Q:

Shear stress in cylindrical coordinates?

In cylindrical coordinates the momentum flux is given by (in the $r$ direction):
$$ \Pi=-\eta \frac{\partial (r\omega)}{\partial r}$$
Where $\eta$ is the viscosity. Therefore one would expect that the shear stress in the $\theta$ direction would be given by:
$$ \tau = \eta \frac{\partial (r\omega)}{\partial r}$$
I have, however, seen is said that in this case the sear stress is instead given by:
$$\tau=\eta r \frac{\partial \omega}{\partial r}$$
This means that there must be another 'pressure' equal to $\eta \omega$ that I have not taken account for in my formula. What is this 'pressure' and why does it arise? Is the same sort of thing true in say spherical coordinates? 

A:

There are two methods for determining the r-$\theta$ shear stress for your situation. Both methods give the same answer. One is to transform the equations for the stress tensor from Cartesian coordinates to cylindrical coordinates.  This method is a little tedious for this problem. 
The other method is to derive the equation for the stress tensor for your situation directly in cylindrical coordinates.  The velocity vector is given here by:  $$\vec{v}=v_{\theta}\vec{i}_{\theta}$$
The gradient (vector) operator is given by:
$$\vec{\nabla}=\vec{i}_r\frac{\partial}{\partial r}+\frac{\vec{i}_{\theta}}{r}\frac{\partial}{\partial \theta}$$
If we evaluate the gradient of the velocity vector, we obtain:
$$\vec{\nabla}\vec{v}=\vec{i}_r\frac{\partial (v_{\theta}\vec{i}_{\theta})}{\partial r}+\frac{\vec{i}_{\theta}}{r}\frac{\partial (v_{\theta}\vec{i}_{\theta})}{\partial \theta}=\vec{i}_r\vec{i}_{\theta}\frac{\partial v_{\theta}}{\partial r}-\vec{i}_{\theta}\vec{i}_r \frac{v_{\theta}}{r}$$
where we have assumed that $ v_{\theta}$ is not a function of $\theta$ and we have differentiated $\vec{i}_{\theta}$ with respect to $\theta$ to obtain $-\vec{i}_r$.  The next step is to determine the rate of deformation tensor $\vec{E}$, which is the symmetric part of the velocity gradient tensor:
$$\vec{E}=(\vec{\nabla}\vec{v}+(\vec{\nabla}\vec{v})^T)/2=\frac{1}{2}\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right](\vec{i}_r\vec{i}_{\theta}+\vec{i}_{\theta}\vec{i}_r )$$
The stress tensor $\vec{\tau}$ is equal to twice the viscosity $\eta$ times the rate of deformation tensor:
$$\vec{\tau}=\tau_{r\theta}\vec{i}_r\vec{i}_{\theta}+\tau_{\theta r}\vec{i}_{\theta}\vec{i}_r =\eta\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right](\vec{i}_r\vec{i}_{\theta}+\vec{i}_{\theta}\vec{i}_r )$$
So, finally, 
$$\tau_{r\theta}=\tau_{\theta r}=\eta\left[\frac{\partial v_{\theta}}{\partial r}- \frac{v_{\theta}}{r}\right]$$

