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I Interpreting the Permittivity of Metals

  1. Jun 27, 2016 #1

    Twigg

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    In a dielectric material, permittivity makes sense, because it's easy enough to interpret the polarization density as the number density of molecules times their induced dipole moment. I'm basing this description off chapter 4 of Griffith's E&M textbook. In this model, the charges are all valence electrons and nuclei. What is the nature of polarization in metals? Clearly light interacts with some kind of miroscopic dipole-oscillators in metals, or else all metals would be silver-colored. It wouldn't make sense for there to be a restoring oscillator force on a conduction band electron, so I assume these dipole-oscillators are dipole moments formed between gold nuclei and valence band electrons.

    What is the criteria for a metal to start behaving more like a dielectric than a conductor in a given wavelength band? Intuitively, I want to say that metals start looking more like dielectrics than conductors when the restoring force on an atomic dipole is less stiff than the Ohmic drag force on conduction electrons, i.e., that the bound charges are less constrained than the conduction charges. If this is the right idea, what's the formal criteria for the threshold wavelength?

    Sorry if this is too open-ended. Any pointers, references, or partial answers would be greatly appreciated.
     
  2. jcsd
  3. Jun 28, 2016 #2
    First off, I will most likely not be able to help you since I know next to nothing about this, but your question sounds interesting.
    I always thought of polarization in metals as a macroscopic effect.Charges, not moving on the scale of molecules but free electrons that actually redistribute themselves.
    Also, I personally find your comment quite confusing.:frown::confused:
    A few reasons for that :

    I assume by silver colored you mean "very good reflectors(which aren't very frequency specific)".
    Why exactly would that have to be ?
    Aren't there higher energy states than the conduction band into wich the electrons could get excited ?
    Even unbound electrons can get excited, so why would the valence electrons have any part in this ?
    Aren't those much harder to influence with electric fields?

    What do you mean by "behaving more like a dielectric" ? Transmitting or absorbing ?
    Since even the tiniest thicknesses of metal are already not transmitting light, I think we can dismiss transmitting as a thought.
    So what about absorbing ? Can't the electrical resistance in the material cause the light to be absorbed rather than reflected ?
    And another way to explain the absorption may be that the electrons relax some of their energy they got from the light into lattice vibrations
    What about absorption would be "dielectric behavior" ?
    PS:To my knowledge, it is very hard to predict the optical properties of a substance just by knowing its "electric" structure.
     
  4. Jun 28, 2016 #3

    Twigg

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    On the contrary, your post is very helpful. I made some overhasty assumptions that I might not have noticed if you hadn't questioned my train of thought. Thank you!

    I'm quite confused myself, and I should have said so more clearly in the original post. To clarify, what I'm writing are just my thoughts and limited understanding. If at any point I'm spouting complete nonsense, please someone speak up.

    To be embarrassingly honest, I hadn't even thought of that. I think you may be right that valence electrons shouldn't be what's responsible for the resonances of metals. Unbound electron transitions sounds way more plausible. What would the modes/states of the unbound electrons be, and in what way do they satisfy the condition that E = 0 in the interior of the metal for incident EM wavelengths far off resonance? How do these states produce microscopic dipoles, with resonance characteristic behavior comparable to the molecular dipoles in a polar dielectric material like quartz? Are they some kind of surface charge distribution that oscillates with the driving field? Is the absorbance of the metal then determined by the inelastic collision of these excited free electrons with the nuclei and/or lattice vibrations? Apologies if this isn't making much sense. I would like to know how much of what I'm thinking is wrong.

    I think I'm beginning to see what you mean. There seem to be non-electric pathways for excited electrons to decay.

    Thank you!
     
  5. Jun 29, 2016 #4
    That is pretty much how I imagined it to this day.

    I guess we'll have to wait for someone who can tell us whether that is a good model or not and how it really works.

    It seems strange to me that "free electrons" would have modes and resonances :confused:
    But how else would you explain the frequency dependencies of the absorption ?
    I am just guessing that the electrons "resonate" pretty well at all frequencies. Imagine a "free sea of charges".
    And when you get closer to reality(by adding in the nuclei, rather than spread out positive charge and lattice vibrations as well as imperfections in the crystal) you must at some point realize that the driving frequency does matter.How you would really go about solving this problem is quite the mystery to me.
    PS: I am surprised but nonetheless happy that I could help you a bit. :)
     
  6. Jul 4, 2016 #5

    Twigg

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    I found a good answer in Hecht's book, section 4.8 of the 4th edition. It suggests we are actually both right. Conduction electrons are the primary mechanism for the light-matter interaction in metals, but a metal's color (if it has one) is due to it's bound electron response.

    "Envision the conductor as an assemblage of driven, damped oscillators. Some correspond to free electrons and will therefore have zero restoring force, whereas others are bound to the atom, [. . .]. The conduction electrons are, however, the predominant contributors to the optical properties of metals. Recall that the displacement of a vibrating electron was given by ## x(t) = \frac{q_{e}/m_{e}}{(\omega_{0}^{2} - \omega^{2})} E(t) ##. With no restoring force, ##\omega_{0} = 0##, the displacement is opposite in sign to the driving force q_{e} E(t) and therefore 180 degrees out of phase with it. This is unlike the situation with for transparent dielectrics, where the resonance frequencies are above the visible and the electrons oscillate in-phase with the driving force. [...] It should be noted that if a metal has a particular color, it indicates that the atoms are partaking of selective absorption by way of the bound electrons, in addition to the general absorption characteristic of the free electrons."

    The equation Hecht gives for the complex index of refraction of metals is:
    ##n^{2}(\omega) = 1 + \frac{Nq_{e}^{2}}{\epsilon_{0}m_{e}}[\frac{f_{e}}{-\omega^{2} + i\gamma_{e} \omega} + \sum_{j} \frac{f_{j}}{\omega_{0j}^{2} - \omega^{2} + i\gamma_{j} \omega}] ##
    where ##f_{e}## is the fraction of free electrons and ##f_{j}## is the fraction of bound electrons with resonance ##\omega_{0j}##. The selective absorption has to come from the latter terms, which are caused by bound electrons, because the conduction electron term has no local extrema, if that makes sense.
     
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