4
$\begingroup$

In my research, I found that my system has the following dispersion relation: $$\omega^2=k^2+k_0^2\ , $$ where $k_0^{-1}$ is an intrinsic lengthscale of the system and the units are chosen so that the wave speed is unity. This is what it's looks like (blue curve):

enter image description here

So for short wavelengths (large $k$) we have standard linear dispersion, and for long wavelengths we have vanishing group velocity and diverging phase velocity.

My question is kind of soft: I can't quite get an intuition about what this does. Do any of you know of a similar situations? Can you describe qualitatively how things behave?

$\endgroup$
3
$\begingroup$

I start my answer with the the second question (similar situations).

This is eerily similar to the free-particle dispersion relation (or the relativistic energy-momentum relation) in natural units ($c = 1$ and $\hbar = 1$), $$\omega^2 = k^2 + m_0^2$$ This system of units is commonly adopted in particle physics and quantum field theory. It can be clearly discerned that in this system of units, the characteristic ''Compton wavelength'' of the the particle would correspond to the inverse of mass ($\lambda_0 \sim 1/m_0$), or strictly speaking have a factor of $2\pi$ in between. If you write in terms of the wave number $k_0 = 2\pi/\lambda_0 = m_0$, we recover exactly the equation you have, $$\omega^2 = k^2 + k_0^2$$, which means that your ''intrinsic length-scale'' is getting decided by the most fundamental ''intrinsic'' property of the particle - its mass.

Of course, then, we expect these systems to have similarities,

  1. When $k >> k_0$, (short wavelength limit), the relation is approximately $\omega^2 =k^2$, since the second term on RHS is negligibly smaller in comparison to the first. Thus, $\omega = k$ (maybe that's what you mean by linear dispersion), hence $v_{\rm phase} = \omega/k = {\ \rm constt}$, and hence $v_{\rm group} = d\omega/dk = 0$, which is what you get.

  2. When $k << k_0$ (long wavelength limit), the relation is approximately $\omega = k_0$. Then, $\omega/k = k_0/k = $ very large, or infinity in the limiting case.

The point here is that large or small depends on how $\lambda$ compares with the intrinsic scale, which is decided by the particle's mass. So, your system is behaving the more or less the same as a free particle. That may or may not be profound, but nothing more can be said unless you mention the context.

Hope this helps :)

$\endgroup$
5
$\begingroup$

This is the generic form of a dispersion relation with a low frequency cutoff. It is a good model for the dispersion relation of electromagnetic waves in a plasma, i.e. the ionosphere, where: $$ \omega^2 = \omega_0^2 + c^2 k^2 $$ Which is the same as your form, though I've put back in $c$ for the nondispersive phase velocity and changed $k_0 \to \omega_0$ to make it a little clearer. For frequencies below the cutoff $\omega_0$ the wave number is imaginary, that is, the waves don't propagate, they are evanescent decaying exponentials. As you point out, the phase velocity $$ v_\phi^2 = \frac{\omega^2}{k^2} = c^2 + \frac{\omega_0^2}{k^2} $$ can get arbitrarily large in the large wavelength limit. The group velocity $$ v_g^2 = \left( \frac{\partial \omega}{\partial k} \right)^2 = \frac{ c^2 }{ 1 + \frac{\omega_0^2}{c^2 k^2} } $$ which vanishes in the long wavelength limit and approaches the nondispersive phase velocity in the short wavelength limit.

Notice that, for the case of plasmas, you might be even more concerned about the diverging phase velocity, since in that case it is quickly exceeding the speed of light. But, it is no bother. The phase velocity is a fairly artificial thing, it just relates the relative phase of different parts of our system after they have reached their steady state.

To get a good physical analogy to reason with, consider a line of pendula of mass $m$ hanging from the wall of length $l$, connected with springs with spring constant $K$ separated by a distance $a$, we get a dispersion relation of the form $$ \omega^2 = \frac{g}{l} + \frac{4K}{m} \sin^2 \frac{ka}{2} $$ which if we take long wavelength limit (i.e. make $ka \ll 1$, so put our pendula close together or consider wavelengths small compared to their separation, we get $$ \omega^2 = \frac{g}{l} + \frac{Ka}{M} k^2 $$ which is the same form as the one under consideration. But now we know how to reason about the situation. First, the presence of a low frequency cutoff makes sense. The pendula want to oscillate at their natural frequency of $\omega = \sqrt{g/l}$ and there is nothing we can do to compel them to oscillate any slower. This will be generic in any system that has some kind of "internal oscillation" aside from the coupling.

This is the large wavelength limit, in which the frequency of our wave is just the frequency of our pendula by themselves. Imagine the whole row of pendula all moving back and forth in perfect synchronization. Equivalently, you can imagine making the coupling between the pendula tend towards zero, which will put us in the same regime. Now all of the pendula move back and forth in perfect synchronization, so if you try to follow the motion of a "wave crest", it will appear to move at an arbitrarily fast speed, approaching infinity in the limit that our wavelength goes to infinity and all of the pendula are truly synchronized (they all have the same phase). Additionally, if we consider this as the limit of vanishing coupling, it makes sense that the group velocity should vanish, as the group velocity gives us the velocity at which a disturbance moves through our system. If the pendula's coupling is arbitrarily small, and you jostled one of them, that disturbance would move arbitrarily slowly to the other pendula.

The other end makes intuitive sense as well. As we increase the frequency of the wave, or lower is wavelength, more and more we are probing the physics of an infinite line of coupled masses with springs, the pendula bit doesn't enter into it, so we recover the physics of a nondispersive medium with constant phase and group velocities, which are equal. (As long as we don't try it with too small of a wavelength so as to ruin our long wavelength approximation in the first place).

$\endgroup$

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.