In my book on fluid mechanics there is an expression
$$
\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}
$$
where
$\boldsymbol{\tau}_{ij}$ is a rank-2 tensor (=matrix). Given that $\boldsymbol\nabla=(\partial_x, \partial_y, \partial_z)$, a vector, what do I get when I dot it with a matrix?
If I was to write $\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}$ in Einstein notation, then how would it look?
The tensor $\boldsymbol{\tau}_{ij}$ is given by
$$
\begin{pmatrix}
\tau_{xx} & \tau_{yx} & \tau_{zx} \\
\tau_{xy} & \tau_{yy} & \tau_{zy} \\
\tau_{xz} & \tau_{yz} & \tau_{zz}
\end{pmatrix}
$$
and the dot-product yields (by comparison with later expressions in the chapter)
$$
\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij} = \mathbf{i}(\partial_x \tau_{xx} + \partial_y \tau_{yx} + \partial_z \tau_{zx})+\mathbf{j}(\partial_x \tau_{xy} + \partial_y \tau_{yy} + \partial_z \tau_{zy}) + \mathbf{k}(\partial_x \tau_{xz} + \partial_y \tau_{yz} + \partial_z \tau_{zz})
$$
However, I don't see how this last expression comes about.
Best Answer
Here's a simple argument that I think agrees with the initial discussion on this question some years ago, to the effect that the stress tensor quoted in the original question is not in agreement with the result and in fact should be the transpose of what's shown.
Inner product of del with stress tensor: $\nabla.$T
$\nabla=(\partial_{x}\mathbf{i}+\partial_{y}\mathbf{j}+\partial_{z}\mathbf{k)}$, and $\textbf{T }$is the second order stress tensor $\tau_{ij}$ with components $\left(\begin{array}{ccc} \tau_{11} & \tau_{12} & \tau_{13}\\ \tau_{21} & \tau_{22} & \tau_{23}\\ \tau_{31} & \tau_{32} & \tau_{33} \end{array}\right)$, which can also be expressed as $\tau_{11}\mathbf{ii}+\tau_{12}\mathbf{ij}+\tau_{13}\mathbf{ik}+\tau_{21}\mathbf{ji}+\tau_{22}\mathbf{jj}+\tau_{23}\mathbf{jk}+\tau_{31}\mathbf{ki}+\tau_{32}\mathbf{kj}+\tau_{33}\mathbf{kk}$
Using the rule that for the vector $\textbf{a }$ and dyad (second order tensor) $\textbf{bc }$(the product of vectors $\textbf{b }$and $\textbf{c}$) we have $\textbf{a.(bc) = (a.b)c}$, then: $$ \nabla.\mathbf{T=}(\partial_{x}\mathbf{i}+\partial_{y}\mathbf{j}+\partial_{z}\mathbf{k)}.\left(\tau_{11}\mathbf{ii}+\tau_{12}\mathbf{ij}+\tau_{13}\mathbf{ik}+\tau_{21}\mathbf{ji}+\tau_{22}\mathbf{jj}+\tau_{23}\mathbf{jk}+\tau_{31}\mathbf{ki}+\tau_{32}\mathbf{kj}+\tau_{33}\mathbf{kk}\right) $$ $$ =\left(\partial_{x}\mathbf{i}.\tau_{11}\mathbf{ii}\right)+\left(\partial_{x}\mathbf{i}.\tau_{12}\mathbf{ij}\right)+...+\left(\partial_{x}\mathbf{i}.\tau_{33}\mathbf{kk}\right)+\left(\partial_{y}\mathbf{j}.\tau_{11}\mathbf{ii}\right)+\left(\partial_{y}\mathbf{j}.\tau_{12}\mathbf{ij}\right)+...+\left(\partial_{y}\mathbf{j}.\tau_{33}\mathbf{kk}\right)+\left(\partial_{z}\mathbf{k}.\tau_{11}\mathbf{ii}\right)+\left(\partial_{z}\mathbf{k}.\tau_{12}\mathbf{ij}\right)+...+\left(\partial_{z}\mathbf{k}.\tau_{33}\mathbf{kk}\right) $$ $$ =\left(\partial_{x}\tau_{11}\mathbf{\left(i.i\right)i}\right)+\left(\partial_{x}\tau_{12}\mathbf{\left(i.i\right)j}\right)+...+\left(\partial_{x}\tau_{33}\mathbf{\left(i.k\right)k}\right)+\left(\partial_{y}\tau_{11}\mathbf{\left(j.i\right)i}\right)+\left(\partial_{y}\tau_{12}\mathbf{\left(j.i\right)j}\right)+...+\left(\partial_{y}\tau_{33}\mathbf{\left(j.k\right)k}\right)+\left(\partial_{z}\tau_{11}\mathbf{\left(k.i\right)i}\right)+\left(\partial_{z}\tau_{12}\mathbf{\left(k.i\right)j}\right)+...+\left(\partial_{z}\tau_{33}\mathbf{\left(k.k\right)k}\right) $$ And all of the inner products are zero apart from $\mathbf{i.i}$, $\mathbf{j.j}$ and $\mathbf{k.k}$ which equal 1, so the above reduces to:
$$ =\partial_{x}\tau_{11}\mathbf{i}+\partial_{x}\tau_{12}\mathbf{j}+\partial_{x}\tau_{13}\mathbf{k}+\partial_{y}\tau_{21}\mathbf{i}+\partial_{y}\tau_{22}\mathbf{j}+\partial_{y}\tau_{23}\mathbf{k}+\partial_{z}\tau_{31}\mathbf{i}+\partial_{z}\tau_{32}\mathbf{j}+\partial_{z}\tau_{33}\mathbf{k} $$ $$ =\left(\partial_{x}\tau_{11}+\partial_{y}\tau_{21}+\partial_{z}\tau_{31}\right)\mathbf{i}+\left(\partial_{x}\tau_{12}+\partial_{y}\tau_{22}+\partial_{z}\tau_{32}\right)\mathbf{j}+\left(\partial_{x}\tau_{13}+\partial_{y}\tau_{23}+\partial_{z}\tau_{33}\right)\mathbf{k} $$