Complex dielectric constant implies absorption?

In summary, dielectric constants describe the ability of a material to impede the propagation of electric and magnetic fields. When the imaginary part of the dielectric function is zero, the material is said to be an ideal medium for wave propagation. The imaginary part of the dielectric function is always positive, meaning that the medium is transferring energy to the wave. References consulted include Omar's "Elementary Solid State Physics", Kittel, and the study guide given to us by our lecturer. Jackson's "Classical Electrodynamics" is also a good reference.
  • #1
adamjford
6
0
Hi everyone,

My very first post after simply using physicsforums as a (very) handy FAQ resource.

Okay, so my question is probably pretty simple, but I've been unable to find anything on here already.

How does a complex dielectric constant imply the absorption of EM radiation? I understand that the frequency dependence of the dielectric constant can be expressed as a complex function, but can't seem to understand mathematically how the imaginary part of the function corresponds to a loss of energy (wave) to the medium (dielectric)... implying absorption.

Three resources I've used seem to make the mathematical and physical arguments pertaining to my question too quickly for me to understand, so any help is greatly appreciated!

Cheers,

Adam.
 
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  • #2
adamjford said:
Hi everyone,

My very first post after simply using physicsforums as a (very) handy FAQ resource.

Okay, so my question is probably pretty simple, but I've been unable to find anything on here already.

How does a complex dielectric constant imply the absorption of EM radiation? I understand that the frequency dependence of the dielectric constant can be expressed as a complex function, but can't seem to understand mathematically how the imaginary part of the function corresponds to a loss of energy (wave) to the medium (dielectric)... implying absorption.

Three resources I've used seem to make the mathematical and physical arguments pertaining to my question too quickly for me to understand, so any help is greatly appreciated!

Cheers,

Adam.

One way to think of it is that the imaginary part of the dielectric function is proportional to the conductivity so that when there is an imaginary part to the dielectric function there is a non-zero current in the bulk. This means that is macroscopic motion of electrons (j~nev) and hence the field does work on them (W~Fd~eEvdt) and loses energy.

Probably a better thing to do is just to write down the expression for energy lost (dQ/dt--follow along with Jackson or some other reference) and see explicitly that dQ/dt is the integral of (|E|^2*Im(\epsilon)) and so is zero if the imaginary part of the dielectric function is zero.

What are the references that you have consulted previously?
 
  • #3
Thanks for the reply,

I've used Omar's "Elementary Solid State Physics", Kittel, and the study guide given to us by our lecturer. Your second method sounds good to me... I think the penny may finally drop if I can firstly show that dQ/dt is the integral you mentioned, and subsequently show that the integral, when evaluated, implies a loss of energy to a wave traveling through the dielectric. (I'm guessing some sort of decaying exponential function maybe?)

Does this sound reasonable?

Also a quick search of the reference you mentioned, Jackson's "Classical Electrodynamics", yeah?

Thanks again!
 
  • #4
adamjford said:
Thanks for the reply,

I've used Omar's "Elementary Solid State Physics", Kittel, and the study guide given to us by our lecturer. Your second method sounds good to me... I think the penny may finally drop if I can firstly show that dQ/dt is the integral you mentioned, and subsequently show that the integral, when evaluated, implies a loss of energy to a wave traveling through the dielectric. (I'm guessing some sort of decaying exponential function maybe?)

Does this sound reasonable?

Also a quick search of the reference you mentioned, Jackson's "Classical Electrodynamics", yeah?

Thanks again!

Yes, that's the Jackson I was thinking of. In the third addition of that book look in section 6.8.

Also, I think a better reference is Landau and Lifgarbagez "electrodynamics of continuous media". In the sencond edition of that book see section 80.
 
  • #5
Thanks!

And my reasoning is sound, yeah?
 
  • #6
Also... will a negative imaginary part correspond to the medium transferring energy to the wave, as opposed to absorbing from it?
 
  • #7
From what I remember from Jackson's text, an EM wave would have form like
[tex]
E(t) = E_0 e^{i\omega t - \vec{k}\cdot\vec{x}}
[/tex]
and in a medium it becomes
[tex]
E(t) = E_0' e^{i\omega t - n \vec{k} \cdot\vec{x}}
[/tex]
because the frequency stays the same at the boundary but the wavelength and speed change. If the index of refraction is complex, then you get an exponentially decaying solution. Actually, there might be an exponentially growing solution to the wave equation but this one probably doesn't conserve energy and is physically not meaningful.

This is all from memory a few years back so I could be wrong.
 
  • #8
I think that the imaginary part of the dielectric constant in condensed matter physics is always positive, so that the structure factor [tex]S(q,\omega) \propto - \textrm{Im} [1/\varepsilon(q,\omega)][/tex] describing dissipation is also positive.
 
  • #9
kanato said:
...Actually, there might be an exponentially growing solution to the wave equation but this one probably doesn't conserve energy and is physically not meaningful.

A growing solution is also physically possible if the medium is unstable or is out of equilibrium, like in lasers, etc.
 
  • #10
adamjford said:
Also... will a negative imaginary part correspond to the medium transferring energy to the wave, as opposed to absorbing from it?

yes. thus for a material in it's ground state the imaginary part of the dielectric function should always be positive.
 
  • #11
Thanks everybody for all your help. Much appreciated!
 

FAQ: Complex dielectric constant implies absorption?

1. What is the complex dielectric constant?

The complex dielectric constant, denoted by ε, is a measure of how a material responds to an electric field. It consists of two components: the real part (ε') which represents the material's ability to store energy, and the imaginary part (ε'') which represents the material's ability to dissipate energy through absorption.

2. How is the complex dielectric constant related to absorption?

The imaginary part of the complex dielectric constant (ε'') is directly related to the absorption of electromagnetic radiation by a material. A higher value of ε'' means that the material is more absorptive, while a lower value means that the material is more transparent to electromagnetic radiation.

3. What factors affect the complex dielectric constant and absorption?

The complex dielectric constant and absorption are affected by several factors, including the material's chemical composition, physical structure, temperature, and the frequency of the applied electric field. In general, materials with higher electron mobility and fewer impurities tend to have higher values of ε'' and therefore exhibit stronger absorption.

4. How is the complex dielectric constant measured?

The complex dielectric constant can be measured experimentally using techniques such as dielectric spectroscopy or impedance spectroscopy. These methods involve applying an alternating electric field to a material and measuring its response, which can then be used to calculate the values of ε' and ε''.

5. What are some practical applications of understanding complex dielectric constant and absorption?

Knowledge of the complex dielectric constant and absorption is essential in various fields, including material science, electrical engineering, and telecommunications. It can be used to design and optimize the performance of electronic devices, such as capacitors, antennas, and solar cells. Additionally, understanding the complex dielectric constant of materials can aid in the development of new technologies, such as microwave heating and medical imaging.

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