# Permittivity of plasma in electromagnetic wave

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.

• can you include said formula in your question?
– JEB
Commented Mar 14, 2023 at 15:44
• $\epsilon=1-\frac{ne^2}{m\epsilon_0 \omega^2}$ Commented Mar 14, 2023 at 16:10
• 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 Commented Mar 14, 2023 at 17:13

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: $$\rho=\rho_b+\rho_f,\mathbf{J}=\mathbf{J}_b+\mathbf{J}_f,\\ \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}\Rightarrow\nabla\cdot\mathbf{P}=\rho_b,\\ \nabla\times\mathbf{B}=\mu_0\left(\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)\Rightarrow \mathbf{J}_b=\nabla\times\mathbf{M}+\frac{\partial \mathbf{P}}{\partial t}$$