Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Dispersion: expansion of wavenumber as function of omega

  1. Apr 13, 2017 #1
    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.
     
  2. jcsd
  3. Apr 13, 2017 #2
    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!?
     
  4. Apr 13, 2017 #3

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [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.
     
  5. Apr 14, 2017 #4
    Ok!

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dispersion: expansion of wavenumber as function of omega
  1. Sign of the wavenumber (Replies: 1)

Loading...