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Divergence of second-order Tensor |
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| Nov30-10, 01:48 PM | #1 |
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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? |
| 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
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| Nov30-10, 02:36 PM | #3 |
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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
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| Nov30-10, 03:10 PM | #5 |
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| divergence, matrix, second-order tensor, tensor |
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