Why isn't this divergence theorem question all equal to zero? I have the following conditions for flow velocity and stress tensor:
$$\nabla \cdot \textbf{u} = 0$$ $$\nabla \cdot \mathbf{\sigma} = 0$$
The flow velocity is a vector field, stress tensor a rank-2 tensor, $\textbf{n}$ a normal vector.
I also have the following surface integral:
$$\int_S \textbf{u} \cdot \sigma\cdot\textbf{n} dS$$
Using the divergence theorem, I get
$$\int_S \textbf{u} \cdot \sigma\cdot\textbf{n} dS = \int_V \nabla \cdot[\textbf{u} \cdot \sigma]dV $$
Applying product rule
$$\int_V \nabla \cdot[\textbf{u} \cdot \sigma]dV = \int_V[(\nabla \cdot \textbf{u}) \cdot \sigma + \textbf{u} \cdot (\nabla \cdot \sigma)] dV $$
Due to the first two conditions I stated, this should reduce to zero. Apparently, it is something non-zero, what am I doing wrong?
Thank you
 A: Your product rule is invalid. $u\cdot\sigma$ is contracting with one "slot" of $\sigma$, so $\nabla\cdot[u\cdot\sigma]$ is contracting with the other slot; you can't magically get a contraction between $u$ and $\nabla$. The correct statement is
$$
  \nabla\cdot[u\cdot\sigma]
    = J_u:\sigma + u\cdot[\nabla\cdot\sigma]
\tag{$*$}
$$
where $J_u$ is the Jacobian of $u$ and the double dot product is indicating a double contraction:
$$
  J_u:\sigma = (J_u)^i_j\sigma^j_i
$$
and the pairing of the indices is irrelevant since $\sigma$ is symmetric. Actually, if $\sigma$ weren't symmetric then ($*$) would be invalid; this is a limitation of the "differentiate to the right" convention. More generally I would write
$$
  \dot u\cdot\dot\sigma\cdot\dot\nabla = \dot u\cdot\sigma\cdot\dot\nabla + u\cdot\dot\sigma\cdot\dot\nabla
$$
where the dots precisely indicate what $\dot\nabla$ is differentiating. Index notation is also unambiguous:
$$
  \dot u\cdot\dot\sigma\cdot\dot\nabla = \partial_j(u^i\sigma_i^j) = (\partial_ju^i)\sigma_i^j + u^i(\partial_j\sigma_i^j).
$$
