# High frequency limit of dielectric permittivity

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.

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.