I am trying to prove that the following fluid flow is incompressible (in other words, its divergence is zero):
$$\textbf{u}( \textbf{r, S}) = \alpha {(\textbf{r}\cdot\textbf{S}\cdot \textbf{r})\textbf{r} \over{r^5}}$$ where $S$ is a trace-free stresslet.
This is my attempt so far:
$$\nabla \cdot \textbf{u}( \textbf{r, S}) = \alpha \nabla \cdot {(\textbf{r}\cdot\textbf{S}\cdot \textbf{r})\textbf{r} \over{r^5}}$$ $$=\alpha {(\textbf{r}\cdot (\nabla \cdot\textbf{S})\cdot \textbf{r})\textbf{r} \over{r^5}}$$ and for some reason, since it is traceless, $\nabla \cdot \textbf{S}=0$?
Furthermore, for the first equation of this question, is the numerator $(\textbf{r}\cdot\textbf{S}\cdot \textbf{r})\textbf{r}$ just the same thing as $(\textbf{r}\cdot\textbf{S}\cdot \textbf{r} \cdot \textbf{r})$? The notation confuses me.
Thank you
I am guessing that the quantity $Q= \vec r\cdot S \cdot\vec r$ is a scalar-valued expression, a quadratic function constructed from a symmetric matrix $S$ that has constant coefficients and no trace. That is, in matrix notation where $\vec r= [x,y,z]$, one can expand $Q= [x,y,z] [S ][x,y,z]^t $.
If so, then the usual product and quotient rules for grad and div imply that
$div (\frac{Q \vec r}{r^5}) = \frac{grad Q\cdot \vec r}{r^5} +\frac{ Q div\vec r}{r^5}- 5 \frac{ Q (\vec r \cdot grad r) }{r^6}$
First term. One can check that for any symmetric quadratic form $Q$, $ grad Q \cdot \vec r = 2 Q$.
Middle Term. Use $div \vec r= 3$ to obtain $3 Q/r^5$.
Last term. Use $grad r= \vec r/r$ to get $-5 Q/r^5$.
The sum of all three terms is zero.
P.S. If the entries of matrix $S$ depend on position then there will be an extra term in the expansion of $grad Q$ where we differentiate the entries of matrix $S$ as well. Maybe then the trace-free condition comes into play. If so, I suggest you explore the simpler case of a $2x2$ matrix first.