Interpret current density in plasma as material property?

[lampCable]

Homework Statement

An electromagnetic wave propagates through a gas of N free electrons per unit volume. Neglecting damping, show that the index of refraction is given by
<br /> n^2 = 1 - \frac{\omega_P^2}{\omega^2},<br />
where the plasma frequency
<br /> \omega_P = \sqrt{\frac{Ne^2}{\epsilon_0m_e}}.\quad(1)<br />
We assume that the incident wave is a plane wave, and that on each electron F = qE.

Homework Equations

Maxwell's fourth equation in vacuum where there exist charges and current:
<br /> \nabla\times\textbf{B} = \mu_0\textbf{J} + \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}.<br />

The Attempt at a Solution

Integrating Newtons second law and neglecting any source velocity we get
<br /> \textbf{v} = \frac{iq}{m\omega}\textbf{E}_0e^{i(kz-\omega t)}.<br />
The current density, then, is
<br /> \textbf{J} = Nq\textbf{v} = -\frac{Nq^2}{m\omega^2}\frac{\partial\textbf{E}}{\partial t}.\quad(2)<br />
Using now Maxwell's fourth equation in vacuum together with (1) and (2) we get
<br /> \nabla\times\textbf{B} = \mu_0\epsilon_0\bigg(1-\frac{\omega_P^2}{\omega^2}\bigg)\frac{\partial\textbf{E}}{\partial t}.<br />

Question: Is it legit to here define \epsilon_r\mu_r = 1 - \frac{\omega_p^2}{\omega^2} for the purpose of calculating the refractive index?

My argument for this is as follows. Since
<br /> n = \sqrt{\epsilon_r\mu_r}<br />
it doesn't matter really (for the purpose of determining the refractive index) how \epsilon_r and \mu_r are chosen per se, so long as their product satisfy the above definition. In this sense it is then possible to convert a case where we have a region with current density, to one where we simply have a material with \epsilon_r\epsilon_r\neq1 instead. But since we are free to choose \epsilon_r and \mu_r, the analogy could not go further to where the two are used separately.

 

[Charles Link]

Homework Statement

An electromagnetic wave propagates through a gas of N free electrons per unit volume. Neglecting damping, show that the index of refraction is given by
<br /> n^2 = 1 - \frac{\omega_P^2}{\omega^2},<br />
where the plasma frequency
<br /> \omega_P = \sqrt{\frac{Ne^2}{\epsilon_0m_e}}.\quad(1)<br />
We assume that the incident wave is a plane wave, and that on each electron F = qE.

Homework Equations

Maxwell's fourth equation in vacuum where there exist charges and current:
<br /> \nabla\times\textbf{B} = \mu_0\textbf{J} + \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}.<br />

The Attempt at a Solution

Integrating Newtons second law and neglecting any source velocity we get
<br /> \textbf{v} = \frac{iq}{m\omega}\textbf{E}_0e^{i(kz-\omega t)}.<br />
The current density, then, is
<br /> \textbf{J} = Nq\textbf{v} = -\frac{Nq^2}{m\omega^2}\frac{\partial\textbf{E}}{\partial t}.\quad(2)<br />
Using now Maxwell's fourth equation in vacuum together with (1) and (2) we get
<br /> \nabla\times\textbf{B} = \mu_0\epsilon_0\bigg(1-\frac{\omega_P^2}{\omega^2}\bigg)\frac{\partial\textbf{E}}{\partial t}.<br />

Question: Is it legit to here define \epsilon_r\mu_r = 1 - \frac{\omega_p^2}{\omega^2} for the purpose of calculating the refractive index?

My argument for this is as follows. Since
<br /> n = \sqrt{\epsilon_r\mu_r}<br />
it doesn't matter really (for the purpose of determining the refractive index) how \epsilon_r and \mu_r are chosen per se, so long as their product satisfy the above definition. In this sense it is then possible to convert a case where we have a region with current density, to one where we simply have a material with \epsilon_r\epsilon_r\neq1 instead. But since we are free to choose \epsilon_r and \mu_r, the analogy could not go further to where the two are used separately.

Your derivation looks quite legitimate. The next step in deriving the E-M wave (at least the $B$ part) is to take the curl of both sides of your equation. With a vector identity $\nabla \times \nabla \times B=\nabla (\nabla \cdot B)-\nabla^2 B$, and using Faraday's law on the other side, you have the wave equation for $B$ and the wave velocity is determined precisely by what you have already presented. editing... Alternatively you could begin with Faraday's law $\nabla \times E=-dB/dt$ and again take the curl of both sides of the equation and proceed in a similar fashion to get the wave equation for the electric field $E$.