Dear friends,
I want to know why electrical susceptibility is a second order tensor instead of third order or higher? Please help me to find an answer.
Dear Renjit,
The topic is discussed in detail in the following most recent handooks:
1. Nicholas D. Spencer,John H. Moore: Encyclopedia of Chemical Physics and Physical Chemistry: Fundamentals, 2001
2. Cécile Malgrange, Christian Ricolleau, Michel Schlenker: Symmetry and Physical Properties of Crystals, 2014
I hope that will help you...
Respectfully yours,
Jewgeni
I am quoting from Wikipedia:
https://en.wikipedia.org/wiki/Crystal_optics#Electric_susceptibility
"In an isotropic and linear medium, this polarization field P is proportional to and parallel to the electric field E:
P = χ ε0E
where χ is the electric susceptibility of the medium.
Now, from https://en.wikipedia.org/wiki/Crystal_optics#Anisotropic_media
In an anisotropic medium, such as a crystal, the polarisation field P is not necessarily aligned with the electric field of the light E. In a physical picture, this can be thought of as the dipoles induced in the medium by the electric field having certain preferred directions, related to the physical structure of the crystal. This can be written as:
P = ε0χ E .
Here χ (chi) is not a number as before but a tensor of rank 2, the electric susceptibility tensor.
Sofia's answer seems very natural and convincing. But note the requirement there for linear medium (anisotropy has no direct relation with linearity). When the material's response to external stimuli appears nonlinear and we want to describe it in terms of polynomials, say P = a*E + b*E2 + c*E3 + ..., then higher rank tensors (here b and c) may and should be introduced. This is indeed necessary when operating with electric fields of extremely high amplitude (high power lasers). But is polynomial description correct? Certainly not for magnetic susceptibility, as the magnetization saturates in high fields, that is remains finite (bounded), while polynomials diverge to infinity as their arguments grow.
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