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Complex dielectric constant implies absorption?

  1. Aug 20, 2009 #1
    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.
     
  2. jcsd
  3. Aug 21, 2009 #2

    olgranpappy

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    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?
     
  4. Aug 21, 2009 #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 travelling 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!
     
  5. Aug 21, 2009 #4

    olgranpappy

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    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.
     
  6. Aug 21, 2009 #5
    Thanks!

    And my reasoning is sound, yeah?
     
  7. Aug 21, 2009 #6
    Also... will a negative imaginary part correspond to the medium transferring energy to the wave, as opposed to absorbing from it?
     
  8. Aug 22, 2009 #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.
     
  9. Aug 22, 2009 #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.
     
  10. Aug 22, 2009 #9
    A growing solution is also physically possible if the medium is unstable or is out of equilibrium, like in lasers, etc.
     
  11. Aug 22, 2009 #10

    olgranpappy

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    yes. thus for a material in it's ground state the imaginary part of the dielectric function should always be positive.
     
  12. Aug 29, 2009 #11
    Thanks everybody for all your help. Much appreciated!
     
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