In a perfect crystal, the scattering between quasiparticles and individual crystal sites is already accounted for in the previous argument, with energy transfer happening via Umklapp scattering. However, crystal in nature are not perfect. The impurities, be it lattice defects or other elements act as individual scattering centres for the quasiparticles.\cite{nozieres1999}
Assuming that the matrix element of quasiparticle–impurity scattering is roughly constant, the scattering cross–section of this process is proportional to the density of defects \(n_\mathrm{d}\). Since the density of defects is weakly temperature dependent (most defects being topologically protected), this process gives a conductivity (and thus resistivity) which is constant in temperature.
Therefore, combining the two mechanisms outlined so far gives a Fermi Liquid resistivity of \[\rho = \rho_0 + A T^2 \mathrm{ ,}\] form which is valid at low temperatures, \(T \ll T_F\). At higher temperatures with a large population of phonons, quasiparticle–phonon scattering starts to become important as well, dominating over the electronic contribution.
From equipartition, at high temperatures there’s an energy proportional to \(k T\) in each phonon mode. Assuming that the scattering cross section of each quasiparticle is proportional to the energy available in each phonon mode, we get a lifetime \(\tau_\mathrm{phonon}\) proportional to \(1 / T\). A more careful calculation\cite{Lawrence_Wilkins_1972} starting from the Boltzmann transport equation gives the same scaling.
Therefore, we can conclude the following: the resistivity, inversely proportional to the conductivity, has the following behaviour: \[\rho = \rho_\mathrm{impurity} + A_\textrm{qp-qp} T^2 \quad T \ll T_F\] \[\rho = \rho_\mathrm{impurity} + B_\textrm{qp-ph} T \quad T \gg T_F\]