ZapperZ said:
It does. Why shouldn't it, since e-e scattering contributes to the scattering rate and quasiparticle lifetime, just like any other mechanism?
Well, as I tried to explain, the e-e interaction is different in that there's no momentum transfer to the lattice, unlike the electron-phonon interaction. So there's no degradation of the electron's (quasi)momentum in the e-e interaction. At least, that's what I thought.
It looks like you have the Matthiesen rule in mind (##\rho = \rho_\text{e-p} +\rho_\text{e-e} ##, using the nomenclature of the paper you link to) when you mention the scattering times, where there's a term from the e-e interaction. If that's the case, then my question is why is that term (##\rho_\text{e-e}##) non zero? Your answer in your first paragraph is "because you have that term".
This is what I'm trying to understand. From a quick search,
https://arxiv.org/abs/1805.03650 mentions
paper 1 said:
Kohn’s theorem states that interactions do not change the cyclotron resonance frequencies in single-band metals with parabolic band dispersion relations strictly obeying translational invari-ance or, if applied in the absence of a magnetic field, that electron-electron interactions cannot degrade the current in the system and therefore cannot affect the dc long-wavelength electrical conductivity. As a result, in a clean material with a parabolic dispersion relation, the con-ductivity cannot be changed except by umklapp scatter-ing (which, arising from an underlying lattice, explicitly breaks translational invariance and momentum conser-vation) or Baber scattering (which involves multiband systems also manifesting a breaking of translational in-variance)
Emphasis mine.
ZapperZ said:
The problem is that this is such a small effect that you need (i) very clean material (very few defects and impurities), and (ii) at very low temperatures to be able to detect the impact of e-e interaction in the metal's resistivity. See, for example,
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.19.6172.
Zz.
Now that's something more tangible. From what I can understand from the paper you link, in compensated systems, the electron-electron interaction (more precisely, the electron-hole scattering) has no impact whatsoever on the total resistivity. It is only in uncompensated systems where the e-h scattering can cause a resistivity impact. That's where Matthiesen rule "fails". But the article makes a clear distinction between e-e scattering and the e-h one, so I'm a bit confused.
As far as I understand, Baber scattering is the process responsible for the ##\rho \sim T^2## behavior, which is hinted by the paper
https://arxiv.org/abs/1206.0623 as
paper2 said:
Therefore the suggestion [3] that the T2-dependence of the DC resistivity in SrTi1−xNbxO3 can be provided by the Baber scattering mechanism is supported by the present calculation.
and from the paper
https://epjb.epj.org/articles/epjb/abs/2017/08/b170367/b170367.html I get the impression that an Umklapp process is required in order to affect the resistivity (so the e-e interaction is discarded). But then it points out that electrons with different effective mass can lead to an impact on ##\rho## (for example with (quasi)electrons from s and d bands).
To add another paper to the pile,
https://arxiv.org/abs/1204.3591 claims
another paper said:
While it is well-known that the electron-electron (ee) interaction cannot affect the resistivity of a Galilean-invariant Fermi liquid (FL), the reverse statement is not necessarily true (...)
Overall, from the little I understand, it is only under very specific conditions that the e-e interaction can have an impact on ##\rho##. And that a scattering time does not necessarily correlates with an impact on ##\rho##. Thus, in general, it is not clear at all to me why the e-e interaction leads to a contribution to ##\rho##. If someone could clarify things, I would be glad.Edit: From another (well cited) recent paper:
https://arxiv.org/abs/1508.07812
the last paper said:
However, electron-electron scattering alone does not generate a finite contribution to resistivity, because such a scattering event would conserve momentum with no decay in the charge current. The presence of an underlying lattice is required in any scenario for generating T2 resistivity from electron-electron scattering.