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How to show that the electrical conductivity tensor is symmetric? (or it's not always symmetric?)

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It is not necessarily symmetric. The conductivity tensor $\boldsymbol \sigma$ is given by:

$$\mathbf J = \boldsymbol \sigma \mathbf E$$

And its inverse $\boldsymbol \sigma^{-1}=\boldsymbol \rho$ is the resistivity tensor. If you use matrix notation you have:

$$\begin{pmatrix} J_1\\ J_2\\J_3\end{pmatrix}=\begin{pmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} & \sigma_{23}\\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{pmatrix}\begin{pmatrix} E_1\\ E_2\\E_3\end{pmatrix}$$

As you can see here, $\sigma_{12}$ tells you how a component of the electric field in the "$2$" direction generates a current in the "$1$" direction, whereas $\sigma_{21}$ tells you how an electric field in the "$1$" direction generates a current in the "$2$" direction (you could think of the $x$ and $y$ directions, in Cartesian coordinates). These two need not be the same, and in a solid they could depend on, for example, the type of anisotropy in the material.

Wikipedia gives us a nice example: "In the Hall effect, due to rotational invariance about the $z$-axis $\rho_{xx}=\rho_{yy}$ and $\rho_{yx}=-\rho_{xy}$". So $\boldsymbol \rho$ is not symmetric, and the inverse of a non-symmetric matrix is also not symmetric, so $\boldsymbol \sigma$ is not symmetric.

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  • $\begingroup$ That's right.but I heard something about onsager's principle and I think maybe we can say good things under some conditions(because in general you said it's not necessarily true), about symmetric or antisymmetric of this matrix. $\endgroup$ – a.p Jun 6 at 21:47
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    $\begingroup$ Well, in an isotropic material it should be symmetric, in fact all the off diagonal components should be zero. The reasoning is as follows: since the material is isotropic, the directions are indistinguishable, suppose, for concreteness, that a field $\mathbf E$ is applied in the coordinate direction $+\mathbf\hat i_1$. If this field generates a current in, again for concreteness, $+\mathbf\hat i_2$ this is a contradiction because a current is being generated in one preferential direction, but we had assumed the material was isotropic. Assuming isotropy for a polycrystal solid is usually safe. $\endgroup$ – S V Jun 7 at 1:36
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    $\begingroup$ I an not convinced by the example of the Hall effect. Conductivity is a material property. In the Hall effect the current is changed by an external magnetic field. This should not be described as a material property. $\endgroup$ – my2cts Jun 7 at 6:18
  • $\begingroup$ @my2cts You can actually get Hall conductance even in the absence of magnetic field -- it's called the "anomalous Hall effect". $\endgroup$ – Dominic Else Jun 7 at 10:34
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    $\begingroup$ @my2cts Material properties need not be independent of an externally applied field. For example, in fluid mechanics non-newtonian fluids have a viscosity that depends on the strain-rate tensor $D_{ij}=\partial_{(i}v_{j)}$. There is an external field applied, in this case stress, that causes the property to change. $\endgroup$ – S V Jun 7 at 14:47

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