# I Dispersion: expansion of wavenumber as function of omega

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1. Apr 13, 2017

### EmilyRuck

Hi!
Dealing about wave propagation in a medium and dispersion, wavenumber $k$ can be considered as a function of $\omega$ (as done in Optics) or vice-versa (as maybe done more often in Quantum Mechanics). In the first case,

$k (\omega) \simeq k(\omega_0) + (\omega - \omega_0) \displaystyle \left. \frac{dk(\omega)}{d \omega} \right|_{\omega = \omega_0} + \frac{1}{2} (\omega - \omega_0)^2 \left. \frac{d^2 k(\omega)}{d\omega^2} \right|_{\omega = \omega_0}$

where first derivative is the inverse of group velocity $1/v_g$ and the second derivative is (maybe?) inverse of group velocity dispersion $1 / \alpha$.

In the second case,

$\omega (k) \simeq \omega(k_0) + (k - k_0) \displaystyle \left. \frac{d\omega(k)}{d k} \right|_{k = k_0} + \frac{1}{2} (k - k_0)^2 \left. \frac{d^2 \omega(k)}{dk^2} \right|_{k = k_0}$

where the first derivative is the group velocity itself $v_g$ and the second derivative is the group velocity dispersion itself $\alpha$.

But what about the limit-cases, when for some reason $v_g = 0$ or $\alpha = 0$? The latter case is maybe more common: it would imply that, in a medium, for a signal whose group velocity doesn't vary with frequency, the second order term in the $k(\omega)$ expansion has an infinite coefficient.

2. Apr 13, 2017

### EmilyRuck

Considering the simplest case, the one regarding plane waves,

$k = \omega / v$

with $v$ constant.

$d\omega/dk = v = v_g$ is the group velocity and $dk/d\omega = 1/v = 1/v_g$ is the reciprocal of the group velocity.

$d^2 \omega/dk^2 = \alpha = 0$ is the group velocity dispersion; so, the reciprocal of the group velocity dispersion should be $1 / \alpha \to \infty$. But also $d^2 k / d \omega^2 = 1/\alpha (?) = 0$.

How is it possible!?

3. Apr 13, 2017

### Meir Achuz

$\frac{d^2\omega}{dk^2}$ does not equal $\left[\frac{d^2 k}{d\omega^2}\right]^{-1}$, as you have demonstrated.
$\omega''$ is the dispersion. In either optics or QM, you can use $\omega(k)$ or $k(\omega)$.
They are just two different ways of expressing the same mathematical function.

4. Apr 14, 2017

### EmilyRuck

Ok!

However, as regards the first derivative, $d\omega / dk = v_g$ and $dk/d\omega = 1/v_g$, so they are exactly reciprocal. If you take the unit measures, they are reciprocal too. So, here is still my doubt.