I Dispersion: expansion of wavenumber as function of omega

[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.

 

[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!?

 

[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.

 

[EmilyRuck]

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.

Ok!

\frac{d^2\omega}{dk^2} does not equal \left[\frac{d^2 k}{d\omega^2}\right]^{-1}, as you have demonstrated.

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.