Polarization density in a dielectric?

[Nikitin]

This is polarization density.

Can somebody explain to me why exactly this is so? My book just says that's how polarization is defined.. Can you guys give me an intuitive understanding of why? Only thing I know is that χ is related to κ, the dielectric constant, and κ is related to how well a dielectric polarizes.

 

[Nikitin]

hmm, isn't the formula above equivalent with P = ε₀(E₀-E), where E = the electric field inside the dielectric, and E₀= the electric field without the dielectric?

Thus P = σ_i = the induced charge on the dielectric?

Wow, why would anyone define something so simple and obvious in such a over-complicated way?

 

[DrDu]

How do you get E0?

 

[Nikitin]

Isn't E=E0/k?

 

[BruceW]

I think most people are more familiar with the term 'electric displacement field' $\vec{D}$, for which $\epsilon_0 \vec{E}_0=\vec{D}$

 

[BruceW]

And strictly, the polarisation density is defined as:
$$\vec{P} = \vec{D} - \epsilon_0 \vec{E}$$
(which is the second equation which you wrote). And the first equation you wrote:
$$\vec{P} = \epsilon_0 \chi_e \vec{E}$$
Is only true in a material which is nice and simple. Since you are just starting to learn about this stuff, you can usually assume that the material is nice and simple. It is only when you've got used to the concepts that you will begin to say "what if the material is not nice and simple", and then the polarisation density will no longer be proportional to the electric field. But for now you can assume it is.

 

[BruceW]

Anyway, the intuitive explanation is that the bound charges in the material react to an electric field in such a way that will reduce the electric field in the material. The most simple model we can use, is to say that the response from the bound charges is the same as the electric field causing it, multiplied by a constant. (And the constant is positive, since the minus sign is already taken into account in the equation which defines the polarisation density).

In fact, the word 'response' is very good, because the electric susceptibility is an example of a response function. Response functions are widely used in physics, and can be very useful.

 

[tiny-tim]

Hi Nikitin!

hmm, isn't the formula above equivalent with P = ε₀(E₀-E), where E = the electric field inside the dielectric, and E₀= the electric field without the dielectric?

Yes, but it's usually written ε_oE = -P + D,

where ε_oE is the total field, made up of -P (the bound field) and D (the free field)

(for historical reasons, P is minus the bound field)

(the ε_o is only there as a conversion factor because for historical reasons the three types of electric field are described by two different electric units: E is defined as force per charge, while D and P are defined as charge per area)

Wow, why would anyone define something so simple and obvious in such a over-complicated way?

i think it's because D is so much easier to calculate (it's just the surface charge density, in coulombs/m²), while P itself is difficult to calculate or measure directly, so it's easier to calculate ε_oE and D and subtract

 

[DrDu]

Isn't E=E0/k?

No, at best E=D/k. But D is usually not the field which would be there if the dielectric were absent.

 

[BruceW]

if the dielectric is the sole contributor of bound charges (and if the dielectric doesn't do anything else), then the D field is the same as the electric field that would be there if the dielectric were absent (except the constant factor of epsilon-0). Right?

Edit: or I guess you mean that the time-evolution of the system would be different if we did remove the dielectric?

 

[Nikitin]

What does polarization, P, equal to? Is it simply equal to induced charges?

 

[DrDu]

if the dielectric is the sole contributor of bound charges (and if the dielectric doesn't do anything else), then the D field is the same as the electric field that would be there if the dielectric were absent (except the constant factor of epsilon-0). Right?

Edit: or I guess you mean that the time-evolution of the system would be different if we did remove the dielectric?

No, that's exactly the point why we introduce all these additional fields like D and H.
The point is that D fulfills different boundary conditions than does E_0. E.g. compare the D field of a sphere which is brought into a constant electric field with the constant field before the dielectric was entered.

 

[DrDu]

What does polarization, P, equal to? Is it simply equal to induced charges?

The definition of P is not unique and different conventions are in use.
For problems which rapidly changing fields one often uses
$P=-\int_{-\infty}^{t}j_\mathrm{int}(t') dt'$, where j is the internal current density (either both bound and unbound current density or only bound current density) in the material.
At lower frequencies it is convenient to split j into a rotational and an irrotational part. The former is taken care off introducing magnetisation, the latter is used in the definition of polarisation.

 

[DrZoidberg]

if the dielectric is the sole contributor of bound charges (and if the dielectric doesn't do anything else), then the D field is the same as the electric field that would be there if the dielectric were absent (except the constant factor of epsilon-0). Right?

Yes, if the field exists in between two parallel metal plates and the charge on those plates stays constant.
Without a dielectric (in a vacuum or air) D and E are the same. I'm omitting e0 here since it just serves as a unit conversion factor.
So since D is equal to the amount of charge on the plates divided by their area, it will not change when the dielectric is removed. And without the dielectric E = D, therefore D is the same as E would be without the dielectric.
By the way, in the (microscopically small) gap between the plates and the dielectric E is also equal to D. That is unless there is electric breakdown occurring in the gap.

 

[BruceW]

No, that's exactly the point why we introduce all these additional fields like D and H.
The point is that D fulfills different boundary conditions than does E_0. E.g. compare the D field of a sphere which is brought into a constant electric field with the constant field before the dielectric was entered.

Ah, yeah of course, you are right, now that I think about it. In the case of electrostatics, the electric field will have zero curl, but the D field will have non-zero curl. So we can only say that the divergence of the D field is the same as the divergence of the E field that would be there if the dielectric were absent (except the constant factor of epsilon-0).

 

[DrZoidberg]

Why would D have a non zero curl in a static situation?

 

[BruceW]

As DrDu says, this is the whole reason for introducing the D and P fields. We want a field P that is going to be non-zero inside the material and zero outside. And generally, this means the curl of P is going to be non-zero. But the electric field doesn't curl in time-independent situations, so when we write the equation relating P,E and D, we know that the curl of D must equal the curl of P, to ensure that E doesn't curl. So then we know D must have non-zero curl.

I'm guessing your next question might be: "Why must P have curl, just because it is non-zero inside the material and zero outside?" Well, because the E field has no curl, this means the tangential component of the E field at the boundary must be continuous. But for the P field, we don't want to make this restriction. We will often want it to be discontinuous at the boundary. For example, for the simple material, if the E field is parallel to the boundary, then we want the P field (inside the material) to be parallel to the boundary too. But outside the material, we want the P field to be zero. So we want the tangential component of the P field at the boundary to be discontinuous. Therefore, we must allow the P field to curl.

 

[Nikitin]

No, at best E=D/k. But D is usually not the field which would be there if the dielectric were absent.

In the case of plate-capacitors, V=V0/k => E=E0/k when q=constant.

The definition of P is not unique and different conventions are in use.
For problems which rapidly changing fields one often uses
$P=-\int_{-\infty}^{t}j_\mathrm{int}(t') dt'$, where j is the internal current density (either both bound and unbound current density or only bound current density) in the material.
At lower frequencies it is convenient to split j into a rotational and an irrotational part. The former is taken care off introducing magnetisation, the latter is used in the definition of polarisation.

How about for static fields? I've just begun learning about condensers.

 

[Jano L.]

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:

$$
\mathbf P = \mathscr{N} \langle \boldsymbol \mu \rangle
$$

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.

 

[jasonRF]

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:

$$
\mathbf P = \mathscr{N} \langle \boldsymbol \mu \rangle
$$

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.

For me this is the most intuitive way to understand this. Hopefuly the OP has learned about simple electic dipoles. Here is the way I think of it:

First a definition. If you have a charge +q and a charge -q, separated by $\mathbf{d}$ (a vector that starts at -q and ends at +q) then the dipole moment is $\mathbf{p} = q \mathbf{d}$. Now when we consider a dielectric media, there are molecules that become distorted when an external field is applied, and the distortion can be modeled by a dipole.

If we look on a microscopic scale, the field and dipoles make a rapidly varying (in the spatial sense) electric field in the vicinity of each molecule, in fact we usually do not care about the fluctuations on these tiny scale sizes. Instead, we are usually interested in macroscopic behavior. So one approach is to average the fields, and the polarization vectors, over volumes that contain many of the elementary dipoles and yield (approximately) continuous vector fields, but are still small enough volumes so that the scale sizes we typically are interested in are preserved. When we do this, we obtain what are usually called the macroscopic Maxwell equations. The terms associated with the dielectric media involve this average polarization density,
$$\mathbf{P}(\mathbf{r}) = \frac{1}{\Delta V} \sum_{\mathbf{r}_i \, in\, \mathbf{r}+\Delta V} \mathbf{p}(\mathbf{r}_i)$$
Note that due to the sign convention for electric dipoles, $\nabla \cdot \mathbf{P} = -\rho_b$ where I have let $\rho_b$ represent bound charge density. There is nothing fictitious about these charges, they are really there. To convince yourself of the sign, consider a bunch of elementary dipoles all with their -q charges at the origin (say), and all of the $\mathbf{d}$ vectors pointing radially outward. Clearly this configuration has positive divergence at the the origin, where there is a bunch of negative charges.

Also, $\frac{\partial \mathbf{P}}{\partial t}$ represents a current. To understand this think about a single dipole. if the time derivative is positive, it means either the charges are increasing (meaning positive charge must be moving from the -q location to the +q location) or the vector $\mathbf{d}$ is increasing, which simply means (for example) the +q charge is moving away from the -q charge, so is a positive current.

Hence, to include dielectric media into our macroscopic Maxwells equations (again, where we usually think of all quantities as having been averaged) we obtain,
$$\begin{eqnarray} \nabla \cdot \epsilon_0 \mathbf{E} & = & \rho - \nabla \cdot \mathbf{P}. \\ \nabla \times \mu_0^{-1} \mathbf{B} & = & \mathbf{J} + \frac{\partial }{\partial t}\epsilon_0 \mathbf{E} + \frac{\partial }{\partial t} \mathbf{P}. \end{eqnarray}$$

If we define $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$ then these become the more familiar,
$$\begin{eqnarray} \nabla \cdot \mathbf{D} & = & \rho. \\ \nabla \times \mu_0^{-1} \mathbf{B} & = & \mathbf{J} + \frac{\partial }{\partial t} \mathbf{D}. \end{eqnarray}$$

I hope that helps a little!

jason

 

[Nikitin]

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:

$$
\mathbf P = \mathscr{N} \langle \boldsymbol \mu \rangle
$$

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.

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?

EDIT: Thanks for a great second-post! I only skimmed thru it for now, but I'll make sure to read it later when I got the time. And yeh, I know what dipoles are.

 

[DrDu]

$$
\mathbf P = \mathscr{N} \langle \boldsymbol \mu \rangle
$$
This applies well to dielectric media (the definition is unambiguous).

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 it is at least as intuitive but much more general to think of P as a time integral over fllux density.

 

[DrDu]

How about for static fields? I've just begun learning about condensers.

That's not a problem at all. Also a static field has to be switched on, or the dielectric is brought slowly into the field. During that switching, a current will flow in the dielectric until the final field strength has been reached.

 

[DrDu]

In the case of plate-capacitors, V=V0/k => E=E0/k when q=constant.

If there were only plate condensers, you would hardly need to introduce all these auxillary fields like D and P.

 

[DrDu]

Hence, to include dielectric media into our macroscopic Maxwells equations (again, where we usually think of all quantities as having been averaged) we obtain,

If you look at your equations you will see that they are perfectly valid also without averaging. In fact, the possibility to introduce polarisation instead of charges and currents is a completely general consequence of charge conservation.

 

[DrZoidberg]

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

What about salt water?

 

[Nikitin]

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?

Can somebody answer the above?

 

[Jano L.]

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?

What do you mean by "essentially classical picture"? And why does that imply that it falls short in many ways?

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.

I think it is at least as intuitive but much more general to think of P as a time integral over fllux density.

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.

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 $\mathbf j$.

 

[Jano L.]

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?

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

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.

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?

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.

 

[DrDu]

Dear Jano,

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.

 

[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?)