Divergence of second-order Tensor

[paccali]

Homework Statement

Calculate the Divergence of a second-order tensor:

\sigma _{ij}(x_{i})=\sigma_{0}x_{i}x_{j}

Homework Equations

\bigtriangledown \cdot \sigma_{ij}=\sigma_{ij'i}

The Attempt at a Solution

\sigma_{ij'i}=\frac{\partial }{\partial x_{i}}\cdot\sigma_{0}x_{i}x_{j}
=\sigma_{0}(x_{j})

I'm not sure if this is correct. When I put it into a matrix form and calculate the divergence, I seem to get:

\sigma_{0}\begin{bmatrix}<br /> x_{1}^{2} &amp; x_{1}x_{2} &amp; x_{1}x_{3}\\ <br /> x_{1}x_{2} &amp; x_{2}^{2} &amp; x_{2}x_{3}\\ <br /> x_{1}x_{3} &amp; x_{2}x_{3} &amp; x_{3}^{2}<br /> \end{bmatrix}

\sigma_{ij&#039;i}=\sigma_{0}\begin{bmatrix}<br /> 2x_{1} &amp; x_{2} &amp; x_{3}\\ <br /> x_{1} &amp; 2x_{2} &amp; x_{3}\\ <br /> x_{1} &amp; x_{2} &amp; 2x_{3}<br /> \end{bmatrix}

Which doesn't equal the partial that wasn't put into matrix form. Any help?

 

[tiny-tim]

hi paccali!

σ_ij'i is a vector, not a tensor …

you haven't summed over i

 

[paccali]

Here is the other part of the problem, and please help me out with this:
\sigma(r)=\sigma_{0}\mathbf{r}\otimes\mathbf{r} where \mathbr{r}=x_{i}i_{i}

So, would this be a correct approach?:

\bigtriangledown \cdot\sigma_{ij}=\frac{\partial }{\partial x_{k}}\sigma_{0}x_{i}x_{j}i_{i}\otimes i_{j}\cdot i_{k}
=(\sigma_{0}x_{i}x_{j})_{&#039;k}i_{i}\delta _{jk}=\sigma_{0}(x_{i}x_{j})_{&#039;j}i_{i}=\sigma_{0}x_{i}i_{i}

 

[tiny-tim]

sorry, not my field … you'd better start a new thread on this one

 

[paccali]

sorry, not my field … you'd better start a new thread on this one

Yeah, sorry, the problem is in tensor notation, which implies summation symbols, but it's a shorthand.