In the semiclassical approximation, at an electron gas temperature T, the electrons can overcome the mutual repulsion energy Е_{Coulomb}=e^{2}/r getting closer to each other up to the minimal distance r=r_{min} at which E_{Coulomb} = T k_{Boltzman} = e^{2}/(r_{min}).
In an electron gas, the electrons move isotropically, repeatedly colliding with each other. However, after turning on the longitudinal external voltage, we increase the scattering energy for each electron by \Delta E.
Could we now write Coulomb's law in the form E_{Coulomb} = e^{2}/(r+r_{cut}), where
r_{cut} = e^{2} / (\Delta E) , if (\Delta E) > (T k_{Boltzman}) or
r_{cut} = e^{2} / (T k_{Boltzman}) , if (\Delta E) < (T k_{Boltzman})?
The short answer is no. To appreciate this point, consider the linear relationship between the applied electric field and the current density, the linear constant of proportionality being the conductivity σ. According to the Drude model for conductivity, which can be deduced from the quantum Boltzmann equation in conjunction with the relaxation-time approximation, for the conductivity σ one has [1]:
(1) σ(ω) = σ0/(1 - iωτ)
where
(2) σ0 = ne2τ/m,
in which ω is the frequency of the external electric field, n the number density, e2 the electron-charge squared, m the electron mass, and τ the relaxation time. In this approximation, all interaction effects, including the electron-electron interaction, electron-phonon interaction, and electron-impurity interaction, are subsumed in the relaxation time τ, which in principle is a function of m, n, and ω.
[1] See, for instance, Eq. (1.29), p. 16, in Solid State Physics, by NW Ashcroft, and ND Mermin (Harcourt, Orlando, 1976).
Dear Behnam Farid ,
Thanks for the detailed explanation.
Vladimir
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