# Accurate derivation of electron-phonon scattering contribution to metal resistivity

My lecture derived the expression for this contribtuin using the collision integral approach but I missed lot of details. He considers the lowest order correction to distribution function $$f=f_0+\delta f_p$$, where $$f_0$$ is the equilibrium function and "guesses" the expression for $$\delta f=(-\partial f_0/\partial\epsilon)T\chi$$. Then he writes down the expression of collision integrals for phonon-electron interaction. After several approximations, he represents collision integral as $$-\hat{\Omega}\chi_p$$ where $$\hat{\Omega}$$ plays the role of operator in "Hilbert space". After that, he expands $$\chi_p$$ by a basis $$\lbrace\phi_l(p)\rbrace$$ and then consider only the one function $$\phi(p)$$ in a basis.

Of course, I lost lot of details but the key idea is to reduce the problem to "eigenvalue" problem and solve it. For me, it seems very unnatural.

Hence, I will be very grateful for how to obtain the expression for $$\rho$$ wich has the correct behavior $$\sim T$$ for high temperatures and $$\sim T^5$$ for low temperatures.

• There are other ways to solve the Boltzmann equation (rewrite as Fokker-Planck, for example), but this is the standard (and most efficient way) to do it, and I'm not aware of a text book that solves the electron-phonon without linearizing the collision operator and solving the eigenvalue problem. Of course, instead of eigenvalues you can use variation functions $\delta f_p$, but this is essentially the same method. – Thomas Apr 21 '19 at 0:13
• Thx You, @Thomas ! Can You please help with some references to the derivation of this contribution using the linearization of collision integral? – Artem Alexandrov Apr 21 '19 at 10:18

I can maybe give a quick summary of the logic behind these calculations. The Boltzmann equation is, schematically, $$lhs=rhs$$ where $$lhs=\partial_0f_p+\ldots$$ are the streaming terms, and $$rhs=C[f_p]$$ is the collision term.
1) We are seeking a solution of the form $$f_p=f^0_p+\delta f_p$$. What is the small parameter in this expansion? There are two length scales in the Boltzmann equation. On the $$lhs$$ we have $$\vec{E}=\vec\nabla\Phi\sim\Phi/L$$. On the $$rhs$$ we have a collision length $$l$$ (roughly, $$C[f_p]\sim \delta f_p/\tau\sim v_p\delta f_p/l$$). The expansion parameter is the ration of these two numbers $$\delta f_p/f_p\sim Kn \sim l/L$$ where $$Kn$$ is the Knudsen number.
2) On the $$lhs$$ we have the Lorentz force acting on particles (this is a force term $$\sim E\cdot\nabla_p f_p$$). We expand and get something proportional to $$v_p\cdot E (\partial f^0_p)/(\partial \epsilon_p)$$
3) On the $$rhs$$ we expand the collision integral $$C[f^0_p+\delta f_p] \sim C_L[\delta f_p]$$ where we have used that $$C[f^0_p]$$ vanishes (equilibrium), and $$C_L$$ is a linearized collision operator.
4) The amusing observation about $$C_L$$ is that it is an anti-hermitean operator with respect to an inner product of the form $$\langle\chi|\phi\rangle \sim \int f^0_p\chi_p\phi_p$$ This makes a certain amount of sense. Hermitean operators (in QM) give time evolution $$\sim \exp(-iE_nt)$$. The collision operator accounts for equilibration $$\sim \exp(-t/\tau_n)$$.
5) To study transport phenomena we have to look for the largest $$t_n$$ (just like in QM we look for the ground state, the lowest $$E_n$$), because this mode will dominate the relaxation process at large $$t$$. The corresponding eigenvector $$\delta f_p\sim f^0_p\chi_p$$ (factoring out $$f^0$$ or $$(\partial f^0_p/\partial \epsilon_p)$$ is a convenient normalization) gives $$rhs = f^0_p\frac{\chi_p}{\tau}$$ and now I can just solve for $$\chi_p$$ (and then go ahead and compute the electric current). Clearly, $$\chi_p\sim \tau/L\sim l/L$$ as advertised.