## Divergence of second-order Tensor

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

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

2. Relevant equations

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

3. 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:

$$\sigma_{0}\begin{bmatrix} x_{1}^{2} & x_{1}x_{2} & x_{1}x_{3}\\ x_{1}x_{2} & x_{2}^{2} & x_{2}x_{3}\\ x_{1}x_{3} & x_{2}x_{3} & x_{3}^{2} \end{bmatrix}$$

$$\sigma_{ij'i}=\sigma_{0}\begin{bmatrix} 2x_{1} & x_{2} & x_{3}\\ x_{1} & 2x_{2} & x_{3}\\ x_{1} & x_{2} & 2x_{3} \end{bmatrix}$$

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

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 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor hi paccali! σij'i is a vector, not a tensor … you haven't summed over i
 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?: $$\bigtriangledown \cdot\sigma_{ij}=\frac{\partial }{\partial x_{k}}\sigma_{0}x_{i}x_{j}i_{i}\otimes i_{j}\cdot i_{k}$$ $$=(\sigma_{0}x_{i}x_{j})_{'k}i_{i}\delta _{jk}=\sigma_{0}(x_{i}x_{j})_{'j}i_{i}=\sigma_{0}x_{i}i_{i}$$

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## Divergence of second-order Tensor

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

 Quote by tiny-tim 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.

 Tags divergence, matrix, second-order tensor, tensor