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Monochromatic waves and Maxwells equations

  1. Nov 25, 2011 #1

    Are there other reasons why monochromatic solutions to Maxwells equations of the form E(z, t) = E(z)exp(-iωt) are good other than its plane wave solutions forming a complete set?

  2. jcsd
  3. Nov 26, 2011 #2


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    The material equations D=εE and B=μH are valid only for stationary or very slowly varying fields. As D=ε0E+P, (P is the polarization of unit volume) D=εE means that the dipole moment of the atoms/molecules follows the electric field instantaneously. It is not the case when the fields change very fast in time. The dipole moment at a given time is determined by the past values of the electric field according to causality.

    [tex] \vec D(t)= \int_{-∞}^0 {K(\tau) \vec E(t-\tau)d\tau} [/tex],

    which is the convolution of the electric field the molecule feels with the function K.
    The Fourier transform of this convolution is the product of the Fourier transforms of the functions K and E. The Fourier transform of K is called epsilon(ω), the frequency-dependent permittivity or dielectric function.
    So the proportionality between P and E or D and E is valid for the Fourier coefficients (Fourier transforms) D(ω)=ε(ω)E(ω). That is why we consider the periodic electric field as a Fourier series, sum of terms Enexp(iωn). Even a monochromatic field has to be written as E0+exp(iωt)+E0-exp(-iωt).

    In case when the time needed for the polarization to reach equilibrium is comparable with the time period of the external electric field, there is a phase difference between D(ω) and E(ω), so ε(ω) is complex. As both E and D are real in the real word, the Fourier components belonging to -ω are equal to the conjugate of the component for ω. E0-=E0+* and D0-=D0+*, so ε(-ω)=ε(ω)*. The complex ε values are different for ω and -ω.

  4. Nov 26, 2011 #3
    Thanks for taking the time to write that. Can I ask which book you use as a reference for these things?

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