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?


2 Answers 2


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".


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.


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