cagonder
- 3
- 0
Can someone please explain the concept of optical losses and its correlation with the imaginary part of the dielectric function in elementary terms. I am confused.
The discussion revolves around the concept of the imaginary part of the dielectric function and its relationship to optical losses, polarization of electromagnetic radiation, and the mathematical modeling of these phenomena. Participants explore theoretical aspects, calculations, and implications in various contexts, including classical models and wave equations.
Participants express varying viewpoints on the relationship between conductivity and complex permittivity, with some suggesting they are equivalent while others explore their simultaneous use. The discussion remains unresolved regarding the best approach to analyze these concepts together.
Some mathematical steps and assumptions regarding the models presented are not fully resolved, particularly concerning the treatment of complex permittivity and its implications for optical losses.
This is a very interesting question. For example, the time-independent wave equation can be expressed as,cagonder said:Thank you all for your responses.
How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?
aabottom said:I never seen an analysis that includes both conductivity and complex permittivity, but my search has not been exhaustive. This treatment has been nagging me for quite some time.
Any comments?
cagonder said:Thank you all for your responses.
How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?
Thank you. That confirms my supposition.DrDu said:Descriptions in terms of conductivity vs. complex permittivity are equivalent alternatives, hence it makes not much sense of using both at the same time.
I like to refer to the Lindhard model of the homogeneous free electron gas as a model of a simple metal. Although the electrons are free, the metal is described completely in terms of epsilon.aabottom said:Thank you. That confirms my supposition.