1. Oct 8, 2015

### vin300

A fluid with forced vortex flow and constant angular velocity is given. Newton's shearing says there must be strain due to differential linear velocity. The problem is, the difference of linear velocity is only visible to the external observer, the fluid particles themselves do not observe relative difference, thus accounting to only centrifuging and nothing else. So is there shear or not?

2. Oct 8, 2015

### Staff: Mentor

The rate of deformation tensor is frame invariant. Please provide the velocity distribution, and we can discuss further.

Chet

3. Oct 8, 2015

### A.T.

If you are on a boat, floating on a lake, and spin around, the water also has differential linear velocity in your rotating frame. Is there strain in the water because of this?

4. Oct 8, 2015

### vin300

. There is nothing extraordinary to provide. I am taking the simplest case of bounded fluid like in a bucket (not a boat in the unending sea), which anybody can imagine to rotate in a similar fashion as a moment arm, and every particle rotates around the same axis making constant angular advances at constant times, with linear variation of linear velocity. I can imagine even though it is frankly speculation, that derivative of (r w) with respect to r would not provide any meaningful "strain" (w is ang. vel.)because there is no tangential strain in a rotating rod either. Clearly there has to be a differential of "angular velocity" in this case for fluid particles to be rubbing past each other and causing distortions.

5. Oct 8, 2015

### Staff: Mentor

Yes, you are correct. The fluid is rotating as a rigid body here, and there is no deformation occurring. To get rates of deformation in a fluid, we need to subtract out the rigid body rotations of the fluid elements. We do this by resolving the velocity gradient tensor into an antisymmetric part (the vorticity tensor, which accounts for rotation) and the symmetric part (the rate of deformation tensor), which accounts for rates of strain. In Newtonian fluid mechanics, it is only the symmetric part of the velocity gradient tensor that determines the stress tensor.

Chet