Divergence of a tensor multiplied by a scalar Just wondering what is the result of the following divergence if we apply the product rule to:
$\nabla \cdot \left(\rho \vec v \otimes \vec v\right)$,
where $\rho$ is a scalar.
Is it correct if I write
$\nabla \cdot \left(\rho \vec v \otimes \vec v\right) = \nabla\rho\cdot\left(\vec v \otimes \vec v\right) + \rho\nabla\cdot\left(\vec v \otimes \vec v\right)$?
I considered another way to approach this, which is to consider $\rho\vec v$ as a vector, therefore
$\nabla\cdot\left[\left(\rho\vec v\right)\otimes \vec v\right] = \left[\nabla\cdot \left(\rho\vec v\right)\right]\vec v + \left[\vec v \cdot \nabla\right]\left(\rho\vec v\right)$.
Are these two equivalent to one another? 
Maybe my method is not correct. If that is the case, can anyone give me a legitimate formula to expand $\nabla \cdot \left(\rho \vec v \otimes \vec v\right)$?
 A: Yes you can consider consider $\rho\vec v$  as a (scaled) vector. But your final result in this case isn't true. The correct answer should be [1]:
$$\nabla .\left( {\rho \vec u\vec u} \right) = \vec u\nabla .\left( {\rho \vec u} \right) + \rho \vec u.\nabla \vec u$$
This result can be proved using index notation as follows:
$$\begin{array}\\\nabla .\left( {\rho \vec u\vec u} \right) = \frac{\partial }{{\partial {x_i}}}\rho {u_j}{u_k}{{\hat e}_i}.{{\hat e}_j}{{\hat e}_k} = \frac{{\partial \rho {u_i}{u_k}}}{{\partial {x_i}}}{{\hat e}_k}
\\\vec u\nabla .\left( {\rho \vec u} \right) + \rho \vec u.\nabla \vec u = {u_k}{{\hat e}_k}\frac{\partial }{{\partial {x_i}}}\rho {u_j}{{\hat e}_i}.{{\hat e}_j} + \rho {u_j}{{\hat e}_j}.\frac{\partial }{{\partial {x_i}}}{{\hat e}_i}{u_k}{{\hat e}_k}
\\\hspace{3.7cm}= {u_k}{{\hat e}_k}\frac{{\partial \rho {u_i}}}{{\partial {x_i}}} + \rho {u_j}{{\hat e}_j}.{{\hat e}_i}\frac{\partial }{{\partial {x_i}}}{u_k}{{\hat e}_k}
\\\hspace{3.7cm}= \left( {{u_k}\frac{{\partial \rho {u_i}}}{{\partial {x_i}}}} \right){{\hat e}_k} + \left( {\rho {u_i}\frac{\partial }{{\partial {x_i}}}{u_k}} \right){{\hat e}_k} = \frac{{\partial \rho {u_i}{u_k}}}{{\partial {x_i}}}{{\hat e}_k}\end{array}$$
