# charges inside a vacuum

#### BillO

Joined Nov 24, 2008
999
It looks like I am being my usual confusing self.

What I mean by a non-conducting sphere is a spherical cavity without a conductive lining. Not a sphere filled with a dielectric.

My original contention could be summed up in that Gauss' law will break down when you use electrons in place of continuous charge. Specifically at the boundary. Gauss' law assumes this nice continuous, uniform distribution of charge when you start talking about the field being zero inside a spherical shell of charge.

Let me pose a question (or two...), to Steve or studiot, or both, or anyone. Lets create a 1 meter radius, spherical universe. Now, put 2 electrons in that universe. As expected, the electrons will move to diametrically opposite points at the outer boundary of the universe. We now have a uniform (so to speak), but discontinuous, distribution of charge in our little sphere. Now, will there be zero field at every point within that universe?

Now what happens if we push another electron in? Will it be able to wander to any point on the outer boundary of this universe, or are there areas it is more likely to head toward? Will the other 2 electrons assert an influence on it? Will they possibly have to move to accommodate it? Does this take some of the energy that was expended in getting the 3rd electron in there?

Now, we can continue to add electrons. At what point does the field absolutely disappear everywhere in this little universe? Does the charge distribution ever truly become continuous? How about arbitrarily close to the boundary layer, full of discrete localized charges??

Do you see where I'm going with this? Am I making any sense?

Last edited:

#### steveb

Joined Jul 3, 2008
2,436
Do you see where I'm going with this? Am I making any sense?
Yes, I think you are making sense. Gauss's law will be valid for discontinuous charge, but I think what you are getting at is that the practical application of Gauss's law is pretty much impossible for a unsymmetrical case and it is clear that the field inside would not be zero for two or three or even quite a few more electrons around a sphere. With just a few electrons, the calculation of field is trivial with Coulomb's law. For 2 or 3 electrons in a 1 m sphere most people would just approximate the solution as zero field over the entire interior, but strictly, you are correct and the field is not zero.

So, perhaps you are implying that there might be an intermediate size for a radius of a sphere that is large enough so that it is a good approximation to use classical theory, but not so large that you can fill it up with enough electrons so that the assumption of symmetry can be made. I guess it's hard to answer without a calculation, but it all comes down to just how accurate you need to be to say you have symmetry and to say that the field is zero.

The extreme of 1 m radius is clearly too large to have any worry. Two or three electrons will provide negligible field inside the cavity, and if you ask the question of how many electrons will fit, we are talking about so many orders of magnitude that symmetry is completely valid as an assumption.

The other extreme might be a radius so small that we are talking about individual atoms of any element. Here quantum mechanics combined with classical electromagnetism provides a pretty good approximation, but the solution is not in any way a case of electrons on the surface of a sphere. In this case we can think of atoms with up to about 92 electrons, by the way.

So, there seems to be some size, larger than atoms, where we will transition over to a case where classical theory will be deemed accurate enough. When we reach that size, how many electrons will be allowed? I don't know, but it is definitely above 100 and is likely to be in the thousands. So we can imagine some intermediate size cavity with the maximum allowed electrons in it, based on the radius and potential strength. I suspect that if we have fully transitioned over to classical theory, there are enough electrons allowed to fill the perimeter sufficiently to provide a good approximation of symmetry. But, I can't prove that without a lot of work. However, if we take this relatively small cavity and put a small number of electrons in it, it is conceivable you might find an example of relatively significant non-zero field in the interior. Again, we are making an assumption of idealized conditions and ignoring a lot or possible interactions, but it never hurts to do thought experiments.

#### BillO

Joined Nov 24, 2008
999
Gauss's law will be valid for discontinuous charge,
Concerning the implications of the classical application of Gauss' law, outside the charge, yes, inside the charge, no. For discrete charges, symmetrical or not, the field inside the charge can never be totally zero. Except in the case where the charge is on a conductor. But real conductors (like metal) do what they do by QM means and ruin my hypothesis and are arguably outside this argument.

For 2 or 3 electrons in a 1 m sphere most people would just approximate the solution as zero field over the entire interior, but strictly, you are correct and the field is not zero.
Here, though, the devil is in the details. To insert each electron in that outer shell is gong to cost. Eventually it will cost more to put it in the outer shell than it does to introduce it to the universe. This is the point at which layers will begin to form.

So, perhaps you are implying that there might be an intermediate size for a radius of a sphere that is large enough so that it is a good approximation to use classical theory, but not so large that you can fill it up with enough electrons so that the assumption of symmetry can be made.
Not quite. Any size will do. The area of interest is at the boundary, not near the center where the symmetry will dominate. At the boundary, where every electron will eventually find itself, there are still discrete charges and associated field. As that outer boundary starts to get well populated, the repulsive forces between the electrons in that layer will get high. High enough that it will be difficult to add further electrons to that layer and they will find a lower energy lair inside it. (sorry for the pun)

...we are talking about so many orders of magnitude that symmetry is completely valid as an assumption.
Not where it counts. Not at that boundary layer.

...but it never hurts to do thought experiments.
On this we can fully agree.

#### steveb

Joined Jul 3, 2008
2,436
Overall, I can't say that I'm following your logic. Let's take just one part of what you are saying, as follows.

Concerning the implications of the classical application of Gauss' law, outside the charge, yes, inside the charge, no. For discrete charges, symmetrical or not, the field inside the charge can never be totally zero. Except in the case where the charge is on a conductor. But real conductors (like metal) do what they do by QM means and ruin my hypothesis and are arguably outside this argument.
I can't say you're wrong here, but only that I'm confused of exactly what you are saying. I don't know if you are saying Gauss' Law is not valid, or that you just can't use it, or if you are just saying that Gauss' Law says that the field is not zero in nonsymetrical cases.

I would say that the integral version of Gauss' law is always valid, but not always practical to use. Here it would only work as a useful tool if you have perfect spherical symmetry. So, perhaps you're just saying that because discrete charges can never give perfect symmetry, you can't use Gauss' law.

If so, then OK, strictly that is true, but then, what are the consequences of that? I think it is just an esoteric point and doesn't really amount to anything practical in the macroscopic case with billions upon billions of electrons in play. The electrons just act like a charge fluid and the discrete nature is not relevant.

But, lets look from the point of view of a small number of electrons. In this case, I agree, we can't use Gauss' law to calculate fields and potentials. Gauss' Law is of course true here, but it doesn't tell you anything useful. However, we can use Coulomb's law to calculate fields and potentials, and then we can look for charge distributions that give stationary equilibrium solutions (whether fully stable of marginally stable). Let's look case by case starting with one electron.

One Electron: The electron can go anywhere inside and be in equilibrium. If it hits the wall, it will just bounce off (note many simplifying assumptions are in play here)

Two Electrons: Both electrons will go to the surface as an equilibrium solution. The electrons can be at any place on the surface as long as each electron is on opposite ends. That is they are 180 degrees apart and a line connecting the two will go through the center.

Three Electrons: There are two solutions (as pointed out by studiot). The stable solution has all three electrons on the surface forming an equalateral triangle whos center is at the center of the sphere. The other solution has one electron at the center and the other two on the surface diametrically opposed as in the case of two electrons. Clearly, this second solution is marginally stable.

Four Electrons: There are is at least one stable solution with all four electrons on the surface forming the vertices of a tetrahedron. There may be other marginally stable solutions. For example, try the solution with three electrons at the surface and then place the fourth in the center. Then ask if this is a stationary solution with marginal stability. I would guess that it is, but it's getting hard for me to visualize now, and it just might be truly unstable (like a ball on the side of a hill, rather than on the top of the hill)

Five Electrons: Well, you can continue finding the stationary solutions. The ones with all electrons on the surface are fully stable equalibrium solutions, and the stationary solutions, with one electron at the center, are marginally stable/unstable. I can visualize that the cases with an odd number of electrons should have a marginally stable solution with one electron at the center. I'm not sure abou the case with an even number of electrons. Does putting one electron at the center allow a stationary solutions? Maybe studiot knows the answer. If not, I can calculate it later.

Number of Electrons Going to Large Numbers: There seems to be no end to this pattern and the limiting case is the one we have been discussing. There is the stable solution with all electrons equally distributed around the surface, and there is the marginally stable case with one electron at the center.

If you want to then say that if we keep going and going and add more and more electrons, eventually something different happens, that's fine. However, any effects would not be predictable by the classical theory, and they probably wouldn't actually happen in the real world anyway because other effects might happen first (such as the container explodes, or the electrons begin to be ejected through the surface). It's hard to say because we are talking about a hypothetical container.

Again, this is all just thought experiment with given simplifying assumptions. Classical EM theory with simplifying assumptions allows some very definitive statements to be made, but such an answer is, of course, not the "be-all and end-all" answer.

Last edited:

#### steveb

Joined Jul 3, 2008
2,436
Not quite. Any size will do. The area of interest is at the boundary, not near the center where the symmetry will dominate. At the boundary, where every electron will eventually find itself, there are still discrete charges and associated field. As that outer boundary starts to get well populated, the repulsive forces between the electrons in that layer will get high. High enough that it will be difficult to add further electrons to that layer and they will find a lower energy lair inside it.
I'm starting to visualize what you are saying. I believe the sticking point is that you are not visualizing that a non-symmetric electron near the surface will distort the surface charge and basically open up it's own vacancy. It can then slip into position and then get into the lowest energy configuration.

I've attached three drawings to help in the visualization. The first two drawings (circle and sphere) try to show why any number of electrons in the outer layer is a stable structure. The last drawing (story) tries to show how the electron moves into position.

First I do a two dimensional case (circle )so that it is easier to visualize how the net force from all electrons is always outward and only the wall can contain the force. Then I do the three dimensional case (sphere). This shows why symmetry is the most stable and lowest energy configuration. The last drawing shows why there is no other local minimum configuration that the system might get stuck in before it gets to the lowest energy.

EDIT: I added a fourth picture which helps to support the third drawing. It answers the question of why a nonsymmetrical electron (shown in green) that gets near the surface, does not get repelled, nor does it find a position of zero force to find a stable resting place. Notice how the closer blue electrons that can even provide a net inward force are few in number and off to the side where they can only give a glancing effect. There are many more far away red electrons that can push outward and many of those are giving a head on force outward, and not a glancing force. You may ask why there is not a blue electron, right in the way of the green electron, but clearly the green electron is going to push anything that gets near it away.

#### Attachments

• 144.1 KB Views: 21
• 67.5 KB Views: 16
• 90.7 KB Views: 19
• 145.4 KB Views: 16
Last edited:

#### wes

Joined Aug 24, 2007
242
Thanks Steveb for the pictures, they really help with explaining why the electrons would always want to move toward the surface, I didn't really think about the fact that even though there may be a electron repelling another electron back towards the center, there is still many many more electrons pushing it back towards the surface, so the electrons pushing it toward the center would be forced to make room. Obviously when they got to like distance's where Quantum mechanics takes over they would probably start to form layers

#### BillO

Joined Nov 24, 2008
999
Overall, I don't know if you are saying Gauss' Law is not valid, or that you just can't use it, or if you are just saying that Gauss' Law says that the field is not zero in nonsymetrical cases.
No. Gauss' law still holds. Both inside the layer of charge and outside. However, the implication/interpretation/assumption (whatever you call it) that there is no field structure within the layer is not right.

So, perhaps you're just saying that because discrete charges can never give perfect symmetry, you can't use Gauss' law.
Not sure what you mean by perfect spherical symmetry, but I consider all the cases where there are an even number of electrons will reach equilibrium in perfect symmetry.

Does putting one electron at the center allow a stationary solutions?
Yes, I believe you can have a single electron at the center with 2 or more electrons on the sphere. Let's call it an equilibrium solution though, and not a stable one.

If you want to then say that if we keep going and going and add more and more electrons, eventually something different happens, that's fine.
I do...

However, any effects would not be predictable by the classical theory...
...and I disagree with this. It would not be easy, but I think it could be done.

... and they probably wouldn't actually happen in the real world anyway because other effects might happen first (such as the container explodes, or the electrons begin to be ejected through the surface). It's hard to say because we are talking about a hypothetical container.
Possibly, but let's assume we build a container strong enough. Hypothetically, of course.

I believe the sticking point is that you are not visualizing that a non-symmetric electron near the surface will distort the surface charge and basically open up it's own vacancy. It can then slip into position and then get into the lowest energy configuration.
In order to get into that outer layer it has to move the electrons already in that layer closer together against their mutual repulsion. Let's call this the boundary effect. Inserting this electron will therefore increase the potential energy in the layer. That increase in potential energy has to come from somewhere. For each electron added the cost of insertion will go up because of the nature of the electric field (inverse square law).

Lets say we start off with an empty sphere and add electrons to a point just off it's center.

We add the first electron. This is a completely unique situation that is never repeated in this sequence as there is no potential yet in the sphere, so no energy is required to put it in, and there will be a field everywhere in the sphere, regardless of where the electron ends up.

Adding the second electron requires a little energy as now there is some potential in the sphere. As you said before, the two electrons will move to points diametrically separated. This is also unique as it is the start of the electron layer. We also have more potential in the sphere and a equilibrium point at it's center.

Adding a third electron requires a little more energy than adding the second one, and this 'adding' energy keeps getting higher as we add more electrons. The two electrons in the outer sphere are easily moved closer together as the third electron is closer to them so the repulsion from it is higher than their original mutual repulsion. Essentially we still have large field gradients in the sphere. Because of this the third electron easily finds a place in the outer shell.

Lets skip ahead until the point where we have many electrons in the sphere. The number is that which creates this particular situation. Now the field gradients in the sphere are getting weaker. This means there are smaller forces on each new electron we insert, driving it to the outside. We add another electron and those weak forces eventually move it close to the outer shell. However, these forces are now no longer strong enough to overcome the boundary effect, which has gotten quite large, and that electron becomes the first one that won't be inserted into the outer layer.

However, if we add another electron, there will probably be enough additional repulsion between it and the last one that our new 2nd layer will get 'pushed' into the outer layer, but we can see that that 2nd layer is now beginning to form...

Last edited:

#### studiot

Joined Nov 9, 2007
4,998
I haven't done the maths on this but I wouldn't be suprised if one could not find a 'shell theory' classical solution, whereby you could have stable concentric shells of electrons, where all the forces balanced. However these shells would require many electrons each so you could not simple introduce the electrones one at a time.

Oh and I'm glad to see you are still with us Bill.

#### BillO

Joined Nov 24, 2008
999
However these shells would require many electrons each so you could not simple introduce the electrones one at a time.
It would indeed be a tad tedious..

Yes, I'm still here, but will be a less frequent visitor and for less time. For a while anyway...

#### steveb

Joined Jul 3, 2008
2,436
I haven't done the maths on this but I wouldn't be suprised if one could not find a 'shell theory' classical solution, whereby you could have stable concentric shells of electrons, where all the forces balanced. .
I wouldn't be surprised to find marginally stable solutions, but I would be very surprised to find actual stable solutions. Just as the case of an electron at the center was not in my thinking zone, so are any other marginally stable solutions. The reason why I don't pay to much attention to these is that they are completely unphysical solutions. In a very real sense, it would be like trying to balance one billion pencils on top of each other, vertically- point to eraser.

But, sure, opening up the marginally stable constraint should allow more solutions in principle. This is easy to see if we think about placing 5 electrons on a linear track with end-stops. The two end electrons will be stopped by the end-stop, and one electron will be in the dead center. The other two electrons would have to be 0.23L from each end, where L is the track length. There is no reason why these types of balanced arrangements could not be introduced into a sphere. But, without a guiding track, how would the house of cards ever have a chance to stand up?

I'm dismissing these types of solutions. In the diagrams I provided, I try to show how a real solution evolves. I even addressed the question of "why isn't there an electron in the way" to the surface. In a classical explanation, there is no way to introduce a balanced set of electrons in the real world. In the real world, a lone electron forges it's own path while seeking the lowest energy.

Anyway, if this is the point, - that marginally stable shell solutions might exist in principle, then I concede that point. What would be interesting is if there were some quantum mechanical mechanism that could lock electrons into this configuration. I wonder if there has ever been a demonstration of anything like this.

Last edited:

#### BillO

Joined Nov 24, 2008
999
Steve, you're not even trying to see what I'm saying. So, dismiss away. The solution I'm talking about will be stable.

I even addressed the question of "why isn't there an electron in the way" to the surface.
Your nice diagrams do not take into account the huge amount of potential energy that will build up in the outer shell and the growing amount of force required to add an electron to it as it gets more populated. While at the same time, the forces inside the shell that could put it there are weakening. The new electron cannot just saunter over to the outer shell and expect the electrons there to just slide out of the way for it. For this to happen, the shell electrons have to get closer together, increasing the potential energy in the shell. Where does this energy come from? Especially considering the case I mentioned above where the internal fields have diminished to the point where they cannot contribute. If this is something you want to dismiss, then there is no possible way to understand what I'm discussing.

So, I don't know where that leaves this. Like I said, I have no time to do the math, which is not trivial (because you have to also solve the QM case to see when it dominates. And the classical case would be no cinch either. Try it...), so I can no further in my argument.

Wes, pick an answer. I say you could, at the very least, build up several layers of electrons. Steve says no.

Got to go catch a flight....be back in about a week to see where this settles.

Last edited:

#### steveb

Joined Jul 3, 2008
2,436
Steve, you're not even trying to see what I'm saying. So, dismiss away. The solution I'm talking about will be stable. ... which is not trivial (because you have to also solve the QM case to see when it dominates. And the classical case would be no cinch either. Try it...), so I can no further in my argument.
Bill,

To be fair, I'm trying very hard to see what you are saying, and I have to say that I'm making progress inching towards narrowing down what you are saying. So I'm not dismissing what you are saying, I'm taking it quite seriously to see if there is something there to learn. I now understand that you are saying there WILL be truly stable solutions, and not just marginal solutions as studiot was (I think) alluding to in his last post. So, I'm one step closer to getting there.

I'm still not completely there because I seem to see some contradictions in what you are saying. Perhaps this is just my own misconceptions, but look at the bold part above. I thought we had narrowed down the discussion to the classical case, but you are saying we need to look at where QM dominates. I don't think this is necessary because if you can show how a classical stable state can exist, even if it somehow contradicts a QM limit, I would be very satisfied that you had proved your point.

At this point I can't see the stability you are talking about. You haven't been able to explain how we can identify the critical number of electrons, or the structure of the arrangement, so it's hard to figure out an analysis that might either prove or disprove your point. But, let's keep thinking and discussing, and let's not worry about rushing to find the answer. You are very busy, and trust me, I am very busy also. We can both think and work on it at a comfortable pace.

#### steveb

Joined Jul 3, 2008
2,436
In order to get into that outer layer it has to move the electrons already in that layer closer together against their mutual repulsion. Let's call this the boundary effect. Inserting this electron will therefore increase the potential energy in the layer. That increase in potential energy has to come from somewhere. For each electron added the cost of insertion will go up because of the nature of the electric field (inverse square law).
Bill,

Please don't feel any rush to respond to this post. I know you are busy and may not get to this for a week or more, but I'd like to make a comment on this point you made, with the primary purpose to show that I'm not dismissing what you are saying. I read your points and consider them carefully, but I find that I'm not in agreement with all points. I originally thought that I wouldn't try to nitpick these various points, but I guess by not addressing anything it makes it seem that I'm "dismissing away". I don't want to dismiss good points, and I dont' want to appear to be dismissing good points.

So I can respond to this good point you made, quoted above, as a representative case of me not agreeing. Note that "not agreeing" is not the same thing as saying you are wrong. Also, note that my debating style is to make points as definitive statements, which can sometimes rub people the wrong way, if they are not aware of that. So, I state it explicitely.

Let's say we have a symmetrical case with of a great number of electrons on the surface. I think we agree that this is a possible condition for a starting state. By symmetrical, I just mean that there are a great many electrons and that they are spread out evenly, in a way that minimizes the energy as much as possible. Let's say we want to add one more electron to the mix. I propose that we do work on the electron to bring the it from infinity to the center of the sphere. This takes work, and that's where the energy comes from. If we gradually bring the electron in, we do positive work until the electron gets to the surface. The electron would very much like to get repelled away to infinity, but if we bring the electron through the wall (magically), it is basically now stuck inside. I think you will agree up to this point.

If we then bring the electron from the surface to the center, then we will need to do even more positive work. You may disagree with this part, or you may agree that this is true near the center, but not near the surface. My claim is that the work is positive at every point along the path. My guess is that you would say that the work dips to a negative value at some point before going positive again. (If I'm wrong about what you think, I apologize, but that's my understanding of your claim right now). During this process of moving the electron, the surface charge was distorted away from it's symmetrical distribution. The surface charge is symmetrical when the new electron is at infinity. It is also symmetrical when the new electron is in the surface layer, and finally, it is symmetrical when the new electron is at the center. Now, once the new electron is at the center, the surface charge will be symmetrical, and hence this is an marginally stable state (like the ball on top of the hill). So, the potential is at least locally maximum with the electron at the center, and forces are balanced.

Now, to my point, I disagree with your quote above because you are claiming that the insertion of the new electron into the surface increases the potential energy, compared to the case of the electron that is just inside the radius of the surface. I claim that the electron in the surface layer is, in fact, the configuration of minimum energy. Now, why is the case of the electron just inside the surface radius not a lower potential energy state? You are claiming that the electrons are closer together when the new electron is in the layer, and farther apart when the new electron is at a radius less than the surface layer, but this is dismissing a key point I made before. An electron that is inside the surface radius DISTORTS the electron distrubution in the surface layer. A distorted surface charge has areas where the electrons are closer together also. So, the situation is not so clear cut as you suggest.

Now, how are you able to claim that one configuration has lower energy than the other without a calculation? Of course, you can turn around and ask the same question to me. My answer is that you are right and I should calculate it, which I'll try to do. But, in the mean time, I would say that intuition is on my side. Symmetry is usually representative of the lower local energy state (or local maximum as we see in the marginal stability cases), and I don't need to do a calculation to claim that a soap bubble represents a local minimum energy configuration. The idea that a lone electron (or even a symmetrical or unsymmetrical group of electrons) sitting just inside the surface radius is at lower energy (at least locally) seems very unintuitive to me. But, my intuition has failed me before, so, who knows. Let's keep thinking, but please don't feel that I'm not paying attention to your points.

Last edited:

#### russ_hensel

Joined Jan 11, 2009
825
If you calculate the pe of a uniformly dense sphere of electrons, the calculation is not too hard as outside the sphere ( and the intermediate ones from which you build the final one ) the sphere acts as a point charge of the total charge of all the electrons in the sphere. This configuation is of course wildly unstable we would need a bunch of maxwell's deamons to do the work and hold the sphere together.

#### steveb

Joined Jul 3, 2008
2,436
So, I'm at work now, and the answer to this paradox just occurred to me. I won't explain it yet because I have a calculation in mind that I need to do later on tonight. I'm quite sure the calculation will confirm the picture I have in mind. Not surprisingly, as I mentioned earlier on, there can be truth on both sides this issue, and I believe this is the case here.

The answer is simple in nature and it answers Bill's good points, while at the same time restores Gauss's law in the continuum limit. As strange as it may seem, there appears to be a way for both Bill and I to be correct, without logical conflicts.

Sorry to leave a cliff-hanger, but a calculation and proper documentation of it are needed to make my argument clear. But, if my viewpoint is right, it will be worth the wait.

... Stay tuned.

#### studiot

Joined Nov 9, 2007
4,998

#### steveb

Joined Jul 3, 2008
2,436
Will this be a digital or analog broadcast?

Not sure, but either method is terribly inefficient in comparison to a live discussion.

It's quite possible the four of us would have had this whole thing figured out with a one hour face-to-face discussion standing at a chalk board.

By the way, I don't consider this question answered until all four of us agree about the answer. So, I have some ideas that I feel strongly about and I will present them over the next couple of weeks. There is no rush because we are all busy, but I eventually want us to come to a common viewpoint.

A lot of work is needed for me to document what I want to present and to make it understandable without a live format. With my workload, I think I need a week to finish what I want to do for calculations and modeling and then put it in a proper document that actually makes sense. It might even be two weeks.

I can give a general idea of what I see as the right answer here. I'm at about 90% confidence level with this, and need to finish some modeling to get to a 99% confidence level. I'm determined to use real calculations now, whether fancy math or brute force simulations, because this question is just too tricky.

The calculations I did last night strongly indicate that what occurred to me yesterday is the right viewpoint. First, Bill's notion that the charges form layers is definitely right. The surprise is just how right he is. This layering occurs almost immediately. With a handful of electrons we get the charges on the surface, and then slightly beyond a handful it seems to form layers. These layers would persist until you get to the continuum limit where Gauss's Law indicates an ultrathin layer at the surface. To me, this seems paradoxical. I think both Bill and I latched on to notions that we felt are right (and i do think they are both right), and then once locked into that viewpoint, we arrived at other ideas that are probably not right.

How do we resolve these opposing viewpoints? What occurred to me yesterday is that this layering will build up as electrons are added, but very quickly a potential well near the surface is created that the charges sit in. This basically a thin valley at the surface. As more charges are added, this valley becomes narrower, but the layering increases. In the continuum limit, the well is infinitely narrow, but this infinitely thin well should be thought of as a shell structure that has been compressed down to nothing in the limit. Of course, reality is that electrons are discrete, so the layer will be finite in thickness and the shell structure will be there.

This can be thought of as a competition between two effects. The electrons want to separate in the volume, but they also don't all want to be at the surface. It seems that in the limit as the number of charges goes to infinity, the shell is driven right to the surface, and so the volume can't fill up. I believe that I can show why this is true mathematically, and then suport it with numerical simulations. But, again this is now only at 90 % confidence level, and I plan to hold myself to the same standards you all will hold me too. That is, prove it and don't use intuition, because the intuition is too nebulous in this case.

I hope you all will be open to consider this possible answer, even if it's clear that it is not proven. I find this possible answer a bit yin-yang in nature.

Last edited:

#### studiot

Joined Nov 9, 2007
4,998
My comment about all this is the need to take into account the nature of the outer shell and its interaction with the electrons. If it provides a lattice or other periodic structure with low energy sites that is where they will go.

#### steveb

Joined Jul 3, 2008
2,436
My comment about all this is the need to take into account the nature of the outer shell and its interaction with the electrons. If it provides a lattice or other periodic structure with low energy sites that is where they will go.
When you say outer shell, are you refering to the outermost shell of electrons that forms, or to the walls of the container? If it is the former, then I am trying to include that, at least under classical assumptions. That is, I would not be trying to include QM.

In essence we are looking for the minimum energy solutions, which one would imagine is some type of lattice structure. This is one of the challenges I've taken on, to find at least some of these lattice-like solutions. One of the methods I'll try is to use is brute force numerical dyanamic simulations and basically let the electrons naturally find their resting places. I had originally thought this would be beyond the capability of my computer power because i thought it might take millions or billions of electrons to bring on the effect. But, now that I've seen that very few electrons can bring on this lattice structure, I see no reason why we can't find at least some of the solutions numerically. Again, only under classical assumptions.

Hmmm, maybe my one or two week estimate was a little optimistic? ... we'll see.

#### studiot

Joined Nov 9, 2007
4,998
I mean whatever keeps the electrons from moving further and further apart by simply expanding the (outer) shell. So yes I mean the 'container' if you like.