Suppose an electromagnetic wave with frequency $\omega$ propagates through plasma with free electron concentration $n$. The formula for the permittivity in terms of the frequency $$ \epsilon = 1 - \frac{ne^2}{m\epsilon_0\omega^2} $$ can be obtained by considering that the divergence of the polarization vector is equal to the free charge density in the plasma.

Why is that?

I understand the fact that plasma is not a linear medium and the polarization cannot be expressed in terms of the displacement vector or the electric field vector.

  • $\begingroup$ can you include said formula in your question? $\endgroup$
    – JEB
    Mar 14, 2023 at 15:44
  • $\begingroup$ $\epsilon=1-\frac{ne^2}{m\epsilon_0 \omega^2}$ $\endgroup$
    – I_am_ant
    Mar 14, 2023 at 16:10
  • $\begingroup$ One thing to notice is that your $\epsilon$ depends on $\omega$ so it is nonlocal in time. So it does not come from the divergence of the polarization, but from its Fourier transform $\endgroup$
    – Mauricio
    Mar 14, 2023 at 17:13

1 Answer 1


Polarization and displacement are quite general concepts of macroscopic electrodynamics. What we cannot do in case of inhomogeneous medium is to describe it by a constant dielectric permittivity and permeability.

More specifically, polarization and magnetization, $\mathbf{P},\mathbf{M}$ are just the fields created by the charges and currents induced in the media, $\rho_b,\mathbf{J}_b$ (up to constant coefficients, dimensional in some systels of units). These are obviously parts of the total electric and magnetic fields, $\mathbf{E},\mathbf{B}$, so we can introduces the difference between them as the fields induced by induced by external charges and currents, i.e., the charges and the currents external to the media, $\rho_f,\mathbf{J}_f$.

Polarization and magnetization are then determined by the part of the Maxwell equations that is due to the induced charges and currents: $$\rho=\rho_b+\rho_f,\mathbf{J}=\mathbf{J}_b+\mathbf{J}_f,\\ \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}\Rightarrow\nabla\cdot\mathbf{P}=\rho_b,\\ \nabla\times\mathbf{B}=\mu_0\left(\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)\Rightarrow \mathbf{J}_b=\nabla\times\mathbf{M}+\frac{\partial \mathbf{P}}{\partial t} $$


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