Polarization density in a dielectric?

[Jano L.]

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?

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.

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.

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.

 

[DrDu]

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?

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.

 

[Jano L.]

Usually, one calculates the polarisability using some band structure calculations with the bands being delocalized over the whole crystal.

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.

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.

 

[DrDu]

My intention was mainly to bring out the point that a description in terms of dipole density, although perpetuated in every introductory textbook, is not always the most appropriate description. Clearly there are situations where it is useful, e.g. in the description of gasses of independent molecules. For this situation I recommend the book
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.

 

[BruceW]

I have seen polarisation density defined in some places as the electric dipole moment per volume, and in some places as simply the polarisation due to all electric moments. Clearly, the latter is more general. It seems that the former is very often used. I guess this is because we only need to consider the dipoles in most cases? (i.e. in simple experiments involving dielectric materials?) (and/or it makes it simpler to put into textbooks, to just consider the dipoles?)

 

[Nikitin]

Guys, OP here. Can somebody please clear this up for me?

I made this 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?

As an answer to Janu's post:

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.

 

[DrDu]

As Jano L has written in #29, $\nabla \cdot P=-\rho$, which is the relation between the polarisation and the induced charge density.

 

[Nikitin]

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?

 

[DrDu]

It's not the gradient of P but its divergence $\nabla\cdot P=\partial P_x/\partial x+\partial P_y/\partial y +\partial P_z/\partial z$. I. e. the difference of the polarization flowing in some volume and the one flowing out is minus the bound charge density.

 

[Nikitin]

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

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

 

[tiny-tim]

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

the difference between the total electric field (E) flowing out and in per volume is the total charge density: divE = ρ_total

for convenience, we can (and do) arbitrarily split the total charge into the free charge plus the bound charge: ρ_total = ρ_free + ρ_bound

we then define the free (D) field as being that produced by the free charge, and the bound (-P) field as being that produced by the bound charge: divD = ρ_free, -divP = ρ_bound

(i've left some constants out)