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In continuum mechanics, the stress vector (see Cauchy stress tensor) $T=\sigma\cdot n$ is the surface density of a force. Forces are covectors, since they map a displacement vector to a scalar energy. Then, why is $T$ not a covector too?

Edit: I talking about the stress vector, defined for a given unit normal vector $n$ by $\sigma\cdot n$, not to be confused with the stress (second-order) tensor $\sigma$.

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    $\begingroup$ Is there some reason you're calling it the stress vector? I feel the issue of vector vs. covector is a red herring, given the more important issue of vector vs. rank-2 tensor. $\endgroup$
    – user10851
    Commented Apr 6, 2015 at 23:55
  • $\begingroup$ @ChrisWhite Well, that's the way everybody calls it (google fight). Red herring for what, for who? This determines if Cauchy tensor is a $(1,1)$ or $(0,2)$ tensor. $\endgroup$
    – anderstood
    Commented Apr 7, 2015 at 0:00
  • $\begingroup$ Are we not working in the presence of a metric? If we are, there's a canonical isomorphism between such tensors, and the distinction between them is utterly without meaning. $\endgroup$
    – Muphrid
    Commented Apr 7, 2015 at 1:39
  • $\begingroup$ @Muphrid There is a canonical isomorphism between the vector space $E$ and its bidual $E^{**}$ but I don't see why this implies that no distinction between $(1,1)$ and $(0,2)$ tensors should be made. Are you talking about a Euclidian space? $\endgroup$
    – anderstood
    Commented Apr 7, 2015 at 1:59
  • $\begingroup$ I'm not. See user1260696's answer. The musical isomorphism makes the distinction between vectors and covectors irrelevant. $\endgroup$
    – Muphrid
    Commented Apr 7, 2015 at 2:05

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When people study continuum mechanics they usually do so at first in $\mathbb{R}^3$ where we have usually implied the usual metric tensor $(g_{ij}) = \operatorname{diag}(1,1,1)$ and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert between vector fields and one-forms by means of raising and lowering indices.

So if $M$ is your space and $(x,U)$ a coordinate system, if $X$ is a vector field, which on $U$ can be written in coordinates as

$$X = X^i\dfrac{\partial}{\partial x^i},$$

then $g$ allows you to build the one-form equivalent to it by setting $\omega = g(X,\cdot)$, that is, in $U$ we can write

$$\omega(Y) = g(X,Y) \Longrightarrow \omega = g_{ij}X^idx^j,$$

where $g_{ij}$ are the components of $g$ on the coordinate system $(x,U)$, that is, functions that allows us to write $g = g_{ij}dx^i\otimes dx^j$.

Now the stress tensor you speak off is usually defined as a linear map that takes vectors into vectors: it is capable of taking one normal and giving back one force. Now linear maps on a vector space $V$ may be identified with the tensor product $V\otimes V^{\ast}$ and so linear maps and tensors of type $(1,1)$ are the same.

In that setting it is best to think about the stress tensor as this $(1,1)$ tensor $\sigma$ which on $(x,U)$ can be written

$$\sigma = \sigma^{i}_{j} \dfrac{\partial}{\partial x^i}\otimes dx^j.$$

Now in the same way such a tensor can map vectors to vectors it can map covectors to covectors. In that way, if you consider force as a covector, $\sigma$ can map it to. Now because you have a metric tensor, those operations can all be "matched" using the musical isomorphism. More importantly, when $g$ is the usual metric tensor of $\mathbb{R}^3$ you see no difference at all.

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  • $\begingroup$ Why is it important in analytical mechanics to make the distinction between vectors and covectors then? Forces (elements of the cotangent bundle of the configuration manifold) and velocities (elements of the tangent bundles) don't play a symmetric role. $\endgroup$
    – anderstood
    Commented Apr 7, 2015 at 2:48
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    $\begingroup$ Because a metric is required to identify vectors and covectors. When you deal with a continuum it is usually in some subset of $\mathbb{R}^3$ or some other space $M$ that naturally has a geometry encoded in a metric tensor, so these identifications are already there. If there's no metric, then there's no way to argue in this way. $\endgroup$
    – Gold
    Commented Apr 7, 2015 at 2:54

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