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


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.