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Meaning of a complex polarizability?

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data

    As I understand it, the polarizability of a material is a measure of the tendency of the material to polarize when an electric field is applied across it. However, the polarizability can be both purely real and complex, and I'm not sure what a complex polarizability means physically. Does the imaginary part have something to do with some kind of damping, perhaps? Anyone who knows? Google didn't help me very much.
     
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  3. Jun 10, 2009 #2
    I had a thought. Since p=aE, where p is the induced dipole moment and a is the polarizability, a complex polarizability means that p is out of phase with E. In other words, there's a slight time delay between the application of the electric field and the response to it, the induced dipole moment. Am I on the right track?
     
    Last edited: Jun 10, 2009
  4. Jun 10, 2009 #3
    good question.

    I know that imaginary permitivity has to do with conductivity.
     
  5. Jun 10, 2009 #4

    gabbagabbahey

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    First off, it's important to note that polarizability is a property of individual atoms or molecules and both the electric field [itex]\mathbf{E}[/itex] and the atom's/molecule's dipole moment [itex]\mathbf{p}[/itex] must be real valued quantities.

    So, for a complex polarizability [itex]\tilde{\alpha}[/itex] to make any sense at all you would need to first define the electric field and dipole moments to be the real part of some complex quantities [itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{p}}[/itex]:

    [tex]\mathbf{E}=\text{Re}[\mathbf{\tilde{E}}], \, \; \mathbf{p}=\text{Re}[\mathbf{\tilde{p}}][/tex]

    And then you would have

    [tex]\mathbf{\tilde{p}}=\tilde{\alpha}\mathbf{\tilde{E}}[/tex]

    The only direct conclusion you can draw from this is that the electric field is out of phase with induced dipole moment.

    Physically, this scenario can occur when an atom is placed in an oscillating electric field (as in the case of an EM-wave incident on an atom) and there is some sort of velocity dependent damping of its electron(s) (the radiation reaction force produces a similar damping proportional to [itex]\mathbf{\ddot{v}}[/itex]). See for example Griffith's Introduction to Electrodynamics 3rd ed. section 9.4.3.

    When there are many atoms//molecules present (such as in any bulk material) that have at least one electron each that is to a large extent free to move about (such as in a conductor) this leads to a complex susceptibility, which results in attenuation/absorbtion of an incident EM-wave.
     
    Last edited: Jun 10, 2009
  6. Jun 11, 2009 #5
    Thank you for a very helpful reply, gabbagabbahey.

    Found this book online. Great resource. However, in the section you mention (below equation 9.158), Griffith states that "the angle arctan(some argument) (...) rises to [tex]\pi[/tex] when [tex]\omega[/tex] is much greater than [tex]\omega_0[/tex]". It makes sense that p and E can be [tex]\pi[/tex] radians out of phase, but the limit as arctan(x) goes to infinity (plus or minus) is [tex]\pi/2[/tex] (again, plus or minus) as far as I know.
     
  7. Jun 12, 2009 #6

    gabbagabbahey

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    Although Griffith's never explicitly says so in his text (IIRC), when he uses [itex]\tan^{-1}[/itex] to represent the inverse tangent, he is using the multivalued version; not [itex]\arctan[/itex] which always returns a value between [itex]\frac{-\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex].

    In other words, when he gives an equation of the form [itex]a=\tan^{-1}(b)[/itex] he means that [itex]b=\tan(a)[/itex] not [itex]a=\arctan(b)[/itex]
     
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