1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    n^2 = 1 - \frac{\omega_P^2}{\omega^2},
    where the plasma frequency
    \omega_P = \sqrt{\frac{Ne^2}{\epsilon_0m_e}}.\quad(1)
    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:
    \nabla\times\textbf{B} = \mu_0\textbf{J} + \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}.

    3. The attempt at a solution

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

    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
    n = \sqrt{\epsilon_r\mu_r}
    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
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Interpret current density in plasma as material property?
  1. Current and density (Replies: 6)

  2. Drift speed for plasma (Replies: 1)