I have asked this question a number of times and am still uneasy with it.

enter image description here

I want to know how somebody can derive the equation (also see above, picture from a publication):

$$\epsilon(\omega)=\epsilon_L + \frac{i\sigma(\omega)}{\omega}$$

The sign is completely wrong in my mind! I am aware that physicists and engineers use different harmonic time dependencies. Such that a time derivative could be either $-i\omega$ or $i\omega$. However, this time dependency should only change the sign of $\epsilon_L$, not the second term.

My derivation is as follows, starting with Maxwell's 4th:

$$\nabla \times H = J + \frac{\partial D}{\partial t}$$ For engineers: $$=\sigma E + i\omega \epsilon_0\epsilon_rE$$ For physicists (notice sign change reflecting difference in harmonic time dependence): $$=\sigma E - i\omega \epsilon_0\epsilon_rE$$

Also, $$\nabla \times H =j\omega \epsilon_0\epsilon^* E$$

Equate both sides and you're left with either

$$\epsilon^*(\omega)=-\epsilon_L + \frac{\sigma(\omega)}{i\omega\epsilon_0}$$ $$\epsilon^*(\omega)=\epsilon_L + \frac{\sigma(\omega)}{i\omega\epsilon_0}$$

I realise there is a possibility I'm misunderstanding the second form of Maxwell's 4th I've used? Even if this would also reflect harmonic time dependence and would therefore be $\nabla \times H =-j\omega \epsilon_0\epsilon^* E$, the derivation still could work out. Deriving it this way, there are four possibilities:

  1. $ - \epsilon_L + \frac{\sigma(\omega)}{i\omega\epsilon_0}$ (wrong, mismatched harmonic time dependencies)
  2. $ \epsilon_L + \frac{\sigma(\omega)}{-i\omega\epsilon_0}$ (correct if you're a phyisicist)
  3. $ - \epsilon_L + \frac{\sigma(\omega)}{-i\omega\epsilon_0}$ (wrong, mismatched harmonic time dependencies)
  4. $ \epsilon_L + \frac{\sigma(\omega)}{i\omega\epsilon_0}$ (correct if you're an engineer)

Really, my question is where does $\nabla \times H =-j\omega \epsilon_0\epsilon^* E$ come from, and does it also exhibit harmonic time dependence? This was answered below, very nicely!

  • $\begingroup$ You may want to check this derivation in Born and Wolf Principles of Optics (Chap 13. p. 612-613: archive.org/details/PrinciplesOfOptics/page/n659/mode/2up ) where they derive results in the optics of metals. It seems to be close to what you are looking for. I hope this helps. $\endgroup$
    – ad2004
    Nov 2, 2020 at 19:21
  • 1
    $\begingroup$ $e^{j\omega t}$ is the engineering convention. $\endgroup$
    – Puk
    Nov 2, 2020 at 20:05
  • $\begingroup$ @Puk noted, thank you :) $\endgroup$
    – smollma
    Nov 2, 2020 at 21:09

1 Answer 1


If we move to Fourier space, we have

$$\mathbf H(\mathbf r,t) = \int_{-\infty}^\infty \hat{\mathbf H}(\mathbf r,\omega) e^{-i\omega t} d\omega$$

$$\mathbf E(\mathbf r,t) = \int_{-\infty}^\infty \hat{\mathbf E}(\mathbf r,\omega) e^{-i\omega t} d\omega$$

$$\mathbf D(\mathbf r,t)= \int_{-\infty}^\infty \epsilon(\omega)\hat{\mathbf E}(\mathbf r,\omega) e^{-i\omega t} d\omega$$

$$\mathbf J(\mathbf r,t) = \int_{-\infty}^\infty \sigma(\omega) \hat{\mathbf E}(\mathbf r,\omega) e^{-i\omega t} d\omega$$

in which case Ampere's law becomes $$\nabla \times \hat{\mathbf H}(\mathbf r,\omega)= \sigma(\omega)\hat{\mathbf E}(\mathbf r,\omega) -i\omega \epsilon(\omega)\hat{\mathbf E}(\mathbf r,\omega)$$ $$=-i\omega\epsilon_0\left[\epsilon_r(\omega) + \frac{i\sigma(\omega)}{\epsilon_0 \omega}\right]\hat{\mathbf E}(\mathbf r,\omega) \equiv -i\omega\epsilon_0 \tilde{\epsilon}_r(\omega)\hat{\mathbf E}(\mathbf r,\omega)$$

This is consistent with $\nabla \times \mathbf H = \frac{\partial}{\partial t} \mathbf D$, where $\epsilon(\omega)$ has acquired an imaginary part. The minus sign on the right hand side reflects the fact that $\frac{\partial}{\partial t}\rightarrow -i\omega$ in this convention.

If you replace all of the $e^{-i\omega t}$'s with $e^{+i\omega t}$'s in the definitions of the Fourier-space quantities, then Ampere's law would become

$$\nabla \times \hat{\mathbf H}(\mathbf r,\omega) = \sigma(\omega)\hat{\mathbf E}(\mathbf r,\omega) +i\omega\epsilon(\omega)\hat{\mathbf E}(\mathbf r,\omega)$$ $$=i\omega\epsilon_0\left[\epsilon_r(\omega) -\frac{i\sigma(\omega)}{\epsilon_0\omega}\right]\hat{\mathbf E}(\mathbf r,\omega) \equiv i\omega \epsilon_0\tilde{\epsilon}_r(\omega) \hat{\mathbf E}(\mathbf r,\omega)$$

This too is consistent with $\nabla\times\mathbf H = \frac{\partial}{\partial t}\mathbf D$, where the lack of a minus sign on the right-hand side reflects that in this choice of convention, $\frac{\partial}{\partial t}\rightarrow i\omega$.

  • $\begingroup$ Really helpful response, thank you! $\endgroup$
    – smollma
    Nov 2, 2020 at 21:08

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