- #1
lampCable
- 22
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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
[tex]
n^2 = 1 - \frac{\omega_P^2}{\omega^2},
[/tex]
where the plasma frequency
[tex]
\omega_P = \sqrt{\frac{Ne^2}{\epsilon_0m_e}}.\quad(1)
[/tex]
We assume that the incident wave is a plane wave, and that on each electron [itex]F = qE[/itex].
Homework Equations
Maxwell's fourth equation in vacuum where there exist charges and current:
[tex]
\nabla\times\textbf{B} = \mu_0\textbf{J} + \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}.
[/tex]
The Attempt at a Solution
Integrating Newtons second law and neglecting any source velocity we get
[tex]
\textbf{v} = \frac{iq}{m\omega}\textbf{E}_0e^{i(kz-\omega t)}.
[/tex]
The current density, then, is
[tex]
\textbf{J} = Nq\textbf{v} = -\frac{Nq^2}{m\omega^2}\frac{\partial\textbf{E}}{\partial t}.\quad(2)
[/tex]
Using now Maxwell's fourth equation in vacuum together with (1) and (2) we get
[tex]
\nabla\times\textbf{B} = \mu_0\epsilon_0\bigg(1-\frac{\omega_P^2}{\omega^2}\bigg)\frac{\partial\textbf{E}}{\partial t}.
[/tex]
Question: Is it legit to here define [tex]\epsilon_r\mu_r = 1 - \frac{\omega_p^2}{\omega^2}[/tex] for the purpose of calculating the refractive index?
My argument for this is as follows. Since
[tex]
n = \sqrt{\epsilon_r\mu_r}
[/tex]
it doesn't matter really (for the purpose of determining the refractive index) how [itex]\epsilon_r[/itex] and [itex]\mu_r[/itex] 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 [itex]\epsilon_r\epsilon_r\neq1[/itex] instead. But since we are free to choose [itex]\epsilon_r[/itex] and [itex]\mu_r[/itex], the analogy could not go further to where the two are used separately.