User talk:Sbyrnes321/Archives/2009

A remark and a request

Hi Steve, I have a remark with respect to your recent edit of the Measurement in quantum mechanics page. You may be right with respect to the content of Griffiths's book (it has been some time since I consulted it, which is used at my University and which I liked pretty much). A different question is whether Griffiths is right about nonlocal realism. So, I don't mind to skip the reference to Griffiths. On the other hand the link [[1]] to the paper by Nikolic might be worth maintaining (although I must confess not to have read the paper carefully) in order to balance the widespread idea that nonlocal realism would be a useful proposition.

My request has been induced by visiting your User page on which I found some nice animations. It must be easy for you to comply with the request I put here. If you could find some time for this purpose I would be very happy.WMdeMuynck (talk) 10:23, 21 January 2009 (UTC)

Hi WMdeMuynck,

The paper by Nikolic is already referenced in that very paragraph, no? It's footnote 2. That paper, like Griffiths, does in fact say that nonlocal realism is a legitimate possibility: "But the fact is that nobody knows with certainty whether the fundamental laws of nature are probabilistic or deterministic...The principle of locality is often used as the crucial argument against hidden variables in QM...However, it is important to emphasize that the principle of locality is an assumption..." and so forth. If you know of any general reason to think that nonlocal realism must be false, you should publish it yourself. For example, Ref. 3 in the article was a recent publication that made it into Nature by just disproving a certain specific class of nonlocal realism theories, far short of a general proof but still regarded as an important publication in that field.

As for your request, I appreciate that you like my previous animation(s), but I'm afraid this is a busy time for me so I won't be able to do it. --Steve (talk) 15:32, 21 January 2009 (UTC)

Thank you for your quick answer. I think we are in agreement if you doubt the relevance of the locality condition for a derivation of the Bell inequalities. You might consult my website [[2]] to find my publications on that issue. I think you refer to Anthony Leggett's article, which I interpret as an attempt on his part to fill in a gap in the proof of impossibility of any hidden variable theory. Unfortunately it seems to be interpreted by many as just a restriction on the possibility of nonlocal theories.

It's a pity that you have no time to deal with my request. Hope someone else will take it up.WMdeMuynck (talk) 16:21, 21 January 2009 (UTC)

Hello! I was referring to the article S. Gröblacher et al., An experimental test of non-local realism, Nature 446, 871 (2007) [3]. To be clear, my position is the same as Griffiths', Nikolic's, and Gröblacher's: Bell's theorem does not disprove all deterministic hidden-variable theories; it only disproves local deterministic hidden-variable theories. Is this something you disagree with? --Steve (talk) 20:44, 21 January 2009 (UTC)

Your position is the one maintained by most physicists engaged in the foundations of quantum mechanics. I happen to disagree because it is not sufficiently taken into account that proofs of Bell's theorem are based on an additional assumption (next to locality), viz. that quantum mechanical measurement results can be conditioned on an instantaneous value of a hidden variable $\scriptstyle \lambda$ (the additional assumption is necessary to warrant the existence of conditional probabilities $\scriptstyle p(a|\lambda )$). I think that this assumption is not justified since quantum measurements, presumably being much slower than subquantum dynamics, average over a trajectory within hidden variable space.

On the quantum mechanical side violation of the Bell inequalities must be related to incompatibility of observables, and, hence, must have a local origin because incompatibility is a local affair (since, as a consequence of the principle of local commutativity only observables corresponding to measurements within causally related regions of spacetime can be incompatible).WMdeMuynck (talk) 23:16, 21 January 2009 (UTC)

OK. I know you've thought about this a lot harder than I have, I haven't read your website, and I have no idea whether you're right or wrong. But the model for Wikipedia (see e.g. WP:V, WP:RS) is that Wikipedia should present what is regarded by reliable sources as accurate information, whether it's actually true or not! See WP:RS for what qualifies as a reliable source...I'm not sure your own book would qualify unless for example it's widely used as a textbook. (Is it? Are there other sources besides that one?) Anyway, it sounds to me (correct me if I'm wrong) like you haven't succeeded in convincing the larger physics community about the impossibility of nonlocal hidden-variable theories. When you do, obviously a bunch of wikipedia pages would have to be changed, not to mention textbooks, etc. In the meantime, I wish you luck. :-) --Steve (talk) 23:46, 21 January 2009 (UTC)

Apart from your reference to "the impossibility of nonlocal hidden-variable theories" rather than "the possibility of local hidden-variable theories" your observations are largely correct.WMdeMuynck (talk) 16:23, 22 January 2009 (UTC)

The Stratton Reference

Steve, I managed to get a look at Stratton. I can't find anything that even remotely resembles the quote which I had purported to have been in Stratton. I was obviously misinformed by a secondary source. In section 23 of chapter 1 there was some related stuff, but it is quite distinct from the point that I was making.

The proof in question appears in a January 1984 paper written by myself, and published in a magazine called the Toth-Maatian Review, and elaborated on in the July 1984 edition. I was quite surprised when I read about Stratton in a recently published paper, and that a line which they quoted as being in Stratton, 1941, was an identical proof to what I had published in the Toth-Matian Review in 1984.

I was misinformed by a recently published article, and I am now investigating the matter. I have written to the authors in order to get them to clarify the reference. The proof in question is original research, but in many respects it's already there in front of us, in that the total time derivative version of Faraday's law contains both the partial time derivative version as per the modern Maxwell's equations, and the vXB term which appeared in Maxwell's original papers.

It can't therefore go in the main article. But I can't see how we can ignore the essence of it when considering the relationship between the total time derivative Faraday's law and the two constituent aspects which have been the cause of allegations of asymetry David Tombe (talk) 04:47, 4 February 2009 (UTC)

Interesting. Oh well, it's not the first time in history that a published article misquotes its sources, and it won't be the last. :-) --Steve (talk) 05:08, 4 February 2009 (UTC)

I'm now curious to know where they did actually get the quote from? I've written to ask them, but so far no reply. David Tombe (talk) 06:04, 4 February 2009 (UTC)

Steve, I've just had a note from FyzixFighter regarding the Stratton reference. You should take a look at the discussion on his talk page because it is relevant to the debate at Faraday's law and it mentions your preference for the term 'flux rule'. David Tombe (talk) 06:44, 4 February 2009 (UTC)

Martin's edits

Hi Steve: Martin Hogbin has removed the paragraph below that was on speed of light claiming without any support that I have misread all the documentation. Here is the paragraph:

Outer space and ultra high vacuum approximate free space, but may have a non-trivial refractive index (that is, an index different from one). Ongoing experimental and theoretical work continues to explore the possibility of small departures of these mediums from free space, which could prove or disprove some theories of quantum gravity, or provide further corroboration of the predictions of quantum electrodynamics.

Personally I cannot see how any misreading has occurred. Here is a verbatim quote from Delphenich:

“Now, by the term "electromagnetic vacuum", what we really intend is not a region of space in there is no energy present, whether in the form of mass or photons, but a region of space in which only an electromagnetic field is present. Hence, there is some justification for treating the electromagnetic vacuum as a polarizable medium in the optical sense, which suggests that treating the electric permittivity and the magnetic permeability of the vacuum as simply constants, ε0 and μ0, is basically a pre-quantum approximation, as well as the constancy of the speed of propagation of electromagnetic waves, c0 = 1/√ε0μ0 , or, equivalently, the index of refraction of the vacuum. We shall regard these constants as asymptotic zero-field values of field dependent functions. The fact that c0 itself might vary with the strength of the field suggests that quantum electrodynamics might even have something deep and subtle to say about causality itself that goes beyond the familiar concepts of special relativity.”

Here's a quote from Moulin:

“This study is motivated by a desire to investigate the possibility of using recently developed powerful ultrashort (femtosecond) laser pulses to demonstrate the existence of nonlinear effects in vacuum, predicted by quantum electrodynamics (QED).”

Here's a quote from Marklund:

“The quantum vacuum constitutes a fascinating medium of study, in particular since near-future laser facilities will be able to probe the nonlinear nature of this vacuum. There has been a large number of proposed tests of the low-energy, high intensity regime of quantum electrodynamics (QED) where the nonlinear aspects of the electromagnetic vacuum comes into play…”

Here's a quote from Mourou et al,:

“The laser fields … will also enable one to access the nonlinear regime of quantum electrodynamics, where the effects of radiative damping are no longer negligible. Furthermore, when the fields are close to the Schwinger value, the vacuum can behave like a nonlinear medium in much the same way as ordinary dielectric matter…”

Here's a quote from Davis et al.:

“The most important magneto-optical interactions that can occur in material media are the Faraday effect, magnetic dichroism, and magnetic birefringence (the Cotton-Mouton effect). Quantum electrodynamics predicts that because of photon-photon interactions even the vacuum becomes birefringent in the presence of a strong magnetic field. …An improved experimental arrangement is needed to pursue vacuum magnetic birefringence and polarization rotation effects. With an improved system, detection of the QED- predicted magnetic birefringence should be possible …”

What is Martin talking about? Maybe I should just do a verbatim quote? Brews ohare (talk) 21:28, 12 February 2009 (UTC)

Hi Brews, I'm having a very busy week! :-) I'll read and respond in a couple days, if you haven't already worked it out by then. Sorry! --Steve (talk) 01:46, 13 February 2009 (UTC)

Steve: Thanks for the interest. I've set up a RfC; when you have time please take a look. Brews ohare (talk) 13:28, 13 February 2009 (UTC)

Thanks

Sorry/thanks for fixing my slip-up on Magnetization; I was merging articles and trying to redirect links late at night, and just was being too much of an idiot to see that I was editing a section on paramagnetism while talking about the Curie temp. Sorry, and thanks! Awickert (talk) 19:13, 8 March 2009 (UTC)

Fermi energy/chemical potential

Hi, regarding your recent changes in Electronic_band_structure: I believe the use of Fermi energy instead of chemical potential is misleading. Of course, they are almost identical numerically in the solid state (which is perhaps the reason to be used interchangeably) - but still, it's wrong. --Evgeny (talk) 23:05, 11 March 2009 (UTC)

Not wrong, not approximate, just a different terminology. For example, here's a quote from Pankove, Optical Processes in Semiconductors: "The Fermi level is the energy at which the expectation of finding a state occupied by an electron is 1/2." (page 6)

He's defining the term "Fermi level" in a certain way (the same definition that you would give to the term "chemical potential"), and then using it consistently with that terminology. This use of terminology is extremely common in the semiconductor materials/device literature, although obviously not universal. I can give arbitrarily many more examples if you think that Pankove is the only one. --Steve (talk) 04:29, 12 March 2009 (UTC)

Quite possible. But again, this is an incorrect use of the term. And the only reason this terminology is used in the semiconductor literature is that the relevant temperatures (or, actually, their variations) are orders of magnitude lower than $E_{F}$. However, Wikipedia articles are not supposed to be read only by semiconductor specialists, so a definition used in general-level of physics textbooks should be more appropriate. It is not even a question of the name. The chemical potential is a function of temperature; the Fermi energy isn't. --Evgeny (talk) 08:26, 12 March 2009 (UTC)

Hi again! I don't think you understand what I'm trying to say. The term "Fermi level" is two words. What is the meaning of these two words? (Keeping in mind, of course, that words can have more than one meaning.) When you read one set of textbooks, you get the answer: "The Fermi level is the energy at which the expectation of finding a state occupied by an electron is 1/2, and is a function of temperature.". When you read a different set of textbooks, you get the answer, "The Fermi level is the highest occupied energy level at zero temperature, and is not a function of temperature". You're saying that everyone using the latter definition is "correct" and everyone using the former definition is "incorrect". On what basis could you say this?? You might as well be arguing about whether meter or metre is the "correct" spelling, or which definition of extinction coefficient is the "correct definition", etc. A terminology in itself cannot be right or wrong, it can only be common or unconventional. Do you see what I mean? --Steve (talk) 15:37, 12 March 2009 (UTC)

Hi. Let's make it straight. The last edition (by you) of the article states verbatim that the chemical potential is sometimes called the Fermi energy. Please quote a textbook (a basic one, not specific to semiconductors) with a similar statement. Note that we talk about the Fermi energy, not Fermi level. See, e.g. the definition in Britannica: [4] --Evgeny (talk) 08:11, 15 March 2009 (UTC)

I'm very familiar with the definition in encyclopedia brittanica, that's the same definition as is given in most intro solid-state physics textbooks, including the one I used, and you'll notice that I'm not trying to push the alternate definition as the "main definition" on this page or any other. I just think that when there's more than one definition for a term, or more than one term for a concept, it's worthwhile to have a little parenthetical mentioning it.

I've never heard of anyone in any subfield making a distinction between "Fermi level" and "Fermi energy"...Have you?

Anyway, here is a reference that I think fits all your criteria: The symbol "E_F" is referred to in the text as both "Fermi level" and "Fermi energy" interchangeably, and the book explicitly says that E_F changes with temperature. ("Thus, when the temperature of n-type or p-type semiconductors increases, the Fermi level E_F moves towards the centre of the band gap".) It's a general textbook on electronics intended for a wide audience, not a technical book for semiconductor experts. Is that OK? I had no trouble finding this, and there's many more where it came from. In fact, this one even has a graph of "Fermi energy E_F" versus temperature! (Figure 6.6.)

Is that OK? Are you on board? :-) --Steve (talk) 04:50, 17 March 2009 (UTC)

You wrote: I've never heard of anyone in any subfield making a distinction between "Fermi level" and "Fermi energy"...Have you? Yes, e.g. see the Britannica article: The value of the Fermi level at absolute zero (−273.15 °C) is called the Fermi energy and is a constant for each solid. There are also others, e.g. [5]. I personally don't use the "Fermi level" term at all - chemical potential is the proper term (and more general, not limited to fermions). Anyway, if you call the Fermi energy anything that depends on T, what's the name for the all-important constant? Again, all your reference books are not general - traditional solid-state physics (of which electronics is a subset) deals with temperatures few orders of magnitude lower than the Fermi energy (by any definition...), which makes the Fermi energy<->chem. potential confusion work for practical purposes in such fields. Say that Fermi energy is a function of T to a specialist in statistical physics, astrophysics, plasma physics - and you'll be ridiculed.

I'm not against mention of possible ambiguity in terms. But let's clearly mark the proper one as such, to be consistent with the definition given in Wikipedia itself. --Evgeny (talk) 08:58, 19 March 2009 (UTC)

Don't pin this on me, it wasn't my idea to have every semiconductor physicist in the world call the chemical potential both "fermi level" and "fermi energy"! I just think that given that they do, Wikipedia should communicate this fact. Wikipedia is supposed to be useful to everyone, including semiconductor physics, which is not an obscure subfield but comprises a sizeable portion of the people interested in the electronic band structure article.

Thank you for the bit of trivia regarding Fermi energy vs. level. Very interesting. I wonder how many practicing physicists/engineers actually would agree with that distinction. I don't think it's in the major textbooks (Kittel, Ashcroft/Mermin, etc.), but don't remember for sure. Certainly not the semiconductor physics-centric textbooks.

If you're not against mention of possible ambiguity in terms, then we're on the same side. Here's my edit, you can see that I privileged "chemical potential" as the main definition, and kept the symbol $\mu$. Can you be more specific about what you would edit in that?

To answer your question, in semiconductor physics, the chemical potential at zero temperature is called "Fermi level/Fermi energy at zero temperature". It actually doesn't come up very often.

Not to nitpick, but as a matter of science, it's incorrect to say "there's essentially no difference between Fermi level at zero temperature and the chemical potential at finite temperatures as long as $\mu <<k_{B}T$". (I assume you mean $\mu$ relative to the ground state, which would be the 1s electrons.) What really matters is how uniform is the density of states in the region ~$\mu \pm k_{B}T$. See here for an example of the chemical potential changing a lot (eV's) at temperatures (as always) far less than $\mu /k_{B}$. (I've done some work in thermopower, where changing of the chemical potential with temperature is very important, even way below room temperature.) :-) --Steve (talk) 13:04, 19 March 2009 (UTC)

OK, thanks for correcting me by pointing to cases of strong dependences of $\mu$ on $T$. You wrote: I assume you mean $\mu$ relative to the ground state. No, in fact, I was thinking about something well approximated by the ideal Fermi gas model (so no discrete states at all). Regarding your edit - I think what triggered my reaction in the first place was the implication (that's how I read it) that the other definitions (why plural, by the way?) are wrong. How about something like (... called the "Fermi energy", and denoted $E_{F}$, although the definition of this term in e.g. statistical physics is different).? --Evgeny (talk) 16:11, 19 March 2009 (UTC)

I tried this for a start...there may be other better ways to communicate this, feel free to edit it yourself. :-) --Steve (talk) 04:04, 20 March 2009 (UTC)

Fermi level

Hi Steve !

I think you did a good job on restructuring the format on the "Fermi level" page - but from my point of view you included a few "errors of incompleteness". I don't disagree that what you write is what some of the textbooks say. The problem is that what's in the solid-state physics textbooks is not the complete story. I guess that the authors of these textbooks never had to apply some of the concepts that they write about to real experimental apparatus and situations as used in surface science and electronic engineering. I will work through these perceived problems with you in due couse - but right now I am a bit busy with other things - including an apparently ongoing discussion about what should and should not be on the vacuum resistivity page. (The content is, I hope, now almost error free).

One point to start off with is that I certainly use a different convention from some solid-state-physics textbooks as to the meaning of the terms "Fermi level" and "Fermi energy". Both these terms refer to the energy level of the "occupancy = 0.5 state" in the "conductor under analysis". The term "Fermi energy" always (tries to) define this as measured from the base of the conduction band. The term "Fermi level" measures this from any suitable reference zero, but it is most convenient to take the reference zero as the Fermi level of a "reference conductor" and it is usually best to take the reference conductor as the Earth. If the temperature of the conductor under analysis is changed (but not the temperature of the Earth) then its Fermi energy changes but the Fermi level does not. What happens is that the band structure of the conductor moves in energy, relative to the Fermi level of the Earth.

Assuming you don't mind having discussions of "errors of incompleteness" in public, I'll repeat this point on the "Fermi level" talk page in due course.

Feel free to delete this message from your talk page,

For info., my first degree is in theoretical physics (from Cambridge Univ., a long time ago), although I am attached to a Department of Electronic Engineering.

Regards

Richard

(RGForbes (talk) 22:48, 29 March 2009 (UTC))

Hi Steve !

After some thought on the matter, I have now concluded that the term "Fermi level" is used in the literature in three related but non-equivalent ways. So I have done a more-or-less complete rewrite of the article - but have continued to include some of the points that you were making at the end. One small issue is that, in my understanding, the term "voltage" always relates to the full electrochemical potential, and should not be used as a term describing differences in purely electrostatic potential.

Regards

Richard Forbes (RGForbes (talk) 21:09, 2 April 2009 (UTC))

Magnetic field

I wonder if you could be interested in looking at Magnetic_field#Magnetic_dipoles and Magnetic moment. These articles could benefit from your attention. Brews ohare (talk) 15:28, 4 April 2009 (UTC)

They don't look so bad to me...what did you have in mind? --Steve (talk) 22:27, 5 April 2009 (UTC)

If you agree with Magnetic_field#Magnetic_dipoles, I'll take a look at making Magnetic moment consistent with it. Brews ohare (talk) 12:35, 6 April 2009 (UTC)

They're inconsistent? Maybe I haven't read carefully enough :-) --Steve (talk)

g-force units

Duh, I'm sorry I was just trying to clean up that section a bit, I should have noted the negative exponent! I'll revert it if it hasn't been already! Thanks!!

Reportingsjr (talk) 22:06, 20 April 2009 (UTC)

Electric dipole

Hi Steve: I wonder if you could help me out with the section: paired charges. This is a very common approach to expressing polarization density, but I'm having trouble demonstrating this connection. Much obliged. Brews ohare (talk) 19:14, 28 April 2009 (UTC)

I happen to know a bit about that, having done work on ferroelectrics. It's a complicated business. For example, if you look at an infinite NaCl crystal along the 111 axis there's a plane of Na+, a plane of Cl-, etc.

 ... +  -  +  -  +  -  + ...

Split it into electric dipoles...

... (+  -)(+  -)(+  -)(+ ...

The polarization vector points to the right! But wait...

 ... +)(-  +)(-  +)(-  +) ...

The polarization vector points to the left!

In fact, it makes more sense physically to say P=0, after all this is a centrosymmetric system. The bulk polarization itself is basically meaningless for this reason.

Changes in the bulk polarization, on the other hand, are very meaningful and easily measured (they look like an electric current). They're also easily calculated (using a Berry's phase method). But the polarization itself is more difficult, and I think still controversial even among experts.

In some situations, e.g. when you take a centrosymmetric material and then slightly distort it, there's an obvious definition of P, namely the convention where P=0 in the centrosymmetric state. Switchable ferroelectrics are in this category, for example, as are centrosymmetric insulators. But in more general cases, it's hard to define P for an extended distribution of point charges (or non-point charges).

Does this make sense? If you want to read scientific papers on this, I have tons. --Steve (talk) 01:44, 29 April 2009 (UTC)

Thanks for the interest. I wonder if you agree with my contributions to Electric dipole moment ? Brews ohare (talk) 02:17, 29 April 2009 (UTC)

Polarization and dipole moment

Hi Steve: Here are some ambiguities that I'd like to set straight in the respective articles. They relate to what seem to be conflicting usages of the term "polarization" to refer both to the P, as in div P = -ρ_b, and to the stretching of atomic dipoles by an electric field, or more generally to the dipole moment of a charge array.

The statement often is found that P is the dipole moment per unit volume. Here, the dipole moment per unit volume is found by adding up all the vector dipoles in a region and diving by the volume. This seems to be the the thing that physically is addressed by the susceptibility or permittivity of a medium, saying that the "polarization is proportional to the field" for example.

However, I'd argue that this dipole moment per unit volume is in fact not the polarization P that satisfies div P = -ρ_b. In fact, for any charge neutral array, the dipole moment is a constant independent of position, whether inside the array or remote from the array. It is an intrinsic property of the array, like its mass or its moment of inertia. Thus, its divergence is identically zero everywhere. The dipole moment of an array is:

${\boldsymbol {p}}({\boldsymbol {r}})=\sum _{i=1}^{N}\,\int q_{i}\left(\delta (\mathbf {r_{0}} -\mathbf {r} _{i}+{\boldsymbol {d}})-\delta (\mathbf {r_{0}} -\mathbf {r} _{i})\right)\,({\boldsymbol {r_{0}}}-{\boldsymbol {r}})\ d^{3}{\boldsymbol {r_{0}}}$ $=\sum _{i=1}^{N}\,q_{i}\left({\boldsymbol {r_{i}+d_{i}}}-{\boldsymbol {r}}-({\boldsymbol {r_{i}}}-{\boldsymbol {r}})\right)=\sum _{i=1}^{N}q_{i}{\boldsymbol {d}}_{i}\,=\sum _{i=1}^{N}{\boldsymbol {p}}_{i}\,$

which is independent of position, but accords with the notion of adding up the individual dipoles.

In contrast the P of Maxwell's equations for an array of paired opposite charges is:

${\boldsymbol {P(r)}}=-\sum _{i=1}^{N}\,{\frac {q_{i}}{4\pi }}\left({\frac {\boldsymbol {r-r_{i}+d}}{|{\boldsymbol {r-r_{i}+d}}|^{3}}}-{\frac {\boldsymbol {r-r_{i}}}{|{\boldsymbol {r-r_{i}}}|^{3}}}\right)\ ,$

which satisfies:

${\boldsymbol {\nabla \cdot P(r)}}=-\sum _{i=1}^{N}\,q_{i}\left(\delta (\mathbf {r} -\mathbf {r} _{i}+{\boldsymbol {d}})-\delta (\mathbf {r} -\mathbf {r} _{i})\right)\ =-\rho _{b},$

the negative of the bound charge density, as required.

Thus, the actual P that satisfies div P = -ρ_b varies with position, and in fact has a rather more complicated dependence upon field than a simple susceptibility inasmuch as a field can rearrange the array changing all its multipole moments and all differently (for example, some moments may be unaffected due to symmetry). Hence, setting P = χ E doesn't seem to make much sense except under some severe approximations (such as replacing the real array by a dipole approximation). The stretching of a dipole being set to χ E, on the other hand, is like applying Hooke's law.

Nonetheless, it is common practice to set P = χ E in Maxwell's equations. This means for example, that if E is some featureless uniform field, so is P, even though the P satisfying div P = -ρ_b has a lot more structure than that, even under gross averaging.

Any comments, references? Thank you. Brews ohare (talk) 19:01, 29 April 2009 (UTC)

Based on your formula, you're saying ${\boldsymbol {P(r)}}$ is -ε0 times the electric field due to the point charges. I don't think this is always true. For example, take a uniformly-polarized sphere, with no external field ([6], this is in Griffiths too.) It's not true that E=-P/ε0 inside the sphere, because D is nonzero...despite the lack of "free" charge or "applied" field! In fact, E=-P/(3ε0), and D=(2/3)P. So your formula based on Gauss's law can't be right...it's a formula for -ε0*E_{due to dipoles}, which has no direct relation to P.

You have the correct $\rho _{b}$ on a microscopic scale. What about when you average over a nm or two? (Approximating the material as a smooth continuum, rather than a bunch of atoms and electrons with wildly-fluctuating charge distributions.) Then since each unit cell is charge-neutral, you expect $\rho _{b}=0$ within the bulk of the material. At the edges of the material, you expect a surface charge, since all the charges of one type are sticking out farther into space than the charges of the other type. I think you'll see both these things explicitly if you blur out $\rho _{b}$, (e.g. convolute with a gaussian). The surface charge should go as the q times the dot-product between the surface normal and the charge-displacement direction.

Your expression ${\boldsymbol {p}}({\boldsymbol {r}})$ is the formula for the total dipole moment of the whole slab. What we want, for smoothed-out polarization density, is

$P(r)={\frac {1}{V}}\int _{V}\rho (r')r'dr'$

over a volume V that's macroscopically small but large compared to a unit cell, and centered at around r. You also need to be sure to choose V carefully, so you don't split up any dipole pairs and get funny edge-effect contributions. Anyway, if you do that right, you'll get a P which is constant (q*d*volume density of dipoles) in the bulk, and fades quickly to zero at all the sample boundaries. And if you compute its divergence, you'll find that it matches $-\rho _{b}$, i.e. zero in the bulk, large at certain surfaces.

Do you agree with that? This is a very interesting discussion!! :-) --Steve (talk) 22:18, 29 April 2009 (UTC)

Yes, the P is just the field of the charges selected to be the "bound" charge, in this case some charge neutral subset of all the charges.

Inevitably, the total charge is then ρ_t = ρ_f + ρ_b and ipso facto div D = ρ_f and div P = -ρ_b. So far this is all fine, although one might ask why do it. One detail is that because the bound selection is charge neutral, it has a unique dipole moment, for example, p = Nqd if all N dipoles are the same and similarly oriented. Otherwise, p is a vector sum, but still a unique vector that characterizes the group regardless of where one looks at it.

It seems to me the averaging you mention is the motivation. The question is: what do you have to do to reduce this P(r) to some very simple result involving the constant p?

One way is the multipole expansion. Because the subset is charge neutral the leading term for large distances from the charges is the dipole term:

${\boldsymbol {P(r)}}=-\sum _{i=1}^{N}\,{\frac {q_{i}}{4\pi }}\left({\frac {\boldsymbol {r-r_{i}+d}}{|{\boldsymbol {r-r_{i}+d}}|^{3}}}-{\frac {\boldsymbol {r-r_{i}}}{|{\boldsymbol {r-r_{i}}}|^{3}}}\right)\ ,$

$\approx -{\frac {q}{4\pi }}\left({\frac {3Nq{\boldsymbol {d\cdot {\hat {r}}}}}{|{\boldsymbol {r}}|^{3}}}{\boldsymbol {\hat {r}}}-{\frac {Nq{\boldsymbol {d}}}{|{\boldsymbol {r}}|^{3}}}\right)\ +\ ...\ ,$

which is the same as the dipole N d at r. Because any charge distribution with this dipole potential is equivalent, if I am so perverse I can replace the array of charges with a sphere with a surface charge on it that gives the same dipole moment. If the array has a dipole moment that satisfies N d = χ E, then my P(r) becomes:

${\boldsymbol {P(r)}}-{\frac {q}{4\pi }}\left({\frac {3\chi {\boldsymbol {E\cdot {\hat {r}}}}}{|{\boldsymbol {r}}|^{3}}}{\boldsymbol {\hat {r}}}-{\frac {\chi {\boldsymbol {E}}}{|{\boldsymbol {r}}|^{3}}}\right)\ +\ ...\ ,$

So far, so good. However, I can't see a methodology that is going to allow this approximate dipole sphere to be used in close to the sphere without abandoning all pretense of reality and just saying "Hey, it's a model; whaddaya expect?"

In many cases (like lattice dynamics, say) it seems the material medium is replaced by atoms as dipole source terms in the Maxwell equations and P(r,t) isn't used. This is kind of interesting: you apply boundary conditions that impose an external E, couple this E to the dipoles on the source side, and find the resulting field including the dipoles. The modified E_eff / E then determines χ.

So maybe the case to focus upon is something like a microwave cavity where you can get away with P = χE and one finds c ≠ c₀? What is your experience? Brews ohare (talk) 00:07, 30 April 2009 (UTC)

I think I have this all straight now. Let me know if you agree. Brews ohare (talk) 20:29, 30 April 2009 (UTC)

Well, one thing I'm asserting is that

${\boldsymbol {P(r)}}=-\sum _{i=1}^{N}\,{\frac {q_{i}}{4\pi }}\left({\frac {\boldsymbol {r-r_{i}+d}}{|{\boldsymbol {r-r_{i}+d}}|^{3}}}-{\frac {\boldsymbol {r-r_{i}}}{|{\boldsymbol {r-r_{i}}}|^{3}}}\right)\ ,$

is not a correct expression for the electric polarization density. The right-hand-side is the Gauss's law expression for the electric field due to the bound charges (times -ε0), which is not the same as P. For example, inside an isolated uniformly-polarized sphere (see above), the electric field due to the bound charges (times -ε0) is one-third the value of P. So this formula would have a factor-of-three error in this case. No matter how you average, that won't get rid of a factor-of-three error.

Instead, I would write

$-\epsilon _{0}{\boldsymbol {E_{DueToBound}(r)}}=-\sum _{i=1}^{N}\,{\frac {q_{i}}{4\pi }}\left({\frac {\boldsymbol {r-r_{i}+d}}{|{\boldsymbol {r-r_{i}+d}}|^{3}}}-{\frac {\boldsymbol {r-r_{i}}}{|{\boldsymbol {r-r_{i}}}|^{3}}}\right)\ ,$

with the understanding that E_DueToBound is the electric field resulting from the bound charge, which has no straightforward relation to the electric polarization density P.

Do you agree with this? --Steve (talk) 05:03, 1 May 2009 (UTC)

I guess not. My tentative view would be that the exact form of P is correct, and the question is really how you wish to approximate this expression (or what kind of accuracy one needs). So for example, if one replaces the actual charge distribution with a constant dipole moment density, what this approximation does is replace the the charges interior to the volume by a vacuum, and instead puts a surface charge on the boundary. This surface charge generates a depolarizing field in the interior vacuum that is accounted for by saying the field in the vacuum is reduced from the applied field by the dielectric constant of the medium. If you choose a spherical surface you'll get the geometric factor of 1/3 you mention at the center of the sphere. However, this geometric correction factor is different for a slab or for positions off-center. In determining the dielectric constant experimentally one must be careful about the experimental geometry.

Fundamentally, one is free to partition the charges into "free" and "bound" in an arbitrary manner, and the the free/bound formulation of Maxwell's equations are perfectly happy with however you do it. However, it is mandatory that div P = -ρ_b. Using the exact form for P is the only way to accomplish this result. Any other method is tantamount to approximating P and using the corresponding approximation for ρ_b. Brews ohare (talk) 06:45, 1 May 2009 (UTC)

Again, take the example of a uniformly-polarized sphere, with no other charge present anywhere in space. Then the only electric field is the field due to the bound charge,

E=E_DueToBound

Inside the sphere, E_DueToBound is E=-P/(3ε0), as stated in Jackson, Griffiths, etc. Outside the sphere, E_DueToBound=E is nonzero even though obviously P=0 in the empty space (the exact formula for E outside the sphere is again in Jackson and Griffiths.

Is ∇ × D = 0 ?

How about this argument?

• In the total charge formulation of Maxwell's equations ∇ × E = 0 and ε₀∇ · E = ρ.
• To set up the "free +bound" picture, we divide the charges into two groups, "free" ρ_f and "bound" ρ_b.
• Using superposition, we solve the two problems:

ε₀∇ · E_b = ρ_b & ε₀∇ · E_f = ρ_f

Apparently ∇ × E_b = 0 & ∇ × E_f = 0

• Next we set ∇ · D = ρ_f
• Apparently this result is satisfied by D = ε₀E_f = ε₀E -ε₀E_b
• We then set P = -ε₀E_b & D = ε₀E + P

Under this scenario, ∇ × D = 0. I'd argue that this scenario is intended by the introduction of the "free/bound" division, with the aim of dividing the methodology up into parts that are to be calculated under different bases.

Tipler "The surface charge on the dielectric is called a bound charge, because the surface charge is bound to the surface molecules of the dielectric and cannot move about like the free charge on the conducting capacitor plates"

Stewart "By bound we mean that the charges are not free to flow"

Fleisch] "bound charge (charge that is displaced by the electric field but does not move freely through the material)"

Vanderbilt A more technical view. Eq. 7 is basically a calculation of dipole moment throughout an entire infinite crystal. As is common in the literature, this item is called the "polarization".

Neaton; end of §II "The total polarization ... can be calculated as the sum of ionic and electronic contributions. The ionic contribution is obtained by summing the product of the position of each ion in the unit cell with the nominal charge of its rigid core. The electronic contribution to P is determined by evaluating the phase of the product of overlaps between cell-periodic Bloch functions... The total polarization is then the sum of the two spin contributions and the ionic contribution."

I argue that it is usual to adopt the view that the polarization originates in a partition of charges as described here, no matter how one wishes to proceed afterward. Brews ohare (talk) 19:52, 3 May 2009 (UTC)

It strikes me that the confusing part of all this is separating dipole moment and polarization, which terms are often used interchangeably. In a model where one calculates the dipole moment as a stretching of the dipole or an expectation value for r, this is where the material behavior comes in. If the view is taken that this is also the Polarization, that view implicitly accepts (i) the point dipole model of the medium, and (ii) the expectation that any surface charges that arise are treated using discontinuity conditions. That is not a general viewpoint, and it doesn't work even for a simple scattering problem off an array of charges. Brews ohare (talk) 23:12, 3 May 2009 (UTC)

If I understand correctly your first paragraph above, you propose to solve the equations:

• ε₀∇ · E_b = ρ_b & ε₀∇ · E_f = ρ_f
• ∇ × E_b = 0 & ∇ × E_f = 0
• ∇ · D = ρ_f
• ∇ · P = -ρ_b
• D = ε₀E + P

The problem is that these equations do not have a unique solution. Yes, I agree that these equations are satisfied by P_Brews=-ε₀E_b, D_Brews=ε₀E_f. But they're also satisfied by other {D,P}: D=D_Brews+A, P=P_Brews+A, where A is any field with zero divergence. What's your basis for saying that {D_Brews, P_Brews} are the appropriate solution, and not one of the infinite set of other {D,P} that satisfy all these equations equally well? --Steve (talk) 23:57, 3 May 2009 (UTC)

Hi Steve: There is no need for uniqueness, as the major requirement is div P = --ρ_b. The rest is just external boundary conditions reflecting the applied fields or image charges or whatever. It is a Laplace solution to patch up the boundary conditions. Brews ohare (talk) 01:04, 4 May 2009 (UTC)

Griffiths page 178:

"The curl of D is not always zero.

$\nabla \times D=\cdots =\nabla \times P$

and there is no reason, in general, to suppose that the curl of P vanishes. Sometimes it does, as in Example 4.4 and Problem 4.15, but more often it does not. The [isolated uniformly-polarized cylinder] is a case in point: here there is no free charge anywhere, so if you really believe that the only source of D is ρ_f, you will be forced to conclude that D=0 everywhere and hence that E=-P/ε₀ inside and E=0 outside the [cylinder], which is obviously wrong. (I leave it for you to find the place where curl P ≠ 0 in this problem.)"

--Steve (talk) 15:02, 4 May 2009 (UTC)

Hi Steve: I'll have to wait a few months till I re-establish connection with my library, unless you have a source accessible on-line. My view at the moment is that you can make curl P non-zero, but probably never have to do it. Can you explain what the purpose of the non-zero curl is here? Possibly it is simply an artifact of some revocable assumption about the behavior of a particular choice for P. I'd say that a curl free P can always be found for a static problem. If \dot P is not zero, things are different. Brews ohare (talk) 15:51, 4 May 2009 (UTC)

In general terms, I'll be surprised if any curl P shows up in the static cylinder problem, since as far as I can see, no matter how complex these classical problems become, the solution all boils down to finding some surface charges and their effects, which ipso facto needs no curl. I don't need P to solve these problems, but once solved, and having found the surface charges, I always can find a non-curl 'P that satisfies div P = surface charge. Any added curl to P never shows up in anything real. Brews ohare (talk) 16:13, 4 May 2009 (UTC)

Sorry about the books!! I'll try to use internet-accessible ones when I can in the future. :-)

You appear to be using the definition "P is any field that satisfies div P = -ρ_b". Is that a correct characterization? If not, how do you define P?

My view is that the definition of P is not "any field with the correct divergence". The definition of P is "polarization density". P is by definition the small-volume average of molecular dipole moments. P is unique (apart from possibly very small corrections based on how fine-toothed your average is), and it's zero in vacuum, and it might or might not have zero curl, but it's not up to us to decide! :-) --Steve (talk) 19:45, 4 May 2009 (UTC)

Steve: I'd say p = Σnd is the dipole moment of the array summed over. For any charge neutral assembly it is a vector without spatial dependence. It says the charge-neutral assembly exhibits such and such polarity, but does not say where the assembly is located. If you add the location info: p is at r, then I have the dipole moment distribution p(r) such that div p(r) = ρ_b. Evidently, the linear response theory etc. finds p = χ E. It doesn't care where the assembly is, only what the field acting upon it is. You then have to solve the EM equations to find the field, and you have to locate the charge assembly to see what field it experiences. That's where the P(r) comes in, and in most of the calculations P(r) = p(r). But not always: for example, some problems require P(r) = p_dipole + p_quadrupole + … , for example. Brews ohare (talk) 20:23, 4 May 2009 (UTC)

D, E, and P

It's true that -ρ_b = div P and -ρ_b = div (-ε0 E). But that doesn't mean that -ε0 E=P! In this case, they differ by D, whose divergence is 0 since there's no free charge. Jackson has a field-line picture of what D and E look like. --Steve (talk) 15:54, 1 May 2009 (UTC)

Hi Steve: Unfortunately I don't have my copy of Jackson with me. Let me try to follow what you are saying. The example is a spherical region containing a uniform dipole moment density. Inside the sphere, the only field is due to the surface charge induced by the constant dipole moment density. Likewise for outside the sphere. There is zero bound charge everywhere except on the surface of the sphere. The E-fields inside and out differ by the surface charge. The polarization P is mathematical distribution such that div P is a delta function at the sphere surface. There is no contradiction here between the (let's call it) exact Polarization density and this model. If the constant dipole moment density model is taken literally true, then the exact Polarization density is the mathematical distribution with delta function divergence. If the constant dipole moment density model is taken as an approximation, and a different model of the dipole moment density is taken as more accurate, the Polarization density will change as well. This refinement can continue until the model is very microscopic, say an array of delta functions. At each step of growing modeling detail for the dipole moment distribution, the Polarization density will follow along. Brews ohare (talk) 17:19, 1 May 2009 (UTC)

You can have more than one vector field with the same divergence. If div A = div B, it does not follow that A=B. The field that you call (incorrectly) the "exact Polarization density" is in fact P-D (in this example), not P. In this example, there's no free charge, so div D = 0, and therefore div P = div (P-D) = -ρ_b. So you see, your field, (P-D), has the right divergence. But it's still the wrong field. :-) --Steve (talk) 20:54, 1 May 2009 (UTC)

This all seems a bit strange to me. I suppose you are saying that P and D can differ by a curl? Or, perhaps by a constant? I don't think that is going to change anything. Put differently, div D = 0 but that doesn't need to mean I've got P-D instead of P, does it? In addition, whatever refinement is selected to define ρ_b, the very same approximations used to find ρ_b cause just the right changes in the "exact P". Here I'm referring to the agreement of the two sections arbitrary choice of bound charge and arbitrary approximation of polarization density. Brews ohare (talk) 21:11, 1 May 2009 (UTC)

P and (P-D) differ by a divergenceless field, namely D. The field lines of D form closed loops that pass through the sphere in straight lines and then twist back around outside the sphere. There's a picture of D in Jackson, sorry you don't have it on hand...

The reason I know you have (P-D) is that your expression is manifestly equal to -ε0 E_DueToBound by Coulomb's law, which in turn is equal to -ε0 E since there are no other electric-field sources in this example, and everyone knows that -ε0 E = P-D. So your expression must be P-D.

Another indication that your expression is not P: Your proposed expression is nonzero in regions of space far outside the sphere, even though everyone knows that in free space D = ε0 E and P = D - ε0 E = 0. :-) --Steve (talk) 04:13, 2 May 2009 (UTC)

Hi Steve: You bring up an interesting point: the difference between D and P is ε0 E. Therefore, curl (D-P) = 0 and D-P = ∇φ_b for some scalar φ_b. Also, ∇²φ_b = -∇·P = ρ_b. On the other hand, if there are no free charges, ε0∇·E = ρ_b too. Hence, P and ε0 E differ only in terms of the boundary conditions on their respective φ's. If we look at a collection of bound charge subject to an external field, for instance, we might take φ_b as the solution for the charges in their polarized positions, but in infinite space with no boundaries and field dropping to zero at large distances. We could take E as having the same φ_b plus a homogeneous solution like E_app z (z the z-coordinate that the applied field points along). Then P and E both have the same inhomogeneous terms, and in particular P can be the form postulated above.

If we focus for the moment on the charged sphere, that is pretty much how the exact solution looks: we have the applied field terms in E and they are supplemented by the surface charge terms representing depolarization by the dipole moment distribution. The approximation is that the medium has no microscopic charge variations, but is a featureless array of end-to-end point dipoles resulting in zero charge everywhere except at the boundaries where their tails stick out. Thus, both P and E agree on the shell of charge at the surface, and upon the lack of charge anywhere else. We can take E as supplementing this picture with the applied E_app, while P just focuses on the contribution of the surface charge.

We're also free to ignore any distinction and just take P and ε0 E as the same, as long as there is no free charge.

I'd take it as strictly a matter of convenience which charges are called "free" and which "bound". Possibly the "free" ones have to be treated more carefully, like with Lorentz force law, say, while the "bound" ones can be dumped into a dielectric constant. Or, the free ones have to be treated on a fine scale, while the bound ones can be averaged a bit. This approximation won't affect the relation between E and P as long as the same approximation to the charge density ρ_b is used to find both.

Please take another look at the article. It has evolved a bit. "Evolved" might not be how you see it. :→§ Brews ohare (talk) 21:35, 2 May 2009 (UTC)

Here are some more differences between P and ε0 E in the isolated-polarized-sphere example:

• E must be nonzero in the free space outside the sphere, since there is, by Coulomb's law, an electric field there.
• P must be zero in the free space outside the sphere, since D = ε0 E and P = D - ε0 E = 0 in free space.
• E has zero curl everywhere
• P might or might not have zero curl. (In fact, in this case it does not.)

This seems to contradict your assertion that one can "take P and ε0 E as the same, as long as there is no free charge". Right? --Steve (talk) 17:01, 3 May 2009 (UTC)

Hi Steve: I'll look at your points one by one.

• E must be nonzero in the free space outside the sphere, since there is, by Coulomb's law, an electric field there.

Yes, outside the sphere E is non-zero, exhibiting both the applied field and the field of the induced surface dipole of charge on the sphere. The existence of E outside the region of the polarized charge doesn't strike me as a difficulty for P, which also can exist outside as long as its got zero divergence outside.

• E has zero curl everywhere

Yes, because there are no magnetic effects

• P might or might not have zero curl. (In fact, in this case it does not.)

If D=κE, then curl D = curl P = ∇κ × E. I don't know what the implications of this are; I've never seen an example where this came up, have you?

• This seems to contradict your assertion that one can "take P and ε0 E as the same, as long as there is no free charge".

In principle, maybe so. I'm wondering if polarization due to an applied E-field can lead to a non-zero ∇κ × E, or if there is a fundamental contradiction at a microscopic level. I don't know of an example where P must have a curl. I do take from our discussion that choosing P = ε0 E is only a possible choice when there is no free charge. We can add to P any grad u that satisfies ∇² u = 0. Brews ohare (talk) 18:33, 3 May 2009 (UTC)

Hmm, let's take this one step at a time.

1. In free space, D=ε0 E

2. It's always true that D=ε0 E+P

3. Therefore, in free space, P=0.

Your proposed P does not satisfy 3: The assumption is that there's a uniformly-polarized sphere, and outside of it is a perfect vacuum (free space). But your expression for P is nonzero in that region. So: Do you disagree with 1, 2, 3, or something else? :-) --Steve (talk) 21:39, 3 May 2009 (UTC)

The first statement would not hold true, in my view, unless free space means there are no charges near enough to influence the field. In other words, div P = –ρ_b does not necessarily imply that P is zero outside the region containing the bound charge. For example, the dipole moment of a charge distribution originates in the region containing the distribution, but the dipole field contribution persists at large remove.Brews ohare (talk) 23:22, 3 May 2009 (UTC)

OK, so you're saying that I can have a region of space at zero kelvin, which is a perfect vacuum, with no particles in it...but D≠ε0 E because 10 meters away, outside the box, on the other side of the room, there's a polarized sphere?

You're welcome to look, but I think you'll find that reliable sources agree that D=ε0 E in a region of space which is a vacuum, regardless of whether or not there's a polarized sphere 10 meters away.

The polarization density is zero in a vacuum, because there's nothing there that can be polarized (neglecting crazy quantum-vacuum effects). The electric susceptibility of a vacuum is zero. I hope we can come to an agreement on this basic premise. :-)

It's true that "div P = –ρ_b does not necessarily imply that P is zero outside the region containing the bound charge". However, "div P = –ρ_b" is not the definition of P. It's a property of P, not the definition. It's not enough information to be a definition, because it contains no info about curl P. --Steve (talk) 00:18, 4 May 2009 (UTC)

Hi Steve: Hey, you've put your finger on the problem here. Can you provide me with some reading that explains what P is outside of the polarizable region?

Bear in mind, I am not suggesting there is anything polarizable in vacuum. That is a thing completely different than saying the Polarization field of Maxwell's div P = –ρ_b is zero in vacuum. I believe that you will find plenty of references that calculate dipole moments or < r > and call it polarization, but that is not to say it's polarization density.

In addition, in the common models (for instance, P = χE), as pointed out in the article, the Maxwell P is the same as the dipole moment density in the polarizable medium. It is not the same outside the polarizable medium. And it is not the same inside the medium either if you use a more sophisticated model than P = χE.

I believe you will be pressed to find someone calculating P outside the polarizable region, because that is not condensed matter physics, it's electromagnetism. You might find it in waveguide theory or in scattering problems. Brews ohare (talk) 01:00, 4 May 2009 (UTC)

OK, here's Jackson, page 152:

The "electric polarization P (dipole moment per unit volume) [is] given by

$\mathbf {P} (\mathbf {x} )=\sum _{i}N_{i}\langle \mathbf {p} _{i}\rangle$

where p_i is the dipole moment of the ith type of molecule in the medium, the average is taken over a small volume centered at x and N_i is the average number per unit volume of the ith type of molecule at the point x."

If you're far from any material, than N_i=0, so P=0, right? And this is the exact same P which he puts into D=ε0 E+P one page later. In fact, next time you have Jackson, you should take a look at Section 6.6, where he very carefully starts with the microscopic Maxwell's equations and derives the macroscopic equations, including what averaging procedure to use, how to define P, etc. The definition of P is eqn (6.89):

P is the macroscopic polarization,

$\mathbf {P} (\mathbf {x} ,t)=\left\langle \sum _{n\;(molecules)}\mathbf {p} _{n}\delta (\mathbf {x} -\mathbf {x} _{n})\right\rangle$

(note: the angle-brackets denote an average over a small volume centered at x).

Anyway, my view is that there is one and only one "P" and it is by definition the dipole moment density. The statement div P = –ρ_b is a consequence of that definition (a straightforward consequence), but is not itself part of the definition. :-) --Steve (talk) 03:47, 4 May 2009 (UTC)

Response

Here are some difficulties:

• "The "electric polarization P (dipole moment per unit volume) [is] given by

$\mathbf {P} (\mathbf {x} )=\sum _{i}N_{i}\langle \mathbf {p} _{i}\rangle$

where p_i is the dipole moment of the ith type of molecule in the medium, the average is taken over a small volume centered at x and N_i is the average number per unit volume of the ith type of molecule at the point x."

If the system of charges is overall neutral, as shown in the article, this sum is a constant independent of position, no matter where you are (inside or outside the charge array). Therefore, its divergence is identically zero throughout all space. (Moreover, it would be sanguine to believe all one needs to know is the dipole moment, a pretty crude low-order opinion of the charge distribution.)

This result is an ineluctable consequence of the very definition mentioned,, viz.:

The definition of P is eqn (6.89):

P is the macroscopic polarization,

$\mathbf {P} (\mathbf {x} ,t)=\left\langle \sum _{n\;(molecules)}\mathbf {p} _{n}\delta (\mathbf {x} -\mathbf {x} _{n})\right\rangle$

(note: the angle-brackets denote an average over a small volume centered at x).

Consequently, your view that P is the dipole moment as presented here cannot be sustained without modification or elaboration.

One way to do this is to introduce spatial dependence. For a truly uniform patch of charge, this is done by specifying how the boundary of the patch is constructed, the simplest notion being a step function. For a less uniform patch, the patch can be broken up into cells. It is the actual averaging procedure used that introduces the specific spatial variation picked up by div D. (so much for uniqueness of D or P - there is a different D or P for every choice of averaging method.) That allows a volume polarization density, but still does not duck the need to spell out the boundary layer. Each cell has a spatially independent dipole moment, no matter where it is located. It is the way the moment transitions from cell to cell that matters and that enters P!

Several time honored methodologies for producing the average are cited in the article. Brews ohare (talk) 04:33, 4 May 2009 (UTC)

P(x) involves an average over a small volume centered at x. (For example, a 2nm-radius sphere centered at x.) When you change x, you're taking an average of dipole moment density over an entirely different small volume. So the sum is not a constant independent of position, whether or not the system is charge-neutral. There are no restrictions on P, it can be any vector field in the world, as long as I set up the appropriate molecular dipole moments at each point. --Steve (talk) 14:55, 4 May 2009 (UTC)

Steve: I get that. The point is that when you divide into cells of dipoles, the definition of dipole moment used provides a p for each cell that is a property of that cell and is not dependent upon position anywhere in space. Thus you get no P(r) from the p 's themselves. Where you do get a P(r) is from the boundary regions where each cell transitions to the next. It is just like the dielectric sphere problem where nothing happens except at the boundary. Brews ohare (talk) 15:08, 4 May 2009 (UTC)

Sorry, I'm confused. :-) Here's the definition, again:

"$\mathbf {P} (\mathbf {x} )=\sum _{i}N_{i}\langle \mathbf {p} _{i}\rangle$

where p_i is the dipole moment of the ith type of molecule in the medium, the average is taken over a small volume centered at x and N_i is the average number per unit volume of the ith type of molecule at the point x."

This is a function of one variable, x (assuming time-independent). What you said above is "this sum is a constant independent of position, no matter where you are (inside or outside the charge array)." It sounds to me like you're saying "P(x) is a constant independent of x". Is that right? If not, what do you mean? What quantity is a constant with respect to what variable? Thanks! :-) --Steve (talk) 19:28, 4 May 2009 (UTC)

What makes p = Σ nd different from p(r)?

A quote from the article:

A very common model is to imagine a medium made up of an assembly of pairs of opposite charges, a typical pair having an individual dipole moment denoted by p _i. For such a medium, the dipole moment of a volume of this material is:

${\boldsymbol {p}}({\boldsymbol {r}})=\sum _{i=1}^{N}\,\int q_{i}\left(\delta (\mathbf {r_{0}} -\mathbf {r} _{i}+{\boldsymbol {d_{i}}})-\delta (\mathbf {r_{0}} -\mathbf {r} _{i})\right)\,({\boldsymbol {r_{0}}}-{\boldsymbol {r}})\ d^{3}{\boldsymbol {r_{0}}}$ $=\sum _{i=1}^{N}\,q_{i}\left({\boldsymbol {r_{i}+d_{i}}}-{\boldsymbol {r}}-({\boldsymbol {r_{i}}}-{\boldsymbol {r}})\right)=\sum _{i=1}^{N}q_{i}{\boldsymbol {d}}_{i}\,=\sum _{i=1}^{N}{\boldsymbol {p}}_{i}\,$

As already noted, because of overall charge neutrality, this dipole moment is independent of the observer's position r.

That is to say, if the array is divided into neutral sub-cells, each sub-cell has a unique p that does not depend on any position variable. All the dipole moments do is provide a table of p-values. You have to add the location info to the dipole moment, it is not included automatically. For example, if you choose to model the array as an assembly of dielectric ellipsoids, that act of confining each cell to a specific ellipsoidal location introduces into the polarization a surface charge with a specific location. Its value depends upon the dipole moment of the cell, but what gives it a non-zero div is the step function at the boundary. Brews ohare (talk) 20:05, 4 May 2009 (UTC)

OK, to rephrase, I think you're saying something like: Dipole moment is

$\int \rho (\mathbf {r} )\mathbf {r} d\mathbf {r}$

As long as $\int \rho (\mathbf {r} )d\mathbf {r} =0$ (the volume is overall charge-neutral), the dipole moment is independent of the origin of the coordinate-system for r. Yes, I agree. Now that we've agreed on that, we can forget about the "origin of the coordinate-system" for the rest of the conversation. We're just talking about charge-neutral dipoles. The only position coordinate that is at all relevant is the position of the dipoles in space. There's no hidden secret extra position coordinate that is involved in how "dipole moment" is defined. So can we drop all mention of this extra position coordinate from now on? It's not relevant, so we can forget it and move on, OK? --Steve (talk) 17:52, 5 May 2009 (UTC)

Is an array of point dipoles all we need? And if so, what array is best?

It isn't just a question of the origin from which the dipole moment is calculated. I am glad we agree about that. The issue is the functional dependence of what you call "the position of the dipoles in space". I get the idea that you just look and see what this is. However, its more complicated than that.

It seems that there is some opinion to the effect that any neutral charge array can be represented by some array of point dipoles. See this. That seems tautological, almost, but how is the array found, and how many do you need? It does seem necessary in general to be able to model details of a charge array beyond the dipole, and some authors don't use dipole arrays, but use expansions of the polarization into multipoles. See Brevet. If an array of point dipoles is always sufficient, another question is whether the array found using D = κε₀ E generates the necessary array. I have tried to write the article to encompass these questions without going into any detail about them. 19:53, 5 May 2009 (UTC)

Multipoles of P

BTW, Jackson is on thin ice here if you interpret him correctly. The standard view is the Maxwell relations stand with P and determine the field behavior, and it is the condensed matter people that have to approximate P properly. Brews ohare (talk) 07:27, 6 May 2009 (UTC)

I find that may authors simply treat P as a dipole density. An extreme case is in Chen & Kotlarchyk, who take as an example:

${\boldsymbol {P(r}},\ t)={\boldsymbol {p}}(t)\delta ({\boldsymbol {r}})\ ,$

for which I'd say div P is not even defined!

More generally, they introduce multipoles of P using the Hertz potential:

$\nabla ^{2}{\boldsymbol {{\Pi }_{e}}}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}{\boldsymbol {{\Pi }_{e}}}=-4\pi {\boldsymbol {P}}$

with solution:

${\boldsymbol {\Pi _{e}(r,}}\ t)=\int _{V}d^{3}r_{o}{\frac {{\boldsymbol {P}}\left({\boldsymbol {r_{o},}}\ t-{\frac {|{\boldsymbol {r-r_{o}}}|}{c}}\right)}{|{\boldsymbol {r-r_{o}}}|}}\ .$

Fourier transforming in time:

${\boldsymbol {\Pi _{e}(r,}}\ \omega )=\int _{V}d^{3}r_{o}{\frac {e^{i\omega |{\boldsymbol {r-r_{o}}}|/c}}{|{\boldsymbol {r-r_{o}}}|}}{\boldsymbol {P(r_{o}}},\ \omega )\ ,$

and a Bessel function expansion of the 1/|r - r_o| introduces the multipoles of P. Brews ohare (talk) 16:52, 6 May 2009 (UTC)

Here's another multipole of P example Huang & Shen, pages 12-13. These authors adopt the view P = χ E, but they confine attention to within the polarizable medium, so we don't know if what happens outside ever is of interest. 18:19, 6 May 2009 (UTC)

Stationary, uniformly-polarized sphere sitting in free space

I feel like you didn't really answer my question. There are infinitely many possible {D,P} consistent with those equations. You picked one, {D_Brews, P_Brews}. Why that one, and not any other?

In the example of the isolated, stationary, uniformly-polarized sphere sitting in free space, I think that's a complete description of the problem and should have a unique solution. How do you know that {D_Brews, P_Brews} is the one? What exactly are the boundary conditions that {D_Brews, P_Brews} satisfies, that every other possible {D,P} does not satisfy?

Thanks! :-) --Steve (talk) 03:14, 4 May 2009 (UTC)

That example is found in the article and in many textbooks (some are cited). The standard solution is that the electric field is the sum of two parts: the applied uniform E-field and the depolarization field due to the polarization charge. This polarization charge is confined to the surface of the sphere, being a thin positive shell on the most distant hemisphere and a thin negative shell on the other. From a strictly simple-minded viewpoint, the dipoles filling the sphere head-to-tail just cancel each other out inside, producing a zero charge region. Their tails stick out on the boundary making the polarization charge. Thus the "bound charge" is this surface charge, as there is no other charge present.

The value for the polarization charge is arrived at several ways.

• Discontinuity condition in the electric field at the surface, arrived at by saying D_n is continuous and κ is not, so E_n is discontinuous. You can view this as saying there is no dipole moment outside and there is inside.
• Using the divergence theorem to convert the integral of div D to a surface integral over p·dA, with p the dipole moment inside the sphere.
• Defining P = χ(r)E throughout all space and using the Heaviside step function in χ(r) to obtain the surface charge. (div P has a δ-function at the surface, resulting in the identification of the surface charge as "bound charge".)

The last mentioned is the approach I've adopted. From its standpoint, the polarization field in Maxwell's equation, P(r), is given by the dipole moment density p(r), which is the same as the spatially constant dipole moment inside the sphere p, but in addition contains the step function confining the dipole moment inside the sphere. It is this step function that yields the surface charge. In a more real model, this step function becomes a real property of the material, describing how it transitions from bulk to vacuum.

All methods are equivalent in mathematical terms. Brews ohare (talk) 03:54, 4 May 2009 (UTC)

If you use P(r) = χ(r)E(r) throughout all space and use the Heaviside step function in χ(r), you'll get P(r)=0 when r is outside the sphere, no? --Steve (talk) 15:06, 4 May 2009 (UTC)

Yes, unless the step is to another material. The fact that P drops to zero in this particular model of medium does not establish that P must be zero outside every medium in every approximation. Brews ohare (talk) 15:46, 4 May 2009 (UTC)

I believe in practical terms what goes on is this: one knows the real charge array has way more detail than you need. You know that one way to filter out the detail is to stand at some distance outside the array and make a multipole expansion. You know that if all you want as detail is the dipole contribution, you can drop the rest of the expansion. The resulting dipole model is great outside the array.

Next you adopt this dipole formulation everywhere. The result inside the region containing the array is problematic, but provides about all the info we can handle anyway, so we assume:

$\phi ({\boldsymbol {r}})={\frac {1}{4\pi \varepsilon _{0}}}\int {\frac {{\boldsymbol {p}}({\boldsymbol {r}}_{o}){\boldsymbol {\cdot (r-r_{0})}}}{|{\boldsymbol {r}}-{\boldsymbol {r}}_{o}|^{3}}}d^{3}{\boldsymbol {r}}_{o},$

everywhere. A supplemental Laplacian φ looks after boundary conditions like applied fields. The dipole moment density p(r) is the dipole moment itself supplemented with location information, for example a step in the case of simple geometry. This density satisfies:

${\boldsymbol {\nabla \cdot p(r)}}=\rho _{b}\ ,$

where ρ_b includes surface charges at interfaces. Outside the array we recover the starting point dipole term in the potential. Inside all we need is the usual linear response theory approach to susceptibility. I believe that encompasses most of what passes for "polarization". Brews ohare (talk) 17:18, 4 May 2009 (UTC)

I'm sorry, can you say again what the difference is between p(r) and P(r)? What's the definition of each? Thanks! :-) --Steve (talk) 19:50, 4 May 2009 (UTC)

Hi Steve: I'm using p(r) to refer to the dipole moment density, which is a combination of the dipole moment of an array Σn_id_i plus information locating that array, which attaches a position to the dipole moment. I'm using P(r) as the entity entering Maxwell's equation when it's time to calculate the electric field throughout the solution region. Depending upon how much detail about the field is needed, greater or lesser information is required in P(r). For example, it may be that all that is needed is p(r), and we can take P = p. (That is not the same as taking P =Σn_id_i, because P must have location information.)

However, it may be that more is demanded of the field calculation, and more detail is needed in P(r). Then P = p is too inaccurate. One might try to spruce things up using P = p_d + p_Q + …. Or, one might go further and introduce arrays of dipoles and quadruples on a lattice of sites. That seems to be the local field theory methodology used in lattice dynamics, for example. Brews ohare (talk) 20:39, 4 May 2009 (UTC)

I understand what you're saying...if P=p, then quadrupoles, etc. are not part of P. Still, you have to agree that all the textbooks define P=p by definition. So where did the quadrupoles go?

OK this is interesting. In Jackson Section 6.6 (mentioned above...sorry you don't have it...this is the section where he starts with the microscopic equations and derives the macroscopic equations), is the following:

"...this means that the macroscopic displacement vector D is defined to have components,

$D_{\alpha }=\epsilon _{0}E_{\alpha }+P_{\alpha }-\sum _{\beta }{\frac {\partial Q_{\alpha \beta }'}{\partial x_{\beta }}}+\cdots$

The first two terms are the familiar result. The third and higher terms are present in principle, but are almost invariably negligible."

(Q is the "macroscopic quadrupole density", which is defined earlier as a small-volume average of molecular quadrupole moment density). So the approach of Jackson is to say P=p by definition, but D=ε0 E+P is merely a very good approximation that neglects the quadrupoles, etc.

On the other hand, Beresnev is a reference that says P=p is merely a very good approximation that neglects the quadrupoles, etc., while D=ε0 E+P is exact.

So apparently the precise notational way to add quadrupole, etc. corrections is not universally agreed.

In the meantime, can we agree that P=0 in vacuum, always? P(x) definitely depends on the average dipole moment density in the immediate vicinity of x, and arguably depends on the average quadrupole, etc. moment density in the immediate vicinity of x. But I haven't seen any definition of P where P(x) depends in any way on the charge-distribution far away from x. --Steve (talk) 06:26, 6 May 2009 (UTC)

Hi Steve: If we look at the dielectric sphere as an example, there is a uniform depolarizing field throughout the inside of the sphere given by :

${\boldsymbol {P}}=-\left({\frac {\kappa -1}{\kappa +2}}\right){\boldsymbol {E}}_{\infty }.$

There is a dipole moment throughout the inner sphere of:

${\frac {\boldsymbol {p}}{V}}={3}\varepsilon _{0}\left({\frac {\kappa -1}{\kappa +2}}\right){\boldsymbol {E_{\infty }}}=-3\varepsilon _{0}{\boldsymbol {P}}\ .$

So one observation is that the polarization inside the polarization medium is closely related to the dipole moment. That agrees with your expectations.

However, because the fields and the dipole moment are constant, their divergences all are zero inside the sphere.

So I'd take it that the polarization P in Maxwell's equations satisfies div P = ρ_b and ρ_b is zero inside the polarization region.

How about outside the sphere? Here the electric field is not uniform because of the surface charge distribution. There is no bound charge, ρ_b is zero. It seems we have two choices:

(i) The field distortion is due to D, which satisfies div D = 0 but has to make up for the boundary conditions. The polarization field P is identically zero outside the sphere. D= ε₀ E outside and E contains the effects upon the exterior region of the bound charge on the sphere surface; div D = ε₀ div E. The charge on the sphere surface might be bound or might be free. It depends on how the handover from P to E is handled.

(ii) The polarization field outside the sphere continues to express the effect of the bound charge, and D = ε₀ E + P, but E contains only the external applied field, so div D = div P. The charge on the surface is "bound" charge and div P ≠ 0 on the surface. The polarization field P does exist in the vacuum because it expresses the effect of the bound charge, wherever that effect may appear.

I prefer version (ii). It seems capricious to me to call the surface charge "free" charge (it is still a charge on the dipoles that are near the surface) and to switch the interpretation of E when you switch from the interior (where it is the applied field) to the exterior (where it suddenly recognizes there is surface charge around).

This example involves a boundary condition, which may muddy the water. But what happens if the transition is gradual at the boundary, so the bound charge just tapers off. I'd say P applies until you leave the bound charge region. Then you have to decide whether you will allow it to continue to express the bound charge outside, or make other arrangements? How close to the edge can you get before P is relieved of duty and E takes over? How is the transitional point established? How do you interpret matters?

It seems a cleaner method to me to use P everywhere to express the bound charge influence. After all, the approximations used in determining P still affect the solution in the exterior region. Why pretend to have switched to a more kosher E approach in the outer region when you really haven't changed you solution methods at all? It also fits in well with superposition: just split the problem up into the free and bound charge problems and add them up at the end. Brews ohare (talk) 07:44, 6 May 2009 (UTC)

Hold on...you wrote that there's a "uniform depolarizing field throughout the inside of the sphere given by

${\boldsymbol {P}}=-\left({\frac {\kappa -1}{\kappa +2}}\right){\boldsymbol {E}}_{\infty }.$"

I agree that there's a uniform depolarizing field, given by the RHS. But I disagree that you can call it P. Are you saying: "P is by definition the depolarization field which arises due to bound charge"? Is that what you believe?

As I'm sure you know, calculations in literature say that the value of P in the sphere is different than this, by a factor of 3. --Steve (talk) 05:24, 7 May 2009 (UTC)

Hi Steve: A factor of three is the least of things. There are always geometric effects, and this may be one of them. The real problem in my concept of things is how to reconcile the notion that P is the consequence of a division into bound and free charge, yet inside the boundary one is happy to take this contribution as P while outside it it looked after by E. Why is the contribution of bound charge to a field limited to inside? Fields in dielectric cross boundaries. Apparently P is not itself a field , although its divergence is a source that gives rise to a field?

$\phi ({\boldsymbol {r}})=-{\frac {1}{4\pi \varepsilon _{0}}}\int {\frac {{\boldsymbol {\nabla _{\boldsymbol {r_{o}}}\cdot }}{\boldsymbol {P}}({\boldsymbol {r}}_{o})}{|{\boldsymbol {r}}-{\boldsymbol {r}}_{O}|}}d^{3}{\boldsymbol {r}}_{o}\ .$

Brews ohare (talk) 05:36, 7 May 2009 (UTC)

You're not worried about a factor of 3, but spend time worrying about infinitesimal quadrupole corrections!? :-)

Yes, in the definition of P that I favor, P is not an electric field, and has no simple relation to any real electric field. P is more or less a description of the bound charge. --Steve (talk) 06:50, 8 May 2009 (UTC)

Polarization and dipole moment

There are many authors (possibly none too definitive) that support your view:

Parker "The polarization P denotes the total dipole moment per unit volume at the position r at time t:

${\boldsymbol {P}}={\frac {no.\ dipoles}{vol}}{\boldsymbol {p}}\ .$

Van Rienen "The polarization describes the vector sum of all dipoles with respect to the volume in the presence of exterior fields (macroscopic dipole density of the material system)"

Wolf "the displacement field D = ε_∞E + 4πneu, where the dipole moment per unit volume associated with the relative displacement of the sublattices is P_L = neu."

Cohen "P= electric dipole density"

However, I am inclined to take Stratton as more influential source. He says: Stratton "We define the electric and magnetic polarization vectors by the equations:

${\boldsymbol {P=D-\varepsilon _{0}E}}\ ,$

The polarization vectors are thus definitely associated with matter and vanish in free space." "the presence of rigid material bodies in an EM field may be completely accounted for by an equivalent distribution of charge of density -∇ · P, and an equivalent distribution of current ..."

I'm prepared to accept all this except the statement about vanishing in vacuum. To me that is a discretionary decision not required by the definition given. Later on p. 184 Stratton says (ω is surface charge):

$\phi ={\frac {1}{4\pi \varepsilon _{0}}}\left(\int dv\ {\frac {\rho }{r}}+\int \ da{\frac {\omega }{r}}+\int \ dv\ {\boldsymbol {P\cdot \nabla }}\left({\frac {1}{r}}\right)\right)\ .$

"the third integral is clearly the potential produced by a continuous distribution of dipole moment and P is, therefore, to be interpreted as the dipole moment per unit volume, or polarization of the dielectric. ..." Brews ohare (talk) 23:13, 6 May 2009 (UTC)

Add Jackson, Griffiths, and Feynman. And what does Brews Ohare believe? You keep referring to "the P in Maxwell's equations". I don't know what that means. Can you write down the definition of P that you think is the correct one? --Steve (talk) 05:32, 7 May 2009 (UTC)

The correct definition is Stratton's

${\boldsymbol {P=D-\varepsilon _{0}E}}\ ,$

One way to phrase the issue is: "Why should P be P= Σq_ip_i?" Among other problems, what prescription is added to avoid div Σq_ip_i ≡ 0, so as to provide a proper spatial dependence? (In the case of the dielectric sphere that seems to be a step function at the sphere surface, not some interior volume dipole distribution, which physically would be a constant.)

I suspect the real origin of this Σq_ip_i connection is in the quantum formulation of Maxwell's equations which never uses P but does introduce a source term due to the interaction of matter and EM field that in some approximation is the dipole moment operator appearing as a source term in Maxwell's equations. Basically, [j, ρ] become expectation values of quantum operators in Maxwell's equations, and these expressions can be reinterpreted using M P by comparing the quantum equations to the classical Maxwell equations. See Agranovitch; Cartwright, Scully, Meystre. What this means is that the classical attempts to "interpret" P are superseded. The "top-down" approach is replaced by a "bottom-up" approach. Brews ohare (talk) 14:38, 7 May 2009 (UTC)

OK, you define P by P=D-ε0 E. How do you define D?

[I'm guessing your answer will be "by Maxwell's equations". The problem is that the Maxwell's equations with E,B,D,H are 8 scalar equations in 12 unknowns, so this is not sufficient to be a definition for D. So again, what is your definition of D?] :-) --Steve (talk) 06:44, 8 May 2009 (UTC)

Hi Steve: Well, I guess D is a garden variety electric field that satisfies div D = ρ_f? If that is so, it's hard to understand how P = D − ε₀E is not also such a field. My view, which seems to be all my own, is that in fact P is an electric field that satisfies div P = −ρ_b. So far as I can determine that is consistent with division into free + bound charge, does not get involved in a ridiculous changing interpretation as you cross a dielectric boundary, very naturally leads to the use of multipoles, and agrees entirely with the customary relation to the dipole moment interior to a polarizable medium. But the standard expressed view is that P is a completely different phenomena in the way of fields that is coextensive with its source, and that may or may not satisfy div P = −ρ_b (at the user's discretion, it seems: e.g., P = pδ(r), which has an undefined divergence, is just fine; a view I find nutso and self-contradictory).

I suspect the customary view is a simple outgrowth of confusing two usages: polarization field P and the dipole moment density Σnd/Ω, which often also is referred to as the polarization density. Sometimes they are the same, but (in my view) not always and not everywhere. Brews ohare (talk) 13:19, 8 May 2009 (UTC)

You say: "D is a garden variety electric field that satisfies div D = ρ_f". This isn't a complete definition. There are infinitely many D that satisfy div D = ρ_f, even with the same boundary condition. So, how do you define D?

Are you unfamiliar with the meaning of a derivative of the delta function? The divergence of pδ(r) is perfectly well-defined, or at any rate it's just as well-defined as pδ(r) itself. It's a mathematical distribution, or a limit of functions, whatever you prefer.

The two usages "polarization field" and "dipole moment density" are confused because they are exactly the same thing, always and everywhere, by definition! :-) Unless you want to include the infinitesimal quadrupole corrections (arguably). You still haven't given any alternative definition of P, except in terms of D which you also haven't defined. --Steve (talk) 15:14, 11 May 2009 (UTC)

Hi Steve: Thanks for your conversations over this topic. So far as I am concerned, it is a dead issue. There are gaps to be filled in from a conceptual standpoint, but from the view of the actual implementation there is unanimity. Brews ohare (talk) 13:54, 12 May 2009 (UTC)

Removed material

I've removed material that does not find general support in the literature, though I am unhappy that I haven't got a clear picture of what should be done. Maybe at a future date a quantum mechanical dipole section could be added. Brews ohare (talk) 03:35, 8 May 2009 (UTC)

"Cauchy-type integral"?

In Sokhatsky–Weierstrass theorem, you linked to "Cauchy-type integral". It's a red link and I don't know what you meant by it. Could you have meant contour integral (line integral in the complex plane)? Or maybe Cauchy principal value? I can't help suspecting there's some reasonable redirect target. Maybe Cauchy-type integral should become a "redirect with possibilities"? Michael Hardy (talk) 21:30, 1 June 2009 (UTC)

I don't remember...if I ever knew in the first place. I may have just copied the phrase out of a book without fully understanding it. Perhaps it should be deleted. Sorry! --Steve (talk) 04:36, 2 June 2009 (UTC)

It all came down to an archaic definition of electromotive force

Steve, I'm sorry to see that you felt the need to speak against me on the administrator's noticeboard. If I remember correctly, our dispute was over the question of the underlying unity of the two aspects of Faraday's law of induction. (the convective aspect and the local aspect).

The essence of that dispute lay in the fact that my argument hinged on using an archaic definition of electromotive force which effectively equated to using the relationship F/q = E for the case of the vXB term in the Lorentz force. That dispute ended with an amicable compromise in which I accepted that such a relationship did not conform to modern textbook usage, and you accepted that the deeper underlying principles of my argument were correct and were contained at equation (D) in Maxwell's 1865 paper. We agreed on a final way to write the equation in the introduction. I thought that both of us learned alot from that debate. I was made more explicitly aware of the fact that terminologies had changed since the time of Maxwell. It's not that I wasn't acquainted with modern terminologies from my university days, but I had been recently very engrossed in Maxwell's original papers.

Once again it was a case of trying to work between different sources which emphasized things differently. David Tombe (talk) 21:05, 14 July 2009 (UTC)

Hi David, I want to make things clearer myself:

• Yes, you are a fringe physicist. A fringe physicist is someone who believes things that essentially every physicist thinks is false, and/or disbelieves things that essentially every physicist thinks is true. For example, you don't believe the Biot-Savart law is true. You don't believe in gauge transformations. You don't believe that planetary orbits can be fully explained and quantified without mentioning the centrifugal force. You do believe in the aether, and don't believe in special relativity. You are a fringe physicist. I suspect you already know this. If you don't, try submitting one of your publications to Physical Review Letters and you'll see what I mean. :-)

• Being fringe doesn't automatically mean being wrong. I know enough history of science to know that lots of true things started out as fringe theories. The first people who said they saw rocks falling from the sky were dismissed as fringe astronomers, but now everyone knows that meteorites are true things. I respect that you're thinking hard about these issues and publishing articles on the internet. Not all fringe physicists are wrong. (Most are, and I happen to think that you are, but who knows really.)

• Being fringe does mean steering clear of editing wikipedia. The philosophy of Wikipedia is that it's not the truth that's posted, it's today's agreed-upon knowledge. For example: I happen to hold a fringe opinion about repetitive strain injury (RSI). I believe, with good justification, that the standard medical understanding of RSI is completely wrong. But I haven't rewritten the wikipedia RSI article, or any related article, to reflect the truth. Why not? Because I think it's unethical to write something I know is false, and that it's equally unethical to write something on wikipedia that disagrees with the current medical understanding. Instead, I've posted my view of RSI on my personal website, and I send the link to people I know whenever possible.

It's not a problem that you have a fringe understanding of electromagnetism, centrifugal force, or whatever else. It is a problem that you have edited physics articles on wikipedia. Everything you read and write about physics is through the lens of your fringe understanding, and of course you're in a terrible position to judge which of your views are fringe views and which aren't. The appropriate solution is to stop editing physics articles on wikipedia. Instead you should keep using other venues to advocate your understanding. In 50 years, maybe everyone will recognize David Tombe as the great physicist who got us out of a dark age in physics. At that point, of course wikipedia could be changed.

The comparison with aliens and JFK may have come across in the wrong way. I didn't mean, David shouldn't edit physics articles because he's a crazy lunatic. You're not a crazy lunatic. I meant, David shouldn't edit physics articles because he has a fringe point-of-view that will inevitably keep him from contributing productively. --Steve (talk) 23:43, 14 July 2009 (UTC)

Steve, Thanks for that advice. Yes, you are correct that I hold some fringe views. That is the consequence of having been heavily involved in physics research for years. But I am fully aware of wikipedia's rules on original research and I have been very careful not to put original ideas on the articles. I was a bit upset when Martin Hogbin tried to suggest that I had an ulterior motive in emphasizing the inverse cube law in the centrifugal force. He drew attention to my paper that attributed this fact to a dipole field. But the truth is that I started the debate on centrifugal force, using IP servers in 2007, before I even hit on that idea.

People might suggest that somebody who is heavily involved in research might be using wikipedia to promote their findings. But that does not necessarily follow. For example, I have a few side interests. Fifteeen years ago I did extensive research on the British Empire and the currency systems in the Empire. I made alot of notes which I have kept. They are badly sorted and I have never published any of my findings in this field. But when I stumbled across some related articles on wikipedia, I felt compelled to correct those articles for the benefit of the readers. The most recent example was my edits on the Unfederated Malay States. If I see something wrong involving specialized knowledge, I want to correct it.

Likewise with centrifugal force. I did alot of research in 2006 which led me to link hydrodynamics to the Lorentz force. This all linked up with centrifugal force and the grad (A.v) term. I studied Maxwell's idea about centrifugal pressure arising between vortices and causing magnetic repulsion. I was fascinated with this because I had studied planetary orbital theory many years earlier and I had always been puzzled about the significance of the terms in the radial equation. In one of my early writings, I explained the repulsion that would arise between two adjacent orbital systems in terms of a positive value for the r(double dot) term if the mutual transverse speed was high enough, but I fell short of explicity seeing it as centrifugal force. Just like you, I had been trained at university to believe that centrifugal force doesn't exist. But with the discoveries in hydrodynamics, it all fell into place and I boldly adopted Maxwell's approach and called that pressure by its real name ie. 'centrifugal force'. Previously I had worried that Maxwell's use of the centrifugal force concept might be a weakness in his arguments. I could always recall my physics lecturer with the circle drawn on the blackboard as he pointed along the tangent explaining that only centripetal force is needed to keep it in a circle. So why was the great Maxwell using a fictitious concept to explain a real magnetic pressure?

At any rate, when I then saw the wikipedia article on centrifugal force, I noticed that it didn't cater for the outward push effect of the rw^2 term in the planetary orbital equation. Hence the edit war began. I felt that the existing article was too incoherent and disjointed. It split the topic into two aspects, neither of which I thought helped to make the concept fully understood. My motive in getting involved was genuinely for the purpose of tidying the article up. The real outward push effect had been sidelined to 'reactive centrifugal force' in the restricted case of circular motion, and the rotating frames approach didn't give any recognition to the real outward push effect. I saw the entire topic as falling under the single effect of that inverse cube law term in the planetary orbital equation. I simply wanted to bring unity and clarity to the readers, and of course I was quite prepared to explain it in my own style as a teacher always does. Sources do exist to back up my viewpoint but unfortunately I didn't have them to hand when I started editing.

I could understand the opposition or the accusations of original research if I had been attempting to claim that space is filled with tiny electric dipoles, and then proceeding to fill up the reference section with my on-line papers. I do do alot of research, but when it all boils down to it, who else ever bothers to edit on a topic that they haven't studied in depth themselves? You are of course correct that somebody involved in research may occasionally step over the line a little bit by accident, having forgotten the difference between their own opinions and what is in the textbooks. But I don't think that I have done that very much.

I have examined alot more wiki physics articles than I have actually edited on. Many of these I have simply walked away from on the grounds that any attempts to fix them would clearly cross the line and result in a futile editor war in which I wouldn't have a leg to stand on under wikipedia's rules. And as such I have no interest in touching topics which I am diametrically opposed to. But I don't think that I crossed the line on centrifugal force. On Faraday's law, it was border line. Sources were just about available. The Maxwell sources were interesting in that they were not primary as such (in relation to Faraday's laws), but they contained ideas that were not contemporary. Nevertheless, the final point gave a perfect unity to Faraday's law, with the singular problem being that we had to write E = F/q. This is a true statement in itself for two of the terms, but not recognized in modern textbooks for the vXB term. So the idea passed for the Lorentz force page but had to be dropped on the Faraday page because the concept of taking the curl of F/q was alien unless we could legitimately write it as E. That's what I call a border line edit war as regards being on the right side of the law.

By the way, as you will recall, I didn't actually put my objections to the Biot-Savart law unto the main article. And while we're on that subject, you might as well get the one that followed directly as a result of our debate on that page [[7]]. That attempts to deal with the dilemma that I had raised regarding the singularities in the solenoidal field lines. Biot-Savart is quite a tricky one. I wouldn't at all be surprised if it turned out to be inverse cube law. But that's only a guess based on other reasons such as the dipole field and the centrifugal pressure. David Tombe (talk) 01:42, 15 July 2009 (UTC)

Invitation

I invite you to read my comment (No. 48) to "Faraday's Law of Induction." Mike La Moreaux (talk) 02:22, 13 September 2009 (UTC)

Speed of light arbitration evidence

In accord with the edit notice ([8]), a portion of your evidence submission has been moved to the talk page.[9][10][ http://en.wikipedia.org/w/index.php?title=Wikipedia_talk:Arbitration/Requests/Case/Speed_of_light/Evidence&diff=313528108&oldid=313528006] None of the information has been deleted. Everything moved from the main evidence page is entirely preserved on the talk page. It is possible that the moved portion may require more supporting links to be appropriate for evidence or that the portion moved is simply more commentary than evidence. Please review the moved portion to decide if it needs revision as an evidence submission or if it should remain on the talk page as commentary. If you feel that this contribution was moved in error, please feel free to contact me to discuss the matter. Thank you for your understanding. Vassyana (talk) 07:48, 13 September 2009 (UTC)

Re topic ban length on arb case workshop

Hey Steve, Though I appreciate your concerns about David Tombe's pattern of pushing original research, Arbcom has traditionally restricted itself to sanctions lasting for a maximum of 1 years, hence the time limit on my proposed remedy. TotientDragooned (talk) 09:55, 13 September 2009 (UTC)

Oh. Thanks for letting me know. I think that tradition doesn't make much sense here. But I guess we'll see what happens. :-) --Steve (talk) 17:31, 13 September 2009 (UTC)

Well said

I don't think we've ever interacted, but after reading this, you have my hearty appreciation and my vote for ArbCom anytime. I think you've encapsulated a vexing issue perfectly. MastCell ^Talk 00:14, 15 September 2009 (UTC)

Gee, thanks. :-) --Steve (talk) 00:22, 15 September 2009 (UTC)

Departure

Hi Steve: I know we haven't agreed on many occasions, but I have always felt that your interest was in correct material and good exposition. Unfortunately, you are an exception, and my experience at Speed of light has introduced me to a scurrilous bunch capable of distortion, gossip, rumor mongering and no interest in WP. So I am out. Thanks for your interaction. Brews ohare (talk) 03:23, 15 September 2009 (UTC)

Well thanks, that really means a lot.

I'm sure you'll accomplish a lot with the countless hours you'll be doing "real world" productive stuff instead of wikipedia. God only knows how productive I would be without wikipedia. In fact I've been trying to move that direction myself! :-) --Steve (talk) 03:55, 15 September 2009 (UTC)

The Biot-Savart Law

Steve, You misrepresented me at the arbitration hearing in relation to the Biot-Savart law. The Biot-Savart law is a tricky one. The physical situation that it is attempting to describe is perhaps too complicated for it to be adequately described mathematically. The result that we have in the textbooks is probably the best that can be done in the circumstances. There is a problem reconciling the singularities with the solenoidalness within the mathematical language that is used.

Taking the curl of the Biot-Savart law leads to Ampère's circuital law, which I have no problem with. The important thing is to see the simplicity of the parallel between the Lorentz force to Faraday's law, and the Biot-Savart law to Ampère's circuital law. Seeing that pattern will help no end in being able to present these articles in a coherent fashion for the benefit of the reader. David Tombe (talk) 10:05, 22 September 2009 (UTC)

Faraday's Law

Steve, you have misstated my original opinion. The version of Faraday's Law which is identical to the one of Maxwell's Laws responsible for transformer emf is, of course, true. The version which I thought that we had been discussing all along, the statement of which is at the beginning of the article, is false. Mike La Moreaux (talk) 20:19, 25 September 2009 (UTC)

The terminology is unfortunate. When I say "Faraday's law", I mean a statement relating the magnetic flux through a circuit to the total EMF around that circuit (including motional EMF). That's what I mean when I say "Faraday's law". I'm not talking about the one of the Maxwell's equations, which doesn't include moving circuits and motional EMF.

I'm also not saying the article does or does not state Faraday's law perfectly correctly. I'm saying it can be stated perfectly correctly. This involves some work to define the term "circuit" in a certain specific way, a way that blocks out Feynman's counterexamples and makes the law amenable to the rigorous mathematical derivations (starting from the Lorentz force and Maxwell's equations) which I have shown you but which I guess you don't believe.

OK, now that I've clarified this, please re-read what I wrote on the talk page, maybe you'll understand what I meant better. Thanks! :-) --Steve (talk) 02:34, 26 September 2009 (UTC)

Steve, you have done a great job of stating your position, and I believe it really clarifies things and was necessary. As I said before, you can define Faraday's Law as properly applying to all cases for which it works. The article does not make this clear at all. I believe that the general definition at the beginning of the article is the only useful one. Any other may well be true and provable but merely of academic interest. Mike La Moreaux (talk) 20:24, 26 September 2009 (UTC)

Speed of Light

Steve: You probably are aware that Speed of light is in turmoil right now with an arbitration request in progress. That might give you pause about contributing to the Talk page for this article. However, I think you might have some good ideas concerning this section, and I hope you might take a look at it. Brews ohare (talk) 15:51, 10 September 2009 (UTC)

In my opinion you didn’t anderstand the Devid arguments. So I remind you:

Resolution 1 of the 17th meeting of the CGPM (1983)

The 1889 definition of the metre, based on the international prototype of platinumiridium, was replaced by the 11th CGPM (1960) using a definition based on the wavelength of krypton 86 radiation. This change was adopted in order to improve the accuracy with which the definition of the metre could be realized, the realization being achieved using an interferometer with a travelling microscope to measure the optical path difference as the fringes were counted. In turn, this was replaced in 1983 by the 17th CGPM (1983, Resolution 1, CR, 97, and Metrologia, 1984, 20, 25) that specified the current definition, as follows: The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. It follows that the speed of light in vacuum is exactly 299 792 458 metres per second, c0 = 299 792 458 m/s.

This EXACT values were “obtained” from the sources:

Comptes Rendus de la 17e CGPM (1983), 1984, 97

Metrologia, 1984, 20(1), 25

speed of light in vacuum Value 299 792 458 m s-1 Standard uncertainty (exact) Relative standard uncertainty (exact)

magnetic constant 4pi x 10-7 = 12.566 370 614... x 10-7 N A-2 Standard uncertainty (exact) Relative standard uncertainty (exact)

electric constant Value 8.854 187 817... x 10-12 F m-1 Standard uncertainty (exact) Relative standard uncertainty (exact)195.47.212.108 (talk) 12:12, 20 October 2009 (UTC)

Resolution 7 of the 9th meeting of the CGPM (1948) defined the Ampere, and therefore the magnetic constant.

Ampere law:

$F_{M}={\frac {\mu _{0}I^{2}l}{2\pi a}}={\frac {\mu _{0}}{2\pi }}=2\cdot 10^{-7}\$N

where $a=1m$ is the distance from conductor wire to the point of measurement, $l=1m$ is is the wire length, and $I=1A$ is the current strength. So, the magnetic constant is:

$\mu _{0}=4\pi \cdot 10^{-7}$ N/A

Note that the relationship

$c={\frac {1}{\sqrt {\mu _{0}\epsilon _{0}}}}$

is dropped up from the blue sky (an AXIOM), since it isn't derived from Maxwell equations.195.47.212.108 (talk) 12:36, 20 October 2009 (UTC)

Therefore, there are too MANY "exact" fundamental constants, which leads to contraditions. For example, in the case of characteristic impedance of free space

$\rho _{0}={\sqrt {\frac {\mu _{0}}{\epsilon _{0}}}}=2\alpha {\frac {h}{e^{2}}}$

where $\alpha$ is the fine structure constant (NOT exact!), $h$ is the Planck constant (NOT exact!), and $e$ is the charge of electron (NOT exact!). Thus, how the sets of NOT-exact values produces the EXACT ones??? Such nonsence could be imagined by METROLOGISTs only...195.47.212.108 (talk) 12:49, 20 October 2009 (UTC)

I don't know why you are telling this to me. I do not work for the CGPM, I didn't come up with definitions of SI terms, and I have no power to change them. Some people do have the power to change them, but they are people who do not learn any physics from wikipedia--they learn it from textbooks, journal articles, and colleagues. Plus, wikipedia itself forbids discussions that advance science as you're trying to do (see this link and this link). You're wasting your time here. If you want to make a difference, you should be spending 100% of your time writing articles for physics and metrology peer-reviewed journals and 0% of your time on wikipedia. I'm sorry. --Steve (talk) 18:42, 20 October 2009 (UTC)

Speed of light arbitration

Preface

You have some thoughtful insights, philosophically and psychologically, on so-called fringe science (I think fringe theories is a misnomer because they are usually even less than a well formed hypothesis), and are remarkably candid about holding an idiosyncratic (not a pejorative term) view yourself. I confess I that have "visualization" of spacetime, that the speed of light is the flow of time, that is not only unconventional, but that I know is wrong mathematically. But there is a kind of "poetic" truth to it (if only in my own mind) that I can't entirely let go of. But, I make sure that it doesn't influence my Wikipedia editing. I wouldn't even self-publish it (and I wouldn't even want to post it, or anything, on a fringe web site) because I can't get past the mathematical objection. What distinguishes me from a crank is that I don't push it, and that I doubt my own view of it. Self-doubt, in my opinion, is the single most important factor in science's success; the rest of scientific methodology flows self-doubt. Descartes was on the track, but he wasn't sufficiently rigorous in applying that doubt to some of his own unstated "postulates".

To the business at hand

Your commentary about Tombe's views, in your statement supporting your proposed principle on Workshop, isn't going to persuade arbitrators because you don't cite supporting examples, and they won't cross-check your Evidence page entries to see what is and isn't supported (nor should they). And what is missing from your Evidence is reliable sources to show that what he says is anti-science. Perhaps, for this purpose, the arbitrators would be satisfied if you pointed them to the correct spot in the Wikipedia articles on the relevant subjects. You could do the arbitrators, and Wikipedia, a great service by filling these gaps. No one else is doing it—we just assert that his views are contra-science because it's so obvious to us—and the arbitrators need it. I think that you are ideal for this task, if you will do it. Meanwhile, we parties have our hands full dealing with the behavioral issues.

By the way, if you look closely enough, I think you will see that Tombe's rejection of relativity (along with most of post-19th-century physics) is the root of his arguments about the speed of light. If the speed of light is constant, as it is according to relativity, then his so-called tautology becomes meaningless physically. Finell (Talk) 16:13, 2 October 2009 (UTC)

First of all, thanks for the compliments! Second, I agree with most everything you say, including how I could help more. But I'm really busy right now. Also I feel that David's said more than enough for the administrators to "get to know him". Finally, I'm hopeful that administrators will find my little section at evidence to be sufficient, even if I don't get a chance to add any more. I do have "Bill Nye the Science Guy" as a reliable source, if nothing else!! :-)

As for you, keep up the good work. I've been checking the pages once in a while, and I'm happy that people like you are doing the hard work on those pages, while I can do other real-life stuff!

As for the philosophy, fringe science is a funny thing. I went to a talk once by a philosopher of science, and she said that before meteorites were accepted, scientists would laugh at the crazy people saying that they saw rocks falling from the sky. Wouldn't you?? I don't think that will be the case with David Tombe. But I just think that's a funny example.

As for editing articles from the neutral point-of-view that you disagree with...I took a philosophy class from Michael Sandel, and he could spend 3 hours arguing convincingly for some philosophy, answering all the counterarguments, and then the next 3 hours arguing the exact opposite point of view, again answering all the counterarguments. Good for him. This is extremely rare though, thanks to the human nature works! Too bad about human nature, and that's why I proposed that "policy".

As for terminology, "science" comes from Latin for "knowledge", so maybe "fringe science" is too kind a term too. Oh well. Sometimes it's impossible to be honest and polite simultaneously!! :-) --18:56, 2 October 2009 (UTC)

Thanks. Somewhere in all these discussions, I wrote, and meant, that Einstein was a fringer, and Wikipedia would have not have published his theories, until science accepted his views as being within legitimate scientific discourse. But, he was accepted pretty quickly (though not universally, and not by those who clung to the past like Tombe does now), because the work was sound on its face. Finell (Talk) 19:18, 2 October 2009 (UTC)

Planck mass

Per this AfD, which you nominated, it seems that the IP has edited Planck mass quite heavily as well. If you can take a look and take out the OR in there, it would be great. Thanks. Tim Song (talk) 07:28, 23 October 2009 (UTC)

Done. :-) --Steve (talk) 08:29, 23 October 2009 (UTC)

Negative Multinomial Distribution Applets

Could you please provide your opinion on this Talk:Negative_multinomial_distribution page? It's regarding the value of an external applet link and a results figure. Thanks. Iwaterpolo (talk) 05:41, 9 November 2009 (UTC)