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Lenard Jones Potential & wavenumber

  1. May 6, 2010 #1
    I was just wondering, given that you have the dispersion relation omega = omega(k) for a Lenard-Jones potential in say a monatomic crystalline solid of lattice spacing a, which is proportional to sin(ka/2) where k is the wavenumber. The group velocity is the derivative of the dispersion relation with respect to k.
    My Question is: Can the wavenumber take negative values in general? (Would it be physically meaningful?). I am thinking it would be okay, as the group velocity would have the same values as if k>0.
     
  2. jcsd
  3. May 7, 2010 #2

    Born2bwire

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    Typically no, with a few exceptions. We generally deal with the wave number in terms of the wave vector. The wave vector can certainly be negative as it is indicative of the direction of the phase and group propagation. However, the negative sign is part of the directional vector and not of the wave number. The only time that you should have a negative wave number is with a left-handed material as the wave number is defined as

    [tex]k = \omega\sqrt{\mu\epsilon}[/tex]

    Still, there are times where we do talk about negative frequencies (which would give us negative k) when we are doing a Fourier analysis. These negative frequencies are generally just the complex conjugate of the positive frequency parts and so you can usually fold over the Fourier domain so that you deal only with frequencies greater or equal to zero.
     
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