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 $$ \begin{align*} \nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\ &= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\ &= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\ &= \jmath\omega\varepsilon_c\mathbf{\tilde{E}} \end{align*} $$ ($\mathbf{\tilde{E}}$ and $\mathbf{\tilde{H}}$ are phasors.)

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

I also have a conceptual question: From what I understand, $\epsilon$ 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 $\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}$ (I figure the loss due to rotating the dipoles should be part of $\Im{\{\varepsilon_c\}}$ (from what I can tell, $\Im{\{\varepsilon_c\}}$ accounts for the loss and $\Re{\{\varepsilon_c\}}$ accounts for the phase delay); however, $\Im{\{\varepsilon_c\}}$ 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 $\sigma$ 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 $\varepsilon$ and $\sigma$ also complex numbers? Or maybe they are...

Thank you.

  • $\begingroup$ Did I not answer your question? All optical constants (conductivity, index of refraction, permeability, permittivity, etc) MUST be complex in the Fourier domain since causality requires they be Hilbert transform pairs. If the real or imaginary part is zero, the other one must be zero too. Maxwell's equations in the time domain are purely real and therefore these quantities are real but non-local (convolutional integrals) in the time domain. Here is a tutorial we wrote: mesoscopic.mines.edu/~jscales/causality.pdf We go to the Fourier domain to simplify convolutions. $\endgroup$
    – user27777
    Aug 16, 2013 at 21:55

2 Answers 2


This is a simple mathematical convenience so that the form of the equation is the same whether or not conductivity is present. The key is to remember the Ampere-Maxwell equation in a homogeneous medium without conductivity: $$ \nabla\times\mathbf{\tilde{H}} = j\omega\varepsilon\mathbf{\tilde{E}} $$

If we add conductivity, we choose to define the new equation such that the form is unchanged: $$ \nabla\times\mathbf{\tilde{H}} = j\omega\varepsilon_c\mathbf{\tilde{E}} $$

But we know adding the conductivity term to the original equation results in:

$$ \nabla\times\mathbf{\tilde{H}} = j\omega\varepsilon\mathbf{\tilde{E}} + \sigma\mathbf{\tilde{E}}= \left(j\omega\varepsilon + \sigma\right)\mathbf{\tilde{E}} $$

Now we have two ways of writing $\nabla\times\mathbf{\tilde{H}}$, one in terms of $\varepsilon_c$, and one in terms of $\varepsilon$ and $\sigma$, so we now equate those two expressions $$ \left(j\omega\varepsilon + \sigma\right)\mathbf{\tilde{E}} = j\omega\varepsilon_c\mathbf{\tilde{E}} $$ This is true iff $$ j\omega\varepsilon + \sigma = j\omega\varepsilon_c $$ Divide by $j\omega$ $$ \frac{j\omega\varepsilon + \sigma}{j\omega} = \varepsilon_c $$ Simplify $$ \varepsilon + \frac{\sigma}{j\omega} = \varepsilon_c $$ And recognize that $\frac{1}{j}=-j$ $$ \varepsilon_c = \varepsilon - j\frac{\sigma}{\omega} $$ So what we discovered is that if we define $\varepsilon_c = \varepsilon - j\frac{\sigma}{\omega}$ and a new equation $\nabla\times\mathbf{\tilde{H}} = j\omega\varepsilon_c\mathbf{\tilde{E}}$, then the result is the correct equation that accounts for conductivity. It is helpful that the new equation has the same form as the old too, because now we can just take one equation, the new one, and allow $\varepsilon_c$ to be purely real to recover the no conductivity case, or we can roll the effect of conductivity into the complex part of the permittivity.

Now, to address your second question: there is indeed loss associated with rotating dipoles in a medium as a wave passes through. You can think of the interaction between the field and dipoles as itself having two parts, a "springy" part, and a "damped" part. If there was no damping, you could apply an impulse to the dipole and start it wiggling, and that wiggle would cause fields to carry away energy, and then the wiggling would eventually stop. The energy carried away would be exactly what was delivered from the impulse, and it would be somewhat delayed from the initial impulse because it takes a finite amount of time for this system to react. This is the normal, lossless dielectric interaction captured in a real dielectric constant. Now, it is possible that as the dipole wiggles, it rubs against other dipoles or atoms in the material, and loses some energy through friction. In this case, some of the energy of the original impulse would be radiated away as EM waves, and some of it would be converted to heat energy in the material. The friction and heating part of the interaction is what I called the "damped" part previously, and indeed causes the EM wave to lose energy as it propagates through such a medium.

We can then say that $\varepsilon=\varepsilon_r-j\varepsilon_\text{heating}$ is itself truly complex to account for this, where the real part describes the "springy" part and the imaginary part describes the lossy dielectric heating piece. Then if we wrap this up into the expression for $\varepsilon_c$, we get the following $$ \varepsilon_c = \varepsilon_r - j\varepsilon_\text{heating} - j\frac{\sigma}{\omega} = \varepsilon_r - j\left(\varepsilon_\text{heating} + \frac{\sigma}{\omega}\right) $$

The net effect is that the complex permittivity has a real part that has to do with the lossless properties of the medium, and a complex part that has to do with losses from both electrons being accelerated by the fields and experiencing resistance, and dipoles being torqued in the medium and experiencing friction.

I'll argue now that the details don't matter, and maybe there are even mechanisms by which the electrons oscillate and re-radiate instead of meeting resistance, contributing to the real part. Sometimes its charged ions in the material that move and meet resistance, contributing again the loss. Indeed, there are many conventions and many mechanism for what gets rolled into the complex permittivity. You've seen some of these conventions and models in the other answers to this question. In practice, however, someone will have measured the attenuation and wavelength of EM waves in a medium, and from the overall attenuation, they can come up with the imagnary part of $\varepsilon_c$ that lumps together all loss mechanisms, and from the wavelength, they will calculate a real part that lumps together all lossless interactions processes. The idea really is that the details of the atomic and molecular physics is not so important to the kinds of questions we ask in a macro sense about EM waves. If I transmit a cellphone signal through a concrete wall and want to know the signal strength on the other side, it's not necessarily important to understand the atomic and molecular physics of the concrete; it is often enough to have characterized the lossy and lossless parts of the dielectric constant, and then simply use those numbers in my calculations.


Have a look at the reference cited in this answer: Can the Kramers–Kronig relation be used to correct transfer function measurements?

In a nutshell here is the point: because of causality, in the time-domain, the permittivity, conductivity, etc must be real but non-local. In the Fourier domain, they must be complex. The nonlocality expresses itself as a convolution integral in the time domain. This is easier to write in the Fourier domain as a multiplication; as a result this often leads to misleading abuses of notation involving time-domain fields in the same equation as Fourier domain optical constants. The answer indicated above has a link to a tutorial on this subject. Oh and the phase delay applies to all material, not just dipoles, this is causality.

  • 1
    $\begingroup$ I really don't see how causality affect whether $\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}$ or $\varepsilon_c \equiv \jmath\varepsilon \omega + \sigma$. They're the same thing as far as I can tell. $\endgroup$
    – user26869
    Aug 17, 2013 at 13:17
  • $\begingroup$ Yes, I know you don't see how it's possible. I gave you a link to a recently published tutorial. My rec is that you have a look. I would be very surprised if you still have questions after reading that. $\endgroup$
    – user27777
    Aug 17, 2013 at 15:20
  • $\begingroup$ The paper is interesting; it basically explains why the relationship between D and E actually involves convolution and that linking them with a constant, epsilon, is really shorthand for performing a convolution. It also says that if there were a purely real constant linking D and E, this would constitute an impossible (non-causal) system. This certainly is interest, but I will admit that I study electrical engineering and am not familiar with a chunk of the physics maths involved. Despite being interesting, I don't really see how it answers my question, it answers why epsilon is complex... $\endgroup$
    – user26869
    Aug 17, 2013 at 16:04
  • 1
    $\begingroup$ All response functions ($\sigma$, $\mu$, are complex. So D = $\epsilon$ E is the def of $\epsilon$ for linear response. You have to do some physics to see how $\epsilon$ is related to $\sigma$. It turns out that the real part of $\epsilon$ is related to the complex part of $\sigma$ and vice versa. The problem is that there are many different loss mechanisms that come in at different frequencies. Thermalization by phonons is the dominant mechanism below 1 THz for semi-conductors. So in this case $\sigma$ does not related to electrical conductivity.It's the total conductivity. $\endgroup$
    – user27777
    Aug 17, 2013 at 18:27
  • 1
    $\begingroup$ If you want to know the form of $\sigma (\epsilon)$ in general, books.google.com/books/about/… is a good place to start. Or Landau and Lifshits ECM. I hope I have not misunderstood your question. cheers, $\endgroup$
    – user27777
    Aug 17, 2013 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.