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Time varying e-field polarization

  1. Oct 8, 2006 #1
    i'm in a bit of a tizzy with an optics class question... it goes like this:

    under the application of a time varying electric field the induced polarization in a dielectric may be described by the equation:
    P = [(Ne^2)/(-m(omega)^2-im(omega)(gamma)+k)][E+(1/3(permittivity))P]
    where the electric field at an electron within the dielectric is seperated into the macroscopic field, E, and the field due to the medium polarization, (1/3(permutivity))P. with reference to vector calculus and the physics involved, describe the origin of the second mentioned term.

    the book mentions nothing on the vector calc involved, and looking online isn't panning out to well.
    cheers for your help!
  2. jcsd
  3. Oct 9, 2006 #2


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    In short,

    [tex]\vec{P} = \epsilon_0\ \tilde{\chi_e}(\omega)\left[\vec{E}_{\mbox{macro}}+\frac{\vec{P}}{3\epsilon_0}}\right][/tex]

    (There is a tilde on the chi, indicating it's complex. The real part of this equation describes how P varies as a function of time under the influence of a single plane wave of frequency [itex]\omega[/itex] but the real part of chi itself is not a susceptibility, because it does not establish a relation of direct proportionality btw P and E. There is a phase difference btw P and E "induced" by the i in the complex susceptibility. P "lag" behind E.)
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3


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    you sure it's a "+" and not a "-"?

    Because the relation is

    [tex]\vec{P}=\epsilon_0\chi_e \vec{E}_{\mbox{total}}[/tex]

    and the total field can be broken into the "macroscopic field", which is the field due to everything besides the atom's own polarization and the "polarization field", which is (in approx.) [itex]-\vec{P}/3\epsilon_0[/itex].

    This is kinda neat. We know that given a sphere of "frozen" polarization P, the field it produces is [itex]-\vec{P}/3\epsilon_0[/itex] inside itself and the field of a perfect dipole outside itself. So say an external field is applied on a material. Every atom in the material then becomes polarized, creating a field of their own, which modifies the polarization of every atom, which again modifies the polarization of every atom, until some kind of equilibrium is attained. And in this equilibirum, the polarization of an atom is a function of the external field + it's own polarization. Of course, self-field in terms of P can be put on the other side of the equation and the total P is then only a function of the external field.
    Last edited: Oct 9, 2006
  5. Oct 10, 2006 #4
    the assignment sheet our prof gave us, as well as our in class notes, have a plus sign between the two terms.
    i'll have to bring that up with him when the next class rolls around.

    so, i can get to this solution by knowing that we induce a polarization by displacing a certain number of charges by some distance, and that during this there would have to be a balance of forces (displacing=restoring)... from which i can get the very basic equation for polarization of:
    P = [Ne^2/k][E]
    then to achieve the expression for the polarization under a time-varying field, we consider the charge exacutes damped simple harmonic motion (therefore giving a new force balance equation to bring in the complex bits) which then gives us the polarization:
    P = [(Ne^2)/(-m(omega)^2-im(omega)(gamma)+k)][E]
    but since thats only the external field we need to add the contribution of the internal field, which is where the (1/3(permittivity))P comes into play.
  6. Sep 22, 2008 #5
    Hi Quasar,

    could you please review that idea of polarization due to total electric field E_tot?

    The external electric field polarizes the material. The materials response cause an induced field (which is proportional to the polarization P) that reduces the externally applied field.
    So E_ext > E_tot since E_tot= E_ext-E_induced.

    The relation P proportional to E_tot. I would say that the bigger P, the smaller E_tot, so the relation should be INVERSELY proportional instead of proportional.
    I know you formula is right, but I cannot find a logic out of it....
    thanksa lot!
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