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Imaginary Part of Dielectric function

  1. Feb 7, 2011 #1
    Can someone please explain the concept of optical losses and its correlation with the imaginary part of the dielectric function in elementary terms. I am confused.
  2. jcsd
  3. Feb 8, 2011 #2


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    An imaginary part of the dielectric constant can only be due to the polarization which, in frequency space, is [tex]P=j(\omega)/(i \omega)[/tex] (please don't rely on my signs and factors) and therefore, [tex] \epsilon=\epsilon_0+ \sigma/i \omega[/tex] where sigma is the conductivity. If epsilon is real, the current is out of phase by 90 degree with the driving field and will do no work. If epsilon has an imaginary component, the current and the driving E field will be partly in phase so that due to ohms law energy is dispersed.
  4. Feb 8, 2011 #3
    How do you calculate the imaginary part of the dielectric function?
  5. Feb 9, 2011 #4


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    Not much differently from the real part. E.g. in a simple classical model of a solid consisting of damped harmonic oscillators, the damping would automatically lead to an imaginary part of polarization.
  6. Feb 10, 2011 #5
    The plane wave solution of the wave equation (from Maxwell's equations) in a homogeneous conducting media is given clearly in Panofsky & Phillips, p 200 [1]. For the plane wave

    E(z,t) = Eo*exp(-beta*z)*exp(i*[alpha*z-omega*t]),

    the real and imaginary parts of the propagation constant

    K = i*(alpha+i*beta) are,

    alpha = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}+1]/2)

    beta = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}-1]/2)

    k = the angular wavenumber = 2*pi/wavelength
    epsilon = permittivity
    sigma = conductivity
    omega =2*pi*f = angular frequency.

    Here, alpha is the real part of the angular wavenumber. beta is the attenuation coefficient and attributes to "optical loss" in terms of the Beer-Lambert Law [2].

    In perfect dielectric, sigma = 0, and
    alpha = k, and
    beta = 0,
    as expected.

    [1] Panofsky & Phillips, Classical Electricity and Magnetism 2nd Ed., Addison-Wesley, 1962.
    [2] http://en.wikipedia.org/wiki/Attenuation_coefficient
    Last edited: Feb 10, 2011
  7. Feb 10, 2011 #6
    Thank you all for your responses.

    How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?
  8. Feb 10, 2011 #7
    This is a very interesting question. For example, the time-independent wave equation can be expressed as,


    But what if the permittivity, epsilon, is complex, thus has real and imaginary parts, epsilonr and epsiloni,

    and and the media is conductive?

    If so then,
    E``(z)+(mu*epsilonr*omega^2+i*mu*omega*(sigma+ epsiloni *omega))*E(z)=0,

    so that the "effective conductivity" becomes,

    sigmaeff = sigma+epsiloni*omega, and

    alpha = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}+1]/2)

    beta = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}-1]/2).

    I never seen an analysis that includes both conductivity and complex permittivity, but my search has not been exhaustive. This treatment has been nagging me for quite some time.

    Any comments?
    Last edited: Feb 11, 2011
  9. Feb 11, 2011 #8


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    Descriptions in terms of conductivity vs. complex permittivity are equivalent alternatives, hence it makes not much sense of using both at the same time.
  10. Feb 11, 2011 #9


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    In general, the index of refraction or equivalently the dielectric function [tex] n(\omega,k)[/tex] is a function of both frequency [tex] \omega[/tex] and wavenumber k. In general, it is also a tensor. While the dielectric function is defined for arbitrary combinations of [tex] \omega[/tex] and k, a propagating wave has to fulfill an equation relating n, [tex]\omega[/tex] and k and the polarization. That's the field of crystal optics.
    See, e.g. Landau Lifgarbagez, Electrodynamics of Continuous media, or, for the specialist, Agranovich and Ginzburg, Crystal Optics with Spatial Dispersion, and Excitons.
  11. Feb 11, 2011 #10
    Thank you. That confirms my supposition.
  12. Feb 11, 2011 #11


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    I like to refer to the Lindhard model of the homogeneous free electron gas as a model of a simple metal. Although the electrons are free, the metal is described completely in terms of epsilon.
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