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Resistance of a two-dimensional sample

+ 6 like - 0 dislike
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In this review of the QHE, Steve Girvin makes the following statement (bottom of pg. 6, beginning of Sec. 1.1.1):

As one learns in the study of scaling in the localization transition, resistivity (which is what theorists calculate) and resistance (which is what experimental- ists measure) for classical systems (in the shape of a [d-dimensional] hypercube) of size L are related by [12,13]

$$ R = \rho L^{(2-d)} $$

where $R$ is the resistance of the hypercube and $\rho$ its resistivity. For $d=2$ it is clear that the resistance and the resistivity of the sample are the same, and as Girvin goes on to say, this fact is crucial for the universality of the quantization of the resistivity plateaus in the QHE.

My question is: Is there a simple physical explanation of the above relationship? Girvin cites two papers. Both are RMP: "Disordered Electronic Systems" by Lee and Ramakrishnan and "Continuous Quantum Phase Transitions" by Sondhi et al. I am hoping that someone can spare me the effort of going through these two papers by providing an answer.

This post imported from StackExchange Physics at 2014-04-01 16:33 (UCT), posted by SE-user user346
asked Jan 26, 2011 in Theoretical Physics by anonymous [ no revision ]

1 Answer

+ 6 like - 0 dislike

It's all about scale/units. Resistivity is, by definition, the resistance you would measure if you had a "wire" of unit length and unit cross-sectional area: $$R = \rho * \textrm{length} / \textrm{cross-sectional area}.$$ For your hypercube of lengths $L$, this would correspond to a length of $L$ and cross-sectional area of $L^{d-1}$: $$R = \rho * L / L^{d-1} = \rho * L^{-(d-2)}.$$

Notice that the key is that resistivity has units which change! In 2D it happens to be exactly Ohms, but that actually hides a bit of "pure geometry". It's really something like "Ohms per square" --- but the same rectangle has a different resistance depending on which way it is...

This post imported from StackExchange Physics at 2014-04-01 16:33 (UCT), posted by SE-user genneth
answered Jan 26, 2011 by genneth (560 points) [ no revision ]
That's embarrassingly simple. Thanks @Genneth.

This post imported from StackExchange Physics at 2014-04-01 16:33 (UCT), posted by SE-user user346
Exactly, Genneth, +1.

This post imported from StackExchange Physics at 2014-04-01 16:33 (UCT), posted by SE-user Luboš Motl

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