Uniform / rigid rotation in fluids, velocity gradients and resulting friction In the derivation of the Navier-Stokes equation e.g. in Landau & Lifshtiz Volume 6 on fluid mechanics it is stated that the viscous stress tensor

$\sigma_{ij}^{\prime}$ must also vanish when the whole fluid is in uniform rotation, since it is clear that in such a motion no internal friction occurs in the fluid.

Moreover, a uniform / rigid rotation leads to velocity gradients due to the change of direction. I fail to understand why these velocity gradients are not leading to frictional forces?
How does angular momentum conservation play in here?
Any pointers to literature / explanation / derivation?
 A: If a flow is consistent with a rigid rotation, then you can always "move" into a (non-inertial) reference frame where you observe the flow to have zero speed everywhere. As a result, even though such a rotating flow can induce changes in the stress state of the fluid, those changes must be consistent with a hydrostatic state of the fluid, which naturally obviates friction.
As a thought experiment, consider a glass of water spinning around at a constant angular velocity—the surface of the water will become a parabola as a result of the rotation, indicating that the rotation is clearly inducing a change in the fluid's stress. If I look at it top-down, I'll find that the flow of the water is consistent with a rigid rotation. If, however, I glue a camera to the glass, I'll observe that the water appears perfectly still—which means that the change in the stress state of the fluid has to be consistent with a hydrostatic (i.e. pressure) change, and not a change induced by friction.
An example of a less elaborate, but effectively equivalent, argument can be found in section 6.9 of Spencer's text: rigid rotations do not induce deformations on small fluid elements, and therefore cannot contribute to "friction".
A: We see this by considering the relative motion between close points (see the standard book from Batchelor from which I took most of the following). Let us consider the velocity $\vec{u}$ at position $\vec{x}$ as well as the neighboring velocity $\vec{u}+\delta \vec{u}$ at $\vec{x}+\vec{r}$ at time $t$. Then to first order we can write
$$
\delta u_i = r_j \frac{\partial u_i}{\partial x_j}\,.
$$
Moreover we remember that we can split the gradient of the velocity field into symmetric and anti-symmetric part. We do this for the difference in velocities between the two neighboring points
$$
\delta u_i 
= \delta u_i^{(s)} + \delta u_i^{(a)}
= r_i e_{ij} + r_i \xi_{ij}
= r_j e_{ij} - \frac{1}{2} r_j \epsilon_{ijk} \omega_k
$$
with vorticity $\vec{\omega}$. In vector notation
$$
\delta\vec{u}^{(a)} = \frac{1}{2} \vec{\omega} \times \vec{r} \,.
$$
Thus, $\delta \vec{u}^{(a)}$ is the velocity due to rigid-body rotation at $\vec{r}$ with angular velocity $\frac{1}{2} \vec{\omega}$.
The crucial point is to look at the temporal change of the displacement vector $\vec{r}$ or its square
$$
\frac{1}{2} \frac{\mathrm{D}}{\mathrm{D}t} r_i^2
= r_i \delta u_i
= r_i \left( r_j e_{ij} - \frac{1}{2} r_j \epsilon_{ijk} \omega_k \right)
= r_i r_j e_{ij}
$$
because the second term vanishes once we expand the sum. Thus rotations do not affect the distance between neighboring points and thus do not cause a relative motion between the.
However relative motion between neighboring fluid particles is needed for friction.
