# Divergence of Cauchy Stress Tensor [closed]

On the wikipedia page for the Cauchy Momementum Equation, it's stated that the equation can be written as

$$\rho \frac{D\,\textbf{v}}{D\,t} = \nabla \cdot \sigma + \textbf{f}$$

Where $\sigma$ is the Cauchy Stress Tensor. I have read countless articles on the stress tensor, but I can't see how the divergence of the stress tensor $\sigma$ is a force per unit volume on a differential unit of fluid.

[EDIT]

Here's one reason why this confuses me. Suppose we are in $\mathbb{R}^3$, and the directions are given by $x$,$y$, and $z$. Taking the Cauchy Momentum Equation in the $x$ direction, we get.

$$\rho \frac{D\,\textbf{v}_x}{D\,t} = \sum_{j\in{x,y,z}}\frac{\partial \sigma_{xj}}{\partial j} + \textbf{f}_x$$

If you look at the sum for the sum of the partial derivatives of the stress tensor terms, we can expand it as:

$$\sum_{j\in{x,y,z}}\frac{\partial \sigma_{xj}}{\partial j} = \frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z}$$

I have yet to come across a good intuitive explanation as to why this is the net force (caused by the terms in the stress tensor) as opposed to

$$\frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial \sigma_{xy}}{\partial x} + \frac{\partial \sigma_{xz}}{\partial x}$$

of course, without just saying "That's just what it is." Even more than that, although I see how the units work out, I don't even understand intuitively how the derivatives of those stress tensor terms even result in a differential force per unit volume.

I just really want some intuition on this whole thing. I'm trying to learn all of this just through wikipedia and online resources.

• Without getting too deep, what are the units of the stress tensor? The units of the divergence? What kind of object results from the divergence of such a tensor? Nov 5 '14 at 1:51
• The units of the stress tensor is in force/area, and the divergence operation would make it in differential force per differential unit of volume, but that still doesn't give me an intuitive feel for why it's so. See the edit. Nov 5 '14 at 2:33
• Are you looking for physical intuition or matrix multiplication rules? Do that operation including unit vectors, and make sure that you include rank in discussions of units - a scalar is not a vector. Nov 5 '14 at 2:40
• What do you mean "do that operation including unit vectors, and make sure that you include rank in discussions of units"? I omitted the unit vector in the expansion of the momentum equation in the $x$ direction because it was assumed. However, I don't get what you mean be the "include rank in discussions of units". Could you explain a bit more? By the way, thank you so very much for replying. I'm extremely grateful for your response(s). Nov 5 '14 at 3:11
• I'm voting to close this question as off-topic because it has been bumped by Community a couple of times; it was cross-posted to math.SE and it already has an answer there. Jul 13 '18 at 13:50

$$\dot{\boldsymbol{\mathcal L}}(t) = \int_{\Omega_t}\rho\frac{D\boldsymbol v}{Dt}\, dv = \int_{\Omega_t}\boldsymbol b_f\, dv + \oint_{\delta\Omega_{t_T}}\boldsymbol t\, da.$$
Next one need to use Cauchy theorem to represent traction via Cauchy tensor. And then get rid of integral sign by saying that we could choose $\Omega_t$ as small as we like, and wherever we like.