# How do I intuit viscosity in a rotating fluid?

Suppose I have two plates with a viscous fluid in between. I slide them in the same direction (a direction in their own plane), one at $5 \,\text{m/s}$ and the other at $6 \,\text{m/s}$. Due to the relative velocity, there will be a viscous damping force tending to equalize the velocities of the plates.

Now suppose I have two concentric cylindrical shells, the second with a radius $6/5$ that of the first. I put viscous fluid between them and rotate the entire system such that the cylinders have tangential velocities $5 \,\text{m/s}$ and $6 \, \text{m/s}$. Because the system is executing rigid rotation, there will be no viscous damping.

Suppose I take a small cube of the viscous fluid in either scenario. In both cases, there's a similar velocity gradient across the cube, but only in one case does the cube dissipate energy. Why?

I can work this out by writing the Navier-Stokes equation and transforming to cylindrical coordinates, but I am still having a hard time visualizing the kinematics such that this becomes clear to me.

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I'm not sure I understand what you mean by "viscous damping." How do you quantify it? –  Christoph B. Apr 17 '13 at 15:34
You quantify it by measuring the force. –  Mark Eichenlaub Apr 17 '13 at 21:01

Let's look at the stationary solutions in 2D for incompressible fluids. I'm using the Wiki Couette flow page as a basis. Let $\boldsymbol{u} = (u,v)$ be the velocity field. The stationary Navier-Stokes equations are given by $$\rho (\boldsymbol{u}\cdot\nabla)\boldsymbol{u} = -\nabla p +\mu \Delta \boldsymbol{u},\\ \nabla \cdot \boldsymbol{u}=0.$$

In the case between two flat plates, I plant my reference frame with the $x$-axis parallel to the the channel, right in the middle. The $y$-axis is then perpendicular to the channel (of width $L=1$), the plates are at $y=0$ and $y=1$. The symmetries of the problem lead me to suppose $v=0~$ and $\partial_x u = 0$. This way, the incompressibility condition is automatically satisfied. Then we are left with (after simplifications), $$0 = \rho(u\partial_xu + v\partial_yu)= -\partial_x p + \mu \partial_y^2 u.$$ Suppose then that we prescribe the pressure gradient as a constant, say $\partial_x p = \delta P$. The solution respecting your boundary conditions is $$u(x,y) = \frac{\delta P}{2\mu} y^2 + \frac{1}{L}(v_1-v_0- \frac{\delta P L^2}{2\mu}) y + v_0.$$ If $\delta P = 0$, the solution is $$u(x,y) =(v_1-v_0) y + v_0.$$ The viscous damping at the top plate per unit area is given by $$\sigma_\mu \cdot \boldsymbol{n} = \mu\begin{pmatrix}2\partial_x u & \partial_y u + \partial_x v\\ \partial_y u + \partial_x v & 2\partial_y v\end{pmatrix}\begin{pmatrix}-1\\0\end{pmatrix} \\ =\mu\begin{pmatrix}0 & \partial_y u \\ \partial_y u & 0\end{pmatrix}\begin{pmatrix}-1\\0\end{pmatrix}= \begin{pmatrix}- \frac{\delta P}{2} - \frac{1}{L}(v_1-v_0- \frac{\delta P L^2}{2})\end{pmatrix}$$ Your intuition is thus correct: the plates experience a viscous damping force.

In the concentric shells scenario, we'll use cylindrical coordinates $(r,\theta)$. Symmetries of the problem leads to $u_r = 0, \partial_\theta u_\theta = 0$ and $\partial_\theta p =0$. Writing the Navier-Stokes equations in cylindrical coordinates and simplifying with the symmetry constraints we get $$\rho u_\theta^2 = r \partial_r p,\\ 0 = \mu(r\partial_r (r\partial_r u_\theta)-u_\theta)$$ The solution to the second equation is of the form $u_\theta = C_1\cdot r + \frac{C_2}{r}$, and the boundary conditions you chose (i.e., the two shells have the same angular velocity) are compatible with that if $C_2=0$ and $C_1$ is the angular velocity of the shells. The viscous damping at the outer shell per unit area is given by (from Batchelor's Introduction to Fluid Dynamics) $$\sigma_\mu \cdot \boldsymbol{n} = \mu \begin{pmatrix}2\partial_r u_r & r\partial_r(\frac{u_\theta}{r}) + \frac{1}{r}\partial_\theta u_r\\ r\partial_r(\frac{u_\theta}{r}) + \frac{1}{r}\partial_\theta u_r & \frac{2}{r}\partial_\theta u_\theta + \frac{u_r}{r}\end{pmatrix}\begin{pmatrix}-1\\0\end{pmatrix} \\ = \mu \begin{pmatrix}0 & r\partial_r(\frac{u_\theta}{r})\\ r\partial_r(\frac{u_\theta}{r}) & 0\end{pmatrix}\begin{pmatrix}-1\\0\end{pmatrix} = \begin{pmatrix}0\\-\mu r\partial_r(\frac{u_\theta}{r})\end{pmatrix}$$ Indeed, if $C_2=0$, then there is no viscous damping. However, this is tied to the fact that both shells have the same angular velocity, otherwise, you get a non-zero force damping force.

As such, zhermes's intuition seems sound to me: in your particular case, the fluid indeed rotates along with the shells in a "rigid body motion" as you say. The velocity gradient in the $y$-axis is non-zero between the two plates, but your choice of boundary conditions in the case of the two shells is precisely the one that makes the gradient along the radius of the shells vanish.

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I think the right thing here is to note that we are dealing here not with some gradient in the sense of vector (covetor, 1-form or whatever) but with a deformation tensor, some higher order object. The flaw in the reasoning that the proper cause of viscosity is not some gradient, but shear strain rate tensor.

Roughly speaking in your both cases you take a single coordinate of a tensor and call it "gradient". However just a single coordinate has no physical meaning. Thus it is illegal even to reason about a single coordinate. With parallel plates we we lucky to take the only non-zero tensor coordinate which represents the whole tensor, but with cylinders we forgot the others.

The proper thermodynamic force which causes dissipation due to shear deformations in Cartesian coordinates is:

$$V_{ij} = \frac{1}{2} \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] - \frac{1}{3} \frac{\partial u_k}{\partial x_k}$$

the last term is $\operatorname{div} \boldsymbol v$, here double $k$ means summation. You can easily see that there are other non-zero components apart from the .

The shear stress tensor is $$P_{ij} = \mu V_{ij}$$, the dissipation is proportional to $$P_{ij} V_{ij}$$.

For those who wonders why was $\operatorname{div} \boldsymbol v$ subtracted --- that's because it is another thermodynamic force and it is corresponded by another coefficient rather than $\mu$.

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I think the answer has to do with the velocity gradient in the radial direction being zero in the rotating frame (i.e. consider a Lagrangian fluid element). Conceptually, it should be clear that two radially separated parcels of fluid have no relative motion in the cylindrical case, while they do in the planar case. In other words, there is no gradient in the angular velocity between cylinders.

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