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I've seen this kind of formula a number of times, in the context of elastic deformations.

$$ -\nabla \sigma = f $$

whete $\sigma$ is "the stress tensor" and $f$ is force. I never understood it even though I've coded finite element simulations several times. To me, it all comes down to defining an elastic energy in terms of positions $E(x)$ and then thinking about the body trying to minimize it and that's it, I can use the computer to take care of all the other details. However, I'm trying to understand the terms used in classical mechanics textbooks. First of all, what is $\sigma$ ? It seems to me, from the descriptions I've read that it's just a matrix. If that's true, then why call it a tensor? And also, what's the best way to interpret $\nabla$? Is it a function acting on a matrix? I guess it's some kind of derivative, but it's not clear to me what it computes exactly. Sometimes it appears as $\nabla.\sigma$ (with a dot in between), is it the same thing as $\nabla \sigma$?

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This equation is just the Cauchy momentum equation under conditions of zero material derivative. It's the condition that arises when all motion has ceased and the body has reached equilibrium.

The nabla is an operator and it should be with a dot in this case, $\nabla\cdot\sigma$, since it's the divergence operator (which decreases the tensor order by 1), whereas without a dot it denotes a gradient (which increases the tensor order by 1). See here for more about tensor derivatives.

Regarding the nature of $\sigma$ it is both a matrix and a tensor. A matrix is just a rectangular array of numbers; a tensor exhibits certain transformation properties. So if a matrix exhibits those same transformation properties then it is also a tensor (of rank 2). The difference between a matrix and a tensor has been discussed here many times before, e.g. see here.

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  • $\begingroup$ For a given mesh, elastic material and nodal positions $x$, I know how to compute the elastic energy $E(x)$. With this information, how do I compute $\sigma$ ? Is it a function of $x$ as well? $\endgroup$
    – yewang
    Jun 18, 2018 at 9:06
  • $\begingroup$ In the context of finite elements, I know how to compute the deformation gradient $F$ for an element. Is there any relation between $F$ and $\sigma$? $\endgroup$
    – yewang
    Jun 18, 2018 at 9:12
  • $\begingroup$ @yewang You can compute the strain tensor from $F$ as follows: $\epsilon=F-I$, and then the stress tensor is related to elastic energy by: $\sigma_{ij}=\partial E/\partial\epsilon_{ij}$. (You should double check that online - I recall a $2$ knocking around somewhere). $\endgroup$
    – lemon
    Jun 18, 2018 at 9:19
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The equation you have written (with the dot properly present) is just a differential force balance on a tiny volume of material within the body. The left hand side of the equation (divergence of the stress tensor) is just the sum of the external forces (imposed by adjacent parcels of material) on the differential volume of material. The right hand side of the equation is just the body force per unit volume (typically, gravity). The stress tensor describes the state of stress at each location within a body, and is thus a function of position within the body. To get an understanding of the stress tensor, why it can be represented by a matrix, the gradient vector operator (nabla), and how the divergence of the stress tensor leads to the sum of the forces on the tiny volume, you need to take a course in strength of materials or theory of elasticity.

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