# Thomas-Fermi approximation and the dielectric function (+ small bit on graphene)

1) With the dielectric function, which is a function of wavenumber and frequency,how is it possible to take the limit of either to zero without changing the other one? I thought that frequency and wavenumber are linked, also am I right in thinking that they are both for the 'probe'?

2)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?

3) 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?

Cheers.

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1. The dielectric function to which you refer describes screening. From a phenomenological point of view, you can imagine the function acting as a damper (or sometimes an enhancer) of momentum and energy transfer. The wave vector $q$ and frequency $\omega$ dependence are these quantities, momentum $\hbar q$ and energy $\hbar\omega$ transfer, respectively. They are not strictly linked. For example, inelastic collisions conserve momentum but not energy.
2. The static limit is the time-averaged quantity. This is perhaps easiest seen by looking at the Fourier transform of the time-dependent dielectric function: $$\epsilon(q,\omega)=\int d\tau\, e^{i\omega \tau} \epsilon(q,t) \\ \epsilon(q,0)=\int d\tau\, \epsilon(q,t) \\$$
3. We've already covered that $q$ and $\omega$ are not the particle's momentum and energy in question 1, so this question should essentially be cleared up.