dot product between vector and matrix

Question

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?

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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.

Answer 1

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:

$$ =\partial_{x}\tau_{11}\mathbf{i}+\partial_{x}\tau_{12}\mathbf{j}+\partial_{x}\tau_{13}\mathbf{k}+\partial_{y}\tau_{21}\mathbf{i}+\partial_{y}\tau_{22}\mathbf{j}+\partial_{y}\tau_{23}\mathbf{k}+\partial_{z}\tau_{31}\mathbf{i}+\partial_{z}\tau_{32}\mathbf{j}+\partial_{z}\tau_{33}\mathbf{k} $$ $$ =\left(\partial_{x}\tau_{11}+\partial_{y}\tau_{21}+\partial_{z}\tau_{31}\right)\mathbf{i}+\left(\partial_{x}\tau_{12}+\partial_{y}\tau_{22}+\partial_{z}\tau_{32}\right)\mathbf{j}+\left(\partial_{x}\tau_{13}+\partial_{y}\tau_{23}+\partial_{z}\tau_{33}\right)\mathbf{k} $$

Answer 2

Let $n$ be any $(1,0)$ - (contravariant) tensor and $\tau$ be $(0,2)$ - (covariant) tensor. Then, $n = n^i e_i$ and $\tau = \tau_{ij}e^i \otimes e^j $ and therefore, we have $$ \begin{align} n\cdot\tau &= \left<n^i e_i,~ \tau_{jk} e^j \otimes e^k \right>\\ &= n^i \tau_{jk} (e_i \cdot e_j) e^k\\ &= g_{ij} n^i \tau_{jk} e^k \end{align} $$ where $\left<\cdot,\cdot\right>$ denote the inner product, $e_i$ is a basis vector, $\otimes$ is a tensor product, and $g_{ij} = e_i \cdot e_j$ is a metric tensor, which is a Kronecker delta $\delta_{ij}$ in the Cartesian coordinate system. That is, $$ \begin{align} n\cdot\tau &= \delta_{ij} n^i \tau_{jk} e^k\\ &= n^i \tau_{ik} e^k \end{align} $$ which is corresponding to your problem. You have to be familiar with tensor algebra to follow this.