deletions | additions       diff --git a/scattering.tex b/scattering.tex  index 2f740f7..feec934 100644  --- a/scattering.tex  +++ b/scattering.tex 
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Therefore, since $0 < \omega < \epsilon$ and $0 < \phi < \omega$, there are $O(\epsilon^2)$ possible decay processes (see Figure~\ref{fig:quasiparticle-lifetime} for an illustration). Assuming each decay process has a similar matrix element, the decay cross section will therefore be proportional to the number of decay processes, therefore proportional to $\epsilon^2$. Thus, since lifetime is inversely proportional to decay cross section, the lifetime $\tau$ of a Fermi Liquid quasiparticle is:  \begin{equation}  \tau \sim \frac{1}{\epsilon} \frac{1}{\epsilon^2}  \end{equation}  At a temperature $T \ll T_F$, the quasiparticles will be excited at an average energy of $\epsilon k_B T$ above the Fermi surface. Therefore, their lifetime will be proportional to $1 / T^2$, meaning that the conductivity of a Fermi Liquid scales as: 
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\sigma_{\mathrm{FL}} \sim \frac{1}{T^2}  \end{equation}  Before concluding that conductivity scales as $1 / T^2$, there's one element left to show. A non-infinite conductivity requires both a decay time and an energy loss mechanism. In the current form, the energy the quasiparticle looses goes back to the Fermi sea, effectively accelerating it. However, in order  Problem: energy conservation of qp-qp scattering. Solution: umklapp processes, whereby energy is transferred to the lattice, therefore you have dissipation via qp-qp scattering alone.