Talk:Earnshaw's theorem

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Moved from article. Left by User:208.239.204.208:

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?

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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.

[edit] 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)

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)

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.

Slice of Potential Surface

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)