I just studied the concept of 'complex permittivity' in time-harmonic field. One of the Maxwell's equations expressed in phasor-domain,

$$\nabla\times\vec H = \vec J + j\omega \epsilon\vec E$$

($\epsilon$ & $\omega$ stand for permittivity & angular freqeuncy),

can be expressed with complex permittivity like following.

$$\nabla\times\vec H = \sigma\vec E + j\omega \epsilon\vec E = (\sigma + j\omega \epsilon)\vec E =j \omega(\epsilon +\sigma/j\omega) \vec E = j\omega\epsilon'\vec E$$

($\sigma$ stands for conductivity)

Then, taking the divergence of both side,

$$\nabla\cdot(\nabla\times \vec H) = 0 = j\omega\epsilon' \nabla\cdot\vec E$$

So, I got the conclusion $\nabla\cdot\vec E = 0$ which means the charge density is zero.

This sounds "Time-Harmonic electromagnetic fields cannot occur in charged matter".

Is this true? or Which point do I miss?

  • $\begingroup$ Welcome to the site! This site supports MathJax. I've typeset this question for you, but in the future it would be good if you learn some basic MathJax and make your equations look nicer. $\endgroup$
    – Chris
    Feb 20, 2018 at 1:20
  • 1
    $\begingroup$ $\nabla \cdot \vec{E} = 0$ in vacuum too. Does that mean that a time-harmonic EM field cannot occur in vacuum? $\endgroup$
    – NickD
    Feb 20, 2018 at 1:27
  • $\begingroup$ Thx Chris! absolutely i didn't know that supporting. And Nick, i think you misunderstand my question a little bit. The vacuum cannot be charged originally. So, the charge density is zero clearly. The point of my question is that no matter which materials EM field occurs in, the time-harmonic condition makes charge density zero. I wonder if time-harmonic EM field cannot be made in charged matter (charge density isn't zero). $\endgroup$
    – JH Lee
    Feb 20, 2018 at 1:47
  • $\begingroup$ The charge density equation is $\textrm{div}\textbf{D}=\rho$. (Only in homogoeneous regions you have $\textrm{div}\textbf{E}=\rho/\epsilon$). Even if the space of interest is made of piecewise homogeneous dielectrics that does not exclude the accumulation and fluctuation of surface charges at the dielectric interfaces. $\endgroup$
    – hyportnex
    Feb 20, 2018 at 15:11

1 Answer 1


No, with the given assumption of field solutions $$\vec E=\vec E_0 \exp j\omega t$$ you can either have a field solution without charge $$\nabla \vec E=0$$ Or with charge $$\nabla \vec E \neq 0$$ which implies $$\sigma + j\omega \epsilon=0$$ and thus $$j\omega =-\frac{\sigma}{\epsilon}$$ This means that you get an arbitrary initial charge distribution $\rho_0 (\vec r)=\epsilon div(\vec E_0)$ decaying with the time constant $$\tau= \frac {\epsilon}{\sigma}$$ $$\rho=\rho_0(\vec r)\exp (-\frac{t}{\tau})$$ The time constant $\tau$ is called the dielectric relaxation time or Maxwell's time. This relaxation time is the characteristic time that any accumulation of charge vanishes in a conductor.

Note: $\nabla \vec E=0$ means that you can only have solenoidal (or homogeneous) fields $\vec E=\nabla \times \vec A$ with a vector potential $\vec A$. You can also have a superposition of these two solenoid (divergence-free) and irrotational (curl-free) field solutions.

  • $\begingroup$ Clearly correct! Thank you for your answer and kind explanation $\endgroup$
    – JH Lee
    Feb 20, 2018 at 2:17

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