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

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 $\sigma_{ij}=\sigma_{ji}$.

• – user4552 Jun 1 '13 at 19:55

The $i$th component of the integral is $\oint_S \epsilon_{ijk} x_j \sigma_{kl} n_l\, dS$ We see that $\epsilon_{ijk} x_j \sigma_{kl}$ has its $l$ index contract with $\hat{n}$. Thus the divergence theorem allows us to convert this integral to $\int_V \partial_l \epsilon_{ijk} x_j \sigma_{kl} \, dV = \int_V \epsilon_{ijk} (\partial_l x_j) \sigma_{kl}\, dV + \int_V \epsilon_{ijk} x_j (\partial_l \sigma_{kl})\, dV = \int_V \epsilon_{ijk} \delta_{lj} \sigma_{kl}\, dV + \int_V \epsilon_{ijk} x_j t_k\, dV = \int_V \epsilon_{ijk} \sigma_{kj}\, dV + \int_V \epsilon_{ijk} x_j t_k\, dV$.