Divergence of second-order Tensor

[paccali]

Homework Statement

Calculate the Divergence of a second-order tensor:

[tex]\sigma _{ij}(x_{i})=\sigma_{0}x_{i}x_{j}[/tex]

Homework Equations

[tex]\bigtriangledown \cdot \sigma_{ij}=\sigma_{ij'i}[/tex]

The Attempt at a Solution

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

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

[tex]\sigma_{0}\begin{bmatrix}
x_{1}^{2} & x_{1}x_{2} & x_{1}x_{3}\\
x_{1}x_{2} & x_{2}^{2} & x_{2}x_{3}\\
x_{1}x_{3} & x_{2}x_{3} & x_{3}^{2}
\end{bmatrix}[/tex]

[tex]\sigma_{ij'i}=\sigma_{0}\begin{bmatrix}
2x_{1} & x_{2} & x_{3}\\
x_{1} & 2x_{2} & x_{3}\\
x_{1} & x_{2} & 2x_{3}
\end{bmatrix}[/tex]

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:
[tex]\sigma(r)=\sigma_{0}\mathbf{r}\otimes\mathbf{r}[/tex] where [tex]\mathbr{r}=x_{i}i_{i}[/tex]

So, would this be a correct approach?:

[tex]\bigtriangledown \cdot\sigma_{ij}=\frac{\partial }{\partial x_{k}}\sigma_{0}x_{i}x_{j}i_{i}\otimes i_{j}\cdot i_{k}[/tex]
[tex]=(\sigma_{0}x_{i}x_{j})_{'k}i_{i}\delta _{jk}=\sigma_{0}(x_{i}x_{j})_{'j}i_{i}=\sigma_{0}x_{i}i_{i}[/tex]

 

[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.