Dielectric susceptibility – classical derivation

Authier, A.

doi:10.1107/97809553602060000569

International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli

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International Tables for Crystallography (2006). Vol. B. ch. 5.1, p. 550

Section A5.1.1.1. Dielectric susceptibility – classical derivation

A. Authier^a ^*

^a Laboratoire de Minéralogie-Cristallographie, Université P. et M. Curie, 4 Place Jussieu, F-75252 Paris CEDEX 05, France
Correspondence e-mail: authier@lmcp.jussieu.fr

A5.1.1.1. Dielectric susceptibility – classical derivation

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Under the influence of the incident electromagnetic radiation, the medium becomes polarized. The dielectric susceptibility, which relates this polarization to the electric field, thus characterizes the interaction of the medium and the electromagnetic wave. The classical derivation of the dielectric susceptibility, χ, which is summarized here is only valid for a very high frequency which is also far from an absorption edge. Let us consider an electromagnetic wave, $[{\bf E} = {\bf E}_{\bf o} \exp 2\pi i (\nu t - {\bf k} \cdot {\bf r}),]$ incident on a bound electron. The electron behaves as if it were held by a spring with a linear restoring force and is an oscillator with a resonant frequency $[\nu_{o}]$ . The equation of its motion is written in the following way: $[m\; \hbox{d}^{2}{\bf a}/\hbox{d}t^{2} = -4 \pi^{2} \nu_{o} m{\bf a} = {\bf F},]$ where the driving force F is due to the electric field of the wave and is equal to $[-e{\bf E}]$ . The magnetic interaction is neglected here.

The solution of the equation of motion is $[{\bf a} = - e{\bf E}/ [4\pi^{2} m(\nu_{o} - \nu^{2})].]$ The resonant frequencies of the electrons in atoms are of the order of the ultraviolet frequencies and are therefore much smaller than X-ray frequencies. They can be neglected and the expression of the amplitude of the electron reduces to $[{\bf a} = e{\bf E}/(4\pi^{2} \nu^{2}).]$ The dipolar moment is therefore $[{\cal M} = -e {\bf a} = -e^{2} {\bf E} / (4 \pi^{2} \nu^{2} m).]$

von Laue assumes that the negative charge is distributed continuously all over space and that the charge of a volume element $[\hbox{d}\tau]$ is $[-e\rho \hbox{ d}\tau]$ , where ρ is the electronic density. The electric moment of the volume element is $[\hbox{d} {\cal M} = -e^{2} \rho {\bf E} \hbox{ d}\tau / (4 \pi^{2} \nu^{2} m).]$ The polarization is equal to the moment per unit volume: $[{\bf P} = \hbox{d} {\cal M} / \hbox{d}\tau = -e^{2} \rho {\bf E} / (4 \pi^{2} \nu^{2} m).]$ It is related to the electric field and electric displacement through $[{\bf D} = \varepsilon_{o} {\bf E} + {\bf P} = \varepsilon_{o} (1 + \chi) {\bf E.} \eqno\hbox{(A5.1.1.1)}]$ We finally obtain the expression of the dielectric susceptibility, $[\chi = -e^{2} \rho / (4 \pi^{2} \varepsilon_{o} \nu^{2} m) = -R\lambda^{2} \rho / \pi, \eqno\hbox{(A5.1.1.2)}]$ where $[R = e^{2} / (4\pi \varepsilon_{o} mc^{2})]$ $[(= 2.81794 \times 10^{-15}\;\hbox{m})]$ is the classical radius of the electron.

References

International Tables for Crystallography (2006). Vol. B. ch. 5.1, p. 550