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Construction of an equivalent dielectric tensor

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Feb 13, 2010 #2
    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!
     
  4. Feb 13, 2010 #3
    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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