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What really is the electric field E appearing in the constitutive relation ? 
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#1
May1212, 12:41 AM

P: 13

I assume that everyone knows the constitutive relation of dielectric material :
D = ε_0E + P where D is the electric field displacement ε_0 is permittivity of vacumm E is the electric field (or may be the electric ) and P is the polarization (density*dipole) I am confusing about the electric field E. Is it the external electric field or the local electric field in the dielectric material. I think that it is the local electric field in the dielectric material. If we apply the external electric field E0 to this material, the local electric field should be changed due to the induced electric field. However, most of classical theories of metal and dielectric materials use E0 in the constitutive relation. It is obviously incorrect. What do you think? 


#2
May1212, 06:31 AM

PF Gold
P: 1,141

Hello tapsanit,
in fact the quantity E in the definition of D is a macroscopic electric field. It can be understood as spatial average of the microscopic electric field: [tex] \mathbf E(\mathbf x) = \frac{1}{V} \int_V \mathbf E_{micro}(\mathbf x + \rho) d^3 \rho [/tex] where the integral is over small domain of volume V centred at x. The macroscopic field is not the same as the external field, because the latter does not include the contribution due to the material. 


#3
May1212, 09:59 AM

P: 13

Thank you Jano L. for replying but I am still confusing. If there is the external field E_{0}, the electric field E in the definition of the displacement field should be the sum of the macroscopic E_{m}, that is, E = E_{0} + E_{m}.
This relation is derived from the Gauss's law as described in Griffith's book page 175. He says that E is the total field, not just that portion generated by polarization. please correct me it I am wrong. I want to understand physics correctly. 


#4
May1212, 10:13 AM

PF Gold
P: 1,141

What really is the electric field E appearing in the constitutive relation ?
I can't figure where you do not understand this. In macroscopic electromagnetism, there is
 macroscopic electric field E  macroscopic polarization P, defined as average dipole moment of the molecules  macroscopic electric displacement D, defined as D = eps_0 E + P. The last quantity is only a shorthand to simplify the equations for nonconducting dielectrics. In general, one can do well just with E and P. The notion of external field, although very useful in other questions, does not enter the definition of these quantities. 


#5
May1212, 10:48 AM

P: 13

I am sorry. Let's me clarify this problem.
Let's consider the classical Drude model. You have known that the metal is absorbing dielectric. That means the dielectric constant is complex. The Drude model can give the complex dielectric constant. It treats the electrons in the metal as free electrons. When the external oscillating electric field is applied to the metal, the free electrons will be driven to oscillate relative to the positively ionic background. From the equation of Drude model, we can derive a dipole moment and then the polarization in term of the external electric field field E0. (I am sorry that I don't know how to write the equation. I hope that you understand what I mean. ) To obtain the dielectric constant as a function of frequency ε(ω), we need the constitutive relation : D = ε_0ε(ω)E = ε_0E + P Then, we can write down the polarization in term of the electric field as P = ε_0(ε(ω)1)E Then, we use the electric field E to be the external field E0 . After comparing this polarization with one derived from the Drude model, the solution of the complex dielectric constant as a function of frequency can be obtained. Why does the electric field E become the external electric field E0? That is actually what I want to understand. Or, I have made the mistake? 


#7
May1212, 11:20 AM

P: 13

May be It is just my misunderstanding, the force acting upon the electron at point r are due to the electric field in the material at point r, F(r) = qE(r).
I had thought that the force is due to the external field, F = qE0 because we always assume the harmonic timedependence of E to be the same as E0. Then, I got confusing. Sorry for such a crazy question. 


#8
May1212, 11:53 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

if you mean E_{o} as the constant in a harmonic E = Re(E_{o}e^{iωt}),
then E_{o} is simply the maximum value of E 


#9
May1212, 12:01 PM

PF Gold
P: 1,141

Aha, I think I understand your questions now. I was thinking of very similar things lately too.
The difficulty is in finding the force acting on the electron [itex]a[/itex]. If we consider only the electric force, this is given by the microscopic field: [tex] \mathbf F = q\mathbf E_{micro} (\mathbf r_a,t), [/tex] which is different from the macroscopic field [itex]\mathbf E[/itex]; it should have more rapid and greater variations on the scale of atoms. The microscopic field can be understood as a sum of the external field and the fields due to all other particles in the medium: [tex] \mathbf E_{micro}(\mathbf r, t) = \mathbf E_0(\mathbf r, t) + \sum_b' \mathbf E_b (\mathbf r_a,t). [/tex] It is hard to find out usable expression for the sum, so usually a different approach is taken. There is a method of Lorentz, which, (if I understand it well) asserts that the microscopic field is approximately equal to the field [itex]\mathbf E_{cav}[/itex] that would be present in a spherical cavity when the macroscopic field in the medium outside the cavity is [itex]\mathbf E[/itex]. This can be calculated to be [tex] \mathbf E_{cav} (\mathbf r_a, t) = \mathbf E +1/(3\epsilon_0) \mathbf P. [/tex] So then the force acting on the electron is estimated as [tex] \mathbf E_{micro} \approx \mathbf E + \frac{1}{3\epsilon_0}\mathbf P. [/tex] The validity of this procedure however probably depends on the density of molecules and frequency of the light, so it should be taken with a grain of salt. I think Onsager has a paper with further development of this theory. EDIT: the resulting force on the electron has the same frequency as the source, but its magnitude is not that of the external field only, because all the other electrons contribute to it. 


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