5
$\begingroup$

1#When an electromagnetic wave propagates in a plasma, with increase in plasma density the phase velocity ($v=\frac{\omega}{k}$) of wave increases. So at cut-off point $k=0$, and phase velocity becomes infinite (wavelength becomes infinite). At resonance phase velocity becomes zero (wavelength becomes zero). Wavelength is distance between two consecutive troughs or crests. How it can be zero or infinite. This is expression for the phase velocity of light wave that is propagating in a plasma \begin{equation}\label{} v=\frac{c}{\sqrt{1-\frac{\omega^{2}_{p}}{\omega^2}}}. \end{equation} I can understand this mathematically but not physically.

$\endgroup$
  • 1
    $\begingroup$ But the cut-off defines the extent of wavelength that can exist in your medium. Longer wavelengths cannot propagate. I think it does not have any physical meaning to have infinite wavelength. $\endgroup$ – Amin R. Dec 6 '15 at 7:53
  • 1
    $\begingroup$ @sara - Not all electromagnetic waves propagate faster when the plasma density increases... The phase velocity of some modes decrease (e.g., Alfvén waves) while others show no differences with changing densities. $\endgroup$ – honeste_vivere Dec 6 '15 at 18:34
2
$\begingroup$

Quick Background

The phase velocity, $\mathbf{V}_{ph}$, is not just $\omega/k$, it is actually the real part of this ratio, or $\Re\left[\omega/k\right]$, since both the frequency and wavenumber can be, in general, complex.

Similarly, the group velocity is defined as: $$ \mathbf{V}_{g} = \frac{ \partial \Re\left[ \omega \right] }{ \partial k } $$

Cutoff Frequency

There is another way to think of the cutoff frequency than just $V_{ph} \sim \infty$, which is actually kind of the inverse of how you should think about it.

The more appropriate way to understand the cutoff frequency is to think of wave propagating from one medium to another with a different index of refraction. If the new medium presents a scenario where the incident wave would experience a cutoff, it is another way of saying that the incident wave would need an infinite phase speed to propagate in that medium. Because an infinite phase speed is physically impossible, the wave does not exist in the second medium.

An actual example that is observed in a plasma is when a Langmuir wave gets caught in a density well (either created by the wave's ponderomotive pressure or propagated into it).

Perhaps an easier concept to grasp is an evanescent wave, which is a similar concept except that the amplitude decays slowly with distance into the second medium rather than not existing at all.

I have another answer about boundary conditions here that shows more math and examples, using strings.

Resonance Frequency

This is slightly more difficult to conceptualize. It is not really that a wave's phase velocity tends to zero at resonance, so much as the growth rate, $\gamma \equiv \Im\left[ \omega \right]$, of the amplitude starts to vastly exceed the wave frequency, $\omega_{r} \equiv \Re\left[ \omega \right]$, or $\gamma \gg \omega_{r}$. The reason I say this is that in a plasma, an electromagnetic wave can be in resonance with some population of charged particles. This does not mean that the wave suddenly has $V_{ph} = 0$ because it resonates with those particles.

More Detailed (Math) Explanation

For example, suppose the source of free energy (e.g., see Gibbs free energy) that is resonating with a normal mode of the system happens to be a particle beam (e.g., field-aligned beams discussed in the arXiv paper 1207.5561). If this is the case, then linear growth theory shows that the parameter which determines interaction between the waves and particles is given by: $$ \zeta_{s}^{n} = \frac{ \omega_{r} - \mathbf{k} \cdot \mathbf{V}_{os} + n \ \Omega_{cs} }{ k \ V_{Ts} } $$ where $\mathbf{V}_{os}$ is the beam velocity relative to the bulk flow rest frame of species $s$, $\Omega_{cs}$ is the cyclotron frequency of species $s$, $\mathbf{V}_{os}$ is the thermal speed of species $s$, $n = 0, \pm 1, \pm 2, \pm 3,$ etc, and $s$ can be $e$(electrons) or $i$(ions).

A mode is said to be resonant with the particle beam when $\lvert \zeta_{s}^{n} \rvert \leq 1$ and non-resonant when $\lvert \zeta_{s}^{n} \rvert \gg 1$. The resonant case is another way of saying that the term in the numerator goes a value smaller than the denomenator, or: $$ \left( \omega_{r} - \mathbf{k} \cdot \mathbf{V}_{os} + n \ \Omega_{cs} \right) \leq \left( k \ V_{Ts} \right) $$

In the "best" case, the wave is completely resonant with the particles when: $$ \left( \omega_{r} - \mathbf{k} \cdot \mathbf{V}_{os} + n \ \Omega_{cs} \right) \rightarrow 0 $$

So you can see that $V_{ph}$ can still be finite when the waves are in resonance with the particles, which is why I argued that resonance does not necessarily imply $V_{ph} = 0$.

Fun Side Note

It is possible for a wave to have $V_{ph} = 0$ and $V_{g} \neq 0$, which are sometimes called purely growing modes since $V_{ph} = 0$ often implies that $\Re\left[ \omega \right] = 0$, but not necessarily mean that $\Im\left[ \omega \right] = 0$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

"Infinite" wavelength just means that the solution does not depend on the position but only on time.

A wave solution has the expression

$sin(kx-\omega t)$

$k$ is the wavevector, and the wavelength is $L=2\pi/k$

If the wave depends on time but not on space, (which might well happen, why not ?)

$sin(-\omega t)$

and this is just what happens at the cutoff frequency, then $k$ is zero and the wavelength is infinite. But the wavelength cannot be zero, because that would mean that the wavevector $k$ should be infinite but then the expression

$sin(kx-\omega t)$

would be meaningless.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.