Is there any intuitive explanation of why resistivity of metal goes as $T^5$ at low temperature?

The Debye theory gives that the phonon distribution goes as $n(\omega)\sim T $ at higher temperature and $n(\omega)\sim T^3$ at low temperature? What is responsible for this extra $T^2$ suppression?

Also, how do I go about rigorously deriving the $T^5$ dependence from scattering theory?

  • $\begingroup$ Please specify what are your variables, as n(w) etc... Also, it is not always the case that we get a power of five. Look for Bloch-Wilson limit $\endgroup$
    – daydreamer
    Commented Oct 13, 2020 at 1:25

1 Answer 1


Following this set of course notes (p. 94-95): the number of phonons scales as $n(\omega)\sim T^3$, but for low temperatures one phonon scattering is not going to knock an electron far along the Fermi surface. So to really randomize the electron direction one needs enough scatterings to make the sum of angular changes of order unity. The scattering angle scales roughly like $\approx \omega^2/2k_F^2 c_{sound}^2 \propto T^2$, so one needs $1/T^2$ extra scatterings. So the total scaling becomes $T^5$... except as the text (and daydreamer's comment) points out, the scaling has often a lower exponent because of the shape of the Fermi surface.


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