8
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ The only reasonable motivation to use a dot here is to indicate $\partial_j \tau_{ij}$ (instead of $\nabla \tau_{ij} = \partial_i \tau_{ij}$), but I have not seen this notation before. $\endgroup$
    – Phira
    Dec 29, 2013 at 21:58
  • 3
    $\begingroup$ I think Phira is correct. But the book really messed up with its notation. I should either write what Phira wrote, or write $\nabla \cdot \tau$. The result should be a vector. $\endgroup$ Dec 29, 2013 at 22:07
  • $\begingroup$ @Phira thanks for the comments, I have added more info to the question. maybe that helps? $\endgroup$
    – BillyJean
    Dec 29, 2013 at 22:14
  • $\begingroup$ Your expression corresponds to what I wrote. It looks swapped because the indices in your matrix are swapped compared to the usual convention. You should imagine the $\nabla$ to be a row vector that is multiplied with the usual dot product with the first row of the matrix to give the first component of the resulting vector (Which is the coefficient of your $\bf i$). $\endgroup$
    – Phira
    Dec 29, 2013 at 22:31
  • $\begingroup$ @Phira thanks -- don't you mean that the row from $\nabla$ should be multiplied by each column of $\tau$? $\endgroup$
    – BillyJean
    Dec 29, 2013 at 22:34

2 Answers 2

4
$\begingroup$

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:

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

$\endgroup$
3
$\begingroup$

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 $$ \begin{align} n\cdot\tau &= \left<n^i e_i,~ \tau_{jk} e^j \otimes e^k \right>\\ &= n^i \tau_{jk} (e_i \cdot e_j) \otimes e^k\\ &= g_{ij} n^i \tau_{jk} e^k \end{align} $$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.