Imaginary Part of Dielectric function

In summary, the imaginary part of the dielectric function is due to the polarization of the EM radiation waves. The effective conductivity is also affected by the polarization.
  • #1
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.
 
Physics news on Phys.org
  • #2
An imaginary part of the dielectric constant can only be due to the polarization which, in frequency space, is [tex]P=j(\omega)/(i \omega)[/tex] (please don't rely on my signs and factors) and therefore, [tex] \epsilon=\epsilon_0+ \sigma/i \omega[/tex] where sigma is the conductivity. If epsilon is real, the current is out of phase by 90 degree with the driving field and will do no work. If epsilon has an imaginary component, the current and the driving E field will be partly in phase so that due to ohms law energy is dispersed.
 
  • #3
How do you calculate the imaginary part of the dielectric function?
 
  • #4
Not much differently from the real part. E.g. in a simple classical model of a solid consisting of damped harmonic oscillators, the damping would automatically lead to an imaginary part of polarization.
 
  • #5
The plane wave solution of the wave equation (from Maxwell's equations) in a homogeneous conducting media is given clearly in Panofsky & Phillips, p 200 [1]. For the plane wave

E(z,t) = Eo*exp(-beta*z)*exp(i*[alpha*z-omega*t]),

the real and imaginary parts of the propagation constant

K = i*(alpha+i*beta) are,

alpha = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}+1]/2)

beta = k*sqrt([sqrt{1+(sigma/[epsilon*omega])^2}-1]/2)

where
k = the angular wavenumber = 2*pi/wavelength
epsilon = permittivity
sigma = conductivity
omega =2*pi*f = angular frequency.

Here, alpha is the real part of the angular wavenumber. beta is the attenuation coefficient and attributes to "optical loss" in terms of the Beer-Lambert Law [2].

In perfect dielectric, sigma = 0, and
alpha = k, and
beta = 0,
as expected.



[1] Panofsky & Phillips, Classical Electricity and Magnetism 2nd Ed., Addison-Wesley, 1962.
[2] http://en.wikipedia.org/wiki/Attenuation_coefficient
 
Last edited:
  • #6
Thank you all for your responses.

How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?
 
  • #7
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?
This is a very interesting question. For example, the time-independent wave equation can be expressed as,

E``(z)+(mu*epsilon*omega^2+i*mu*omega*sigma)*E(z)=0.

But what if the permittivity, epsilon, is complex, thus has real and imaginary parts, epsilonr and epsiloni,
epsilon=epsilonr+i*epsiloni,

and and the media is conductive?

If so then,
E``(z)+(mu*epsilonr*omega^2+i*mu*omega*(sigma+ epsiloni *omega))*E(z)=0,

so that the "effective conductivity" becomes,

sigmaeff = sigma+epsiloni*omega, and

alpha = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}+1]/2)

beta = k*sqrt([sqrt{1+(sigma/[epsilonr*omega]+epsiloni/epsilonr)^2}-1]/2).

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?
 
Last edited:
  • #8
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?

Descriptions in terms of conductivity vs. complex permittivity are equivalent alternatives, hence it makes not much sense of using both at the same time.
 
  • #9
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?

In general, the index of refraction or equivalently the dielectric function [tex] n(\omega,k)[/tex] is a function of both frequency [tex] \omega[/tex] and wavenumber k. In general, it is also a tensor. While the dielectric function is defined for arbitrary combinations of [tex] \omega[/tex] and k, a propagating wave has to fulfill an equation relating n, [tex]\omega[/tex] and k and the polarization. That's the field of crystal optics.
See, e.g. Landau Lifgarbagez, Electrodynamics of Continuous media, or, for the specialist, Agranovich and Ginzburg, Crystal Optics with Spatial Dispersion, and Excitons.
 
  • #10
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.
Thank you. That confirms my supposition.
 
  • #11
aabottom said:
Thank you. That confirms my supposition.
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.
 

What is the imaginary part of the dielectric function?

The imaginary part of the dielectric function, also known as the absorption coefficient, is a measure of how much light is absorbed by a material as it passes through it. It is related to the material's ability to store and dissipate energy, and is an important parameter in understanding the optical properties of a material.

How is the imaginary part of the dielectric function calculated?

The imaginary part of the dielectric function is calculated using a mathematical formula that takes into account the material's refractive index, absorption coefficient, and wavelength of light. It is typically calculated using experimental data obtained from spectroscopic techniques such as ellipsometry or reflectance spectroscopy.

What factors affect the value of the imaginary part of the dielectric function?

The value of the imaginary part of the dielectric function is affected by various factors, including the type of material, its composition, and the wavelength of light. Additionally, the presence of impurities, defects, and surface roughness can also impact the value of the imaginary part of the dielectric function.

Why is the imaginary part of the dielectric function important in materials science?

The imaginary part of the dielectric function is important in materials science because it provides valuable information about a material's optical properties, which can be used to understand its electronic and structural properties. It is also crucial in the design and development of new materials for various applications, such as solar cells, optical sensors, and electronic devices.

Can the imaginary part of the dielectric function be used to determine a material's quality?

Yes, the imaginary part of the dielectric function can be used as a measure of a material's quality. A high absorption coefficient in a material can indicate the presence of defects or impurities, while a low absorption coefficient can indicate a high-quality, defect-free material. However, other factors such as surface roughness and sample preparation must also be considered when using the imaginary part of the dielectric function as a measure of material quality.

Suggested for: Imaginary Part of Dielectric function

Replies
3
Views
2K
Replies
1
Views
624
Replies
1
Views
895
Replies
0
Views
1K
Replies
8
Views
2K
Replies
10
Views
3K
Replies
1
Views
1K
Back
Top