# What prevents this magnetic perpetuum mobile from working?

As a child, I imagined this device, which may seem to rotate indefinitely. I have two questions.

1. Is this perpetual motion machine already known? If it is, could you please give some references?
2. What is the exact mechanism that makes it stop? By this, I mean an explanation, not simply "because it would break law of energy conservation". Of course energy conservation is true and this machine cannot work indefinitely. But for any known (presumed) perpetual mobile, there was an explanation, usually based on showing that the force generating the motion is balanced by another force. For example, Stevin obtained the laws of the inclined plane from the perpetual motion device

I would like to make some observations.

• I don't try to convince anyone that it will move forever because I don't believe in breaking the energy conservation. It is true that great physicists like Bohr, Kramers and Slater admitted the possibility, and nowadays some who think that there may be energy exchanges between parallel worlds in MWI believe, but I don't.
• But I don't consider an enough explanation simply to refer to the energy conservation. I am interested in an explanation showing exactly how the magnetic forces making it rotate, are balanced.
• If the forces are balanced, only then friction will make it slow down and stop. I don't think that we can explain only by referring to friction, which in principle can be made as small as needed. There has to be a balance of forces.
• Why spending time trying to understand or explain something that admittedly can't work? Well, even though perpetual motion machines cannot actually work, I think they may be interesting as puzzles.
• What are the red and blue surfaces? Electric charges (of opposite signs)? Magnetic poles? The exact explanation will depend on this. Aug 27 '13 at 22:22
• @Colin McFaul: Magnets. I thought it was clear, because I did use the word in title, text, and tags. But, if they would be electric charges, would this make a difference? I would be interested in this too. Aug 28 '13 at 5:12
• @CristiStoica, electric charges rotating in such a way would create a magnetic field, which might affect the motion in a way to oppose it. I think the same might be happening with the magnets rotating, but I can't seem to formalise it through equations. Aug 28 '13 at 5:21
• @udiboy: Maxwell's equations are invariant at interchange of electric and magnetic fields, provided there are no sources. If we consider electric dipoles instead of magnets, there will be no difference. In particular, if the rotating electric dipoles create magnetic fields, the converse is true as well: rotating magnets create electric fields. Aug 28 '13 at 7:50
• Nov 16 '14 at 21:58

You do not need to invoke friction. The magnetic forces are in equilibrium by themselves so if you place the magnets in that configuration, they will not spontaneously begin to move. The reason is that there is a corresponding force on the magnets when they are vertical that matches the ones you've already drawn.

Let me make a simple model. First of all, start by upping the game and including two big magnets, which can only make it better:

If red is a north pole, then each rotating magnet, when horizontal, has a north pole repelling its backside north pole and a north pole attracting its front south pole. Focusing on the big magnets for the moment, the fact that their north poles face each other suggests that we can trade them for a pair of anti-Helmholtz coils. This means the important character of their field is its quadrupolar nature, and we can approximate the magnetic field as $$\mathbf B=\frac{B_0}a\begin{pmatrix}x\\y\\-2z\end{pmatrix},$$ where the $z$ axis goes from one big magnet to the other, $a$ is some characteristic length, and $B_0$ is some characteristic field strength.

Now, for the little magnets, I think it is uncontroversial to model them as point dipoles. If $\theta$ is the angle the wheel spoke makes with the $x$ axis (with the wheel in the $x,z$ plane) then each magnet is a dipole with moment $$\mathbf m=m \begin{pmatrix}-\sin(\theta)\\0\\\cos(\theta)\end{pmatrix} \text{ at } \mathbf r=R \begin{pmatrix}\cos(\theta)\\0\\\sin(\theta)\end{pmatrix}.$$ With this, the potential energy of each spoke magnet is $$U=-\mathbf m\cdot\mathbf B=3\frac R a mB_0\sin(\theta)\cos(\theta)=\frac32\frac Ra mB_0\sin(2\theta).$$

To see how this behaves, here is a colour plot of the energy, with negative energy in red and positive energy in blue.

You can see there is a gradient pointing up on the right and down on the left. However, these are matched by clockwise gradients when the spokes are vertical. A single magnet will settle on the lower left or the upper right; a pair of magnets will settle on that diagonal. For a symmetrical wheel with three or more magnets, the total potential energy is flat at zero, $$U=\sum_{k=1}^n\frac32\frac Ra mB_0\sin\left(2\left(\theta_0+\frac{2\pi}{n}k\right)\right) =\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta}\sum_{k=1}^n(e^{2\pi i/n})^k\right] =\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta+2\pi i/n}\frac{1-(e^{2\pi i/n})^n}{1-e^{2\pi i/n}}\right] =0$$ and there is no resultant magnetic force.

• If anyone wants to prettify the images, they're welcome. Aug 28 '13 at 15:34
• I am very happy with this answer. I was thinking at the same settlement positions as those you found, but I used very rudimentary reasoning, while your answer is rigorous. I will accept your answer. If you don't mind, I will come with a follow up question, perhaps tomorrow (because now here is late after midnight, and I just returned from Roger Waters's concert). Aug 28 '13 at 21:42
• Well, it is tomorrow anyway:) Here is my follow up question. Consider my version with belt, as in my answer. Let's consider 1. the ambient magnetic field is nearly uniform, and 2. there is only one small magnet. In the regions A-B, C-D, the magnet is accelerated, and in the regions B-C, D-A, it is slowed down and perhaps stopped. After giving that answer, I thought: what if the linear region is long enough, so that the magnet is accelerated well enough to go through the brake regions? Would it move forever? This shouldn't be possible. Do you think your answer can be adapted to this case too? Aug 28 '13 at 21:56
• +1, very nice. I guess an intuitive version of this answer is that although the magnets are being accelerated in the way shown in the question, there is also a torque on the magnets, making them want to twist around to have their south poles facing the big magnet's north pole. Aug 29 '13 at 4:07
• what if you give them an initial rotational motion? Aug 29 '13 at 5:43

I will try to explain what I think is the solution. It is obviously far from complete, it is just an idea, a "progress report". So I will not select it as the correct answer, unless I will be able to finish it.

To explain the solution, I will refer to the following image, which is a simplified version of the original device from the question. It looks more complicated, but I think it makes the main idea simpler to understand.

Each magnet moves along the band ABCD, cyclically. Each magnet, when moving between A and B, and between C and D, doesn't rotate, but it is accelerated by the magnetic force. This is obvious, as anyone who played with magnets knows. Therefore, if the movement is slowed down somewhere, it has to be when moving between B and C, and between D and A, as seen in the picture below.

Between the points B-C, respectively D-A, the magnet rotates. Some of the field lines joining it with the large magnet are broken, and others are created. If we put a small magnet near a large one, we see that it changes its orientation, and locks itself along a preferred orientation. We also know this by looking at the needle of a compass in the magnetic field of the Earth. So, my guess is that it is there where the kinetic energy accumulated when moving along AB and CD is consumed. But I don't know how to calculate the balance. Anyway, my guess is that the system will find a preferred position and lock into it, before being able to make at least a complete rotation.

• The belt mechanism does not explain something. The magnets on the belt when running over the pulleys do the same as they do in the original wheel. So the belt machine is just the superposition of the wheel plus the linear extension. (Btw the I see energy loss/gain in the linear yellow section different) Aug 28 '13 at 14:15
• @Georg: My point with the belt is that on the yellow region there is linear acceleration, and on the gray region rotation. So, somehow, the rotation should balance the kinetic energy gained in the yellow region. I am interested if you could detail your "btw..." :) Aug 28 '13 at 14:22
• ""there is linear acceleration,"" more nonsense! There is linear movement! Re the yellow region: one side ("up") will loose energy, the other down will gain. Please learn basic mechanics and the wording of it. Aug 28 '13 at 14:26
• @Georg: 1. If you say "more nonsense", please tell first where the other nonsense was. 2. "Linear acceleration", yes, it is linear movement, but accelerated. 3. Look at the way magnets are oriented when go up and when go down. In both A-B and C-D the movement is accelerated, hence kinetic energy is gained both times. Do you still believe that going up "loose" energy? You refer to potential energy in gravitational field, which may even not be present, and I was talking about kinetic energy, as you can see if you read again. Please read carefully and don't insult because you don't understand. Aug 29 '13 at 4:19
• in effect as Emilio's answer illustrates, your attempt is correct, the little magnets when rotated at the pivot points (BC, DA) have an exact opposite force (to the one on AB, CD) so the whole system is under equilibrium. One can break equilibrium and make the rotations, but then also introduces asymetry which itself introduces an efficiency $\eta < 1$ Nov 8 '14 at 1:24

The simple answer is the magnetic forces are at equilibrium. The north poles are attracted to the south pole of the fixed magnet with the exact force that the south poles are repelled from it, and the same with the fixed magnet's north pole.

• is the energy required to disturb the equilibrium higher or equal than the energy created from the movement? May 5 at 14:45
• There would only be kinetic energy loss from friction and drag forces. May 5 at 20:45