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Dielectric function

  1. Jul 2, 2007 #1
    Hi guys:

    Could someone kindly explain the relation between dielectric funciton (e) and frequency, wavevetor? What is the condition for wavevector e(k=0, w), e(k, w=0) and why?

    Thanks a lot :smile:
  2. jcsd
  3. Jul 2, 2007 #2


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    the dielectric function depends on both the frequency and wave-vector independently because I get to choose with what frequency and with what wave-vector I probe the system. The system will respond and the response is given by the dielectric function. For example, if I wanted to know how the system responds to a very long wavelength (k->0) probe I would care about [tex]\epsilon(k=0,\omega)[/tex]. If, on the other hand, I rather care about the response to time-independent fields of finite wave-length I would instead care about [tex]\epsilon(k,\omega=0)[/tex].
  4. Jul 2, 2007 #3
    Are we talking probing with light here? In that case the frequency and the wavevector are not independent but related by [tex]\omega/k=c[/tex], right?
  5. Jul 2, 2007 #4


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    not necessarily. the response of the system is a property *of the system* and is independent of the probe used be it light or electrons or neutrons or etc.
  6. Jul 2, 2007 #5

    Claude Bile

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    Only in a vacuum. Inside a medium, particularly media whose permittivity varies as a function of position (waveguides, photonic crystals for example), the dispersion relation can get quite complex.

    Sinayu71 - There is no simple way to obtain how the permittivity changes with frequency, since this function ultimately depends on the band structure of the medium.

    For a regular, isotropic medium however, the permittivity does have a characteristic shape (minus the kinks and other perturbations), approximated in the optical region of the spectrum by the Selmier (sp?) equations.

  7. Jul 3, 2007 #6
    in the solid state physics book, when determine the plasma ocillation, k =0 is chosed in the dielectric funtion. However, the w=0 is chosed when discuss the potential screening. Can someone explain it?

    thank you :shy:
  8. Jul 3, 2007 #7


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    It's a pedagogical choice and nothing more. Those two simple cases are presented *because* they are simple and well known limits; a general study of the dielectric function is not at the level of any textbook.

    In the first example you give (plasma oscillation) one thinks of all the electrons moving together in the whole macroscopic metal--this response is obviously of macroscopic wavelength ([tex]k \to 0[/tex]) but finite frequency. For finite wavelength the plasma frequency is not the same as the k=0 case discussed in textbooks and is much more difficult to determine.

    In the second example, one considers the static ([tex]\omega = 0[/tex]) screening of a point charge. If the point charge were not at rest in the solid the problem would again be more difficult.
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