0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Alex Parker Oct 24 '18 at 0:57
  • 1
    $\begingroup$ Once you notice that $x_i^2=r^2$ (since there is a summation rule) you're done! $\endgroup$ – Hamed Oct 24 '18 at 1:00
  • $\begingroup$ 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? $\endgroup$ – Alex Parker Oct 24 '18 at 1:09
  • 1
    $\begingroup$ Your final answer is indeed $2\vec{r}/r$ (not 2). $\endgroup$ – Hamed Oct 24 '18 at 1:10
  • 1
    $\begingroup$ 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). $\endgroup$ – Hamed Oct 24 '18 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.