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- Thread starter cagonder
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DrDu

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How do you calculate the imaginary part of the dielectric function?

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DrDu

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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

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

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How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?

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This is a very interesting question. For example, the time-independent wave equation can be expressed as,

How is the polarization of the EM radiation waves relevant to the imaginary part of the dielectric function?

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?

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DrDu

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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.

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DrDu

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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.

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Thank you. That confirms my supposition.Descriptions in terms of conductivity vs. complex permittivity are equivalent alternatives, hence it makes not much sense of using both at the same time.

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DrDu

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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.Thank you. That confirms my supposition.

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