# Talk:Polarization density

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## Relation between bound charge and bound current

I added a new section to indicate that the simple velocity equation that relates free charge density and free-charge current density does not hold for bound change density and bound current density. If anyone has any comments, please let me know. Elee1l5 (talk) 18:13, 30 September 2008 (UTC)

Sorry, your addition is just wrong. Even for free charge, it is perfectly possible to have a nonzero free current and a zero charge density. You just need to have equal and opposite densities of positive and negative charges moving at different velocities, and there will be a net current. The only relationship between the free current density and the free charge density is the continuity equation, which is equally true for free or total (free+bound) charge. —Steven G. Johnson (talk) 03:44, 1 October 2008 (UTC)

Thank you for the comments. I agree with what you are saying. However, I am not sure why you are saying that the additional comments that I added were wrong. It seems to me that what you are saying actually supports what I was saying. Of course it is true that in a neutral conductor there can be a current density without a charge density. On the other hand, if there is a charge density and it is moving, then this must correspond to a current density (according to the usual velocity equation). Now, inside of an insulating dielectric that is homogeneous, there is no bound charge density. There is still a bound current density. Physically, this bound current density is arising from the rotation of the molecular dipoles. As the dipoles rotate, the two opposite charges at either end of each dipole will have opposite velocity vectors. Thus, just as you have stated in your example above, there will be a net current, in this case corresponding to the bound current density. Thus, the bound current density is arising from the rotation of the molecules, NOT from the motion of any bound charge density. This is precisely the point that I was trying to emphasize. Perhaps you consider it to be obvious and not worth mentioning, but I thought it might be, especially since bound current density and bound charge density are discussed together in the same article, and one might naturally ask if there in any simple relationship between them. The answer is that they are related by a continuity equation, but not by a simple velocity equation. Elee1l5 (talk) 00:40, 2 October 2008 (UTC)

Your addition incorrectly stated the relationship between free charge density and current density, as I explained; expressing the free current density as the free charge density times a velocity is only true under very restricted circumstances. Nor is it true that the bound charge density can never be associated with a velocity. For example, if you have relatively immobile ions that are polarizable by displacing the valence electrons with an applied field, then the bound current density in an oscillating field can indeed be related to a velocity of the (oscillating) bound electrons. Your attempt to draw a firm distinction here is well-intended, but wrongheaded. —Steven G. Johnson (talk) 02:11, 2 October 2008 (UTC)

For free charge density, it is always true that the current density is equal to a charge density times a velocity vector, if we interpret this properly. For example, in a neutral wire, the charge density would be the charge density of the electrons (the charge carriers that are moving), not the total charge density (which would be zero). If we have positive and negative charges moving in opposite directions, as you originally stated, then we have two different velocity vectors, and two different charge densities. Hence we would have two current densities (both pointing in the same direction), and the total current density would be the sum of the two. For the polarization current, the principle is exactly the same except that the current is physically due to the motion of charges that are bound to the molecules or the atoms. The charge velocity is now due to either the molecules rotating from the applied field (Debye relaxation effects) or the electrons being elastically stretched from the nucleus by the applied field (electronic, atomic, and ionic effects). The polarization current DOES correspond to the motion of charges that are bound to the atoms and molecules. The polarization current DOES NOT correspond to the motion of “bound charge density” as defined by the divergence of the polarization vector. I maintain the correctness of this distinction. Elee1l5 (talk) 04:49, 6 October 2008 (UTC)

If you define the bound charge density as the divergence of the polarization density, then you should define the free charge density as the divergence of the electric field---in which case it is zero in a neutral wire. If you are willing to break the free charge density up into a sum of components moving in different directions, then there is no reason not to do the same for the bound charge. Ergo, you are still wrong. Anyway, to avoid wasting more time on discussion here, go find an authoritative reference that you think backs you up before arguing more. —Steven G. Johnson (talk) 02:05, 24 October 2008 (UTC)
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## Free charge compared to bound charge

How exactly are free and bound charges distinguished? Are free charges those with untrammeled movement while bound charges are those with restricted movement? These questions are natural but misplaced. I'd suggest that an example would greatly add to the clarity of the discussion. A possible addition builds upon the article's statement that bound charge is due to rearrangement as follows:

.................

The idea of bound charge is defined by:

$\rho_b = -\nabla\cdot\mathbf{P} \ .$

This bound charge is due to a "rearrangement" of microscopic charges so, it would seem, the bound charge is an artificial charge introduced for purposes of computation or visualization, not necessarily equal to any actual charge, but expressing an effect of rearrangement of actual charge. Consequently, it is not possible to use the equation above to find P from ρb, which is itself unknown.

Instead, we have to find ρb by other means. How is that done? Ostensibly, at bottom, one can determine for any external electric field E how the actual identifiable microscopic charges respond to E, the simplest case being that the + and − charges separate by an amount proportional to E, imagined as though the + and − charges are connected with a simple spring that stretches until the restoring force of the stretched spring balances the external stretching force due to E. The resulting P is then:

${\mathbf P}=\varepsilon_0\chi{\mathbf E},$

so the resulting ρb is:

$\rho_b = -\varepsilon_0\nabla\cdot\left(\chi\mathbf{ E}\right)\ .$

Imagine a surface enclosing an array of atoms, forming what might be called a 'lattice' of atoms. Inasmuch as there is no charge rearrangement well outside the lattice of atoms because there are no atoms out there, and there definitely is rearrangement inside the lattice, one can approximate the effect of E by taking χ to be a step function at the surface. So, for simplicity, if the solid is a planar layer, making the problem one-dimensional:

$\rho_b = -\varepsilon_0\frac{d}{dx}\left(\chi E \right)$
$= -\varepsilon_0\left(\frac{d\chi}{dx} E +\chi\frac{d E}{dx} \right)\ .$

In this simple case the external field E is independent of distance, so the result is:

$\rho_b = -\varepsilon_0\left(\frac{d\chi}{dx} E \right)\ ,$

which is a nonzero bound charge only at the surfaces where χ steps up from zero to its nonzero interior value χ and down from χ to zero. Thus:

$\rho_b = -\varepsilon_0 \chi \left(\delta(x-x_-) - \delta(x-x_+)\right )\ ,$

corresponding to a sheet of negative charge on the left surface and a sheet of positive charge on the right surface, and no bound charge inside the layer. Thus, in this idealization, the bound charge can be calculated for a hypothetical real charge rearrangement. Of course, in reality, the rearrangement of charge very near the surface is not accurately described by a step function in χ, and instead χ = χ(r), a function of position. The calculation of χ(r) is no longer something easily dealt with, and must be found using a quantum mechanical calculation of the atomic responses according to exactly how they are placed on the lattice, including the interactions between the atoms. In some solids, the rearrangement of actual charges will be nearly independent of position except very close to the interfaces, and the use of a step function (or the equivalent surface layers of bound charge) will prove adequate if one is not interested in details very near the interfaces.

.............

Just a suggestion to answer the questions posed at the outset, which I find to be natural queries by a reader. Brews ohare (talk) 14:01, 22 November 2011 (UTC)

Brews, according to Wikipedia policy, you should follow descriptions in standard textbooks and other widely accepted review material, and should be cautious about proposing your own personal "spin" on things.
The bound charge is indeed a real, measurable, physical charge density that arises in a polarizable object in response to an applied field, not merely an artificial mathematical construct. (Note the distinction between "charges" and "charge density". It is perfectly possible to have a body in which there is no net macroscopic charge density even though it is filled with charges.) And the reason that you can't use the divergence equation to solve for P is not that it is "artificial", but rather that you also need to know the curl for P to be determined.
And your description of how P is determined is misleading. One doesn't simply determine P by taking an applied E field and multiplying it by the susceptibility. The reason is that P then produces a bound charge density that produces electric fields modifying E, which modifies P, so you need to solve the problem self-consistently.
Finally, the whole picture of a susceptibility inherently implies a macroscopic average of the fields and charge density. (Hence they are called the "macroscopic" Maxwell equations.) At a microscopic (atomic) level, you wouldn't use χ or P at all.
— Steven G. Johnson (talk) 15:08, 22 November 2011 (UTC)
Hi Steven: Thanks for the observations. I am aware that this example is a simple one. My reason for suggesting that "bound" charge is best conceived using the divergence of polarization is that I cannot find another definition of bound charge that always works. For example, at one time I simply took "bound" charge to be a subset of charges that were treated in a macroscopic way, while free charges were treated microscopically. That idea seems to run afoul of the usual definition of polarization. The article does say, I think correctly, that bound charge is an expression of charge rearrangement, which does not preclude identification of 'bound' charge with real charge, but doesn't seem to require it either. My memory is not sharp, but it strikes me that quantum-mechanical approaches to the dielectric behavior of materials commonly proceed by calculation of a response function, for example, a susceptibility calculation. In any event, it would be helpful to elucidate the definition of "bound" charge one way or another. How would you go about it? Brews ohare (talk) 15:50, 22 November 2011 (UTC)
The -divergence of P is indeed the bound charge density. I'm not disagreeing with that standard definition of bound charge density (whose integral gives the amount of bound charge in any given volume). I was objecting to your claim that this is "artificial charge". (And also to your careless conflation of macroscopic electromagnetism's net "charge" and "charge density" with "charges" in the sense of particles. The bound charge density is unquestionably a physical net charge density; whether it is associated with "particles of bound charge" is beside the point in macroscopic electromagnetism.)
(And your example is not just simple, it is wrong. P is not given by the susceptibility multiplied by only the external E field, because you must include the field of the bound charges.)
People attempting accurate ab initio calculations of susceptibility and polarizability do indeed need quantum-mechanical methods that operate at an atomic scale. However, they use these methods to predict the macroscopic susceptibility, which is not the same thing as trying to define an atomic-scale "susceptibility". At a microscopic level, you (ideally) compute the actual quantum probability distributions of the electron wavefunctions and other charges in response to applied fields, not a "susceptibility". The susceptibility comes from an average of these quantities (e.g. over a unit cell for a periodic medium), and it is only useful for describing electromagnetism at a macroscopic scale (averaged over many atoms) where the atomic-scale field and charge-density variations are irrelevant. (Almost invariably, however, macroscopic electromagnetic calculations use susceptibilities that were measured in experiments rather than computed ab initio.) — Steven G. Johnson (talk) 16:10, 22 November 2011 (UTC)

outdent Steven: You haven't provided your idea of what 'bound' charge is. Is it in your view simply a set of charges set aside for some reason to treat macroscopically, for example, by simply averaging out the short-wavelength components of the charge density originating in the selected set of microscopic charges? Such a definition is incomplete because it leaves out the concept of rearrangement of the charge by the applied field. Somehow we have to include the rearrangement of the charge by the applied field in the definition of bound charge. If our selected charges don't respond to an applied electric field with some particular wavelength, they are not a component of the bound charge at that wavelength. Do you agree? Brews ohare (talk) 17:33, 22 November 2011 (UTC)

The ultimate point here is how to amplify discussion of the role of rearrangement in defining bound charge in order to clarify the article. Perhaps this point is best pursued by tracking down a source. Most probably this source will be found in condensed matter physics, probably involving linear response theory, and it may become a bit complex for the article if presented in detail. I'd hope that we might agree upon some simplified presentation of such work that avoids a lot of mathematics. That simplification requires some agreement upon the principles involved. Brews ohare (talk) 18:02, 22 November 2011 (UTC)

My thoughts: Probably worthwhile to say "polarization in solids almost always arises from pulling 'bound electrons' (electrons attached to specific nuclei) a bit to the side of the nuclei they orbit. Bound charge is the macroscopic charge imbalance that can result from this microscopic displacements." Or something like that.
Deriving the formula $\sigma_b = \mathbf{P}\cdot\mathbf{\hat n}_\mathrm{out}$ from the formula $\rho_b = -\nabla\cdot\mathbf{P}$, using delta-functions and all that, as Brews suggests, seems unhelpful and distracting IMO. I guess in a show/hide box it would be OK. :-) --Steve (talk) 22:20, 22 November 2011 (UTC)
Steve: I'm not averse to a show-hide presentation of the example. However, from Stevenj's comments, I'd guess the point of the example is not clear. Because bound charge is directly related to polarization, it doesn't exist unless polarization takes place. To state matters in a very striking and somewhat alarming fashion, polarization creates bound charge. In other words, if an atom is rigid (a very hypothetical case to be sure) despite having electrons and protons, it doesn't contribute to bound charge. But if it polarizes in response to the applied field, then it does contribute to bound charge. This peculiarity of bound charge is contained in the div P definition, I'd say. I'd guess that a div P analysis relating "bound" charge to a quantum-mechanical linear-response analysis of polarization would support this picture. Maybe a simple ball-and-spring example would work just as well. Do you find merit in this viewpoint? Brews ohare (talk) 17:42, 23 November 2011 (UTC)
One way to look at the "creation" of bound charge by polarization is to think of coextensive plus and minus charge clouds on a fixed site. The overall charge is zero. But under the electric field the two charge clouds separate, and now we have a certain amount of plus and minus charge exposed, an amount that increases as the separation increases and the overlap of the charge clouds reduces. The amount of "bound" charge is a function of the "spring constant" that resists their separation when subject to the applied field. Brews ohare (talk) 18:02, 23 November 2011 (UTC)
The terminology "bound charge" is confusing; some authors call it "polarization charge" and some call it "structural charge", but "bound charge" is the most common name. The problem with the "bound charge" terminology is that it suggests a charge that exists independent of any applied field, as a constituent of the matter itself even when there is no field, when in fact it comes into existence only when rearrangement occurs. That is the point I think the article needs to emphasize more. Brews ohare (talk) 23:46, 23 November 2011 (UTC)
At the moment the leading sentence of the article states: polarization density "expresses the density of permanent or induced electric dipole moments in a dielectric material". I am questioning the underlined reference. For example, in a single crystal of a ferroelectric substance, does one include the permanent electric dipoles in P, which are intrinsic to the material and generated by internal cooperation among the atoms, and so do not depend upon any external field, or does one consider P to involve only the field-dependent re-orientation or field-dependent change in strength of these dipoles? Brews ohare (talk) 15:42, 24 November 2011 (UTC)
It appears that the literature on ferroelectricity describes the spontaneous dipoles of ferroelecrics in terms of polarization in a close analogy with ferromagnetism. The connection to the paraelectric case is made by positing a coercive field that resembles the Weiss field for ferromagnets. This field acts like an external field but is internally generated. Apparently this old fashioned idea is now replaced by the Berry phase theory of ferroelectricity, which is unfamiliar to me. In any event, it would appear that the notion of rearrangement of charge by an external field can be circumvented to include rearrangement by internal fields. According to this source polarization is defined as dipole moment per unit volume, which would seem to suggest that rearrangement is not required at all. I am left uncertain about how the matter of bound charge should be presented. Brews ohare (talk) 17:12, 25 November 2011 (UTC)
Yes, P is the density of electric dipole moment per unit volume. That's the definition already in the article and the definition everybody uses. Certainly it includes both permanent and induced. The Berry phase thing you're referring to is not a different definition, it's a calculation method, for calculating how much P changes when the crystal structure gradually changes from one arrangement to a different arrangement (for example the distortion that occurs as you cool a single-domain ferroelectric through its curie point).
Bound charge is the net charge of the electrons and protons that makes up all those dipoles in P. Key word is NET. Obviously, it's possible to have bound charges but still zero net bound charge, if the electrons and protons are in the same places. I don't think it's necessary to go on and on about the difference between charge and net charge, in this particular article: One imagines that most readers will be reasonably familiar with that distinction. :-) --Steve (talk) 17:33, 25 November 2011 (UTC)

outdent Hi Steve: Thanks for engaging and helping me sort this through. Of course a dipole has zero net charge, and yet contributes to the dipole moment per unit volume, and hence to the polarization density. For the sake of argument, lets suppose an imaginary solid made up of closely spaced pairs of + & − charges. Suppose for discussion that these dipoles are all alike and randomly oriented, so the polarization density is zero. Now we apply a field and the randomness is reduced so there is some net polarization. For the sake of argument again, lets suppose the randomness is due to temperature, and the temperature has a spatial dependence. So the applied field is more successful in overcoming the randomness in a spatially varying fashion. What is the bound charge in such a case? Evidently it is div P, and will relate to the temperature distribution that governs the reorientation. The net charge of course is zero regardless of the dipole orientation, so that is not a factor in deciding the bound charge, and the higher the temperature or the smaller the field the smaller the bound charge will be.

In a case of permanent dipoles, the dipole moment sets the torque experienced by the "molecule" in the electric field, the polarizability decides the amount of rotation realized, and the bound charge will be zero regardless of the reorientation (except possibly at the bounding surface) unless the polarization is spatially varying (for example, due to the nonuniform temperature distribution imagined in the example) because div P=0.

Calling the "bound charge" the net charge constituting these dipoles is not going to clarify matters, I'd guess? Am I missing something here? Brews ohare (talk) 14:41, 26 November 2011 (UTC)

Steve, some authors follow your lead and refer to bound charge as "charges of equal magnitude but of opposite signs that are held in close proximity and are free to move only atomic distances (roughly 1Å or less)." See Brian J. Kirby (2010). Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices. Cambridge University Press. p. 97. One may reasonably inquire whether this definition is equivalent to the div P definition. Looking at the definition of polarization, a material with composed of only a spatially independent fixed array of dipoles {dj}, the same in every sub-volume ΔV will have a spatially independent polarization P, the same in every sub-volume:

$\mathbf{P} =\frac{1}{\Delta V} \sum_{j\in \Delta V}\ \mathbf{d}_j \ ,$

and hence a polarization with zero divergence and zero bound charge, incompatible with this author's definition of bound charge. In fact, even if the dipoles change strength or orientation under the influence of an applied electric field, but all shift by the same amount so the dipole moments are still distributed uniformly in space, there will be zero bound charge. That is the case, for example, when a dielectric sphere is introduced into a uniform electric field: the bound charge is everywhere zero except at the surface of the sphere where a surface charge appears because of the step in dipole moment on leaving the sphere for classical vacuum. Brews ohare (talk) 17:31, 26 November 2011 (UTC)

I'd be inclined to change this author's definition from a definition to only an association, as in "bound charges are related to charges of equal magnitude but of opposite signs that are held in close proximity and are free to move only atomic distances (roughly 1Å or less)." Brews ohare (talk) 18:11, 26 November 2011 (UTC)

As I've repeatedly tried to explain to you, you are once again confusing the net bound charge density (macroscopically averaged), div P, with the microscopic "bound" charges that physically constitute a polarizable medium. There is no contradiction between the fact that an infinite polarizable medium in a uniform E field has a uniform P field and hence no net macroscopic charge density div P, and yet it still contains positive and negative charges at a microscopic level. (Nor is there any contradiction in the fact that div P shows up only as a surface charge in finite objects containing no free charge.) — Steven G. Johnson (talk) 01:21, 27 November 2011 (UTC)
Stevenj: I haven't made the claim that zero div P contradicts the presence of microscopic charge in the sample, and I do not understand where that misconception comes from. I have not said there is no microscopic charge when div P =0. What I did say is that there is no bound charge when div P=0, and you agreed with me. My view is that the term "bound charge" is in fact an expression of the spatial inhomogeneity in dipole distribution (on whatever scale you wish to average), and that this interpretation is poorly conveyed by the term "bound charge" and so deserves some explanation in the article. Brews ohare (talk) 04:02, 27 November 2011 (UTC)
I'd say that when you coin the term ‘microscopic "bound" charges that physically constitute a polarizable medium’ and put quotes around "bound" to indicate (I'd guess) that these bound charges are not div P bound charges that you are implicitly saying the same thing. Brews ohare (talk) 04:11, 27 November 2011 (UTC)
The words "bound charge" are used in two senses. First, there is the macroscopically averaged net bound charge density, div P. Second, there are the microscopic bound charges that constitute a physical polarizable medium. When one talks about plural "bound charges", implying discrete particles, one is normally talking about the latter. When one talks about "bound charge density" or uses "bound charge" as a mass noun, one is normally talking about the former. You are confusing matters because you aren't differentiating between the two usages. — Steven G. Johnson (talk) 14:09, 27 November 2011 (UTC)
Hi Stevenj: I had not heard the term "mass noun" before. It'll take a while to enter my vocabulary.
I'd like to discuss the idea of "net" charge. In my vocabulary the "net" charge of an array of charges {qj} is the sum with signs attached of all the qj. Although not a universal choice, most authors in referring to dipole moments or to polarization restrict themselves to arrays {qj} with a net charge in this sense given by Σqj=0. An example would be an array of dipoles, or an array of paired equal and opposite charges as in "charges of equal magnitude but of opposite signs that are held in close proximity and are free to move only atomic distances (roughly 1Å or less)." See Brian J. Kirby (2010). Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices. Cambridge University Press. p. 97.
On this basis the "net" charge contributing to div P is exactly zero. That is the case regardless of whether one counts the charges individually on a microscopic level of individual dipoles or macroscopically averaging over sub-volumes of larger size containing multiple dipoles. Thus, if the dipole density is nonuniformly distributed inside a dielectric sphere, due say to a spatially varying strain field, div P is nonzero and there is bound charge, even though the net charge Σqj in any selected sub-volume is zero provided the sub-volumes are designed so any chosen dipole populates only one sub-volume. For example, there is bound charge if every sub-volume has 1000 dipoles but in adjacent sub-volumes on average they are oriented slightly differently.
Apparently the word "net charge" has a different meaning for yourself. So some semantic confusion still clogs my arteries here. Can you clear this up? Brews ohare (talk) 16:41, 27 November 2011 (UTC)
If you could choose subvolumes only to contain an integral number of dipoles, then every subvolume would have no net bound charge. However, such a thought experiment fundamentally conflicts with the way that P is defined as a macroscopic average of the microscopic quantities, so as to give a piecewise-continuous dipole moment density. Effectively, one has to average over all possible shifts of the microscopic dipoles within their unit cells to obtain a piecewise-continuous density rather than delta functions, and when you do this you will find that a P with nonzero divergence indeed gives nonzero average net charge within subvolumes. Another way of putting it is that if you have atomic-scale resolution available to you in picking your subvolumes, then you should be using the microscopic Maxwell equations and not the macroscopically averaged ones with P and D etcetera. (There are many textbooks that discuss the distinction between the microscopic and macroscopic Maxwell equations in more detail, e.g. Jackson or Landau.) — Steven G. Johnson (talk) 05:01, 28 November 2011 (UTC)
Hi Stevenj: Apparently you have in mind a particular way to construct sub-volumes, and have found the requirements of continuity of P requires that not every dipole can be contained within a single sub-volume. You suggest that Jackson or Landau support this notion. I haven't been successful in locating such an argument, and I do not see the necessity of div P ≠ 0 implying that one cannot simultaneously allow Σqj=0. In fact, if div P ≠ 0 requires that some dipoles must appear in multiple sub-volumes, as required to allow Σqj≠0, it implies surface charge at the interfaces between sub-volumes, which appears disastrous for continuity of P, don't you think? Brews ohare (talk) 06:34, 28 November 2011 (UTC)
Another difficulty in using sub-volumes for which Σqj≠0 is that the dipole moment of a sub-volume Σrjqj becomes dependent upon the origin of coordinates selected for the location vectors {rj}, making the dipole moment of the sub-volume no longer a simple property of the sub-volume alone. Brews ohare (talk) 16:04, 28 November 2011 (UTC)
I've located a discussion in Stratton, which indicates that continuity of potential requires piecewise continuity of P, which allows steps in P with P taking the value at the step of the average of its values on either side.
A discussion by Clemmow explicitly introduces an averaging procedure using a continuously differentiable weighting function W(r) that is shaped so the first two terms of its Taylor expansion are an adequate approximation over dimensions comparable to a molecule. In particular, W doesn't exhibit any steps. He determines the polarization as P = ΣmpmW(rrm), with m the index of each molecule in an assembly of molecules. Then div P is related to grad W, which exists as W is continuously differentiable. The bound charge would then be given by −div P = ρb = −Σmpm∘gradr W(rrm) and the charge on a molecule at rm is given by ρ(rrm)=qm W(rrm)-pm∘gradrW(rrm). The first term is zero if the molecules are charge neutral making qm=0. Brews ohare (talk) 17:40, 28 November 2011 (UTC)
I like the weighting function approach, or else (what amounts to the same thing) replacing each point charge by a smeared-out charge distribution with the same total charge. But whatever, any reasonable procedure gives the same result.
An unreasonable procedure is to try to choose a subvolume where each dipole is either totally inside or totally outside the subvolume. Each dipole is (in general)) spatially spread-out, so you would have to carefully gerrymander the boundaries of the subvolume to not cut through any dipole. Even worse, the dipoles may overlap and fill space. In that case, it would be totally impossible to draw a subvolume that didn't cut through any dipoles. --Steve (talk) 20:28, 28 November 2011 (UTC)
Steve: about gerrymandering, you might notice that the molecules above have been replaced by a point charge that equals their net charge, and by a point dipole that equals their dipole moment, and then the W-function proceeds to smear those items out. Thus, the Σqj inside any sub-volume is zero because all the {qj} are zero for assemblies of neutral molecules. There are some unexplained points here about how well this procedure represents any particular real assembly of molecules. It seems likely that for molecules small enough compared to the averaging range, you won't be able to tell the difference, because the higher order multipoles all are negligible. But given the procedure, for neutral molecules it leads to zero net charge in every sub-volume and non-zero bound charge wherever a dipole moment exists.
The underlying question for the article remains, in my opinion, just how to broach this topic in the article. I feel that the "bound charge" concept is a treacherous one that needs more explanation. Brews ohare (talk) 21:10, 28 November 2011 (UTC)

Sure, here W is just a smoothing kernel that you are convolving with the implicit delta functions in the dipole density (from the point dipoles assumed in this model). Convolving with a smoothing kernel is what "averaging" means. (Technically W has to have support comparable to but somewhat greater than the atom-atom separations if you want to smooth out the atomic-scale "bumpiness" to apply a macroscopic theory.) And note that div P = Σ pm ∇W is in general nonzero.

It's odd that you evaluate a charge density and call it a "charge", and your math from that point seems confused. Suppose that we actually evaluate the net bound charge in a vicinity of rn: i.e. you integrate -div P in an atomic-scale volume ΔV surrounding rn (e.g. one unit cell in a crystalline medium); by the divergence theorem you can write this as a surface integral of P. Even if W and ΔV are symmetrical around rn so that the n-th term in the Σ cancels out, by inspection you will still get a nonzero net charge contribution from any unequal adjacent dipoles, exactly as I said. You seem to be neglecting these contributions in your final conclusion, which would only be true if the support of W were too small to smooth out the atomic-scale variations (i.e. if you are not looking at a macroscopic P).

On a separate note, there are no steps in P or delta functions in div P if you resolve things at a near-atomic scale. However, at an interface between two media P changes over an atomic-scale length (the lengthscale of the support of the W function in this model). In the common case where you aren't interested in (or capable of) resolving things at a near-atomic scale, it is convenient to approximate this by a step function in P and hence a surface-charge density. Indeed, avoiding irrelevant atomic-scale features is the whole point of the macroscopic averaged theory. — Steven G. Johnson (talk) 21:55, 28 November 2011 (UTC)

Hi Stevenj: It seems that any attempt to formulate some text for the article has been forgotten.
As for the approximation using W(r), I'd understand it roughly as follows. For an individual molecule, write out its potential in multipoles. Establish a distance at which only the molecular charge qm=Σj∈mqj and molecular dipole moment pm=Σj∈mrjqj are significant. Then drop the higher multipoles and replace the real molecule by some diffuse charge distribution that has the same effect at that distance. Now, since the charge of the neutral molecule qm is zero, this approach leads only to a diffuse dipole distribution pmW(rrm) with the same dipole moment. We now place these clouds at every molecular site and take the molecules away. Now no matter what surface you draw around whatever location in the array of molecules, you will get zero Σqm, because the point charges qm that have been diffused are nonexistent, but you will get a div P bound charge. Do we agree about that? It may be that this approach works only for some types of charge distributions – for example, it seems likely to work for a gas of molecules, say, but I doubt it works in a polar solid where there is long-range order.
Do you think anything has to be added to the article? Brews ohare (talk) 22:56, 28 November 2011 (UTC)
You seem to be artificially distinguishing between -div P and the "clouds of charge" from the averaged charged densities, and are apparently claiming that they give a different result; if this is what you mean to say, it is wrong. (qm is zero for neutral molecules, so forget about it; including it here just muddies the waters.)
Mathematically, you can write a point dipole directly as a charge density given by the derivative of a delta function. When you convolve this with W to get a smoothed charge distribution, you obtain precisely -p⋅∇W terms (that's what convolving with the derivative of a delta does). So, the "diffuse charge distribution with the same dipole moment" is exactly -div P.
Regarding potential improvements to the article, I'm sure there are plenty, but I don't think it's worthwhile to discuss them in the context of this thread, which mainly seems to be about clearing up your personal confusions. — Steven G. Johnson (talk) 23:08, 28 November 2011 (UTC)
Hi Stevenj: I see nothing added by your recasting of the matter in terms of gradients of delta functions. The fact remains that the bound charge exists as div P, and there is zero Σmqm inside every closed surface whatsoever. Sorry you find this thread useless; I find that unfortunate. Brews ohare (talk) 23:34, 28 November 2011 (UTC)
The latter Σ statement is almost content-free since qm (as you defined it) is zero even before talking about sums or closed surfaces. But have you finally acknowledged that macroscopically averaged spatially varying dipoles have a bona-fide nonzero charge density -div P and hence nonzero net charge when integrated over subvolumes? Or are you persisting in your confused claim that only Σmqm represents "net charge"? — Steven G. Johnson (talk) 00:15, 29 November 2011 (UTC)
Stevenj: I have never said anything different from div P existing and non-zero whenever there is spatially varying dipole moment, and the entire point of the qm of a molecule being identically zero is that this means the spatially distinct and somewhat separated pairs of opposite charges constituting a molecule and forming the dipole can be considered as populating only one sub-volume however that is drawn, as otherwise one of the two charges constituting the dipole would be outside the volume and the sum of charges inside the sub-volume would not be zero. I guess I have to claim a complete disconnect between the two of us on this topic. Sorry we've been unable to compare notes. Brews ohare (talk) 01:19, 29 November 2011 (UTC)
Yup, complete disconnect. You still apparently believe that you can construct "one molecule" subvolumes with zero net charge in matter with a spatially varying (nonzero divergence) dipole density, even though this is no longer true once you macroscopically average the dipoles (as is normally assumed whenever one talks about P). — Steven G. Johnson (talk) 02:24, 29 November 2011 (UTC)
About gerrymandering: Using point dipoles does not help the gerrymandering problem. (i.e., the problem that, unless you mischievously gerrymander the subvolume boundary, the boundary will cut some dipoles in half.) With point dipoles, the boundary will be only infinitesimally likely to cut a dipole in half...but when it does, it's a doozy! You get an infinite charge in the subvolume! When you average over the different possible places to draw the subvolume boundary, you get the appropriate finite value. Even though the gerrymandering is easy with point dipoles, it is still an unreasonable way to proceed. It gives a different answer than the correct answer you get with a "typical" (average) boundary or a smeared-out boundary.
Brews, the start of your setup is OK: "replace the real molecule by some diffuse charge distribution that has the same [dipole moment]". Sure: This would be a charge density ρ(r). You seem to have the idea that the charge density is constant and zero everywhere: ρ(r)=0. But of course, the charge distribution ρ(r)=0 has the wrong dipole moment. So ρ(r)≠0, instead ρ(r) needs to be positive in some places and negative in other places, with an overall integral of zero. The subvolume boundary can include regions where ρ(r)>0 but cut off regions with ρ(r)<0. So the sum of charges inside the sub-volume would not be zero. It would be proportional to -div P. --Steve (talk) 03:23, 29 November 2011 (UTC)

outdent Steve: Thanks for the thoughts. I don't have exactly that view. What I imagine is a molecule with a potential expanded in multipoles, and replaced by (i) a point charge with charge the same as the molecule (zero if the molecule is neutral) and (ii) also a point dipole, the same size as the dipole moment of the molecule. Thus, although the molecule itself, as you say, has a separated plus and minus charge density, the substitute doesn't. So as you point out at the outset, drawing a surface at random through the replacement system runs the hazard of actually hitting a point dipole. If the replacement system is smeared using a dipole density p = pmW(rrm) the likelihood of intersecting the smeared dipole becomes significant, I imagine, but intersection actually doesn't have the consequence of putting some charge inside the volume and some outside, because there isn't any charge at all (assuming a neutral molecule). Instead, what intersection does is put some dipole moment outside and some inside. Brews ohare (talk) 03:52, 29 November 2011 (UTC)

It's totally false that a smeared out dipole does "not have any charge at all" in any subvolume. The charge density of a point dipole is a delta derivative, and the charge density of a smeared dipole is proportional to -∇W, which is indeed a cloud of nonzero charge density (even though its integral over all space is zero, this is not true for subvolumes). You don't understand point dipoles or what happens when you convolve them with a smoothing kernel; I tried to explain this earlier but you brushed it off.
Once you smear out the dipoles enough so that they are all overlapping (which you must do if you want a macroscopic theory that smooths the atomic-scale variations), then every subvolume will have surfaces that hit the dipoles. The "gerrymandering" problem disappears, and you get a smoothly varying macroscopic charge density -div P (with the exception of surface charge densities at interfaces, which one normally approximates as a discontinuity in P with a delta function in div P). (Note that smearing out the dipoles is mathematically equivalent to averaging over positions of the subvolumes or smearing out the subvolume interfaces.)
— Steven G. Johnson (talk) 15:59, 29 November 2011 (UTC)
Stevenj: Yes, I agree there is a div P or "bound charge" density. The point of this discussion is to separate that from the notion of "bound" charge as simply charge or pairs of charges tied to the vicinity of a site. Brews ohare (talk) 17:14, 29 November 2011 (UTC)
No, the bound charge density really truly comes from real charges tied to the vicinity of a site. The problem is that you want to smear out the dipole moment but without introducing any charge anywhere. This is impossible. It is mathematically necessary that a smeared-out dipole moment brings positive charge to some areas and negative charge to other areas. Let's do the 1D case: A point dipole has charge distribution ρ(x) = P δ'(x). We smear it out to a dipole distribution P(y), for example P(y)=P0 e-y^2. Then the overall charge distribution is
$\rho(x)=\int dy \, P(y) \delta'(x-y) = -P'(x)$
(integration by parts), i.e., a finite charge which is positive in some places and negative in other places. :-) --Steve (talk) 19:51, 29 November 2011 (UTC)
Hi Steve: I've no doubt, of course, that a microscopic dipole involves charge separation. However, my notion was that one could define a point dipole simply as an idealized object that has a dipole form for its potential, even though such idealized objects are a pure abstraction. Using this idealized object to model a real dipole with charge separation would be indistinguishable at distances far enough away that the real dipole lost all its multipole contributions. In any event, the whole subject of approximating charge distributions by some kind of averaging is pretty complicated, and possibly not the most direct route to interpreting div P charge. Thank you so much for taking the time to discuss this subject with me. I'll return to it later when I have studied the matter more. Brews ohare (talk) 14:29, 30 November 2011 (UTC)
The "idealized object" that represents a "point dipole" is precisely a charge distribution that is the derivative of a delta function. As we have both tried to explain to you, many times. — Steven G. Johnson (talk) 02:18, 2 December 2011 (UTC)
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