# $\nabla\cdot(\vec{a}\otimes\vec{b})$ 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 Commented Mar 5, 2019 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? Commented Mar 6, 2019 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. Commented Mar 6, 2019 at 2:14
• @Gilbert Do you mean that interpreting vectors as matrix is not correct way of doing physicst? Commented Mar 6, 2019 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
Commented Mar 6, 2019 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 $${0}$$.