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- Thread starter Tosh5457
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In matter, the electric and magnetic fields cause polarization and magnetization of the material. These effects cause, for example, neutral objects to separate opposite charges. The charge densities and currents created entirely from these effects are said to be

Free charges and currents are those that you can imagine existed even before the medium was polarized or magnetized. It is often the case that one knows what free charges are present in a physical system, and as such, defining ##D## such that ##\nabla \cdot D = \rho_f## is useful to help ignore the effects of bound charges (or, even to help calculate what the bound charges are, if you know the electric field and, hence, the total charge density).

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And how to arrive at the expression:

?

And why does D appear in the first macroscopic Maxwell's equation instead of E? And why does E appear in the 3rd equation, and not D?

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- #4

jtbell

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And how to arrive at the expression:

See the section from equations 806 to 808 here:

http://farside.ph.utexas.edu/teaching/em/lectures/node69.html

You might find the rest of those lecture notes useful:

http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html

But what you really need is a real textbook. You can't expect lecture notes or an Internet forum to substitute for a textbook.

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- #5

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Electric and magnetic fields behave in a perfectly symmetrical way as long as no matter is involved. It's the properties of matter that introduce asymmetries. That's because matter contains electric charges but no magnetic charges(magnetic monopoles). Also in the SI system asymmetric units are used. If you use the cgs system the units are symmetrical.

Look here

http://en.wikipedia.org/wiki/Magnetic_monopole#In_Gaussian_cgs_units

Look at the maxwell equations with cgs units and with monopoles. You will notice that they are completely symmetrical. So the asymmetries in the "normal" form of the maxwell equations are due to the fact that the properties of matter were incorporated into the equations and because of the way the SI system is defined.

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I find that line of reasoning somewhat misleading. An understanding of special relativity tells us that stationary "electric" charges are in fact currents. It's not a property of matter so much as a statement that the EM field has a single vector field as its source.

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If someone wants to avoid the notion of D and H ,then it is completely possible.They were defined before quantum mechanics formulated so as to avoid the machinery of microscopic world in terms of electrons and others.As feynman has pointed out(vol. 2) that ε,μ etc can be explained properly only by using quantum mechanics.here it is stressed that one can avoid the notion of D and H if one wants and treat everything in terms of E and B.

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Thanks for your help, I get it now

- #9

DrDu

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Today the dielectric function can be calculated from the microscopic properties of the medium.

A distinction into bound and free charges is not made any more and would also be difficult due to the quantum nature of the electrons which makes it impossible to ascribe some electron to be bound to a specific atom.

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"Bound charges" means that they cannot travel freely through the volume of the sample, as is the case for a conductor. Instead, they respond to the external electric field by displacing from the equilibrium positions by a microscopic distance. The macroscopic effect of this displacement is the appearance of an electric dipole moment in every

Now, imagine you have a point dipole with an electric dipole moment

[tex]

\rho(\mathbf{x}) = -\mathbf{d} \cdot \nabla \delta( \mathbf{x} - \mathbf{x}_0 )

[/tex]

For a continuous distribution, then, we would get:

[tex]

\rho(\mathbf{x}) = \int d\mathbf{x}' \, (-\mathbf{P}(\mathbf{x}') \cdot \nabla) \delta(\mathbf{x} - \mathbf{x}')

[/tex]

[tex]

= -\nabla \cdot \int d\mathbf{x}' \, \mathbf{P}(\mathbf{x}')\delta(\mathbf{x} - \mathbf{x}')

[/tex]

[tex]

= -\nabla \cdot \mathbf{P}(\mathbf{x})

[/tex]

Notice that, because the bound charge density is given by the (negative) divergence of the polarization, there would be an ambiguity in the definition of polarization up to a curl of an arbitrary vector field. Namely:

[tex]

\tilde{\mathbf{P}}(\mathbf{x}) = \mathbf{P}(\mathbf{x}) + \nabla \times \mathbf{K}(\mathbf{x})

[/tex]

would give the same charge density.

However, if integrate any over any volume to find the total dipole moment, we would get:

[tex]

\tilde{\mathbf{d}} = \mathbf{d} + \oint_{S} da' \hat{\mathbf{n}}' \times \mathbf{K}(\mathbf{x}')

[/tex]

where we used Gauss's Theorem. If we want the dipole moment to be unchanged, we must make the tangential component of

A similar consideration for magnetic materials leads to the notion of an induced magnetic dipole moment density -

[tex]

\mathbf{J}(\mathbf{x}) = \nabla \times \mathbf{M}(\mathbf{x})

[/tex]

Again, adding a gradient of a scalar to the magnetization [itex]\tilde{\mathbf{M}} = \mathbf{M} + \nabla F[/itex] gives the same current density. Thus, specifying only the current density does not uniquely determine the magnetization. The total magnetic moment goes to:

[tex]

\tilde{\mathbf{\mu}} = \mathbf{\mu} + \oint_{S} da' \hat{\mathbf{n}}' F(\mathbf{x}')

[/tex]

The requirement that the magnetic moment is unaltered for an arbitrary surface fixes F = 0.

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- #11

DrDu

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Dickfore, I think this whole description of polarization in terms of dipole densities of bound charges is very outdated. The polarisation can be defined as [itex] P(t)=\int_{-\infty}^{t}j(t) [/itex] where j(t) is the microscopic charge density. "Macroscopic" values of P or D or epsilon are simply obtained considering the low wavenumber components of its Fourier transform.

- #12

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Well, your "definition" has several issues:

- What is your integration variable t?
- As is writen, your formula is dimensionally inconsistent
- The polarization is a vector, whereas charge density is a scalar. So, the formula is also rotationally inconsistent.

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One has to make a distinction between the definition of electric displacement:

[tex]

\mathbf{D} = \varepsilon_{0} \mathbf{E} + \mathbf{P}

[/tex]

and the

[tex]

\mathbf{D} = \varepsilon_{0} \, \varepsilon \, \mathbf{E}

[/tex]

which is valid only for

My previous post was concerned with expressing a charge and current density if a polarization and magnetization are given. It was not concerned with any microscopic details of the material, and is a simple mathematical statement following from the definitions of the quantities involved.

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DrDu

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Well, your "definition" has several issues:

- What is your integration variable t?
- As is writen, your formula is dimensionally inconsistent
- The polarization is a vector, whereas charge density is a scalar. So, the formula is also rotationally inconsistent.

t is time, and j is the vector of electric current. So no inconsistency.

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t is time, and j is the vector of electric current. So no inconsistency.

Because you said j is the charge density. But, your formula follows from mine through the continuity equation:

[tex]

\frac{\partial \rho'}{\partial t} + \nabla \cdot \mathbf{J} = 0

[/tex]

assuming no external free charges are present (notice that bound currents are solenoidal, i.e. their divergence is zero).

EDIT:

There needs to be a correction to the formula for the bound current density as the curl of magnetization. Namely, it is only valid if the polarization is time-independent. For time-dependent polarization, we have:

[tex]

\mathbf{J}_{\mathrm{bound}} = \nabla \times \mathbf{M} + \frac{\partial \mathbf{P}}{\partial t}

[/tex]

This ensures the law of conservation of bound charges (continuity eqn. for bound charge and current density), which is valid if there is no ionization of the material, when some of the bound charges break apart from their bonds in the molecules, and become free to move.

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- #16

DrDu

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Oh yes, this was a typo. I wanted to say current density.

EDIT:

There needs to be a correction to the formula for the bound current density as the curl of magnetization. Namely, it is only valid if the polarization is time-independent.

Yes, that is true, but I am more used to time dependent phenomena. Electrostatics and magnetostatics are for engineers :-)

My point is that it is somewhat artificial to distinguish between bound and free charges. Even relatively localized bonds, like e.g. in silicon form bands which are spread out over the whole solid and this delocalization effects the dielectric function.

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My point is that it is somewhat artificial to distinguish between bound and free charges. Even relatively localized bonds, like e.g. in silicon form bands which are spread out over the whole solid and this delocalization effects the dielectric function.

This is wrong. Bands are spread out in energy, not position. The important thing is that there is a gap between the highest occupied energy level and the next free energy level (at zero temperature). Thus, silicon is a band insulator. However, since this gap is rather small, it is characterized as a semiconductor, since there is an appriciable number of electrons than can become conduction due to thermal excitations at room temperature. But, that is another story, that is derailing this thread.

Bound charges are a good abstraction when one considers macroscopic phenomena. The macroscopic Maxwell's equations are obtained by an averaging procedure over volumes that contain a large number of structural units (atoms/molecules), and bound charges are charges that get displaced by essentially confined to interatomic distances. Thus, it makes sense to speak of surface charges and currents.

I once again want to emphasize that introducing the D and H fields does not solve a problem, and that it is solely based on the distinction between bound and free charges. One needs to close the system of equations by assigning relationships between these and the E and B foelds. This gives the response functions of the medium, which depend on the atomic structure of the material.

But, I would further argue that the distinction between bound and free charges fails in the same situation where the dielectric constant looses its meaning. For example, a conductor with conductance [itex]\sigma[/itex] formally has a dielectric function that is frequency dependent:

[tex]

\tilde{\varepsilon}(\omega) = i \frac{\sigma}{\varepsilon_0 \, \omega}

[/tex]

First of all, this is a purely imaginary response function, which indicates that the processes involved are dissipative. Second, it has a simple pole at zero frequency, which indicates the inability of existence of zero electric field in the static limit.

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DrDu

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This is wrong. Bands are spread out in energy, not position.

What I meant is that the orbitals forming the band are spread out over the whole solid. Furthermore orbitals with different k vector have different energies.

A classical example where completely free charges are described in terms of the dielectric function is the Lindhard dielectric function:

http://iopscience.iop.org/1402-4896/28/3/005/pdf/physscr_28_3_005.pdf

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