# Are these forms for effective permittivity equivalent?

I was browsing through a book and noticed the following form for effective permittivity (Eq. 1.14.2 of this book):

$$\epsilon(\omega) = \epsilon_d(\omega) + \frac{\sigma_c(\omega)}{j\omega}$$

Though, I am only familiar with this form:

$$\epsilon(\omega) = \epsilon_r(\omega)\epsilon_0 - j\frac{\sigma_c(\omega)}{\omega\epsilon_0}$$

Are these two forms equivelant? It looks like the book author has absorbed $$\epsilon_0$$ into $$\epsilon_d(\omega)$$ for $$\epsilon'$$. However, I cannot see how the $$\epsilon''$$ parts are equivelant, can somebody help me understand?

The latter form is more familiar to me and can be found on the wikipedia page for relative permittivity (Lossy materials). It also appears more frequently in academic articles than the first form, one such example is linked here as Equation 2.

• It seems that in the second equation, the $\epsilon_0$ in the second term should not be there. Do you have a source for this equation? – EigenDavid Dec 5 '19 at 8:22
• @David the latter for is just a general form: see the wiki. It also appears in numerous academic articles (Eq. 2). – smollma Dec 5 '19 at 14:33
• The thing with your second formula is that either it expresses the absolute permittivity and the $\epsilon_0$ in the denominator is wrong, or it expresses a relative permittivity and then the $\epsilon_0$ in the first term is wrong. In both Wikipedia and the article you gave, the formulas give a relative permittivity hence the absence of $\epsilon_0$ in the first term. The sign of the second terms changes because of the different conventions for the harmonic timedependence. – EigenDavid Dec 5 '19 at 14:42
• @David the author of the first formula calls it an "effective permittivity" which is the same as relative permittivity, isn't it? He has no $\epsilon_0$ in the denominator. The sign change makes sense mathematically, since $j^{-1}=-j$. The academic paper's equation doesn't reflect the 'correct' sign change, though. – smollma Dec 5 '19 at 15:00
• No; effective here is in the sense that we have a model for the permittivity and that the material acts as if the permittivity is the one given by the model. For the sign, it comes from the choice of the implicit harmonic dependence of the quantities, either exp($-j\omega$t) or exp($j\omega$t). These different conventions lead to a different sign of the "$j$"' that appears due to the properties of the Fourier transform. For the first convention we have $\partial_t \leftrightarrow -j \omega$, for the second $\partial_t \leftrightarrow j \omega$, hence the difference in sign in different sources – EigenDavid Dec 5 '19 at 15:06

The two forms are not equivalent; the second is wrong.

In the book you refer too, "effective" does not mean relative. Effective here is in the sense that we have a model for the permittivity and that the material acts as if the permittivity is the one given by the model. As it can be seen from the previous equation 1.14.1, the author talks about the absolute permittivity; This can also be checked by a dimensional analysis.

In your second equation, you have two possibilities to correct it. Either it expresses an absolute permittivity (as does the first one) and the $$\epsilon_0$$ in the denominator is extra, or it expresses a relative permittivity and then the $$\epsilon_0$$ in the first term is extra: $$$$\epsilon_{abs}^{eff}=\epsilon_0\epsilon_{r}+\frac{\sigma}{j\omega} \,.$$$$ The corresponding relative permittivity is simply obtained by dividing the absolute one by $$\epsilon_0$$: $$$$\epsilon_{rel}^{eff}=\epsilon_{r}+\frac{\sigma}{j\omega\epsilon_0} \,.$$$$

Concerning the different signs of the second term in different sources, namely $$\frac{\sigma}{j\omega}=-j\frac{\sigma}{\omega}$$ vs. $$-\frac{\sigma}{j\omega}=j\frac{\sigma}{\omega}$$, this comes from the different sign of the harmonic time dependence chosen. When working with harmonically oscillating quantities, we usually assume that all the quantities have a time dependence described by a complex valued exponential. However, we are free to choose $$\mathrm{exp}(j\omega t)$$ or $$\mathrm{exp}(-j\omega t)$$ without meaningful physical consequences. The problem is that when applying time-derivatives to the quantities, those two conventions lead to different results. The first one gives $$\partial_t\leftrightarrow j\omega$$ whereas the second one gives $$\partial_t\leftrightarrow -j\omega$$. Thus the imaginary parts of the equation have a different sign in the resulting equations. See https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-007-electromagnetic-energy-from-motors-to-lasers-spring-2011/readings/MIT6_007S11_sign.pdf for additional details.

The important thing is to be careful when picking formulas in different sources and be sure to know the convention used. There seems to be roughly an engineer ($$\mathrm{exp}(j\omega t)$$) vs. physicist ($$\mathrm{exp}(-j\omega t)$$) type of convention. Finally, the choice of convention leads to some choice of the sign of some physical quantities. The simplest example being the imaginary part of the refractive index (linked to the permittivity) that needs to be positive or negative in order to translate as losses in the medium; the wrong choice of sign or convention would result in a gain. Conversely, for example, gain in e.g. lasers can indeed be described by an imaginary part of the refractive index with the correct sign.

Here (https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/lecture09and10.pdf) you can find a slightly different derivation of the effective permittivity you encountered; I think it shows more clearly why the conductivity appears in the effective permittivity.

• David, I've just looked at this again and if you derive the equation in question from Maxwell's fourth you get two answers and the sign of the first term changes, not the second. I think I'll make a new question for that. – smollma Nov 2 '20 at 18:34