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Magnetoresistance and Fermi surfaces 
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#1
Nov813, 01:53 PM

P: 928

There seems to be relation between Fermi surfaces and magnetoresistance, but I guess because I don't have a clear picture of fermi surfaces,I have problem understanding this relationship.
Also I have heard about open and close fermi surfaces and saturation of magnetoresistance which I can't find enough explanation about. I will appreciate any clarification. Thanks 


#2
Nov1213, 11:32 AM

P: 928

I've been searching different books and sites on the subject but non of them give a satisfactory proof of the saturation of magnetoresistance in the directions that the fermi surface are closed.Can someone explain?
Thanks 


#3
Nov1213, 11:59 AM

Sci Advisor
P: 3,628

Try to explain me how you think this all works and you get my comments for free!



#4
Nov1213, 08:26 PM

P: 928

Magnetoresistance and Fermi surfaces
I can only suggest you to take a look at chapters 11 and 12 of:http://www2.physics.ox.ac.uk/sites/d...ompleteSet.pdf It helps you into the subject. There is also chapter 12 of solid state physics by Ashcroft and Mermin. You can find explanations about the subject in other books and websites too,but non that I found had a full and satisfactory proof of magnetoresistance saturation in the directions that the Fermi surface is closed. Also,these texts tell you what I know about it and how I think. 


#5
Nov1313, 02:00 AM

Sci Advisor
P: 3,628

Have a look at:
arXiv:0907.2021v1 [condmat.meshall] They discuss the high and low field limit. 


#6
Nov1313, 03:45 AM

P: 928

In the high field limit,it gives the current density by the following formula: [itex] \vec{J}=e \frac{\vec{E}\times\vec{B}}{B^2}n [/itex] To make things simple,let's have [itex] \vec{E}\cdot\vec{B}=0 [/itex],So we will have [itex] J=\frac{ne}{B}E \Rightarrow \sigma=\frac{ne}{B} \Rightarrow \rho=\frac{B}{ne} [/itex] which is proportional to B and so doesn't approach a constant as [itex] B\rightarrow \infty [/itex] I have the same problem with the argument presented in Ashcroft and mermin. But in http://www2.physics.ox.ac.uk/sites/d...ompleteSet.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 [itex] \rho_{xx}=\rho_{yy}=\rho_0 [/itex] and [itex] \rho_{yx}=\rho_{xy}=\frac{B}{ne} [/itex]. Then in section 11.3.3,[itex] \rho_{xx}[/itex] is said to tend to a constant as [itex] B \rightarrow \infty [/itex]!!! Also,[itex] \sigma_{xx}=\sigma_{yy}=\frac{\sigma_0}{1+\omega_c^2 \tau^2} (\omega_c=\frac{eB}{m})[/itex] 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. 


#7
Nov1313, 06:19 AM

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P: 3,628

Well, I suppose, the question of million is what is the form of the cyclotron frequency in the case of an open fermi surface?
For an open surface, the electron will not perform a cyclic motion at all but will be driven along a, maybe wiggly line. Hence the frequency must be 0. I suppose this is formally due to the effective mass  when averaged over the orbit  being infinite. 


#8
Nov1313, 09:14 AM

P: 928

And in the first paragraph of section 11.3.3,it is said that [itex] \sigma_{yy} [/itex] and [itex] \sigma_{xx} [/itex] both vary as [itex]B^{2}[/itex].This is OK but if we use previous formulas,then we're not talking only about closed orbits.So the derivation of such a behavior should use something else which in that part of the text,seems to be the zero average of the velocity in the plane perpendicular to B and equations 11.27 and 11.28 in the high magnetic field limit.But that just suggests that they should vary as [itex] B^{n}(n>0) [/itex] and there should be another way for determining n. To be honest,I lost my hope on this paper! 


#9
Nov1413, 03:29 AM

Sci Advisor
P: 3,628

I am not sure either. However the equality of sigma_xx and sigma_yy is derived in 11.1 under the explicit assumption of an isotropic and energy independent effective mass.
I think it is this assumption which breaks down when you have an open FS. 


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