# $\nabla\cdot(\vec{a}\otimes\vec{b})$ being a row vector

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. 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. Another problem is that I don't know how to derive them.

Anybody to help me?

• Honestly this is really more of a mathematics question than physics, even if it does arise from a physical system, the motivation is purely mathematical. Also is there some reason to avoid index notation? It normally makes life much easier to do that – Triatticus Mar 5 at 23:47
• I thought if I posted this to math SE they would say not to abuse notation. (like because del is not a real vector and divergence is not del dot)And index notation is not elegant in my opinion... For final result they usually recombine it to vector/matrix notation then why don't use it consistently? – Septacle Mar 6 at 1:46
• When you use index notation, you realize that whether a vector is a row vs column vector is just a notational convenience for matrix multiplication. If you write out their indices, you find that they can, in fact, be added, for example. And furthermore, the math notation is simply conveying a message about the physical system. Finally, the complicated identities are all derived with indices. – Gilbert Mar 6 at 2:14
• @Gilbert Do you mean that interpreting vectors as matrix is not correct way of doing physicst? – Septacle Mar 6 at 7:19
• This is indices notation \begin{aligned}\nabla \left( u\oplus v\right) =\\ \nabla ^{i}u_{i}v_{j}=\left( \nabla ^{i}u_{i}\right) v_{j}+u_{i}\left( \nabla ^{i}v_{j}\right) \end{aligned} – Eli Mar 6 at 7:42

$$\vec{a}\otimes\vec{b}$$ is the dyadic product - or 2nd order tensor.
Assuming $$b$$ is a 3x1 vector, then the result is a 3x3 dyadic.
And $$u\oplus v$$ is the direct sum of two vector spaces $$u$$ and $$v$$ - it doesn't make sense unless the intersection is $${0}$$.