0
$\begingroup$

I have a $2D$ fluid parcel with coordinates $(0.5,-0.5), (-0.5,-0.5), (0.5,0.5)$ and $(-0.5,0.5)$ and this parcel is deformed by a steady flow field of $u=ay$ and $v=0$, defined on the basis ${(1,0), (0,1)}$. I tried to calculate the velocity gradient tensor $\tau_{ij}$ given by the matrix $ \left( \begin{array}{ccc} 0 & a \\ 0 & 0 \end{array} \right) $. I now need to decompose this into the symmetric strain rate tensor and the antisymmetric rotation tensor. However, this matrix isn't diagonalizable. Am I missing something here?


I need to use this part to "solve" the transformation equation of a point that is given by $$x_i(t+dt)=x_i+(u_i+du_i)dt$$where $du_i = \tau_{ij}dx_j$. Then, I must remove the rotation rate tensor from the velocity gradient tensor (which basically means I have the strain rate tensor left, if I am not mistaken) and use that to show that the transformation equation then becomes $$x_i(t+dt)=x_i+(u_i+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})\partial x_j)dt$$ and finally determine the transformation equations for $x(t+dt), y(t+dt)$.

$\endgroup$
1
$\begingroup$

Any tensor $A_{ij}$ can be decomposed into symmetric and antisymmetric parts, regardless of whether or not it is diagonalizable.

$$ \begin{align} S_{ij} & = \frac{1}{2}\left(A_{ij} + A_{ji}\right) \\ \Omega_{ij} & = \frac{1}{2}\left(A_{ij} - A_{ji}\right) \end{align} $$ so that $$ A_{ij} = S_{ij} + \Omega_{ij} $$

$\endgroup$
  • $\begingroup$ But is my velocity gradient tensor correct? I tried proceeding but it made no sense. $\endgroup$ – Artemisia Sep 3 '14 at 6:03
  • $\begingroup$ Your velocity gradient tensor looks consistent with the velocity fields you gave. Where did it start not making sense? $\endgroup$ – Darwin Sep 3 '14 at 7:41
  • $\begingroup$ Haha oops well I tried it once again and I think I have it this time :) thank you. $\endgroup$ – Artemisia Sep 3 '14 at 10:29

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.