DrDu said:
Have a look at:
arXiv:0907.2021v1 [cond-mat.mes-hall]
They discuss the high and low field limit.
Thanks for introducing the paper,it seems so useful...But it still doesn't give me the thing I want.
In the high field limit,it gives the current density by the following formula:
<br />
\vec{J}=-e \frac{\vec{E}\times\vec{B}}{B^2}n<br />
To make things simple,let's have \vec{E}\cdot\vec{B}=0,So we will have J=-\frac{ne}{B}E \Rightarrow |\sigma|=\frac{ne}{B} \Rightarrow |\rho|=\frac{B}{ne} which is proportional to B and so doesn't approach a constant as B\rightarrow \infty
I have the same problem with the argument presented in Ashcroft and mermin.
But in
http://www2.physics.ox.ac.uk/sites/default/files/BandMT1112_CompleteSet.pdf ,there is another problem.
First,in section 11.3.1,the conductivity tensor is derived and then inverted to give the resistivity tensor.It turns out that \rho_{xx}=\rho_{yy}=\rho_0 and \rho_{yx}=-\rho_{xy}=-\frac{B}{ne}.
Then in section 11.3.3,\rho_{xx} is said to tend to a constant as B \rightarrow \infty!
Also,\sigma_{xx}=\sigma_{yy}=\frac{\sigma_0}{1+\omega_c^2 \tau^2} (\omega_c=\frac{eB}{m}) is derived in section 11.3.1 regardless of the shape of the fermi surface.But in section 11.3.3 it is argued that it behaves differently for closed and open fermi surfaces.
If only the inconsistencies are cured and the explanations are clarified a little,then this last paper may help.
