# Torque in circular fluid motion

1. Aug 4, 2007

### da_willem

I'm reading a book (intro to, by Davidson) about MHD now, but found I'm a bit rusty on tensors and curvilinear coordinates. It is written that for a circular flow the azimuthal component of the NS equations in the steady state gives (with F some body force)

$$\tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr$$

Shouldn't this read, for axisymmetric flow, without the square on both r's? I would argue that the remaining terms in the NS equations

$$\sigma _{ij,i}+F_j=0$$

would yield for the azimuthal ($\theta$) component (any suggestions welcome if the notation is obscure):

$$F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta})$$

Now the author continues,

$$\tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r})$$

In which he says he used Newtons law of viscosity, which I think one can write

$$\tau_{ij}=\mu u_{i,j}$$

(Is it by the way ok to write this as $$\overline{\overline{\tau}}=\mu \nabla \vec{u}$$?)

But how does one come from that to the (r, theta) component?

Last edited: Aug 5, 2007
2. Aug 5, 2007

### da_willem

I managed to get the r,theta component of the stress tensor as it is written in the book now. But I still can't see why there would be squares on the r's in the first formula I posted; anyone?