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I have a neutral plasma and I need to solve Maxwell equations given the charge and current densities on the plasma. In order to do it I need to know the electrical permittivity $\varepsilon$, I've seen in a lot of books that this quantity is given by:

$$ \varepsilon_r = 1- \left(\frac{\omega_p}{\omega}\right)^2 $$

Where $\omega_p$ and $\omega$ are the plasma and external wave frequencies, however the Lorentz-Drude model to the first order (Lorentz's result) shows us a slightly different result:

$$\varepsilon_r = 1 + \frac{\omega_p^2}{ (\omega_0^{ 2}-\omega^2) + 4i \Gamma \omega}$$

My questions are three:

  1. What are the assumptions and approximations to get the first equation, do they hold in the optical region?
  2. Could the damping factor $\Gamma$ be related somehow to the collision frequency in a plasma?
  3. Is the approximation $\mu = \mu_0$ correct?

Thank you in advance!

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1 Answer 1

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In the Drude model, the electrons are bound to the material by a Hooke's law restoring force, and this coupling with the material provides the damping. In the usual plasma model, this is not the case. Rather, the electrons are free to move. They respond to the electric field in the plasma by producing currents, and the changing currents produce time-varying magnetic fields that self-consistently induce electric fields (and hence currents). To answer your questions specifically:

  1. The assumptions are: (a) Fixed ions and mobile electrons in the plasma; (b) negligible mean magnetic field; (c) current density is $\vec{J} = -en_e\vec{v}_e$, where $n_e$ is the number density of electrons and $\vec{v}_e$ is their velocity; (d) the only force on the electrons is the Lorentz force, so $m_e\frac{\partial \vec{v}_e}{\partial t} = -e\vec{E}$ (note that Ohm's law does not apply); and (e) Maxwell's equations.

  2. $\Gamma$ is not directly related to the collision frequency in the plasma, since the analysis goes through even for a collisionless plasma.

  3. Yes, this derivation is for an unmagnetized plasma, and $\mu = \mu_\circ$. Hence $\epsilon_r = \frac{c^2}{v_\phi^2}$, where $v_\phi = \frac{\omega}{k}$ is the phase velocity of traveling em waves.

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  • $\begingroup$ Thank you, what would be the meaning of $\Gamma$ then?, is it just a quantity imposed by the model or does it have a physical explanation? $\endgroup$
    – manuel91
    Jan 19, 2015 at 22:56
  • $\begingroup$ $\Gamma$ in the Drude model reflects damping that corresponds to transfer of energy from the wave to the material. If you imagine the electrons bound to the lattice by springs, $\Gamma$ is the damping coefficient of those springs. Physically, the damping could be resistive or due to radiation losses. In any case, it is not there when deriving your first equation. $\endgroup$
    – pwf
    Jan 19, 2015 at 23:10
  • $\begingroup$ Why does Ohm's law not apply? Because it is a plasma? I have found derivations where it is used, but also some where your result is found. This would probably solve my question. $\endgroup$
    – Lagrangian
    Jan 2, 2016 at 11:42
  • $\begingroup$ @Lagrangian Ohm's law sometimes applies in a plasma, but not in the derivation of the first equation. In this case, the plasma is assumed to be "collisionless," i.e. collisions of electrons on ions are so infrequent (compared with frequencies of interest, like wave frequencies) that they may be neglected in the force balance equation. $\endgroup$
    – pwf
    Jan 4, 2016 at 17:59
  • $\begingroup$ I do have a derivation of the Drude dielectric function that does use Ohm's law and obtains $\epsilon_r = 1 - \omega_p^2/omega^2$. Isn't it weird that there is one derivation using a collisionless electron gas or plasma and one using ohm's law that obtain the same result? $\endgroup$
    – Lagrangian
    Jan 5, 2016 at 9:54

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