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Can you point out physical interpretation of the complex permittivity?

  1. Apr 6, 2009 #1
    Hi,

    Can u point out physical interpretation of the complex permittivity? (ie. what can u expect from a material with [tex]\epsilon[/tex]' = x and [tex]\epsilon[/tex]'' = y ? what happens if y is very large etc.

    [tex]\epsilon[/tex] = [tex]\epsilon[/tex]' -j [tex]\epsilon[/tex]''

    thanks

    Krindik
     
  2. jcsd
  3. Apr 6, 2009 #2

    Born2bwire

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    The imaginary part can model gain/loss of a medium. Normally we model the imaginary part to be equal to

    [tex]\frac{\sigma}{\omega}[/tex]

    [tex]\sigma[/tex] is the conductivity of the medium and has been defined for a myriad of materials. What happens in the material is that the conductivity allows for the movement of electrons, excited by incident electromagnetic waves. The moving electrons create eddy currents which in turn create secondary electromagnetic waves. These secondary waves cancel the incident wave. The result is that a conductive medium will attenuate waves as they travel through the medium. A perfect conductor, one that has infinite conductivity, will suppress all waves from traveling in the conductor. The result here is that you have currents confined to the surface of the conductor that perfectly cancel any incident wave.

    If we make the imaginary part of epsilon to be positive, using the j convention, then the fields will increase as the travel through the medium. I'm not aware of any physical simulation where you would do this though the real part of epsilon can certainly become negative (like in a plasma or left-handed material).
     
  4. Apr 7, 2009 #3
    Thanks.

    As I read elsewhere
    [tex]{\epsilon'' > 0}[/tex] - passive medium(as you said where the wave is attenuated)
    [tex]{\epsilon'' < 0}[/tex] - active medium where energy is supplied to the wave and amplification happens as in laser/maser (this is what I read)


    So, [tex]{\epsilon'}[/tex] does mean what it represents in simple media?
    ie. [tex]{\epsilon'} = {\epsilon_0}\cdot{\epsilon_r}[/tex] ?
     
  5. Apr 7, 2009 #4

    Born2bwire

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    Ah, I don't do optics at all so I didn't think of modeling laser's as having an active medium but I guess that certainly would fit the bill. I always work with passive mediums.

    The real part describes the behavior of the medium as a dielectric and is indicative of how well the medium can polarize. It can also be abstractly thought of as relating to capacitance, and just like a capacitor a higher real permittivity allows for more energy to be stored in a medium given a constant electric field amplitude. When we have an electric field in a medium, the field exerts a force on charges in the medium. This causes a separation, the negative charges move opposite to the field while the positive charges move with the field, to result in polarization of the medium. The polarization is a local effect, the overall charge doesn't change, but it allows us to store energy in the medium.
     
    Last edited: Apr 7, 2009
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