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Thomas-Fermi approximation and the dielectric function

  1. Nov 14, 2012 #1
    1)What exactly is meant by the 'static limit' where the frequency is taken to zero, but the wavenumber is finite? I am getting confused because if the frequency is zero, then surely the probing electrons/photons/whatever have no wavelength, so how can the wavenumber be finite and non-zero?

    2) Regarding the Thomas-Fermi approximation, in my textbook (Kittel) it says that it is valid for electron wavenumbers much smaller than the fermi wavevector - so larger wavelengths than the fermi wavelength. If I am looking at impurity scattering in a metal, then surely you cannot apply the TF approximation since the electrons will all be at the Fermi level and so the wavenumber of the scattered electrons will equal that of the fermi wavevector. However I have seen the TF used for graphene particularly, so how is that a valid assumption?

  2. jcsd
  3. Nov 14, 2012 #2


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    Consider the potential of a static charge (a Coulomb potential in vacuum). Although it is static, the Fourier transform of the spatial distibution will contain all values of k.
    The distinctive point with respect to photons is that you need a source for a Coulomb potential while photons are source free solutions of the Maxwell equations. The latter are only possible for special relations (dispersion) of k on omega.

    As regards to question 2 I suppose (although I am not sure) that it is sufficient in scattering that the change in wavenumber is much smaller than the Fermi wavenumber.
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