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Interpret current density in plasma as material property?

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    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].

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Oct 10, 2016 #2

    Charles Link

    User Avatar
    Homework Helper

    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 ##.
     
    Last edited: Oct 10, 2016
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