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Divergence of second-order Tensor

 
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Nov30-10, 01:48 PM   #1
 

Divergence of second-order Tensor


1. The problem statement, all variables and given/known data
Calculate the Divergence of a second-order tensor:

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

2. Relevant equations

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

3. 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?
 
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Nov30-10, 02:15 PM   #2
 
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hi paccali!

σij'i is a vector, not a tensor …

you haven't summed over i
 
Nov30-10, 02:36 PM   #3
 
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]
 
Nov30-10, 02:47 PM   #4
 
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Divergence of second-order Tensor


sorry, not my field … you'd better start a new thread on this one
 
Nov30-10, 03:10 PM   #5
 
Quote by tiny-tim View Post
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.
 
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divergence, matrix, second-order tensor, tensor
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