Construction of an equivalent dielectric tensor

  • #1

Homework Statement



A medium is described by the response functions [tex]\varepsilon (\omega )[/tex] and [tex]{\mu ^{ - 1}}(\omega )[/tex] in

[tex]\textbf{D} = \varepsilon (\omega )\textbf{E}, \textbf{H} = {\mu ^{ - 1}}(\omega )\textbf{B}.[/tex]

Construct the equivalent dielectric tensor [tex]{K_{ij}}(\omega ,\textbf{k})[/tex] in terms of [tex]\varepsilon (\omega )[/tex] and [tex]{\mu ^{ - 1}}(\omega )[/tex]

Homework Equations



[tex]\textbf{D} = \varepsilon (\omega )\textbf{E}[/tex]

[tex]\textbf{H} = {\mu ^{ - 1}}(\omega )\textbf{B}[/tex]

[tex]\textbf{P} = {\varepsilon _0}{\chi ^e}\textbf{E}[/tex]

[tex]\textbf{M} = {\chi ^m}\textbf{B}/{\mu _0}[/tex]

[tex]{P_i} = {\varepsilon _0}\chi _{ij}^e{E_j}[/tex]

[tex]{M_i} = \chi _{ij}^m{B_j}/{\mu _0}[/tex]

The Attempt at a Solution



I seriously have no idea. I know what the answer should be but I only need a push in the right direction. Dont know where to start or how to attack the problem. All help is appreciated.
 

Answers and Replies

  • #2
14
0
Hi! I actually did this one just a moment ago. I guess you are also solving exercise 6.1 in Melrose, McPhedran's book "Electromagnetic processes in dispersive media". :) You should use the following equations:

[tex]
K_{i,j}(\omega, \textbf{k}) = \delta_{ij} + \frac{i}{\omega \epsilon_0}\sigma_{ij}(\omega, \textbf{k})[/tex]

[tex]
\left(\textbf{J}_{ind}\right)_i(\omega, \textbf{k}) = \sigma_{ij}(\omega, \textbf{k})E_j(\omega, \textbf{k})[/tex]

[tex]
\left(\textbf{J}_{ind}\right)_i(\omega, \textbf{k}) = -i\omega P_i(\omega, \textbf{k}) + i \epsilon_{ijk}k_j M_k(\omega, \textbf{k})[/tex]

[tex]
P_i(\omega, \textbf{k}) = D_i(\omega, \textbf{k}) - \epsilon_0 E_i(\omega, \textbf{k})[/tex]

[tex]
D_i(\omega, \textbf{k}) = \epsilon(\omega)E_i(\omega, \textbf{k})[/tex]

[tex]
M_i(\omega, \textbf{k}) = \frac{1}{\mu_0}B_i(\omega, \textbf{k}) - H_i(\omega, \textbf{k})[/tex]

[tex]
H_i(\omega, \textbf{k}) = \mu^{-1}(\omega)B_i(\omega, \textbf{k})[/tex]

[tex]
B_i(\omega, \textbf{k}) = \frac{1}{\omega}\epsilon_{ijk}k_j E_k(\omega, \textbf{k})[/tex]

Then I guess the rest is straightforward. Good luck!
 
  • #3
14
0
Btw, is it possible that you also take T. Hellsten's course at KTH and have this exercise as a deadline until next thursday? Just wondering. :)
 

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