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I Time harmonic case of Gauss's Law

  1. Sep 24, 2018 #1
    In a chapter building up to the theory of plane waves my book starts by introducing
    time harmonic electric fields and defines a special case of Gauss's law.


    curl(H) = J + dD/dt

    curl(H) = sigma * E + epsilon * dE/dt

    if E is time harmonic and spacially dependent... E(x,y,z,t) let E' represent the phasor form

    curl(H) = sigma * E' + epsilon * j * w * E'

    curl(H) = (sigma + epsilon*j*w) E'

    of curl(H) = jw(epsilon - j*sigma/w) E'
    where epsilon - j*sigma/w = epsilon_c (complex permittivity)


    given that... divergence(curl(H)) = 0....

    divergence( jw * epsilon_c * E') = 0

    therefore divergence(E) = 0

    so pv (volume charge density) = 0 by Gauss's law


    I am very confused why a time harmonic E field can never bound a charge source and why it's divergence is always zero as my book seems to suggest.
    I am guessing of have missed a major assumption and or am misinterpreting something? Looking for some guidance. Thanks!
     
  2. jcsd
  3. Sep 24, 2018 #2

    Charles Link

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    Let's take the divergence of both sides of this equation, but let's assume ## J=J_{free} =0 ##, and let's look at ## 0=\frac{d \nabla \cdot D}{dt}=\frac{d (\epsilon_o \nabla \cdot E+\nabla \cdot P)}{dt} ##. If we have a single material, and a single frequency, we can write ## P(\omega)=\epsilon_o \, \chi(\omega) E(\omega) ##, with ## E(t)=E(\omega)e^{i \omega t} ## and ## P(t)=P(\omega) e^{i \omega t} ##. Having a single homogeneous material means we get no polarization charges on any surface interface, (because there are no surface interfaces), and with the equation as we have it, it shows that we must have ## \nabla \cdot E=0 ##. We won't get any polarization charge inside the single uniform material. ## \\ ## I think a similar argument could be applied to the ## \nabla \cdot J_{free}=\nabla \cdot (\sigma E ) ## term. If the conductor is homogeneous, and responds linearly with ## J=\sigma E ##, so that ## J(\omega)=\sigma(\omega) \, E(\omega)##, there is no charge build-up anywhere. (If the conductor has a boundary, so that ## \sigma ## is not constant, then you will get charge build-up, and ## \nabla \cdot E \neq 0 ## ). ## \\ ## I don't know that what is found in your textbook is saying anything of any more significance than what I have just shown.
     
    Last edited: Sep 24, 2018
  4. Sep 24, 2018 #3
    Thank you for the response, I think this helps me narrow down my confusion a bit more. This makes sense for a 'charge and current' free region as you are showing, that an externally produced E field would not result in any long term charge in a homogeneous material.
    I think my confusion is that my book doesn't seem to make this assumption (maybe I missed this?) of being in a current free charge free region.
    If there was some theoretical time harmonic source of charge wouldn't it produce a time harmonic E field and then wouldn't the volume charge density have nonzero value at various points in time in that region?
     
  5. Sep 24, 2018 #4

    Charles Link

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    I only assumed ## J_{free} =0 ## ,(and presumably ## \rho_{free}=0 ##), for the very first paragraph. After that, the more general case applies, but again, the assumption of a homogeneous material still applies. ## \\ ## The assumption of a single homogeneous material seems to be the important one in all cases here, rather than a harmonic time dependence. ## \\ ## Note: In the equation ## \nabla \times H=J+\frac{\partial{D}}{\partial{t}} ##, the ## J ## here is ## J_{free} ##. ## \\ ## The ## J_m =\nabla \times M ## and ## J_p=\dot{P} ## are not part of ## J ## here, in this equation. ## \\ ## And also notice if ## \sigma ## and ## \epsilon ## are spatially dependent, i.e. a non-homogeneous material, then ## \nabla \sigma \neq 0 ## and ## \nabla \epsilon \neq 0 ##, so that ## \nabla \cdot ( \sigma E) \neq \sigma \nabla \cdot E ##, and ## \nabla \cdot (\epsilon E) \neq \epsilon \nabla \cdot E ##, so that an algebraic step that was assumed would not be permissible.## \\ ## (Note: ## \nabla \cdot (\sigma E)=(\nabla \sigma) \cdot E+\sigma \nabla \cdot E ## ).
     
    Last edited: Sep 24, 2018
  6. Sep 25, 2018 #5
    Thanks! The spatially dependent case clears up my confusion.
     
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