I'm physics student, so please forgive my abuse of notation. I know some of my notation are not rigorous, like $\nabla$ is not a really vector. It's not wholly my fault, because I learned that way from physics textbooks. And that's one of the reason I came here today.
I'm learning fluid physics, and I see a lot of $\nabla\cdot(\vec{a}\otimes\vec{b})$ (sometimes they just write it $\nabla\cdot(\vec{a}\vec{b})$ but I somehow don't like it)
I'm proving everything in matrix notation, not index notation, but I'm stuck with this thing. So if we assume a vector $\vec{a}$ is a 3*1 matrix a,
$\vec{a}\cdot\vec{b}=a^Tb$
$\nabla\cdot\vec{a}=\nabla^T a$
$(\vec{a}\otimes\vec{b})\cdot\vec{c}=ab^Tc$
So far it's fine. (Well actually the third equation is suspicious, because in first eq. we transposed a but in third we didn't. But I just followed what wikipedia (and many other textbooks) says about dyadics.(https://en.wikipedia.org/wiki/Dyadics#Product_of_dyadic_and_vector) But now
$\vec{a}\cdot(\vec{b}\otimes\vec{c})=a^Tbc^T$
Now RHS is row vector. But my textbook mix it with other vectors.
And when $\nabla$ comes into play,
$\nabla\cdot(\vec{b}\otimes\vec{c})=\nabla^Tbc^T$
So this is also definitely a row vector, but in textbooks it's mixed with other (column) vector. This is a problem, because $\nabla$ always come to left. My solution is that when I do inner product vector$\cdot$2nd rank tensor, make it transpose, like
$\vec{a}\cdot(\vec{b}\otimes\vec{c})\equiv (a^Tbc^T)^T$
$\nabla\cdot(\vec{b}\otimes\vec{c})\equiv(\nabla^Tbc^T)^T$
It works, but with so many T's it looks annoying. (Though still better than struggling with index) I feel like this to be related to co/contravariant vector but not sure.
I searched about this online but couldn't find a satisfying answer. They always explain it in index notation... One thing I found is https://biomechanics.stanford.edu/me337/me337_s03.pdf
In slide 16 of the link it defines $\nabla\cdot F(x)=tr(\nabla F(x))$ (here F is 2nd rank tensor) I know that $\nabla F(x)$ is 3rd rank tensor but then how you define trace there? And the link also has many fancy formulae in slide 18, including
$\nabla\cdot(\vec{u}\otimes\vec{v})=\vec{u}\nabla\cdot\vec{v}+\vec{v}\cdot\nabla u^t$
but here also RHS they're adding column vector with row vector. And another problem is that I don't know how to derive them.
Anybody to help me?
I am not very fond of this notation, but by the wikipedia page you linked the correct "rule" is \begin{equation} \vec{a}\cdot(\vec{b}\otimes\vec{c})=a^Tbc \end{equation} which is indeed a (column) vector. I find the notation with the $\cdot$ product misleading since the product of a dyad with a vector is not symmetric: $\vec{a}\cdot(\vec{b}\otimes\vec{c})=a^Tbc \neq (\vec{b}\otimes\vec{c})\cdot a=b c ^T a$.
Your equation involving $\nabla$ is therefore wrong: it should be \begin{equation} \nabla \cdot (b\otimes c)= (\nabla^T b ) c=\mathrm{div}(b) c \end{equation} which again is a column vector.