# Why do Fermi liquids have T^2 resistivity?

I have often read that metals that are Fermi liquids should have a resistivity that varies with temperature like $\rho(T) = \rho(0) + a T^2$.

I guess the $T^2$ part is the resistance due to electron-electron interactions and the constant term is due to impurity scattering.

Is there a simple argument to show this? Or maybe you could point me to a nice reference?

Also, it seems that for electron-electron interactions to introduce a finite resistivity, some umklapp scattering is necessary (to break Galilean and translational invariance). Is this correct? Which of these symmetries (Galilean or translational) has to be broken?

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I am looking for a better answer, but my simple understanding is as follows: $\rho\sim\Im\Sigma\sim\omega^2\sim T^2$. And $\Im\Sigma\sim\omega^2$ is what that defines the Fermi liquid behavior. –  Everett You Apr 3 '13 at 12:03
The $T^2$ scaling needs both Umklapp and electron-electron scattering. Effectively, a $O(kT)$ vicinity of the Fermi surface for quasiparticles participates in the interactions which implies the scaling, arxiv.org/abs/1204.3591 . –  Luboš Motl Apr 3 '13 at 12:04
@EverettYou: That's what I was thinking too, but where does the umklapp come in? –  jjj Apr 3 '13 at 12:45
@LubošMotl: Thanks for the reference! –  jjj Apr 3 '13 at 13:39
Has someone some good references about the computation of umklapp effect in the Fermi liquid theory? –  Andrea Amoretti Jan 31 at 11:17