In the context of anisotropy, I have often read that the use of a rank 2 tensor is "a model". But what is the idea behind this choice? Can anyone describe in what sense the use of tensor in this context is a "model"?


3 Answers 3


The tensor itself is not the model, but the the tensor is used to model (one could also say describe or quantify) the anisotropy.

One example is an anisotropic electric conductor. The conductivity $\sigma$ describes which current occurs in response to an electric field $\vec j = \sigma \vec E$. In isotropic materials (e.g. glass, microcrystalline metals when averaged), this quantity is scalar, that means that the current points in the same direction as the electric field and how much current density is generated by the electric field does not depend on the direction.

In general, however, the conductivity is a tensor of rank 2. For example in a graphite monocrystal (which consists of loosely coupled layers) the conductivity in the layers is much higher, than the conductivity perpendicular to the layers. So in a coordinate system where the layers are stacked along the $z$-direction we will have a conductivity tensor of the form (assuming conductivity is approximately isotropic within the layers): \begin{align*} \sigma &= \begin{pmatrix} \sigma_l & 0 & 0 \\ 0 & \sigma_l & 0 \\ 0 & 0 & \sigma_p \end{pmatrix} \end{align*} Here $\sigma_l \gg \sigma_p$. So we can compute the current that occurs if we apply an arbitrary electric field with this tensor, and the current will be larger along the layers than perpendicular to the layers. The nice thing about this relation is, that $\vec j = \sigma \vec E$ holds in any coordinate system, we just have to transform the components of $\sigma$ accordingly. (We can even determine the orientation of the layers from a macroscopic measurement of the conductivity tensor by computing the main axes of the tensor).

Note, that there are anisotropies that must be described by tensors of higher rank. For example, the mechanical stress in a material is a rank 2 tensor and the elasticity tensor (relating the stress to the strain) is a rank 4 tensor.

That tensors are the objects that occur here has two reasons:

  1. The equations must be invariant under the choice of the coordinate systems and tensors are the natural objects when we seek equations invariant under rotations (or more general under arbitrary coordinate transformations).

  2. We often work in linearized theories and the equation $A^{(n)} = B^{(n+m)} C^{(m)}$ (where the juxtaposition denotes contraction and the upper indices denote the tensor rank) is the most general linear relation between the tensors $A$ and $C$.


I would expand upon Sebastian's nice answer to point out that any orientation-sensitive quantity $f(\hat v)$ may be expanded by spherical harmonics and the symmetric rank-2 tensor can be often used to represent the first nonzero term.

To understand this, start by noting that all of these spherical harmonics come with a polynomial structure factor. You may recognized them from atomic orbital theory (from wikipedia):

enter image description here

In this case we are trying to represent the wavefunction $\psi(\vec r)$.

Notice the third row there are these polynomials that label the different orbitals. For a fixed angular momentum (l), a state with that momentum may be labeled by a degree l polynomial with complex coefficients, where we consider the "trace polynomial" $x^l + y^l + z^l$ to be equivalent to zero (this just contributes to the isotropic ($l = 0$) response).

In particular, for $l = 2$ (the quadropole moment), we're talking about quadratic polynomials, and we can represent these as a traceless symmetric matrix $A$ by

$$v^T A v,$$

where $v$ is the vector $v^T = (x,y,z)$. This traceless symmetric matrix $A$, once we let it depend on spatial position, becomes our rank-2 symmetric tensor. For many applications, this is the first multipole moment that is nonzero (because generic potentials are quadratic near equilibrium). In those cases, to first order, this matrix is the object of interest.

In general, however, the spherical harmonics are labelled by symmetric tensors of all rank, and this decomposition has to do with the representation theory of $SO(3)$. A more advanced model may have to include higher multipole moments to capture the orientation-dependence.

I think Sebastian's example of the conductivity matrix is actually a bit confusing, since there we're talking about a vector-valued quantity that also depends on orientation, namely $\vec j(\vec E)$. In this case, the conductivity matrix really comes from the $l = 1$ moment (it's rank (1,1), not rank (2,0)). The symmetry of this matrix is not guaranteed by representation theory but by Onsager reciprocity. Higher moments would come from nonlinear corrections to Ohm's law but this symmetry relation would still hold near equilibrium!

Similarly, the elastic tensor is a rank (2,2) tensor, symmetric in each factor ($l=2$) and actually also symmetric between the two, but the reason for this last symmetry is mysterious to me...


In the context of anisotropy, I often read that the use of a tank 2 tensor is "a model". But what is the idea behind this choice? Can anyone describe in which sense the use of tensor in this context is a "model"?

Without context, it's difficult to guess at what exactly someone might mean by referring to a second-order tensor as "a model", as it could be a reference to a bunch of different observations.

Personally, I think that I've complained most about it for the typical presumption of locality that goes into it. For a simple early example, there's the Cauchy stress tensor from fluid mechanics:
I can completely understand why folks would think that this is pretty general since it seems to avoid assumptions about dynamics, at any given point.

However, the tensor reflects the implicit assumption that mechanical interactions are local, i.e. that there're fully captured at any given point. And perhaps this seems like a defensible enough approximation for a crystalline solids, but even in that best-case scenario, it's merely an approximation.

Then on the opposite end of the spectrum, e.g. low-pressure gases, obviously there're significant free paths, such that mechanical interactions of a fluid are poorly described by assuming the local model implied by the tensor.

This isn't necessarily what a speaker means in any given context, though it's one possible type of modeling assumption represented by a tensor that someone might be referring to.

Found a Wikipedia article that discusses a relaxation:

In continuum mechanics, the finite strain theory—also called large strain theory, or large deformation theory—deals with deformations in which strains and/or rotations are large enough to invalidate assumptions inherent in infinitesimal strain theory.

"Finite strain theory", Wikipedia

Though in general, if you're working with second-order tensors, you're probably working with a model that's using those tensors to capture infinitely localized interactions, which is an extremely-convenient-but-obviously-flawed model in any real-world physics situation.

  • $\begingroup$ Modelling using tensors does not imply locality: You can have a tensorial integral kernel that is convoluted with the input quantity. This convolution decouples to a point-wise product in Fourier space and you arrive at $\vec j(\omega, \vec k) = \sigma(\omega, \vec k) \vec E(\omega, \vec k)$, so interpreted right, $j = \sigma E$ does cover non-local and time dependent linear response! $\endgroup$ Commented May 8, 2018 at 20:54

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