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Flaviu Cipcigan edited magnetoresistance.tex
over 10 years ago
Commit id: 512b0ad6e679ee83c7544d3a01b4f82bead59165
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However, this derivation hinges on a crucial assumption: \emph{all} charge carriers have the same decay time. We've seen that not to be true in the case of the Landau quasiparticles -- their decay time depends on how far away their energy is to the Fermi energy.
Therefore, in a Fermi Liquid, there will be a population of decay timescales $\tau_k$ and conductivities $\sigma_0^{(k)}$. Without a magnetic field, these populations will just average out to result in an average timescale $\overline{\tau}$ and conductivity $\overline{\sigma_0^{(k)}}$. Thus, in the absence of a magnetic field, a distribution of decay timescales will not affect the conductivity. However, in the presence of a magnetic field, the picture changes. Let's assume we have $N$ different decay times. Let\footnote{The way I see these equations is as a weighted average, with a weight of
{\sigma_0} $\sigma_0 / {(1 + (\omega_\textrm{c} \tau_k
)^2)}} )^2)$}
\begin{equation}
\begin{split}
\gamma(1) &:= \frac{1}{N} \sum_{k=1}^{N} \frac{\sigma_0^{(k)}}{1 + (\omega_\textrm{c} \tau_k)^2} \\