Question

Derive the relation between electrical susceptibility and the atomic polarizablity

Answer 1

Answer:

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Answer 2

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They are very similar, dealing with almost the same physics, but are slightly different in definition due to the units they take and when they are used.

➤ Electric Susceptibility (χeχe)

• Electric susceptibility is the dimensionless proportionality factor that relates the electric field in a material to the polarization density, as follows:

P⃗ =ϵ0χeE⃗ P→=ϵ0χeE→

• For non-homogeneous systems, this factor doesn’t need to be a single number, but rather a 3x3 matrix which describes the willingness for electric dipoles to align with an external electric field, on the microscopic scale. However, this is a macroscopic, averaged value of these microscopic situations. It’s also really helpful for describing the displacement field, as follows:

D⃗ =ϵ0(1+χe)E⃗ D→=ϵ0(1+χe)E→

➤ Electric Polarizability (αα)

Electric polarizability speaks more to the understanding of how atoms or dipoles interact with an electric field on the microscopic scale.

p⃗ =αElocal→p→=αElocal→

I make the distinction between E⃗ E→ and Elocal→Elocal→by saying that the first is what you would measure to be the electric field; whereas, the second is the electric field minus that produced by the atomic dipole.

What’s their relation?

Because they both talk about the same kind of physics, there are derivations (See Griffiths “Introduction to Electrodynamics” 3rd edition, Problem 4.38) that relate the two in what is known as the “Clausius-Mossotti relation”, as follows:

χeχe+3=nα3χeχe+3=nα3

where nn is the number density of molecules in a material.