same, hence the positively charged cat would not get attracted.

But is that really the case?

The cat, the protons and the electrons are at rest to each other.
For the sake of argument, we assume that we can accelerate the protons along with the cat, while leaving the electrons "untouched".

From the perspective of the cat, the protons retain the same density as they remain at rest relative to the cat. The electrons however would be denser now (lorentz contraction) and closer to each other.
So contrary to the video explanation, it seems to me that the positively charged cat at rest relative to the wire, should be attracted to the wire when a current flows through it.

Last edited: Oct 28, 2016
7. Oct 28, 2016

### vanhees71

It's strictly speaking not the case, because the current in the wire leads to a magnetic field, which itself influences the moving charges (in usual metallic conductors the electrons) via the full Lorentz force (including the magnetic part). That leads to the buildup of a surface charge and an additional electric field around the wire (Hall effect). The reason why that's negelected in the usual electrodynamics textbooks is that they use the non-relativistic approximation of Ohm's Law. The full constitutive relation for the conduction current is (in Heaviside-Lorentz units)
$$\vec{j}_{\text{cond}}=\sigma (\vec{E}+\vec{v}/c \times \vec{B}).$$

However, for usual household currents the speed of the electrons is so tiny that you can forget about it. The drift velocity of the electrons in that case is of the order of 1mm/s (millimeters per second) only! So you can very savely forget the magnetic part in Ohm's law and are back at the usual textbook treatment.

8. Oct 29, 2016

### Jeronimus

So basically. Special Relativity does not really explain how magnets work. You can explain why the cat gets repelled from the cat's perspective, but when looking at it from the perspective of an observer at rest relative to the wire, you are back to magnetic fields.
Also, even from the perspective of the cat, there is a magnetic field involved still. As the protons moving away from the cat should be generating a magnetic field.

I often read "magnetism is electrostatics combined with special relativity" but that does not seem to be the case. Even when using SR you seemingly have to deal with magnetic fields still and while with electric fields we have electrons and protons among other particles as the emitting source of such fields, with magnetic fields it seems a little bit stranger. There are no magnetons. They just seemingly pop out of nowhere whenever there is a current flowing to ensure that everything stays consistent.

9. Oct 29, 2016

### Staff: Mentor

Maxwell's Equations explain how magnets work. Maxwell's Equations are Lorentz invariant, so they are compatible with SR.

I'm not sure where you've read that, but it's not correct. I've read (in one of Feynman's books, can't remember which one) that magnetism is a sort of "relativistic correction to electricity" (not those exact words, but the key is that "static" is left out), which is a better way of saying it, but still somewhat vague. This is why physicists use math when they want to be precise.

10. Oct 29, 2016

### jartsa

Yes that is the case.

Two electrons one meter apart, starting to move at the same high velocity at the same time, stay one meter apart. That's just a very simple fact.

Now let's say there is a magnetometer nearby when many electrons start moving simultaneously at the same velocity. The magnetometer is just a wire loop with electric current. The magnetometer will tell us that there is a magnetic field, and then we will think that there is a magnetic field.

How does the loop measure a magnetic field? Electrons on one side of the loop think that the electron formation contracted when it started moving, while electrons on the other side expanded. That causes the loop to point to the direction of the magnetic field.

11. Oct 29, 2016

### Jeronimus

Yes, that is indeed a fact, if from the perspective of an observer at rest relative to the wire, the electrons would keep the distance when accelerated. But that is not what happens. The distance between the electrons shortens and the electrons themselves get denser.

You would certainly agree, that instead of accelerating the electrons, you would accelerate the cat in the opposite direction, according to SR the cat would be observing compressed electrons which are closer to each other. Length contraction according to SR.

If you accelerate the electrons inside an electric field instead however, you claim that they would stay at the same distance apart from each other. One could certainly device such a field for this to be the case i assume, but you would have to show mathematically that this is the case within a standard field.

12. Oct 29, 2016

### Ibix

@jartsa is correct. From the point of view of an observer at rest with respect to the wire the number density of electrons is unchanged. It has to be, otherwise the wire would acquire a charge - and where would the extra electrons come from? The point is that there is nothing requiring the electrons to maintain a constant separation in their rest frame. This is different from a cat (or indeed the wire) which has internal forces that insist that its atoms maintain a constant average separation in their rest frame.

13. Oct 29, 2016

### Jeronimus

Yes it would.

In the scenario where there is a current flowing through the wire, and the cat accelerates to be at rest relative to the electrons, the protons get closer to each other and become denser from the perspective of the cat.
From the perspective of the cat, the wire acquires a positive charge and that is why the cat is repelled.

I don't see why the wire acquiring a negative charge would be a problem at all.

That is a good question. But in the opposite case, where the cat gets repelled because of the denser and closer packed protons, where do the extra protons come from? It is still the same amount of electrons and protons, yet the wire is charged seen from the perspective of the cat, and this is why the cat gets repelled.

At least that is my understanding.

14. Oct 29, 2016

### Ibix

In the rest frame of the wire we observe that the wire does not acquire a net charge; if your explanation says it does then your explanation is wrong.

Let's imagine a current loop consisting of two straight wires connected by negligible short wires at either end - a very long, very narrow rectangle. In the rest frame of the electrons on one side of the loop the wire is length contracted, which is why the proton density rises (no more extra protons - just less length). But the electrons in the other side of the loop are moving faster than the protons so are even further contracted. So there is a positive charge on the side of the loop where the electrons are at rest and a negative charge on the other side - for a net charge of zero.

15. Oct 29, 2016

### Jeronimus

Let's imagine one wire which contains only protons and another wire which contains only negative charged particles glued/confined within the wire just like the protons are (ignoring if such a wire can be produced). The wires are very long and confined within tubes to not smash into each other, floating in empty space.

Now imagine we accelerate the negatively charged wire.

The positively charged cat remains at rest relative to the positive charged wire.

From the perspective of the cat, the negatively charged wire would contract in its length and the negative charges would become denser.

Would in such a case the cat get attracted towards the system of positive and negative charged wires?

16. Oct 29, 2016

### Ibix

That's a different scenario. In that case your negative charges are constrained to have identical separations in their rest frame before and after the acceleration. So yes, in that case I'd expect there to be a net negative charge (plus a magnetic field).

Incidentally "cat" is usually associated with a negative charge, except in solution.

17. Oct 29, 2016

### Jeronimus

Glad i got this right at least.

Now my question would be, related to your question "where would the extra electrons come from?" - What exactly would stop more electrons from entering the wire?

And finally. Which mechanism decides the distance between electrons at rest within the wire. Hence, no current flowing/at rest relative to the protons. By which mechanism would we calculate the distance between the electrons (in the electrons' rest frame) when a current is flowing, hence the electrons are moving in relation to the protons?

18. Oct 29, 2016

### Ibix

If there are extra electrons entering the wire then your battery must be acquiring a positive charge - in which case there is a force drawing electrons back into the battery.

In a conductor the electrons are free to move. They aren't tied together so they don't have to be the same distance apart. But you can note that there was (for the sake of argument) one electron per proton before you turned the current on so there must be one per proton afterwards.

In your second example you stipulated that the negative charges were glued down. So that one was easy to understand.

19. Oct 29, 2016