What does "non-dispersive" mean in terms of waves and group velocity?

Question

I'm confused about the term wave group velocity: It is usually explained in terms of a superposition of harmonic waves with very closely spaced wave vectors and frequencies. It is then easily shown, that the envelope of such packet moves with group velocity

$$v_g=\frac{d \omega(k)}{dk}$$

On the other hand, the individual wave components move with its phase velocity

$$v_p=\omega/k$$

This means, that although the envelope moves without changing its shape, the sinusoidal components inside change with regard to the envelope:

Is this what we call "dispersion" ?

In my opinion, the only way to keep the detailed form constant, we must have

$$v_p=v_g \rightarrow \frac{\omega(k)}{k} = \frac{d \omega(k)}{dk}$$

In such case we get:

I would guess, this is a case without dispersion – right?

So if there is a wave vector $k$ for which $v_g(k)=v_p(k)$ and my wave packet is closely centered around that k, there is no dispersion. However, this doesn't imply, that $d \omega(k)/dk = const=c$, as it would be for completely non-dispersive waves.

So what does "non-dispersive" mean? Is it with regard to a particular wave vector or for all wave vectors (like sound or free plane EM-waves)?

Answer 1

Yes, it is for all $k$, which only works if $\omega=ck$:

$$ \frac{\omega}k = c =v_p$$ $$ \frac{d\omega}{dk} = c = v_g $$

Anything else is dispersive.

BTW, you can do the following with fully formed wave packets from Fourier transforms of spectral distributions, but for now, consider the equal sum of two nearby monochromic waves:

$$ A(x, t) = e^{i(k_1x-w_1t)} + e^{i(k_2x-w_2t)} $$

and consider the time derivative:

$$ \frac{dA(x, t)}{dt} =-i(\omega_1 e^{i(k_1x-w_1t)} + \omega_2e^{i(k_2x-w_2t)} )$$

Now if the waves are close: $$k \equiv k_1$$ $$k_2 =k+dk $$

with

$$\omega \equiv \omega_1$$ $$\omega_2 = \omega + d\omega = \omega + \frac{d\omega}{dk}dk = \omega(1 + v_{gr}dk) $$

so

$$ \frac{dA(x, t)}{dt} =-i\omega[e^{i(kx-\omega t)} + (1+v_{gr}dk)e^{i(kx-(\omega+v_{gr}dkt))}] $$

$$ \frac{dA(x, t)}{dt} =-i\omega e^{i(kx-\omega t)}[1 + (1+v_{gr}dk/\omega)e^{-iv_{gr}dkt))}] $$

So the phase factor out front is still moving at $v_{ph}$, and there should be some steps to convince you that the exponential in the $[,]$ is a translation operator moving the envelope at $v_{gr}$