- #26

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What about salt water?In case the medium is conducting, like metals, this concept of polarization does not apply, because there are no neutral molecules.

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- #26

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What about salt water?In case the medium is conducting, like metals, this concept of polarization does not apply, because there are no neutral molecules.

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Can somebody answer the above?Electric dipole moment it measures the strength of the dipole, right?

But what practical conclusion, in this case, can you draw from multiplying the dipole moment with the particle density, other than measuring how polarized the dielectric is?

In some cases the polarization of the dielectric equals its induced charge density. Can you please explain to me why this is so, and in what situations it is so?

- #28

Jano L.

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What do you mean by "essentially classical picture"? And why does that imply that it falls short in many ways?Although this is sometimes a useful model picture, it falls short in many ways,as it is essentially a classical picture. How about higher moments? What do you do e.g. in the case of quartz which is not made up of individual molecules?

I think the common concept of "polarization" is applicable to any model where charges are bound to some immovable centers. I do not know what problem is there with quartz. If it is non-conducting quartz, I do not see a problem - we should be able to refer the charge positions with respect to the centers of the crystalline lattice, introduce dipole moment of some neutral region etc.

But in thinking this, you are introducing new definition for the quantity denoted by symbol P, which does not necessarily give the same value as the above definition. I believe that this new quantity is called also Hertz vector, or polarization potential. It is a bit different thing than the polarization - they can be different when both are applicable, since the polarization potential depends on how the macroscopic current density is defined in the model, which is somewhat ambiguous, while the average dipole moment of the medium is unambiguous.I think it is at least as intuitive but much more general to think of P as a time integral over fllux density.

The polarization potential is indeed useful in spectroscopy and in antenna theory. But the definition through the integral you indicated requires knowledge and integration of the macroscopic current density ##\mathbf j## over infinite past time. Even if we knew it for that time, it is not clear that it is integrable in that way. As far as I know, the only important property that motivates this concept is

$$

\partial_t \mathbf P = \mathbf j.

$$

This ca be taken as the definition. It leaves some ambiguity in the value of ##\mathbf P##, but it usually does not matter since the important thing is the current density ##\mathbf j##.

- #29

Jano L.

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The quantity ##\mathbf P## defined above is useful macroscopic representation of the microscopic electric state of the medium. It allows us to express bound charges in a mathematically convenient wayBut what practical conclusion, in this case, can you draw from multiplying the dipole moment with the particle density, other than measuring how polarized the dielectric is?

$$

\nabla \cdot \mathbf P = -\rho.

$$

Often it is in simple linear relation to the electric field and due to this it can be used to find the electric field in the dielectric medium.

It is easy to understand this qualitatively. When there is polarization in some direction, from the definition it follows that the molecules have average dipole moment in that direction. This is only possible if the charges get displaced along that direction, as compared to the neutral state. When this happens on the surface of a body which is perpendicular to that direction, some electric charge traverses this surface from one half space to another. This transferred charge is the surface charge, and is proportional to the magnitude of the displacement, hence to polarization.In some cases the polarization of the dielectric equals its induced charge density. Can you please explain to me why this is so, and in what situations it is so?

- #30

DrDu

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The distinction between bound and free charges has become obsolete with the introduction of quantum mechanics. That's what I meant with it being a classical picture.

The definition in terms of dipole moments contains many assumptions. You first stated it for neutral molecules, that's why I asked for quartz which is a macromolecule.

So you have to break down the molecule into neutral regions (with the boundaries of these regions being time dependent for non-static processes), determine their dipole moments and finally do a macroscopic averaging.

Even this definition is not complete as it does not contain contributions from e.g. quadrupole moments and, also in static situations, it is ambiguous:

http://en.wikipedia.org/wiki/Polarization_density (see polarization ambiguity)

If you have a look at the article cited there

http://inside.mines.edu/~zhiwu/research/papers/E04_berry2.pdf

you will find that after many complicated considerations he uses (after eq. 58) ##\delta P=j/(i\omega)## which is the time Fourier transform of the equation I used.

Therefore modern solid state physics book usually give the definition I used in terms of current density.

- #31

Jano L.

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can you explain why the distinction between bound and free charges has become obsolete with the introduction of quantum mechanics ? Do you agree that there is quite a distinction in the behavior of electric charges in say oil and copper?

Yes, only the "macroscopic averaging" is simply average of many dipoles, or expected average calculated from some assumed probability distribution. Charges move, but this is alright, it could not be simpler.So you have to break down the molecule into neutral regions (with the boundaries of these regions being time dependent for non-static processes), determine their dipole moments and finally do a macroscopic averaging.

The example of polarization ambiguity on Wiki is flawed, since in (c) the grouping of charges left one charge on the edge ungrouped. The proper way is to group charges into neutral groups, so that their dipole moment is the first important term in expansion of their field - otherwise the electrostatic field due to ungrouped charges will be strong. The polarization has a good meaning for neutral medium composed of neutral constituents, provided that in the definition no charges are left unaccounted.

Also, the unfortunate mistake on wiki could arise only be because it is 1D chain of charges. Consider what would happen if one forgot to group first layer of cubic lattice of NaCl.

Polarization is defined based on dipole moment and it is supposed to give just that. It is not supposed to give you density of quadrupole moment or exact information on the electric field - there are other quantities to do that.

- #32

DrDu

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Usually, one calculates the polarisability using some band structure calculations with the bands being delocalized over the whole crystal. Use of localized orbitals is possible if bands are full but does not yield much additional insight. In the case of metals, you find that polarizability has a pole at ##\omega=0## and some effects of spatial dispersion. Not a problem. The calculation of the dielectric function (and thus also polarisability) of metals in the Lindhard approximation is standard in most solid state curriculums.Dear DrDu,

can you explain why the distinction between bound and free charges has become obsolete with the introduction of quantum mechanics ? Do you agree that there is quite a distinction in the behavior of electric charges in say oil and copper?

- #33

Jano L.

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This is one respectable way to calculate response function of crystals. Perhaps you use this method usually, but I would like to remind you that there are other models where it is not calculated this way, i.e. liquids. Anyway, the way how it is calculated is not that important. We surely agree that the electrons in the molecules of oil are usually bonded to molecules, while those in the metals are free to move around.Usually, one calculates the polarisability using some band structure calculations with the bands being delocalized over the whole crystal.

It seems to me that we are talking about different meanings of the term "polarization" used in different areas of physics. I think both are rightful concepts in their own setting.

- #34

DrDu

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Craig, David Parker, and T. Thirunamachandran. Molecular quantum electrodynamics. Dover Publications, 1998

which among other very interesting insights contains a correct formula for the polarisation beyond the dipole approximation (here for an atom with nuclear charge Ne at position R),

namely

##p(r)=-e\sum_i^N(q_i-R)\int_0^1 \delta(r-R-\lambda(q_i-R))d\lambda##

first derived by Wooley in 1971.

This expression at least fulfills the relation ##\nabla P=\-\rho=e\sum_i \delta(q_i-R)## without further approximations. An expansion in terms of ##\lambda## yields the contribution of multipoles beyond the dipole.

- #35

BruceW

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- #36

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I made this post:

As an answer to Janu's post:Electric dipole moment it measures the strength of the dipole, right?

But what practical conclusion, in this case, can you draw from multiplying the dipole moment with the particle density, other than measuring how polarized the dielectric is?

In some cases the polarization of the dielectric equals its induced charge density. Can you please explain to me why this is so, and in what situations it is so?

One important definition of polarization which was not mentioned so far is that it is the total electric dipole moment of neutral set of molecules divided by the volume they occupy, or, which is the same thing, number density of molecules times their average dipole moment:

P=N⟨μ⟩

This applies well to dielectric media (the definition is unambiguous).

In case the medium is conducting, like metals, this concept of polarization does not apply, because there are no neutral molecules.

- #37

DrDu

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Ah, thanks. I appreciate your help, but what exactly does the formula mean?

I learned from Calculus 2 recently that the gradient of a function F(x,y,z) is the direction in a 3D-area in which F(x,y,z) grows the most. How does this tie with P? Geometrically, when is ∇P = ρ? Note: I haven't learned about vector-fields yet.

Could you please explain this to me?

I learned from Calculus 2 recently that the gradient of a function F(x,y,z) is the direction in a 3D-area in which F(x,y,z) grows the most. How does this tie with P? Geometrically, when is ∇P = ρ? Note: I haven't learned about vector-fields yet.

Could you please explain this to me?

Last edited:

- #39

DrDu

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Sorry for being a bit incompetent in the math department - we have just started on gradients, and have not learned about vector fields yet.

PS: Since you're quite skilled in physics, would you mind assisting me here, too? https://www.physicsforums.com/showthread.php?t=676590

- #41

tiny-tim

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the difference between theSo the difference between the polarization flowing out and in per volume is the bound charge density? Is there an intuitive explanation for this?

for convenience, we can (and do) arbitrarily split the total charge into the free charge plus the bound charge: ρ

we then

(i've left some constants out)

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