# Dot product of the vector r and second order tensor

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 (dot product) $$\vec r.\overline{\overline{T}}$$ 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 $$\vec r.\overline{\overline{T}}=\vec r.\overline{\overline{T_1}}+\vec r.\overline{\overline{T_2}}$$ or not.

If a rank 2 tensor $$T$$ is thought of as a matrix $$M$$ (in certain standard basis), $$T\cdot v$$, for $$v$$ a (column) vector, is simply $$Mv$$. However, $$v\cdot T$$ is $$v^tM$$ with $$t$$ denoting transpose.
In your index notation, if $$T=T_{ij}e_ie_j$$ and $$r=x_ke_k$$, then \begin{aligned} r\cdot T &= x_kT_{ij} (e_k\cdot e_i)e_j = x_kT_{ij}\delta_{ik}e_j = x_iT_{ij}\\ T\cdot r &= x_kT_{ij} e_i(e_j\cdot e_k) = x_kT_{ij}\delta_{jk}e_i = T_{ij}x_j \end{aligned} Generally, dotting from right or left matters. However, if your rank 2 tensor is symmetric, i.e. $$T_{ij}=T_{ji}$$, then the left and right does not matter. Your tensor $$T=\frac{r^2\delta_{ij}+x_ix_j}{r^3}e_ie_j$$ is symmetric.
For a rank $$n$$ tensor $$T$$, the situation is even more complicated. Because now the notion of $$T\cdot v$$ needs extra clarification. It is a good idea to write $$T\cdot_m v$$, meaning the dot product is done over the $$m$$th component. Or better yet, avoid using dot products in this form altogether. Either stick to indexes from start to finish or use the relevant abstract notations related to tensors.
• By doing the dot product, I got $\frac{x_j}{r}(1+\frac{{x_i}^2}{r^2})e_j$. Is this correct or I do need to simplify it more? – Alex Parker Oct 24 '18 at 0:57
• Once you notice that $x_i^2=r^2$ (since there is a summation rule) you're done! – Hamed Oct 24 '18 at 1:00
• So I can write as $\frac{2x_je_j}{r}$ or $\frac{2\vec r}{r}$ and if we have $r=\vert {\vec r}\vert$, then we get 2 as the answer? – Alex Parker Oct 24 '18 at 1:09
• Your final answer is indeed $2\vec{r}/r$ (not 2). – Hamed Oct 24 '18 at 1:10
• It is true that $|\vec{r}|=r$, but there is no way $\vec{r}/|\vec{r}|=1$. In general, the dot product of a (rank 2) tensor and a vector is another vector (not a scalar). – Hamed Oct 24 '18 at 1:23