If we have a tensor A = $\vec {u}\otimes\vec v$ then you can rewrite it to be A = $A_{ij}(\hat e_i\otimes\hat e_j)$, where $A_{ij}$ denotes the components of the tensor. If $(i,j = 1,2,3)$ then you can put it into a matrix $$ (\mathbf A)_{ij} = \mathcal A = \begin{bmatrix}A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}$$Does this mean that $A_{11}, A_{21}$ and $A_{31}$ are all acting in the $e_1$ direction? Is there any significance in direction between them?

I'm just overall a bit confused on what all this means when applied to something.


The simplest and most intuitive case is that of (1,1)-rank tensors, which are linear transformations between vectors. Such tensors can be faithfully represented by matrices.

For example, Ohm's law says that $$ \vec J = \sigma \vec E $$

where $\vec J$ is the current density and $\sigma$ is the conductivity tensor for the material. In simple cases, the conductivity of a material is a scalar, so the current flows in precisely the same direction as the applied electric field. However, there's no reason why that has to be the case; the weird crystalline structure of the atoms and molecules in a material may cause the current to flow at an oblique angle.

In such a case, we would have that

$$ \vec J=\pmatrix{J_1 \\J_2\\J_3} = \pmatrix{\sigma_{11}E_1+\sigma_{12}E_2+\sigma_{13}E_3\\\sigma_{21}E_1+\sigma_{22}E_2+\sigma_{23}E_3\\\sigma_{31}E_1+\sigma_{32}E_2+\sigma_{33}E_3}= \sigma \vec E$$

In colloquial terms, all three components of the electric field may contribute to the each component of the resulting current. For instance, $\sigma_{21}$ tells us how much of $E_1$ goes into $J_2$.

In the simple, isotropic case we learn in first-year electromagnetism, we have that $\sigma_{11}=\sigma_{22}=\sigma_{33}\equiv \sigma_0$, and $\sigma_{ij}=\sigma_0 \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta. In such a case, we can treat the conductivity as a scalar, but it's worth remembering that it's actually a tensor.

  • $\begingroup$ Thanks very much that's cleared it up nicely with a neat example too :) $\endgroup$ – CooperCape Aug 17 '17 at 18:45
  • $\begingroup$ @CooperCape No problem. I should add, however, that you should think of matrices only as a special type of tensor. You can, of course, always lay out the components of a tensor in a matrix-like grid, but you need to be very careful if the tensor is not rank (1,1). $\endgroup$ – J. Murray Aug 17 '17 at 18:50

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