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Bound Charges - Polarized Distributions

  1. Jul 25, 2013 #1
    The potential due to a polorized distribution is given by:
    [tex]V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int _{V} \frac{ \hat{r} \cdot \vec{P} ( \vec{r}')}{r^{2}} dV [/tex]

    After working some voodoo math, this is worked into the form [tex]V = \frac{1}{4 \pi \epsilon _{0}} \oint _{S} \frac{1}{r} \vec{P} \cdot d \vec{a} - \frac{1}{4 \pi \epsilon _{0}} \int _{V} \frac{1}{r}( \nabla \cdot \vec{P}) dV [/tex]

    the dipole moment per unit length is [itex]\vec{p}= \vec{P} dV [/itex]

    What exactly does P mean? I'm confused about the meaning of P, which seems like a good place to start asking questions. I feel the rest of my misunderstandings stem from not really "getting" what P actually IS.

    It also describes the potential of the surface charge as [itex]\sigma = \vec{P} \cdot \hat{n}[/itex] and [itex] \rho _b = - \nabla \cdot \vec{P}[/itex] which I have little clue where it comes from.


    As an example of how befuddled I am, take the problem "Find the electric field produced by a uniformly polarized sphere of radius R". The answer is [itex]\frac{P}{3 \epsilon _{0}} r cos \theta [/itex] but when I set the integral up like this: [tex]\frac{1}{4 \pi \epsilon _{0}} \int ^{2 \pi} _{0} \int ^{\pi} _{0} \frac{P R^{2} cos \theta sin \theta}{ \sqrt{R^{2}+z^{2}-2Rz cos\theta}} d \theta d \phi [/tex]

    This integral is probably the most hideous thing I've ever seen (used wolfram).

    I figure a "bound charge" is each term of the final integral [itex]V = \frac{1}{4 \pi \epsilon _{0}} \oint _{S} \frac{1}{r} \vec{P} \cdot d \vec{a} - \frac{1}{4 \pi \epsilon _{0}} \int _{V} \frac{1}{r}( \nabla \cdot \vec{P}) dV [/itex], the first being the surface charge, and the second being the volume charge, but this doesn't really help me understand what it actually means.

    NB sorry for any mistakes in the math text, it's a laborious process to write this out on a computer, and I'm prone to making mistakes as it is.

    Is there a program that will simulate electric/magnetic fields? Visual as well as numerical, it would be interesting to see the electric field generated by different objects and such, which would be impossible to solve analytically.
     
    Last edited: Jul 25, 2013
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  3. Jul 26, 2013 #2

    WannabeNewton

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    First orient the coordinate system so that ##P## is along the ##z##-axis and the observation point ##r## is arbitrary. Then rotate the coordinate system about the ##z##-axis until ##r## is on the ##x-z## plane (this leaves ##P## unchanged of course). Now rotate the coordinate system until ##r## is along the ##z##-axis in which case ##P## lies in some direction on the ##x-z## plane.

    Hence ##P\cdot \hat{n} = P_x \sin
    \theta \cos\varphi + P_{z}\cos\theta## but notice that the ##P_{x}## term isn't going to survive the integral so all we're left with is ##V = \frac{P \cos \Psi R^{2}}{4\pi \epsilon_{0}} \int _{S^{2}}\frac{\cos\theta \sin\theta d\theta d\varphi}{\sqrt{z^{2} + R^{2} - 2Rz\cos\theta}}## where ##\Psi## is the angle ##P## makes with the ##z##-axis in this coordinate system. If we integrate this out, apply the bounds, and simplify we will get (for the potential inside the sphere) ##V = \frac{P z\cos \Psi }{ 3\epsilon_{0}} = \frac{P\cdot r}{3\epsilon_{0}}## where the last part comes about because in this coordinate system the observation point ##r## is along the ##z##-axis and ##P## is on the ##x-z## plane so since ##\Psi## is the angle ##P## makes with the ##z##-axis, ##Pz\cos\Psi = P\cdot r##. Now ##V = \frac{P\cdot r}{3\epsilon_{0}}## is a coordinate independent result so it must hold in all coordinate systems. Orienting our coordinate system back to the original orientation (in which ##P## was completely along the ##z##-axis but ##r## was arbitrary), we see that ##V = \frac{Pr\cos\theta}{3\epsilon_{0}}## as desired. The potential outside the sphere can be obtained in an identical manner.
     
  4. Jul 26, 2013 #3

    DrDu

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    I argued here already several times that a definition of polarization in terms of a dipole moment density is purely classical (and even there incomplete as there are also contributions from quadrupole moments etc.) and becomes unfeasible when the actual distributions of electronic densities are considered.
    A more useful definition which is used in present day calculations is [itex] P=-\int_{-\infty}^t dt' j_\mathrm{int}(t') [/itex] in terms of the internal current density ##j_\mathrm{int}##. It is easy to show that this P fulfills al the Maxwell equations.
     
  5. Jul 26, 2013 #4

    ShayanJ

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    About the meaning of [itex] \vec{P} [/itex]:Consider a dielectric material placed in an electric field.The molecules of the dielectric material,as a result of the presence of the electric field,rotate to align in a new direction and what that direction is,depends on the properties of the material and the direction of the electric field.For a linear dielectric,that direction is the same as the direction of the electric field.
    But to make it quantitative and take out [itex] \vec{P} [/itex]from it.When the molecules of the dielectric are aligned to a new direction,they each have a dipole moment [itex] \vec{p}_i [/itex] in that direction.And so if you take a portion of the dielectric,it may have a net dipole moment.The dipole moment of this portion of the dielectric is the sum of the moments of its constituent molecules,i.e. [itex] \vec{\Pi}_{portion}=\sum_{molecules} \vec{p}_i [/itex]. But the quantity [itex] \vec{\Pi} [/itex] is dependent on the volume and isn't proper for general studies about the polarization of dielectrics,so physicists define the dipole moment density [itex] \vec{P}=\large{\frac{d \ \vec{\Pi}}{dV}} [/itex].
    About the formulas [itex] \sigma_b=\vec{P}\cdot\vec{n}_{out} [/itex] and [itex] \rho_b=-\vec{\nabla}\cdot\vec{P} [/itex].They arise from the "voodoo math" you explained at the beginning of your post.The point is,when people do that voodoo math and turn that one integral into two integrals,they find two quantities with dimensions of surface([itex]\sigma_b[/itex]) and volume([itex] \rho_b [/itex]) charge densities and the potential takes the form of a potential produced by ordinary charge distributions.
    But maybe you have problem with interpreting [itex]\sigma_b[/itex] and [itex] \rho_b [/itex]. They are simply providing information about the net charge residing on infinitesimal portions of the dielectric resulting from its polarization.
    Take an initially unpolarized dielectric and focus on an infinitesimal portion of it.Because the molecules of this portion have no net charge and are not polarized,this portion has no net charge.But when the electric field is applied,the molecules become polarized and so some parts of them may go out of the portion we're considering.Now in this portion,we have the positive part of some molecules and the negative part of some others,and so this portion may have a net charge.This net charge is called bound charge.You can apply a similar argument to the surface of the dielectric and so find an interpretation for [itex] \sigma_b [/itex]
     
  6. Jul 26, 2013 #5

    ShayanJ

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    mmm...interesting...never heard of that.Could you introduce some references?
     
  7. Jul 26, 2013 #6

    Jano L.

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    The symbol ##\mathbf P## and also the term "polarization" is used in several different meanings. The basic one from electrostatics, which is probably the one you are interested in, is that it is a vector that quantifies average dipole moment of a large set of neutral molecules around some point in a dielectric. Choose some region ##\Delta V## around the point ##\mathbf r##. Divide all neutral molecules into two groups: those that belong to the volume and those that do not (if some molecule is only partly in the volume, take it in or out, does not matter).

    The polarization at ##\mathbf r## is

    $$
    \mathbf P \approx \frac{total ~electric ~dipole ~moment ~of ~all~ neutral ~molecules ~belonging~to~the~volume ~\Delta V}{\Delta V}
    $$

    This is the classic meaning of the word "polarization" in macroscopic electrostatics.

    What DrDu is referring to is a different quantity, which can better be called by other name, for example "electric polarization potential". Let us denote it by ##\tilde{\mathbf P}## (not a standard notation). It behaves similar mathematically and is often named and denoted in the same way as the above polarization, but it serves a different purpose: namely that it is the quantity that in oscillating electromagnetic field gives the electric current density:

    $$
    \mathbf j = \partial_t \tilde{\mathbf P}
    $$

    It belongs to and is very useful in time-dependent theory (spectroscopy), not electrostatics.
     
  8. Jul 26, 2013 #7

    DrDu

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    That's not true. You can calculate the static polarisation using the formula I have given when you interprete the time integral for a path where the external field which induces the current and polarisation is swiched on adiabatically. In the (purely hypothetical) cases where the polarisation can really be described in terms of dipole densities, the two formulas can be shown to coincide.
    The formula requires some modifications to describe magnetostatic phenomena, i.e. only the transversal part of the current density should be used.
     
  9. Jul 26, 2013 #8

    DrDu

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  10. Jul 26, 2013 #9

    Jano L.

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    You obviously want to assign the name polarization to a different concept than I. In my view, the defining property of polarization is that it ##is## density of electric dipole moment. I have a feeling that it will be hard for us to agree on common terminology. I do not see any problem with my definition, it is simple and uses only instantaneous quantities. Your definition requires integration over infinite time, which is odd for a quantity that should characterize instantaneous state of the medium. That's like definition of radius vector as an integral of velocity.


    In your formula, what do you mean by internal current density ##j_{\text{int}}##?
     
  11. Jul 26, 2013 #10

    DrDu

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    I think polarisation and magentisation, or alternatively dielectric displacement and corresponding magnetic properties were concepts introduced already by Maxwell without a microscopic interpretation, although I am also not too familiar with the historical development.
    However Maxwells equations fix the analytic properties of P and M which we like to explain.
    The interpretation in terms of dipole moment densities goes back to Lorentz. As it stems from the pre-quantum era, in my point of view it is a nice toy model nowadays but little more.
    What weights even more in my eyes is that you need to make, as Astrum called it - some Vodoo math approximations which are hard to justify to show that the dipole moment density approximately fulfills Maxwell's equations.
    What is even more serious a problem and is pointed out in the article I cited is the fact that the definition of polarisation in terms of moment densities cannot be translated into a quantum mechanical operator.
    Schemes to calculate the polarisation based on the formula I gave without having to explicitly perform a time integral have only been obtained in the last 25 years or so but seem to be the standard in quantum solid state programs today.

    With ##j_\mathrm{int}## I mean the microscopic currents induced in the medium, these are all the currents not explicitly included in the Maxwell equations. The external currents are usually due to external macroscopic currents.
     
  12. Jul 26, 2013 #11

    Jano L.

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    I am not aware of any need of voodoo approximations. The only approximation is that the expressions

    $$
    -\nabla \cdot \mathbf P
    $$
    $$
    \partial_t \mathbf P
    $$
    in electrostatics and
    $$
    \mathbf j =\nabla \times \mathbf M
    $$
    in magnetostatics give total charge and current only approximately. This can be shown from microscopic particle model of matter. It is not a problem, it only signifies that ##\mathbf P,\mathbf M## are concepts from quasistatic theory limited in usefulness and applications.

    For fast processes (spectroscopy) the moment densities are not that useful. It is much easier to work directly with the current density or ##\tilde{\mathbf P}##. But there is no need to call ##\tilde{\mathbf P}## polarization. It is just an auxiliary variable. The distinction seems important, since the polarization ##\mathbf P## may differ from the polarization potential ##\tilde{\mathbf P}##.

    For neutral molecule, let ##\psi(\mathbf r_a, \mathbf R_A)## be the ##\psi## function. The electric moment operator is
    $$
    \hat{\mathbf p} = \sum_a q_a \mathbf r_a + \sum_A Q_A \mathbf R_A
    $$

    where ##a## indexes electrons, ##A## indexes nuclei. The expected average value of dipole moment is
    $$
    \int \psi^* \hat{\mathbf p} \psi d\tau
    $$

    I asked you a question in my above post. I will try again: what exactly is ##j_{\text{int}}## in your definition?
     
  13. Jul 29, 2013 #12

    DrDu

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    The electric moment operator is fine for a molecule, but we are talking about a macroscopic sample of matter. Many substances like e.g. SiO2 aren't made up of isolated molecules. And even for molecular crystals you will have to define somehow boundaries between the molecules. I don't see a way to take care of this consistently.

    With ##j_\mathrm{int}## i mean the currents produced by all the electrons in the sample.
     
  14. Jul 29, 2013 #13

    Jano L.

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    I did not say that the molecule has to be isolated. All that is needed is a set of particles whose total charge vanishes.
    This set does not need to be "one small molecule". We can take a lot of nuclei and electrons from SiO2 and define its electric moment via the mentioned expression. Boundaries between molecules are not necessary.

    By currents produced by electrons you mean the microscopic current density? But then your ##\tilde{\mathbf{P}}## will fluctuate in space on the atomic scale, so it will not be useful in macroscopic Maxwell's equations (no linear relation to macroscopic electric field, etc.)
     
  15. Jul 29, 2013 #14

    DrDu

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    Of course you can define the total dipole moment like this, say, for a macroscopic crystal. But this does not yield the polarisation density which is usually not homogeneous over the macroscopic sample.






    Yes, I mean the microscopic density. It is rather trivial to separate a slowly varying part, e.g. taking only the small k components in a Fourier transform. But this is not conceptually necessary (my solid state prof. used to say: Blos net mitteln! (Never take averages) in his Viennan accent).
    The linear relation between polarisation and a macroscopic external field is another question and not directly related to the definition of the polarisation. It is nicely worked out in the book by Max Born and Kun Huang "Dynamical theory of crystal lattices".
    An explicit method to calculate dielectric constants is the work of Adler from 1962
    http://prola.aps.org/abstract/PR/v126/i2/p413_1
    which forms the basis for any ab initio calculation of dielectric constants used today.

    I just found an article which contains much of the stuff I tried to explain, like the definition of the microscopic polarisation P in terms of the current density j, and how to include local field effects to calculate the macoscopic dielectric constant:
    http://www.tandfonline.com/doi/abs/10.1080/00018737800101384#.UfaIRKzwvRk
     
    Last edited: Jul 29, 2013
  16. Jul 29, 2013 #15

    Jano L.

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    Choose a point ##\mathbf x## inside a dielectric. Take a ball centred at ##\mathbf x## with radius larger than 10 A, but small enough to allow us to treat the ball like a point in macroscopic theory. Say ##R = 100 ##A. Take any neutral set of particles ##S## whose majority is inside the ball and define the polarization density by
    $$
    \mathbf P(\mathbf x) = \frac{1}{V} E \left\{ \sum_{a\in S} q_a \mathbf r_a\right\}
    $$
    where E denotes expected average value.

    This does give density of electric moment for the purposes of macroscopic theory provided the macroscopic field does not vary significantly over distances ##< R##.
     
  17. Jul 29, 2013 #16

    DrDu

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    Ok, mostly we are interested in the change of polarisation with an applied field. How do you ascertain that the ball remains charge neutral?
     
  18. Jul 29, 2013 #17

    Jano L.

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    It is not the ball that is neutral. It is the set S of particles chosen in such way. The summation is over the particles of the set, not over the particles in the ball.
     
  19. Jul 29, 2013 #18

    DrDu

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    Electrons are indistinguishable. I do not see how to define such a set.
     
  20. Jul 29, 2013 #19

    Jano L.

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    You can begin with a model where the electrons are labeled by some index ##a## and have positions ##\mathbf r_a##. In writing the Schroedinger equation one does exactly that. This does not prevent us from using symmetrized or anti-symmetrized ##\psi## functions later, which describe the "indistinguishable" character of statistics of the electrons.
     
  21. Jul 29, 2013 #20

    DrDu

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    Yes, I am with you now. If you want to do this quantum mechanically, you have to use Wannier functions
    http://en.wikipedia.org/wiki/Wannier_function
    see especially the last paragraph "Modern theory of polarization".
    I still think this is only possible in true isolators where bands are either full or empty. This corresponds to the restriction to bound charges in the classical formulation.
     
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