Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E-M waves in dielectrics

  1. Aug 27, 2006 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Let's see if I understand this correctly... In an isotropic material of conductivity [itex]\sigma \neq 0[/itex] (and charge density [itex]\rho=0[/itex], the 4th Maxwell equation has a non-null current term [itex]\vec{J}=\sigma \vec{E}[/itex] so the resulting wave equations for [itex]\vec{E}[/itex] and [itex]\vec{H}[/itex] take the form

    [tex]\nabla ^2\vec{E}=\frac{\epsilon \mu}{c^2}\frac{\partial^2}{\partial t^2}\vec{E}+\frac{4\pi \mu \sigma}{c^2}\frac{\partial}{\partial t}\vec{E}[/tex]

    and the same thing for [itex]\vec{H}[/itex]. (I am using the same unit convention as in Greiner's 'Classical Electrodynamics' book)

    If we try a solution of the form

    [tex]\vec{E}=\vec{E}_0e^{i(\vec{k}\cdot \vec{r}-\omega t + \phi)}[/tex]

    (the monochromatic plane wave solution), we find that it is a solution provided that the wave number vector satisfies the following complex dispersion relation:

    [tex]|\vec{k}|^2=k^2=\epsilon \mu\frac{\omega^2}{c^2}\left( 1+\frac{4\pi i\sigma}{\epsilon \omega}\right)[/tex]

    We know that if a complex function satisfies a linear diff. equ. such as the above "wave equation", its real and imaginary part taken separately are also solutions. In vacuum, separating the real and imaginary part of the sine wave solution was easy because [itex]\vec{k}[/itex] was real. Now, [itex]\vec{k}[/itex] is non-real as soon as the conductivity is non-zero, and to get the real part of [itex]\vec{E}_0e^{i(\vec{k}\cdot \vec{r}-\omega t + \phi)}[/itex], we must decompose the exponential according to Euler's formula and then expand the sine and cosine in their Taylor series, and finally group together the real parts and the imaginary parts.

    At this point I don't know what to think. This seems rather impractical. Does it turn out in the end that a monochromatic plane wave solutions (with real wave vector [itex]\mathcal{K}[/itex]) exists? Please comment!
     
    Last edited: Aug 27, 2006
  2. jcsd
  3. Aug 27, 2006 #2

    Claude Bile

    User Avatar
    Science Advisor

    Pure plane wave solutions do not exist because of the attenuation that inevitably occurs as a wave propagates through a dielectric. The degree of attenuation is determined by the imaginary component of the refractive index.

    If the attenuation is small, then a plane wave solution can be an excellent approximation to the full solution. More often though when modelling, theorists neglect the attenuation terms altogether, preferring instead to tack it on later, as loss is a figure that is typically measured rather than worked out using first principles, in the optical region of the EM spectrum at least.

    (On a side note with regard to your second last paragraph - Typically the real terms are brought out into a second exponential so you get two exponentials multiplied by one another, one representing the attenuation and one representing the oscillation.)

    Claude.
     
    Last edited: Aug 27, 2006
  4. Aug 28, 2006 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah, of course!
     
  5. Aug 28, 2006 #4

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The wave propagates like exp(ikz). You calculate k as
    k=k_r+k_i=[k^2]^1/2. This is a bilt complicated, but is done in EM textbooks. Then the proagation is exp[ik_r] exp[-k_i],
    showing the attenuation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: E-M waves in dielectrics
  1. E-M waves detail (Replies: 3)

Loading...