How come group and phase velocity can be different in a non-dispersive media?

Question

I'm a (ex-)physicist working in the applied-physics world and was digging into group vs phase velocities.

There is one thing that confuses me : when would that be the case where group and phase velocity actually differ but the media is still non-dispersive ?

One can see an animation of this on the wiki page for phase velocity: https://en.wikipedia.org/wiki/Phase_velocity

There is a Gaussian Pulse where group velocity is less than phase velocity, and yet, the media is "non-dispersive" !

I checked the notebook that built this animation (available here https://gitlab.com/Carlson/dispersion-relation-sim) and it looks like in the Gaussian pulse example, the author actually use $\omega = c \times k + \rm constant$

This confuses me a little because phase velocity now equals $v_\phi = \frac{\omega}{k} = c + \frac{\mathrm{constant}}{k}$ In other words, now the phase velocity depends on $k$.

But I thought that since $k = \frac{\omega}{c}$, now we have: $v_\phi = c + \frac{\mathrm{constant} \times c}{\omega} = c(1 + \frac{\mathrm{constant}}{\omega})$, which means $v_\phi$ depends on the frequency! Which is, BY DEFINITION, a dispersive media.

So the author says it's non-dispersive, but to me it looks like it is dispersive, but at the same time, the packet travels un-distorted so it looks like it is not dispersive.

Ugh...

Is it just a special case of a non-dispersive media, where both velocities differ? And because it is non-dispersive, then, automatically, the wave packet will NOT deform?

Maybe (Probably) I'm wrong when I assume $k = \frac{\omega}{c}$, because then $v_\phi$ is simply equal to... $c$ ?! Which is in contradiction with the formula above $v_\phi = c(1 + \frac{\mathrm{constant}}{\omega})$

I'm obviously doing something wrong here, but not sure what.

Answer 1

In short, I think the claim made in the Wikipedia example must be incorrect.

Phase velocity, by definition, is $v_p = \frac{\omega_p (k)}{k}$.

Group velocity, by definition, is $v_g = \frac{d \omega_p (k)}{dk}$.

In his book Wave Propagation and Group Velocity (1960), Léon Brillouin defines a dispersive medium in the very first chapter as follows:

A medium exhibiting a wave velocity W(k) is called a dispersive medium

He uses the symbol $W(k)$ for phase velocity, so he is saying that phase velocity that varies with frequency is indicative of a dispersive medium. Otherwise, the medium is non-dispersive.

Clearly, in a non-dispersive medium, the phase velocity $v_p$ is constant, and that forces $v_g = \frac{d \omega_p (k)}{dk} = v_p$.

I am citing Brillouin's book as an authoritative source, but I can see many other search results on Google (e.g. Quora, Harvard Physics class, etc.) that use the same definition of a non-dispersive medium and hence conclude $v_p = v_g$.

N.B: I haven't read Brillouin's book in detail, but I learned of the book when I took a class in nanophotonics, and the instructor introduced it as a reference that clarifies many confusions regarding the definition of energy density in dispersive media. Apparently there was a flurry of papers claiming superluminal transport in nanophotonic structures many years ago, but they all had to do with incorrect notions of energy density in dispersive media, while Brillouin already set down what it is in 1960.

Answer 2

In essence, the answer is more or less derived here:

https://en.wikipedia.org/wiki/Group_velocity#Derivation

The phase velocity does not contribute anywhere inside the integral, it is outside of the integral. Inside the integral is only the omega prime (that is, the group velocity).

The derivation above shows that the group velocity ($\omega_0^\prime$) and phase velocity ($\omega_0$) are decoupled, and both are tied to the enveloppe and the carrier respectively and the latter is monochromatic, therefore, even though we have a 'dispersive' media because 'phase velocity depends on frequency', there's only one frequency that is affected by that (the monochromatic, central frequency)

In fact, one can show that, if we limit the Taylor expension of $\omega$ to first order, called linearization, as done in the wiki, it is impossible to explain a packet distortion, because both zero-th and first order terms are constant, and they define phase and group velocity respectively. So, by linearizing, we "force" the group velocity to be constant, which is equivalent to saying that the packet travels undistorted. One has to push the expension to at least the second order to mathematically explain packet distortion.

In addition to that, if one really does this second-order expension, it turns out that the case of a Gaussian pulse is particular : the second-order term only affects the variance of the pulse, which will become larger as time goes. Consequently, the higher the second-order term, the higher the Gaussian pulse dispersion will be. But if the second-order term is zero, then the Gaussian Pulse will travel freely, even though the zero-th order (phase velocity) is a function of frequency via the dispersion relation, because, again, only one frequency is 'affected' by this dispersion relation.

In the wiki picture, the dispersion formula is $\omega = c\cdot k + \alpha$ where $\alpha$ is a constant. (Note: The wiki author provides links to a very nice jupyter notebook that can help reproduce the animation and see what the dispersion formula is ($\alpha=4$ in his notebook))

The consequence of this is that if you derive it once (w.r.t to $k$), you get the group velocity $c$, but if you derive it a second time, it is zero, hence the second-order term is zero, and this is another (necessary) reason for why the Gaussian pulse travels undistorted.

You can try making the dispersion relation second-order using e.g.

def dr(k):
    return pow(k, 2.) + 4

in the jupyter notebook and you will see the pulse shape distort.

[below is a tentative visual interpretation to get a sense of why a linear dispersion relation with a constant term doesn't really affect wave packets]

An important point to realize is that with a first-order dispersion relation like the one used here, the constant $\alpha$ merely shifts the spectrum in Fourier space. To illustrate with a "visual interpretation", the fact that the sum of monochromatic waves 'forms' a nice packet with a constant shape stems from the fact that a higher frequency wave have a higher wavenumber $k$ than one with lower frequency, and that their relation is some constant factor ($c$). Because it is constant, all their 'interferences' will stay the same. For a given time interval, all waves (irrespective of their frequency) will 'move' by their wavenumber proportionnal to some velocity ($c$ here). Adding a constant to the dipsersion formula doesn't change that : it simply 'shifts the origin' but for a given time interval, all waves will move in the same manner than without the constant. Only non-linear relationships will distort the signal. Indeed, if there is a quadratic dependency ($k^2$) in the dispersion relation, now in a given time interval, higher-frequency waves will 'move' by their wavenumber times their wavenumber times some velocity ($c$ here), so different-frequency sub-packets will 'move' at different speed and thus distort the packet.

In summary, a monochromatic-modulated Gaussian pulse with a first-order-only dispersion relation is kind of a particular case, where you technically have (phase) dispersion but no (group) distortion.