%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\sect{Einstein relation for the conductivity of electrons}

Given a metal design. We mark with ${\varphi\left(x\right)}$ the
electrical potential in the sample and with ${N\left(x\right)}$ the
spacial density of the electrons in the design. According to the
kinetic theory s
%
%
\[\vec{J}\left(x\right)=-\sigma\nabla\varrho-eD\nabla N\]
%
%
${\sigma}$ is the conductivity and ${D}$ is the diffusion
coefficient. In an equilibrium state  ${\vec{J}\left(x\right)\equiv
0}$, especially in a state of equilibrium that we get in the
presence of outer field  ${\varphi \left(x\right)\neq const}$ and
therefore has to : ${\frac{\sigma}{eD}=\frac{\nabla N}{\nabla
\varphi}}$.

Use the principles of the statistical mechanics to show that from
here derives
%
%
\[\frac{\sigma}{D}=-e^{2}\int dE \ g(E)
f'\left(E-\mu\right)\]
%
%
low temperatures ${\frac{\sigma}{D}=e^{2}g\left(E_{r}\right)}$

High temperatures ${\frac{\sigma}{D}=N\frac{e^{2}}{T}}$

${g\left(E\right)}$ is the uniparticles states density per volume
unit. Hint - notice that
%
%
\[N\left(x\right)=\int g\left(E -eV(x)\right) dE f\left(E-\mu\right)\]






