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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!
[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!
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