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The electric susceptibility $\chi_e(t)$ of a material is defined via a convolution relation between the polarization field $P$ and the electric field $E$: $$P(t) = \epsilon_0 (\chi_e * E)(t) = \epsilon_0 \int_{-\infty}^t \chi_e(t-\tau) E(\tau) d\tau,$$

where $\epsilon_0$ is the electric permittivity of free space. In particular $\chi_e$ vanishes on the negative semiaxis $\mathbb{R}_-$. Taking Fourier transform this relation becomes $$\hat{P}(\omega) = \epsilon_0 \widehat{\chi_e}(\omega) \hat{E}(\omega)$$

Then one defines the dielectric permittivity of a material as $$\epsilon(\omega) = \epsilon_0(1+\widehat{\chi_e}(\omega))$$

My question regards to the high frequency limit of this quantity, what I've seen so far is that $\epsilon(\infty) = \epsilon_0$ (e.g. in the plasma limit). This will follow for example when $\chi_e$ is an integrable function then its Fourier Transform has to vanishes as $\omega \to \infty$.

Sorry if my question is trivial, due to my lack of knowledge in physics, but I want to ask is this the only physically meaningful assumption, namely that $\epsilon(\infty) = \epsilon_0$? Does it make sense at all if for example $\epsilon(\infty)=0$? Intuitively I can understand that when the frequency is infinite, the wave probably goes through the material as if it goes through vacuum. I'd appreciate more insight into this.

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Electric permittivity is a multiresonant property. Whether there exists something that can reply to infinite frequency I cannot tell, but if you look at a dipole oscillator subjected to an electric field of frequency $\omega$ it yields in the limit of $\omega\to\infty$ the value $0$, which as a result gives $\epsilon = \epsilon_0 (1 + 0) = \epsilon_0$, identical to vacuum.

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  • $\begingroup$ Sorry about the wrong formatting. I was using mobile version of SE where formatting helper is not visible :) $\endgroup$
    – Radek D
    Sep 14 at 22:06

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