# Polarizaition and susceptibility

 P: 369 In some unit, the relation of (linear) polarization and susceptibility can be written of $$P(t) = \chi E(t)$$ but I also read some expression in other text reads $$P(\omega) = \chi(\omega) E(\omega)$$ why change the time to frequency? Why polarization depends on frequency?
 Sci Advisor P: 5,517 Your equations are not really written correctly. The first one, the time dependent one, should really be written as a convolution: The polarization of a linear isotropic medium with a local but noninstantaneous response (but still independent of time) is: P(t)=$\int \chi(t-\tau)E(\tau)d\tau$ And taking the Fourier transform of this equation provides your second expression. If the material responds instantaneously and has no memory[$\chi(t-\tau) = \chi\delta(t-\tau)$], then the convolution integral reduces to your first expression. Having a frequency-dependent susceptibility is simply dispersion.
 Quote by Andy Resnick Your equations are not really written correctly. The first one, the time dependent one, should really be written as a convolution: The polarization of a linear isotropic medium with a local but noninstantaneous response (but still independent of time) is: P(t)=$\int \chi(t-\tau)E(\tau)d\tau$ And taking the Fourier transform of this equation provides your second expression. If the material responds instantaneously and has no memory[$\chi(t-\tau) = \chi\delta(t-\tau)$], then the convolution integral reduces to your first expression. Having a frequency-dependent susceptibility is simply dispersion.