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Understanding complex permittivity

  1. Aug 15, 2013 #1
    I'm not sure if this is the appropriate forum for my question as I actually am studying this as part of electrical engineering and I don't actually study physics. Nonetheless, I shall ask and if need be, move my question to another venue.

    My question is with regard to how complex permittivity is defined. According to my book
    $$
    \begin{align*}
    \nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\
    &= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\
    &= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\
    &= \jmath\omega\varepsilon_c\mathbf{\tilde{E}}
    \end{align*}
    $$
    ([itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{H}}[/itex] are phasors.)

    I really do not understand why [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] and not [itex]\varepsilon_c \equiv \sigma + \jmath\omega\varepsilon[/itex]. What is the sense in creating a complex value, [itex]\varepsilon_c[/itex], and then multiplying by [itex]\jmath\omega[/itex] when you could just modify the definition of [itex]\varepsilon_c[/itex] such that [itex]\nabla \times \mathbf{\tilde{H}} = \varepsilon_c \mathbf{\tilde{E}}[/itex]?

    I also have a conceptual question: From what I understand, [itex]\varepsilon[/itex] determines the phase delay between the H and E fields. This phase delay, as far as I know, comes from the finite speed involved in 'rotating' the dipoles in the medium. When these dipoles are 'rotated' though, since they take a finite time to rotate, that implies to me that there are some sort of losses involved in rotating the dipoles. These losses, though, as far as I can tell, are not accounted for in [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] (I figure the loss due to rotating the dipoles should be part of [itex]\Im{\{\varepsilon_c\}}[/itex] (from what I can tell, [itex]\Im{\{\varepsilon_c\}}[/itex] accounts for the loss and [itex]\Re{\{\varepsilon_c\}}[/itex] accounts for the phase delay); however, [itex]\Im{\{\varepsilon_c\}}[/itex] only seems to take into account frequency and loss from electorns crashing into atoms).

    Similarily, I would've thought that the loss that comes from electrons crashing into atoms (which [itex]\sigma[/itex] looks after), would also have the affect of at least somewhat slowing down the wave and causing lag.

    Basically, what I'm saying is, why aren't [itex]\varepsilon[/itex] and [itex]\sigma[/itex] also complex numbers? Or maybe they are...

    Thank you.
     
  2. jcsd
  3. Aug 15, 2013 #2

    Philip Wood

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    Gold Member

    Perhaps not my place to say so, but I think your question is perfectly appropriate on the Physics forum. The equation you quote :
    is a development of one of Maxwell's equations - the one embodying Ampère's Law and 'the Maxwell term'.
    [tex]\nabla \times\textbf{H} = \textbf{J} + \frac{\partial \textbf{D}}{\partial t}[/tex]
    Your version expresses J, the current density, and D, the electric flux density, in terms of E using [itex]\textbf{J}= \sigma \textbf{E}[/itex] and [itex]\textbf{D}= \epsilon \textbf{E}[/itex]. These assume a homogeneous medium. Replacing [itex]\frac{\partial}{\partial t}[/itex] by [itex]j\omega[/itex] assumes that E is of the form [itex]\textbf{E} = \textbf{E}_0 e^{j \omega t}.[/itex] [It also assumes that [itex]\epsilon[/itex] is a constant throughout the oscillation cycle, so it passes unchanged through the differentiation.]

    I'll now try to answer your first question. Observe that ε, the permittivity, appears because of the basic relationship [itex]\textbf{D}= \epsilon \textbf{E}[/itex], and that permittivity was first used as a real number describing the dielectric properties of an insulator. Thus we define complex permittivity for any (homogeneous) medium, such that its value collapses to the real permittivity when [itex]\sigma[/itex] = 0, that is when the medium is an insulating, lossless one. Your [itex]\epsilon _c[/itex] does just this, collapsing to the real coefficient of [itex]j \omega \textbf{E} [/itex] when [itex]\sigma[/itex] = 0.

    Your suggested simplified definition of [itex]\epsilon_c[/itex] would be fine for defining [itex]\sigma_c[/itex], in other words it would define a complex conductivity. Some time over a century ago, I imagine, the convention was adopted to take care of lossiness by the imaginary part of epsilon, rather than taking care of electric flux density by an imaginary part of conductivity. Presumably this was because the theory was mainly applied to dielectrics which may or may not be significantly lossy.

    I'll just make one remark about your next batch of questions: don't assume that a finite rotation /vibration time for dipoles itself implies a lossy (dissipative) process. Consider a pendulum: its finite period of oscillation is determined almost exclusively by inertia and restoring force, neither of which is dissipative (i.e. converting energy to random). Losses do usually need to be considered, of course. They can be added to the model easily enough.
     
    Last edited: Aug 15, 2013
  4. Aug 15, 2013 #3

    Philip Wood

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    Your suggested simplified definition of [itex]\epsilon _c[/itex] would be fine for defining [itex]\sigma _c[/itex], in other words it would define a complex conductivity. Some time over a century ago, I imagine, the convention was adopted to take care of lossiness by the imaginary part of epsilon, rather than taking care of electric flux density by an imaginary part of conductivity. Presumably this was because the theory was mainly applied to dielectrics which may or may not be significantly lossy.
     
  5. Aug 15, 2013 #4
    I'm a little confused with what all the parameters are modeling. As I understand it, sigma models the loss due to electrons crashing into other atoms when they get accelerated by the E-field (I know that's not 'really' what happens but it's good enough for me). Epsilon accounts for the fact that there are dipoles in the material that cause a delay because they cannot move infinitely quickly.

    I imagine that when the electrons are accelerated and 'crash' into other electrons, this would not only cause a loss but also a delay (because surely crashing into something will make you go slower...). What is it that models that? I would've thought it'd be something in the real part of epsilon_c; however, that only consists of permittivity. It seems as though there should be a complex conductance to model the lag caused by crashing into electrons.
     
  6. Aug 15, 2013 #5

    Philip Wood

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    Gold Member

    (1) The equation describes how magnetic field arises both due to moving charges (first term on right) and/or due to rate of change of electric field (second term on right). The second term alone will be there - with its [itex]j\omega[/itex] - even in a vacuum, when [itex]\epsilon = 1 (\times \epsilon_0)[/itex]

    (2) The average effect of the 'crashing into' forces is quite well modelled by a force proportional to an electron's velocity and in the opposite direction to its velocity.

    (3) But the phase of the electron's oscillation compared to that of the electric field is mainly determined by restoring 'elastic' forces on the electron, due to its being bound to a nucleus, and its inertia. We're in the realm of forced oscillations. The phase of the oscillations depends on the frequency of E relative to the natural frequency of the electron, and the phase determines whether the relative permittivity is made more or less than 1. The presence of resistive forces makes the change of phase - and hence of [itex]\epsilon[/itex] - with frequency less sharp and than it would be with no damping.

    In an attempt to be brief, I've hugely oversimplified what is actually a complicated business. Hope I've been of some help.
     
  7. Aug 16, 2013 #6

    Claude Bile

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    The factor of j(omega) is preserved as it comes from taking the derivative of E, assuming a time dependence of exp(-j(omega)t). I can't say for sure off the top of my head, but it is a good chance it is written this way to preserve consistency with other forms of this equation (e.g as applied to dielectrics).

    The sign of the complex part is arbitrary, and dictated by convention.

    I hope this contributes something....

    Claude.
     
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