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Position vector r is given as $\vec r=x_i\hat e_i$ and the second order tensor T is given as: $\overline{\overline{T}}=\frac{\delta_{ij}\hat e_i\hat e_j}{r}+\frac{x_ix_j \hat e_i\hat e_j}{r^3}$.

How to determine (cross product) $\vec r\times\overline{\overline{T}}$ by using index notation rules? I would appreciate any suggestions as I don't know whether I can dot the vector into the two parts of the tensor separately as $\vec r\times\overline{\overline{T}}=\vec r\times\overline{\overline{T_1}}+\vec r\times\overline{\overline{T_2}}$ or not.

Actually I have get this answer, but I don't know how I can simplify it more. $\vec r\times\overline{\overline{T}}=(\delta_{ij}+\frac{x_ix_j}{r^2})\frac{x_m}{r}\epsilon_{mik}\hat e_k\hat e_j$.

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Let's use Greek letters for scalars, lowercase Latin for vectors, and uppercase Latin for matrices.
And for ease of typing, let's avoid decorating variables with things like arrows, hats, overbars, etc.

The main variables are $$\eqalign{ \rho &= \|r\| \cr n &= \rho^{-1}r \,\,\,\,\,\,\,\,\,\,{\rm [\,unit\,vector\,]} \cr r &= \rho n \cr T &= \rho^{-1}(I + nn^T) \cr }$$ and we'd like to evaluate the matrix $\,\,N = r\times T$

Since the cross product distributes over addition, we can use the fact that $n\times n=0$ to simplify the calculation to $$N = n\times(I + nn^T) = n\times I$$ Switching to index notation $$N_{jp} = -n_{i}\varepsilon_{ijk}\delta_{kp} = -n_{i}\varepsilon_{ijp}$$ or in matrix notation $$N =\begin{bmatrix}0&-n_3&n_2\\n_3&0&-n_1\\-n_2&n_1&0\end{bmatrix} =\frac{1}{\rho}\begin{bmatrix}0&-r_3&r_2\\r_3&0&-r_1\\-r_2&r_1&0\end{bmatrix} $$

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  • $\begingroup$ Thanks, but based on what I wrote above, how we can write your final answer in that way? $\endgroup$ – Alex Parker Oct 24 '18 at 5:30

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