Dielectric Function: Relation to Frequency & Wavevector

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SUMMARY

The discussion centers on the relationship between the dielectric function (ε) and its dependence on frequency (ω) and wavevector (k). It is established that ε(k=0, ω) pertains to responses to long-wavelength probes, while ε(k, ω=0) relates to time-independent fields. The complexity of the dielectric function arises from the medium's band structure, particularly in non-isotropic materials. The choice of k=0 for plasma oscillation and ω=0 for potential screening is a pedagogical simplification, highlighting well-known limits in solid-state physics.

PREREQUISITES
  • Understanding of dielectric function and its significance in solid-state physics
  • Familiarity with wavevector (k) and frequency (ω) concepts
  • Knowledge of plasma oscillations and potential screening in materials
  • Basic principles of electromagnetic wave propagation in different media
NEXT STEPS
  • Study the Selmier equations for approximating permittivity in optical regions
  • Explore the band structure of various materials and its effect on dielectric properties
  • Investigate the dispersion relations in complex media like waveguides and photonic crystals
  • Learn about the mathematical modeling of plasma oscillations in solid-state physics
USEFUL FOR

Physicists, materials scientists, and electrical engineers interested in the properties of dielectric materials and their applications in optics and electronics.

sinayu71
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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:
 
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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 \epsilon(k=0,\omega). If, on the other hand, I rather care about the response to time-independent fields of finite wave-length I would instead care about \epsilon(k,\omega=0).
 
Are we talking probing with light here? In that case the frequency and the wavevector are not independent but related by \omega/k=c, right?
 
Repetit said:
Are we talking probing with light here?

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.
 
Repetit said:
Are we talking probing with light here? In that case the frequency and the wavevector are not independent but related by \omega/k=c, right?
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.

Claude.
 
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:
 
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 (k \to 0) 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 (\omega = 0) 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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