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Speed of light in lossy dielectric medium

  1. Jan 31, 2013 #1
    It is usually written that the speed of light in a dielectric medium is ##v=\frac{c}{\sqrt{\epsilon_r}}##, where ##c## is the speed of light in vacuum and ##\epsilon_r## is the relative permittivity. But, how can it be calculated for lossy and not necessarily low-loss dielectrics, i.e. those with a complex permittivity ##\epsilon_r=\epsilon'-j\epsilon"##?

    a) ##v=\frac{c}{\sqrt{\epsilon'}}##?
    b) ##v=\frac{c}{\sqrt{|\epsilon_r|}}##?
    c) None of above?

    Related to previous question: If the natural resonance frequency of a resonant mode in an empty (vacuum) microwave cavity is ##f_0##, which is the natural resonance frequency for the same mode, but with the cavity completely-filled with the previous lossy dielectric material?

    a) ##f=\frac{f_o}{\sqrt{\epsilon'}}##?
    b) ##f=\frac{f_o}{\Re{(\sqrt{\epsilon'-j\epsilon"})}}##?
    b) ##f=\frac{f_o}{\sqrt{|\epsilon_r|}}##?
    c) None of above?

    All of the textbooks and webs (that I've found) about this topic, consider only lossless dielectrics or approximations for low-loss dielectrics, but no general expressions.
     
  2. jcsd
  3. Feb 1, 2013 #2

    K^2

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    Science Advisor

    It helps to understand the equation, not just know it. Ignoring, for the moment, frequency dependence of permittivity and permeability, electromagnetic wave is described by the following equation.

    [tex]\nabla^2 E = \mu \epsilon \frac{\partial^2 E}{\partial t^2}[/tex]

    In your case, ##\small \mu = \mu_0## and ##\small \epsilon = \epsilon_r \epsilon_0##. And to make things easier, lets take solution of the form ##E = E_0 e^{i(k\cdot x - \omega t)}##.

    [tex]- k^2 E = - \mu_0 \epsilon_r \epsilon_0 \omega^2 E[/tex]

    Since ##\small c^2 = 1/\mu_0\epsilon_0##, the above gives you ##k^2 = \frac{\epsilon_r \omega^2}{c^2}##.

    Now we can look at velocity. Since you seem to be interested in the phase velocity, we are looking at points in space and time where ##\small E## has same phase as ##\small E_0##. In other words, ##\small k \cdot x - \omega t## is purely imaginary. Realizing that ##\small x## and ##\small t## are always real, that gives you a simple enough formula.

    [tex]Re(k)\cdot x = Re(\omega)t[/tex]

    Differentiating both sides, we get the equation for phase velocity.

    [tex]v = \frac{dx}{dt} = Re\left(\frac{\omega}{k}\right)[/tex]

    From earlier, we have ##\frac{\omega^2}{k^2} = \frac{c^2}{\epsilon_r}##. So we have to take the square root here, keeping in mind that we are working with complex numbers.

    [tex]v = Re\left(\frac{c}{\sqrt{\epsilon_r}}\right)[/tex]

    If ##\small \epsilon_r## is real, that trivially simplifies to your formula. It is, however, not. We need to take square root of a complex number. In other words, we are looking for some ##\small z = a + ib## such that ##\small z^2 = \epsilon_r##.

    [tex]a^2 - b^2 + 2iab = \epsilon' + i\epsilon''[/tex]

    This is a system of equations with four possible solutions, but only one is physical. It's a bit messy, so I only give the result for a, which is all we need for the real part above.

    [tex]a = \frac{\sqrt{\epsilon' + \sqrt{\epsilon'^2 + \epsilon''^2}}}{\sqrt{2}}[/tex]

    So the correct formula for phase velocity is as follows.

    [tex]v = \frac{\sqrt{2}c}{\sqrt{\epsilon' + \sqrt{\epsilon'^2 + \epsilon''^2}}}[/tex]

    Which again simplifies to formula you have whenever ##\small \epsilon'' = 0##.
     
  4. Feb 4, 2013 #3
    Thank you very much.
     
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