# Showing symmetry of the stress tensor by applying divergence theorem to $\int\int_{\delta V(t)} \vec{x}\times \vec{t} dS$

I'm currently working through the symmetry of the stress tensor, in relation to viscous flow. I am looking at this by examining the conservation of angular momentum equation for a material volume $V(t)$ with unit normal $\vec{n}=(n_1,n_2,n_3)$. I am having issue with applying the divergence theorem to this term

$$\int\int_{\delta V(t)} \vec{x}\times \vec{t} dS$$

Where $\vec{x}=(x_1,x_2,x_3)$ and $\vec{t}$ is the stress vector where $\vec{t}=\vec{e}_i\sigma_{ij}n_j$, using the summation convenction, where $\sigma_{ij}$ is stress vector.

If I can extract a normal from this expression I can use the divergence theorem to convert to a volume integral and combine with the other terms of the conservation of angular momentum equation, which are volume integrals, this will lead to showing $\sigma_{ij}=\sigma_{ji}$.

Many thanks to anyone who could help.

EDIT: Under angular momentum on this page is basically doing what i'm looking for, but can't for the life of me see how they do it -or what their notation relates to http://bobbyness.net/NerdyStuff/Navier%20Stokes%20Equations/Navier%20Stokes.html

EDIT2: Here is a link to the notes i'm learning from, page 14 http://www.maths.ox.ac.uk/system/files/coursematerial/2012/2386/9/B6aLectureNotes_img.pdf

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Is it fair to say that $\vec{e}_i$ are tangent to the surface $\delta V(t)$? I need a few more details on the conventions of your notation. Interesting problem. I have a derivation of the inertia tensor from KE of a rigid body, the inertia tensor naturally arises a symmetric tensor which gives the components of the quadratic form in the angular velocity... but your problem is fluid physics so perhaps my starting point is wrong. –  James S. Cook Dec 6 '12 at 18:25
@JamesS.Cook: At sorry, I tried to include as many possible but I see that now could be confusing, $\vec{e_1}=\vec{i}, \vec{e_2}=\vec{j}, \vec{e_3}=\vec{k}$ –  Freeman Dec 6 '12 at 19:02

Sorry, this is a bit too late but here goes.

The first thing is to rewrite the cross product using indices and the summation convention

$\mathbf{x} \times \mathbf{t}|_r = \epsilon_{rmn} x_m t_n$

where $\epsilon_{rmn}$ is the Levi-Civita symbol.

Now rewrite the traction vector $t_n$ as $\sigma_{jn} n_j$ which is how you sneak a normal into the equation.

Before you can apply the Divergence theorem to the resulting integral, ie.

$\int_{S} \epsilon_{rmn} x_m \sigma_{jn} n_j dS$

you have to understand that the integrand without the $n_j$ is an object $T$ with two free indices. So you can think of this integrand as

$\int_{S} T_{rj} n_j dS$

By the Divergence theorem this is

$$\int_{S} T_{rj} n_j dS = \int_{V} \frac{\partial T_{rj}}{\partial x_j} dV$$

Now you can resub for $T_{rj}$, so the integral is

$$\int_{V} \frac{\partial ({\epsilon_{rmn} x_m \sigma_{jn}})}{\partial x_j} dV$$

Take the constant $\epsilon$ out, apply partial derivatives and simplify; you get

$$\int_{V} \epsilon_{rmn} \left ( \sigma_{mn} + x_m \frac{\partial {\sigma_{jn}}}{\partial x_j} \right) dV$$

You can proceed from here. You'll see the momentum balance appear if you group the other terms, and you'll finally just have

$$\int_{V} \epsilon_{rmn} \sigma_{mn} dV = 0$$

which yields the result because $V$ is an arbitrary fluid volume and the Levi-Civita symbol is skew-symmetric.

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