# What is the physics behind the plasma frequency?

I can't seem to find a satisfactory, intuitive explanation for the way electromagnetic waves behave when they encounter a cold plasma (that is, one where $\omega^2 = \omega_p^2 + c^2 k^2$). My intuition for the behavior below the plasma frequency is that the incoming wave makes the electrons in the plasma oscillate in such a way that they generate radiations that cancel the transmitted wave and create a reflected wave in the opposite direction (is that right?). I also understand that at very high frequency the electrons don't respond fast enough making the plasma behave like a vacuum. However the existence of this very sharp change in behavior, and what happens beyond the plasma frequency, stumps me.

• I'm not an expert, so I'm making this a comment, but I'd bet that the transition is not particularly sharp. – Sean E. Lake Nov 4 '17 at 1:43
• Well the wave vector does vary smoothly as a function of the frequency, but the transmission and reflection coefficients respectively increase and decrease quite sharply just after the plasma frequency -- or at least that's what my textbook wants me to believe. – user115153 Nov 4 '17 at 1:57
• That sounds like it's based on a harmonic oscillator model. If you graph the response of a simple harmonic oscillator as a function of frequency, the phase has a sharp discontinuity at the resonant frequency. Adding realistic damping broadens that a bit. See Wikipedia for SHO details: en.wikipedia.org/wiki/Harmonic_oscillator#Steady-state_solution – Sean E. Lake Nov 4 '17 at 2:03
• Right, so in this model there is a transition! And I'm inquiring as to how it works – user115153 Nov 4 '17 at 14:47
• Isn't it simply that the plasma frequency is the free oscillation frequency - the plasma will wobble at this frequency undriven, and so you get the same behavioral transitions and resonances with changing frequency as you do for any other driven harmonic oscillator behavior? – Selene Routley Nov 9 '17 at 1:12

Your understanding is basically right. A helpful way to understand the boundary where $\omega = \omega_p$ is to look at the refractive index,
$$n=\frac{c}{v_{p}}=\sqrt{\frac{\epsilon \mu}{\epsilon_0 \mu_0}}\approx\sqrt{\epsilon_r}\,,$$ where $v_p$ is the phase velocity and I have taken $\mu_0\approx \mu$.

For electromagnetic (EM) waves incident on a cold unmagnetized plasma the ions can be assumed to be at rest relative to the electrons due to their inertia. The much lighter electrons move more readily in response to the electric field of an EM wave. The displacement $x$ between an electron and ion due to the EM wave creates an electric dipole moment, which affects the plasma's relative permittivity $\epsilon_r$,

$$\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}= 1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\,,$$ where $E= E_0 e^{-i\omega t}$ is the electric field of the EM wave, $P= n_e p_e$ is the polarization density (which aligns with the electric field in a simple medium), $n_e$ is the electron number density and $p_e =e x$ is the electric dipole moment.

The displacement $x$ is given by the equation of motion $\ddot{x} = \frac{e}{m_e} E$ and also follows simple harmonic motion as Sean has suggested, $\ddot{x} = -\omega^2 x$, so $$x = -\frac{\ddot{x}}{\omega^2}= -\frac{e E}{m_e \omega^2}\,.$$ Plugging this into our equation for the relative permittivity gives $$\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2}=1 - \frac{\omega_p^2}{\omega^2}\,.$$ Therefore, the refractive index becomes zero at $\omega=\omega_p$ and is imaginary for $\omega<\omega_p$.

Now we consider the various cases depending on the frequency of the EM wave. Generally, in a cold plasma the electrons are able to respond to the EM wave on a characteristic timescale of $1/\omega_p$:

1. For $\omega<\omega_p$, the math tells us that waves cannot propagate in a medium with imaginary refractive index (the solution to the wave equation falls off exponentially and the field becomes evanescent). Physically, the electrons are able to cancel the electric field of the EM wave within the period of oscillation of the wave ($\sim 1/\omega$). Thus, the cold plasma behaves like a metal, which has essentially free electrons that cancel fields tangential to the surface. You are therefore correct to say that the electrons establish a reflected field of radiation emanating from the plasma's surface. The majority of the wave energy bounces off the plasma and does not enter it (e.g. radio waves bouncing off ionosphere).
2. For $\omega=\omega_p$, the wave energy gets absorbed by the plasma with little reflection.
3. For $\omega>\omega_p$, the electrons are unable to completely cancel the electric field of the EM wave, so the wave is able to propagate in the plasma (see figure below). The collective electron motion is still able to cancel some of the electric field within one period of oscillation of the wave and establish a reflected radiation field as well.
4. In the limit $\omega\gg \omega_p$, you are right that even the electrons hardly respond due to their inertia and establish only a weak radiation field. Thus the plasma behaves much like a vacuum with only very little reflection.

I hope this explains the sharp frequency dependence of the reflection and transmission coefficients. For a good description of radiation in a dielectric, I recommend reading this answer.

Plot of the dispersion relation for EM waves in a plasma:

• Thank you very much for your detailed explanation however what I was looking for is indeed a physical interpretation of what happens beyond the plasma frequency; for instance the maths foresee that below it the incoming wave is reflected and an interpretation I found (that may be incorrect) is that the oscillating electrons themselves radiate, which creates the pattern we know. However this can't happen beyond $\omega_p$ because there's no reflected wave while an oscillating dipole radiates in all directions. What then, beyond the plasma frequency,causes the 'slowing down' of the incoming... – user115153 Nov 4 '17 at 10:29
• ...waves? Incidentally, too, why don't the electrons radiate anymore? – user115153 Nov 4 '17 at 10:31
• Thank you for the comments, I have edited my answer to address your questions (e.g. there is still a reflected wave for $\omega>\omega_p$). I have included a link to a good physical explanation for EM waves traveling in a dielectric medium. – Tom Neiser Nov 8 '17 at 22:29