1
$\begingroup$

I have a lecture on solid mechanics that introduced the stress tensor as a bunch of indices squeezed into a matrix (my point isn't that the indices aren't justified).

What I'm trying to understand is what kind of geometrical object the stress tensor is.

What I've understood so far is that no one really cares in engineering whether the stress tensor is contravariant/covariant/mixed since we almost always work with an orthonormal basis.

What I also did understand is that the stress tensor may be used to find the force acting on a plane by applying it to a vector normal to that plane. That would indicate the stress tensor would be a linear map, so it would be a (1,1) tensor.

I understand that in a orthonormal basis, you can apply the metric tensor and get a (2,0) tensor or a (0,2) with exactly the same value for the indices. So it would be pretty much a convention to chose which one of the object you to use. But what would be the geometrical meaning of the stress tensor viewed as a bilinear form, or as a (0,2) tensor?

Lastly, are there application where you would want to use the stress tensor in curvilinear coordinate system, and therefore have to know what transformation rule to use?

$\endgroup$
3
  • $\begingroup$ To figure out which tensor it is, write down in which equations it appears. The co-/contra-variant indices can then be defined based on requirement that equation as a whole is independent of the coordinates and basis. $\endgroup$
    – Cryo
    Jan 18, 2021 at 19:37
  • $\begingroup$ I was more thinking about a visual/geometrical intuition. The stress tensor indeed seems to be (1,1) in most case, but apparently, not always, and I was wondering why and what it would represent geometrically. $\endgroup$ Jan 18, 2021 at 19:42
  • $\begingroup$ I was partially asking for myself since I don't work with stress tensors. I would probably proceed like this. From (Canonical coordinates)[en.wikipedia.org/wiki/Canonical_coordinates] one should be able to work out that momentum is co-variant. Which the allows you to work out that force would also be co-variant etc. $\endgroup$
    – Cryo
    Jan 18, 2021 at 19:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.