# Talk:Earnshaw's theorem

## Contents

## More precise language in introductory paragraphs

I'm leaning toward reworking the first few paragraphs of the article to be more precise with respect to certain concepts: force versus potential energy, electric fields versus magnetic fields and the need to invoke other properties of these fields besides zero divergence. In particular, the article should be clear that magnetic fields are not inverse square force fields (or even force fields, per se - a magnetic field vector gives the potential energy of the reference dipole rather than the force on the reference dipole). In addition to zero divergence, it is also necessary to invoke either conservation of energy or zero curl (or possibly both for magnetic fields). If anyone is interested we can work together to craft some changes. Otherwise, I'll just have a go myself sometime in the next couple weeks. Compbiowes (talk) 05:27, 19 April 2008 (UTC)

## Why a ferromagnet cannot be levitated if it is aligned anti-parellel to the external field initially

In the article the extrema of the potential of a ferromagnet aligned parallel or anti-parallel to the external field are discussed. The dipole moment stated in the article is:

According to the calculation made there it seems as if levitation can be possible if the magnet is aligned anti-parallel. Then there is this sentence:

It should be noted, however, there are no known configurations of permanent magnets that stably levitate so there may be other reasons not discussed here why it is not possible to maintain permanent magnets in orientations anti-parallel to magnetic fields (at least not without rotational motion - see Levitron).

There are no "other reasons" necessary to explain this. If the magnet is aligned anti-parallel at **any time** (as required by the used magnetic moment) levitation is possible. This is exactly the principle of the Levitron (compare [1]).

I think what is meant by the ansatz for the magnetic moment above is something different. If the magnet is aligned anti-parallel initially and then pertubed a little bit we may claim that the orientation of the magnet does not change, but this does not mean that is is still aligned anti-parallel to the field. The field might change! So a correct ansatz for a initially anti-aligned magnet whos orientation is constant would be:

This case is allready discussed in the article and does not lead to stable levitation.

So i think we should clarify that with the ansatz in the article the magnet is aligned parallel or anti-parallel with the field at **any time** and then the sentence i quoted above should be unnecessary.

- I'm not sure I understand the objection to the sentence in question but since I wrote the sentence in question I'll try to explain what I meant. Ordinarily, a permanent magnet will align itself
*parallel*to the field lines of the external magnetic field. In particular, if we try to align a permanent magnet*anti-parallel*to the external field lines then, as soon as we "let go", the permanent magnet will flip around to align itself*parallel*to the external fields lines. Suppose, however, that we have not just one permanent magnet but a whole system of permanent magnets all connected together in a flexible (and possibly very complicated) way - for example, with a system of gears and rods and pulleys. Suppose also that the static external magnetic field that will levitate this system of permanent magnets is also arbitrarily complex - for example, involving many magnetic poles of varying strengths. Can we be sure that there is no way for the system of permanent magnets to hold (at least some of) each other*anti-parallel*to the external field lines? Note that such a system would be very difficult to design because translational movement of the system would, in general, cause the orientation of the field lines to change relative to the permanent magnets. A successful system would have to be able to "detect" the translational movement and change the orientation of the permanent magnets accordingly. Earnshaw's theorem might be disproved if such a system was possible. That is, for the proof given on the Wikipedia page to be completely rigorous, we must prove that such a system is not possible - but I'm not sure how to do that. For that matter, somewhere out there in the vast body of scientific literature there may be a different more general proof of Earnshaw's theorem that doesn't have this limitation. Compbiowes 00:09, 4 May 2007 (UTC)

- I think the system you described above is not possible. A (somewhat lengthy) argument why it is not possible was given by Werner Braunbek in 1939 (Werner Braunbek: Freischwebende Körper im elektrischen und magnetischen Feld, Zeitschrift für Physik 112 (1939), p. 753-763). Unfortunately i don't know an online source for this article (and an english version ...). Braunbek concludes that it is not possible to levitate an object using
**any**static combination of magnetical, electrical and gravitational fields as far as no diamagnetic substances are involved. However there is one objection against this argument. He only considers translational movements and concludes that if a system is unstable regarding translations it has to be unstable regarding any movement. This conclusion is proven to be wrong by the levitron ([2], p. 6).

- A while back, I took the trouble to get a copy of the Braunbek article myself but I only got as far as the translating the abstract into English and I didn't get around to understanding his proofs in detail. Strictly speaking, Earnshaw's Theorem in its original form only dealt with electric charges and other inverse square law forces (magnetic forces are
*not*inverse square). The main motivation I had for including my proofs for certain special cases of generalizing Earnshaw's Theorem to magnets was to show that the magnetic case is much more complicated and difficult to prove. One problem with the magnetic case is that it's not clear what is meant by a "static" arrangement of magnets. Is the arrangement of magnets still "static" if the magnets are allowed to rotate with respect to each other (or even just relative to the external field)? If Braunbek's proofs are more general than than my proofs then it could be good to replace my proofs with Braunbek's proofs. On the other hand, if neither proofs are able to treat rotation of the magnets in a fully general way then perhaps we could merely add a more precise qualifier as to what is actually being proved. Compbiowes 18:16, 7 May 2007 (UTC)

—as far as no diamagnetic *or* diaelèctric materials or interactions are involved. One could reach static levitation of a magnèt with help of a wall (as in one toy) or a string (cheap homebrew), but this contact would count as diaelèctric stability. Also, rotation in the Levitron is *not* static levitation, as it's acceleration—where it over the full flipover span sets up a energetic barrier which is pseudodiamagnetic against the magnetic potential minimum. The levitation of charges or planèts against their attractors are also such inertial pseudodiamagnetism. -lysdexia 04:01, 23 March 2008 (UTC) —Preceding unsigned comment added by 68.123.4.199 (talk)

## All inverse square law forces are divergenceless

I am confused as to your statement that all inverse square law forces are divergenceless. We know from Maxwells equations that Div(E)=charge density/epsilon naught or in a vacuum, Div (E)=0. We know that Div (B)=0 because magnetic field lines are circmferential, and do not diverge, but electric field lines do. Where have I misunderstood?

<email address can be found in history>

Check Electricity and Magnetism by Griffiths, a physics college undergraduate standard.

Any field that falls off like one over r squared does not necessarily have zero divergence. The Electric field is an example: Maxwells equations tell us that Div(E)=[charge density/permittivity of free space]. Magnetic fields are circumferential and therefore always have zero divergence. I think this is an important point to make, as the divergence of the E field only vanishes in a vacuum, where there are no charges located. The character of the field, being a one over r squared field allows us to do some mathematical tricks, but it does not exclude a field from having a divergence. The aspect of the magnetic field which makes its divergence vanish is the fact that there have been no scientifically observed MAGNETIC MONOPOLES. Thus, magnetic field lines close in on themselves. Electric field lines do not have this quality, as point charges do exist. Hence please be sure to differentiate between fields where one can apply similar mathematical formalism versus the physical characteristics of the field (e.g. whether the fields have a divergence or a curl).

- [Also posted to their talk page:]
- Thank you for your comments on Earnshaw's theorem. I have moved them to the talk page of the article instead, where comments belong. It's been a while since I've done vector calc, but I think the only discrepancy here is that these fields are divergenceless
*in free space*. Yes there are maxima and minima at point charges, but you can't create a minima in free space for an object to "fall into", so it is impossible to levitate something stably.

- Feel free to make more comments on the talk page or edit the article if you think it needs expanding.

- You said "as the divergence of the E field only vanishes in a vacuum, where there are no charges located". I think that is exactly the point. The only way to create a local minima in the presence of gravity is if the local minima is at the center of a charged/magnetic object. You can't create one in free space without violating his assumptions. - Omegatron 18:09, August 10, 2005 (UTC)

It should be noted that this theorem is valid in 3 dimensions but not in 2 dimensions.

## a contradiction

I have a question: consider this problem: Four poin charges q are placed on the corners of a square and one poin charge q is placed on the center of the square. We want to consider the eqiblirium of centeral point charge in the direction of one diagonals of the square. If you investigate low displasment, you can see that this charge has stabel equiblirium! What is the connection of this problem with the Earnshaw's theorem ?

- That wouldn't be a stable configuration. — Omegatron 14:41, 12 January 2006 (UTC)

This is a stable configuration if you are only considering the plane. If you consider 3 dimensional space, it is obvious the configuration is unstable. Possibly also of interest, four charges seems to be the least you must have to create a stable equilibrium in the center; with three charges, the center is a saddle-point and thus unstable.

- It's not stable in a plane either. The test charge will shoot out in between two of the fixed charges, if it is the same polarity, or be attracted to one of them, if it's the opposite polarity. It's an equilibrium, but it's not a stable equilibrium. — Omegatron 16:19, 22 January 2006 (UTC)

- You can visualize the potential in 2 dimensions as an infinite rubber sheet. The 4 charges are vertical sticks underneath, pushing up the sheet to make peaks. The saddle point between the sticks will be higher than the plane far away, so a marble placed there will tend to roll through one of the 4 valleys between the sticks.

You can generalize this image to any number of charges/sticks, with negative charges being sticks pushing the sheet down to make a hole. There will be saddle points between the sticks, where a marble could sit stationary, but the slightest disturbance will cause it to either roll into one of the holes, or roll out into the plane far away. As Omegatron says, unstable equilibria. -- Chetvorno 03:16, 5 March 2006 (UTC)

- You can visualize the potential in 2 dimensions as an infinite rubber sheet. The 4 charges are vertical sticks underneath, pushing up the sheet to make peaks. The saddle point between the sticks will be higher than the plane far away, so a marble placed there will tend to roll through one of the 4 valleys between the sticks.

- Continuing the rubber sheet analogy, imagine connecting the tops of the sticks with pieces of string so the strings hold up ridges in the rubber sheet. If the strings are connected diagonally then the surface of the rubber sheet will not have any depressions (a marble would be unstable). If the strings are connected along the sides and if the rubber sheet has weight then the rubber sheet will sag in the middle forming a depression (where a marble would be stable). Basically, either stability or instability is possible depending on the precise form of the potential. Playing around with MatLab and a point charge potential, I get the result that for this two dimensional case there
**will**be a depression where a marble would be stable. Compbiowes 20:50, 20 October 2006 (UTC)

- Continuing the rubber sheet analogy, imagine connecting the tops of the sticks with pieces of string so the strings hold up ridges in the rubber sheet. If the strings are connected diagonally then the surface of the rubber sheet will not have any depressions (a marble would be unstable). If the strings are connected along the sides and if the rubber sheet has weight then the rubber sheet will sag in the middle forming a depression (where a marble would be stable). Basically, either stability or instability is possible depending on the precise form of the potential. Playing around with MatLab and a point charge potential, I get the result that for this two dimensional case there

- Possibly your introduction of weight in the rubber sheet, and/or strings, enabled you to create a depression. To be analogous to a potential, the rubber sheet has to be uniform and of negligible weight itself, stretched tight, like a trampoline. Under those conditions it shouldn't be possible to create a stable point with sticks. --Chetvorno 01:50, 24 December 2006 (UTC)

Your examples could really use some images. ;-) — Omegatron 21:16, 20 October 2006 (UTC)

- I calculated this gray-scale image of a slice of the potential surface around four point charges. Compbiowes 17:28, 22 October 2006 (UTC)

On the subject of generalizing this to larger numbers of charges, the requirement for zero divergence implies that there will be a 2-diminsional point of stability at the center of the ring. For any point not in the plane of the ring, the force will be directed away from the plane of the ring. Because of the symmetry of the ring, zero divergence requires that the forces in the plane of the ring will be pointing inward near the center of the ring. Compbiowes 17:40, 22 October 2006 (UTC)