# Independency of the frame of reference of the strain rate tensor

1. Jul 12, 2013

### Idontknow84

I've got a problem regarding tensors.

Premise: we are considering a fluid particle with a velocity $\mathbf{u}$ and a position vector $\mathbf{x}$; $S_{ij}$ is the strain rate tensor, defined in this way:

$\displaystyle{S_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} +\frac{\partial u_j}{\partial x_i}\right)}.$

OK, the problem is in this paragraph taken from Fluid Mechanics. Fifth Edition, P. K. Kundu, I. M. Cohen, D. R. Dowling, 2011, p. 78:

My question is: why is $S_{ij}$ independent of the frame of reference in which it is observed? Sure, it is zero in every frame in which the fluid particle translates with constant linear-velocity $\mathbf{U}$ and rotates with constant angular-velocity $\mathbf{\Omega}$, but this doesn't explain why it should be the case "even if $\mathbf{U}$ depends on time and the frame of reference is rotating."

[*] This is the Exercise 3.17:

[$R_{ij}$ is the rotation tensor: $\displaystyle{\frac{\partial u_i}{\partial x_j} -\frac{\partial u_j}{\partial x_i}}$.]

Last edited: Jul 12, 2013
2. Jul 12, 2013

### Staff: Mentor

The rate of deformation tensor is a linear function of the velocity field. Therefore, if you superimpose two velocity fields, the rate of deformation tensor for the combined velocity field is the linear sum of the rates of deformation tensors for the individual velocity fields. If one of these velocity fields is a rigid body translation and rotation, its rate of deformation tensor is zero. Now, changing the frame of reference of the observer is mathematically equivalent to superimposing a rigid body translation and rotation. Therefore, the rate of deformation tensor is not affected by the frame of reference of the observer.

Actually, this is not precisely correct. The rate of deformation tensor as reckoned from the frame of reference of the rotating observer is equal to the rate of deformation tensor as reckoned from the frame of reference of the non-rotating observer, pre-dotted by the transpose of the rotation tensor, and post-dotted by the rotation tensor. (The rotation tensor as used here is not the same as the rotation tensor you defined, which I usually call two times the vorticity tensor).

3. Jul 13, 2013

### Idontknow84

Thanks for the reply, but I'm not so much convinced by your argumentation. In fact, at page 80 it is stated:
All in all, on one hand "$S_{ij}$ is independent of the frame of reference in which it is observed"; on the other hand, instead, "$S_{ij}$ change as the coordinate system is rotated".
I am a bit confused.

4. Jul 13, 2013

### Staff: Mentor

You are confused because there is a difference between the components of a tensor expressed with respect to a particular coordinate system, and the tensor itself. If you change the coordinate system, the components of the tensor change, but the tensor itself is independent (invariant) of the coordinate system you use. This is analogous to the case of a vector. Here is an example. Consider a vector $\vec{V}$ expressed with respect to a particular cartesian coordinate system:
$$\vec{V}=V_x\vec{i_x}+V_y\vec{i_y}+V_z\vec{i_z}$$
where $V_x$, $V_y$, and $V_z$ are the components of $\vec{V}$ with respect to the x-y-z coordinate system, and $\vec{i_x}$, $\vec{i_y}$, and $\vec{i_z}$ are the unit vectors for the coordinate directions. Now consider the exact same vector $\vec{V}$ expressed with respect to a second cartesian coordinate system x'-y'-z':
$$\vec{V}=V_{x'}\vec{i_{x'}}+V_{y'}\vec{i_{y'}}+V_{z'}\vec{i_{z'}}$$
Notice that, even though the vector $\vec{V}$ is exactly the same, its components with respect to the two different cartesian coordinate systems are different (because the coordinate axes and unit vectors are pointing in different directions). The x'-y'-z' coordinate system can be obtained from the x-y-z coordinate system by a sequence of rigid rotations.

The exact same thing happens with second order tensors, such as the rate of deformation tensor and the stress tensor, although the transformation law for relating the components of the tensor between the two coordinate systems is a little more complicated.

5. Jul 14, 2013

### Idontknow84

The problem is the book isn't referring to the strain rate tensor itself (whose nature is expressed by writing $\mathbf{S}$), but to its components (so it would be an objective tensor): otherwise, it would have been trivial to state its independency of the frame of reference in which it is observed, since every tensor possesses this feature. In addition, carefully look at the following paragraph:

The keyword in here is thus, which means: "as a result or consequence of this"; in other words, the fact "$S_{ij}$ is independent of the frame of reference" has to be a consequence of the preceding sentence.
Moreover, on page 79 it is stated:

So, in the above mentioned case "$R_{ij}$" must clearly represent the components of the tensor $\mathbf{R}$ in a particular basis (as well as $\mathbf{\omega}$—so it should be written $\omega_i$ for consistency), whereas, according to your reasoning, $S_{ij}$ is the tensor $\mathbf{S}$.
But then again:
• Why the "Thus"?
• Why to state such an obvious truth about the tensors themselves (about their invariance of the coordinate system)?
• And why, lastly, on page 80 is it written "the components $S_{ij}$ change as the coordinate system is rotated" and not "the components of $S_{ij}$ change as the coordinate system is rotated", if we even assume "$S_{ij}$" is another way of writing the tensor $\mathbf{S}$ itself?

That's why I'm confused.

6. Jul 14, 2013

### Staff: Mentor

It is no wonder that you are confused. Your assessment of the situation is completely correct, and the statement in the book beginning Thus is incorrect. The components Sij will certainly depend on the rotation of the frame of reference (coordinate system) of the observer. I alluded to this in my first reply. If S is the matrix of components for the non-rotating observer, and SR is the matrix of components for the rotating observer, then

SR=QT S Q

where Q is an orthogonal matrix related to the amount of rotation that has taken place between the coordinate axes of the two frames of reference. This is just a straight coordinate transformation.

Chet

Last edited: Jul 14, 2013