Dispersion: expansion of wavenumber as function of omega

  • #1
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Hi!
Dealing about wave propagation in a medium and dispersion, wavenumber [itex]k[/itex] can be considered as a function of [itex]\omega[/itex] (as done in Optics) or vice-versa (as maybe done more often in Quantum Mechanics). In the first case,

[itex]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}[/itex]

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

In the second case,

[itex]\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}[/itex]

where the first derivative is the group velocity itself [itex]v_g[/itex] and the second derivative is the group velocity dispersion itself [itex]\alpha[/itex].

But what about the limit-cases, when for some reason [itex]v_g = 0[/itex] or [itex]\alpha = 0[/itex]? 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 [itex]k(\omega)[/itex] expansion has an infinite coefficient.
 

Answers and Replies

  • #2
136
6
Considering the simplest case, the one regarding plane waves,

[itex]k = \omega / v[/itex]

with [itex]v[/itex] constant.

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

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

How is it possible!?
 
  • #3
Meir Achuz
Science Advisor
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[itex]\frac{d^2\omega}{dk^2}[/itex] does not equal [itex]\left[\frac{d^2 k}{d\omega^2}\right]^{-1}[/itex], as you have demonstrated.
[itex]\omega''[/itex] is the dispersion. In either optics or QM, you can use [itex]\omega(k)[/itex] or [itex]k(\omega)[/itex].
They are just two different ways of expressing the same mathematical function.
 
  • #4
136
6
In either optics or QM, you can use [itex]\omega(k)[/itex] or [itex]k(\omega)[/itex].
They are just two different ways of expressing the same mathematical function.
Ok!

[itex]\frac{d^2\omega}{dk^2}[/itex] does not equal [itex]\left[\frac{d^2 k}{d\omega^2}\right]^{-1}[/itex], as you have demonstrated.
However, as regards the first derivative, [itex]d\omega / dk = v_g[/itex] and [itex]dk/d\omega = 1/v_g[/itex], so they are exactly reciprocal. If you take the unit measures, they are reciprocal too. So, here is still my doubt.
 

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