Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.