Can optical conductivity be thought of as a frequency-dependent tensor form of the reciprocal of resistivity?

Question

Before reading Phys.org's Study finds billions of quantum entangled electrons in 'strange metal' I'd never heard the term optical conductivity.

When I think of an EM wave interacting with electrons in a surface and reflecting from it I think of the skin effect formula which depends on a resistivity. Electrons respond to the electric field from incident light, and the longer their mean free path (lower resistivity or $\rho$) the smaller the depth necessary to radiate the reflected wave and cancel the incident wave.

The linked Wikipedia article on optical conductivity points out that materials may be insulators against static electric fields and yet have substantial reflectivity at optical frequencies. It then goes on to say:

Only in the simplest case (coarse-graining, long-wavelength limit, cubic symmetry of the material), these properties can be considered as (complex-valued) scalar functions of the frequency only. Then, the electric current density $\mathbf {J}$ (a three-dimensional vector), the scalar optical conductivity $\sigma$ and the electric field vector $mathbf {E}$ are linked by the equation

$$\mathbf{J}(\omega) = \sigma(\omega) \mathbf{E}(\omega)$$

while the dielectric function $\varepsilon$ relates the electrical displacement to the electric field:

$$\mathbf{D}(\omega) = \varepsilon(\omega) \mathbf{E}(\omega)$$

so I can see that optical conductivity is at least somewhat analogous to the reciprocal of a resistivity.

Question: Can optical conductivity be thought of as a frequency-dependent tensor form of the reciprocal of resistivity?

While that might sound awkward, that's the way I'm going to approach the concept unless told otherwise.

The right hand side of your first equation is wrong. By the way resistivity is defined as the matrix inverse of the conductivity. — KF Gauss, Jan 17 '20 at 02:50

score 1 · Accepted Answer · edited Jan 17 '20 at 13:58

1

Can optical conductivity be thought of as a frequency-dependent tensor form of the reciprocal of resistivity?

Yes.

The conductivity and resistivity are matrix inverses by definition.

Recall the definitions of the two

$$J_a(\omega)=\sigma_{ab}(\omega) E_b(\omega)$$

And

$$E_a(\omega)=\rho_{ab}(\omega) J_b(\omega)$$

This latter is Ohm's law in terms of the electric field accounting for local variations, more general than $V=IR$ which is spatially integrated.

This means that the conductivity and resistivity matrices are inverses by very definition.

$$J_a(\omega)=\sigma_{ab}(\omega) E_b(\omega) \\=\sigma_{ab}(\omega) \rho_{bc}(\omega) J_c(\omega)\equiv \delta_{ac} J_c(\omega)= J_a(\omega)$$

edited Jan 17 '20 at 13:58

uhoh

6,308

answered Jan 17 '20 at 09:29

KF Gauss

7,823

Thank you! To the question itself, is the answer closer to a "yes" or a "no"? – uhoh Jan 17 '20 at 12:33
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Not only is it "yes", but that's its very definition. It's somewhat akin to asking whether an apple can be thought of as a fruit. – KF Gauss Jan 17 '20 at 13:11
Feel free to adjust the edit, thanks! – uhoh Jan 17 '20 at 13:59

Can optical conductivity be thought of as a frequency-dependent tensor form of the reciprocal of resistivity?

1 Answers1