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Background: In Electrodynamics, the scalar permittivity $\epsilon(\omega)$ relates the Electric displacement field $\vec D$ to the electric field $\vec E$ as $\vec D=\epsilon\vec E$ when assuming a linear, isotropic medium. Causality requires that in time-space the Fourier transform $\tilde\epsilon(t)$ must vanish for $t<0$ (i.e. the future cannot influence $\vec D$), resulting in $\epsilon(\omega)$ being analytic and that the Kramers-Kronig relation (basically using the Hilbert transform) can be used to relate the real and imaginary parts of $\epsilon$. The imaginary part describes absorption and must thus not be negative1, does this put any additional restraints on $\Re \epsilon$?

So in summary:

For $\epsilon(\omega)$ analytical (because the Fourier transform $\tilde\epsilon(t)$ vanishes for $t<0$) and $\forall\omega>0:\Im\epsilon(\omega)\ge0$, what are properties of $\Re\epsilon(\omega)$ for $\omega>0$?

1) or positive, depending on convention as in whether the time dependence is $e^{+i\omega t}$ or $e^{-i\omega t}$

edit also, since $\vec E$ and $\vec D$ are real valued in time-space, $\epsilon(-\omega)=\epsilon(\omega)^*$

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When you write $\epsilon(-w) = \epsilon(w)^*$, does $z^*$ denote complex conjugation? for what values of $w$ does that hold? Finally, do you know anything about the growth/decay/asymptotic expansion of $\epsilon(w)$ as $|w| \rightarrow \infty$? – Sam Lisi Mar 30 '11 at 13:17
@Sam yes, I mean complex conjugation. Is there some different convention I should use? This hold for all real $\omega$ and there is no a priori knowledge about complex $\omega$ involved. For $\omega\to\infty$, $\epsilon \to \epsilon_\infty$ being a constant. – Tobias Kienzler Mar 30 '11 at 13:23
somehow I didn't see your follow-up comment. If $\epsilon$ is entire (analytic over the whole complex plane), and $|\epsilon(w)| \rightarrow \epsilon_\infty$ as $|w| \rightarrow \infty$, then by Liouville's theorem, $\epsilon(w) = \epsilon_\infty$. Or did you mean that $\epsilon(w) \rightarrow \epsilon_\infty$ only as $|w| \rightarrow \infty$ along the real axis? (I normally use $\bar z$ to denote the complex conjugate of $z$, but I have seen $z^*$ also used on occasion.) – Sam Lisi Apr 9 '11 at 0:16
@SamLisi: I only mean real valued positive $\omega\to\infty$. A constant $\epsilon(\omega)$ is definitely not the case, so nothing about $|\omega|\to\infty$ in general can be stated – Tobias Kienzler Apr 12 '11 at 15:31

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