Are the stress and strain tensor covariant or contravariant?

Question

My question is related to this question but I don't find the answer there to be completely satisfactory.

The displacement of an elastic medium is a contravariant quantity, which I think is pretty uncontroversial, so we'd write it as $u^i$ if we cared about the difference between upper and lower indices. When we want to form the infinitesimal strain tensor, we want to take a symmetrized gradient of the displacement, but we have a choice to make about whether to lower the indices of the displacement or raise the indices of the differential. I've always thought of it as $$\epsilon^{ij} = \frac{1}{2}\left(\nabla^iu^j + \nabla^ju^i\right)$$ with all upper indices. Likewise I've always thought of the elasticity tensor $\mathscr C$ as having all lower indices, which makes the stress also a covariant tensor: $$\sigma_{ij} = \mathscr{C}_{ijkl}\epsilon^{kl}.$$ This is nice because we can then form the strain energy density $E = \frac{1}{2}\epsilon^{ij}\sigma_{ij}$ without having to raise or lower any indices. It's nicer yet when you want to take the Legendre transformation use the Hellinger-Reissner form of the elasticity problem.

Is there a compelling reason to prefer the stress / strain tensors to be covariant / contravariant? Same question could be asked for viscous fluid flow with the strain replaced by the rate of strain.

My guess is that the answer is no, there is no reason to prefer one or the other. It's purely a matter of taste or aesthetics. The usual musical isomorphism lets you change between them however you like. The important questions (e.g., does the stress on this structure under this load exceed the fracture toughness of the material?) don't depend on the answer to this incredibly pedantic question.

On the other hand, I can think of cases where there really does appear to be a preference to use vectors for certain quantities and covectors for others. For example, when you start doing classical mechanics on manifolds, you tend to think of velocities as vectors but momenta as covectors due to the fact that the Hamiltonian arises through a Legendre transformation of the Lagrangian. There really is a preference in this case for one type of index or the other because the argument to the Legendre transformation of a convex functional lives in the dual space. I tend to think of "forces" in the most general sense as living in the dual space (i.e. lower indices) and "displacements" in the primal space (i.e. upper indices). I don't know whether this is a general preference in physics or just me. The problem becomes arguably more interesting for field theories. The solution you're looking for lives in some Sobolev space, and the dual of a Sobolev space looks very different from the primal space unlike in finite dimensions.

Answer 1

The stress and strain tensors are neither covariant nor contra-variant. They are independent of the basis vectors used to express their components. Their components are considered covariant or contravariant (or a mixture of the two) depending on how they are expressed in terms of covariant and contravariant basis vectors.

Answer 2

I'm not sure if this is what you had in mind, but coordinate-like degrees of freedom are usually thought of as transforming contravariantly for consistency with function composition: i.e. if $x' = \phi(x)$ is a new set of coordinates for some coordinate patch $U_i$, then functions $f(x)$ expressed in the new coordinates are $f'(x') = f(\phi^{-1}(x'))$. Thus, references to $x$ are replaced in the new frame by references to $\phi^{-1}(x')$ in the new frame. The reason functions are defined to transform this way (rather than "actively", where $f'(x') = f(\phi(x'))$) is that the contravariant transformation rule dovetails nicely with function composition, and therefore doesn't need to be imposed "manually" (as it is for active transformations): that is, if $x'' = \psi\circ\phi(x)$, then $f''(x'') = f((\psi\circ\phi)^{-1}(x'')) = f(\phi^{-1}\circ\psi^{-1}(x'')) = f'(\psi^{-1}(x''))$, and this isn't true if instead $f'(x')$ were defined as $f(\phi(x))$ (because $f(\psi(\phi(x)))\neq f(\phi(\psi(x)))$). The (conventional) contravariant transformation law emerges naturally from familiar properties of functions.

Typically the energy densities (or Lagrangian densities) in which a "covarying" (either contra- or co- variant) strain tensor would appear transform as "weighted scalars" under diffeomorphisms: they would be multiplied by an appropriate inverse power of the determinant of the metric (i.e. 1/2) so that they could be integrated "covaryingly" to obtain scalar quantities, but would otherwise transform like ordinary scalar fields. Hence, in some ways it might be clearer to think of the strain (and "inverse stress") as "coordinates" of these density fields that transform contravariantly, but obviously the math should be the same whether you use contravariant or covariant forms (i.e. you should know how to go from one to the other.)

To some extent the "covariant" and "contravariant" terminology is unfortunate, because the word covariant is often used in two senses: firstly to convey that a quantity is frame-dependent (i.e. it 'co-varies' with the reference frame) and secondly in the more specific sense (i.e. that the tensor or component is multiplied by a positive power of the Jacobian rather than a negative power.) So from a certain, possibly impressionistic, perspective, a contravariant vector is always "covariant", but obviously not a "covariant vector."