Determinant of the strain tensor?

Question

In fluid mechanics, we often decompose the tensor $\nabla\boldsymbol u$ into its symmetric and antisymmetric parts, called the strain rate tensor and the vorticity tensor respectively, $$\nabla\boldsymbol u=\underbrace{\frac{1}{2}\big(\nabla \boldsymbol u+(\nabla\boldsymbol u)^\intercal\big)}_{:=\boldsymbol \varepsilon}+\underbrace{\frac{1}{2}\big(\nabla \boldsymbol u-(\nabla\boldsymbol u)^\intercal\big)}_{:=\boldsymbol \Omega}$$ These tensors are all of shape $(1,1)$ and so are bona-fide matrices, so their determinant makes sense.

Clearly, $\det(\nabla\boldsymbol u)$ is just the Jacobian determinant of $\boldsymbol u$. The relevance of this is obvious. It is also obvious that $\operatorname{tr}\nabla\boldsymbol u=\operatorname{tr}\boldsymbol \varepsilon=\operatorname{div}\boldsymbol u$. The significance of this is also clear.

However, what can we say about $\det\boldsymbol \varepsilon$ and $\det\boldsymbol \Omega$? Do these have any physical significance? Knowing these quantities, what can we say, qualitatively and quantitatively, about the flow field $\boldsymbol u$? I tried computing these quantities in general but I just ended up with a complete mess of indices. I do know that in an odd number of dimensions, the determinant of an antisymmetric matrix is zero. But what about two dimensional flows? What then?

Motivation:

In my fluid mechanics course, I have solved for the two dimensional incompressible steady flow field $$\boldsymbol{u}\left(\begin{bmatrix} x\\ y \end{bmatrix}\right)=\begin{bmatrix} ( E+\Omega) y\\ ( E-\Omega) x \end{bmatrix}$$ I am asked to:

Compute the rate-of-strain tensor $\boldsymbol \varepsilon$ and the vorticity tensor $\boldsymbol \Omega$. Comment on the results in the five cases

1: $\Omega=0, E\neq 0$ 2: $\Omega\neq 0, E=0$ 3: $\Omega=E$ 4: $\Omega<E$ 5: $\Omega>E$

I have computed the strain rate and vorticity tensors, $$\boldsymbol{\varepsilon} =\begin{bmatrix} 0 & E\\ E & 0 \end{bmatrix} \ \ ;\ \ \boldsymbol{\Omega} =\begin{bmatrix} 0 & \Omega\\ -\Omega & 0 \end{bmatrix}$$

But I'm having problems seeing the physical significance of the five different cases. Clearly when $\Omega=0$ then $\operatorname{curl}\boldsymbol u=0$, but what else?

Answer 1

This should give you a start.

To understand the nature of these flows, you should try to plot streamlines. It is also helpful to compute the eigenvalues and eigenvectors of the rate-of-strain tensor which are the principal strain rates and principal axes of strain.

(1) If $\Omega = 0$ and $E > 0$, then the principal strain rates (eigenvalues of $\boldsymbol\varepsilon$) are $\pm E$ with corresponding eigenvectors $(1,1)^T$ and $(1,-1)^T$. Hence, the principal axes of strain are oriented at $+45$ degrees and $-45$ degrees with respect to the $x$-axis. This is an example of biaxial extensional flow where the streamlines are hyperbolas with these principal strain axes as asymptotes. Fluid particles are stretched and compressed in directions parallel to the principal strain axes oriented at $+45$ degrees and $-45$ degrees, respectively.

(2) If $\Omega \neq 0$ and $E = 0$ we have a rigid rotation with circular streamlines.

(3) If $\Omega = E$ we have a simple unidirectional shear flow with $u = 2Ey$ and $v = 0$.