Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In a question, I have to find the acceleration of a fluid parcel in a steady line vortex. I am given that $u_\theta=\frac{A_0}{r}$. So for a steady line vortex, the parcels are following circular paths therefore in cylindrical coordinates, $u_r=u_z=0$ and the acceleration comes from the Lagrangian derivative

$\frac{D\vec{u}}{Dt}=\frac{\partial\vec{u}}{\partial{t}}+(\vec{u}\dot{}\vec{\nabla})\vec{u}$

evaluating this for the velocity given above gives a zero acceleration as the only term from $\vec{u}\dot{}\vec{\nabla}$ that gives a non zero derivative is $\partial/\partial{r}$ but then $u_r=0$ which kills it. However this doesnt ring true as from just considering circular motion, $\vec{a}=-\frac{u_\theta^2}{r}\hat{r}=-\frac{A_0^2}{r^3}\hat{r}$.

If I use the identity $(\vec{u}\dot{}\vec{\nabla})\vec{u}=-\vec{\omega}\times\vec{u}+\vec{\nabla}(\frac{\vec{u}\dot{}\vec{u}}{2})$ then it can be shown that the vorticity is zero and the expected acceleration is obtained. How can this not work when evaluating the acceleration before the identity is used!? Three of us have puzzled over this and got nowhere!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

You need to remember to differentiate your unit vectors. In cylindrical coordinates the unit derivatives of the unit vectors are not simply zero as they are in cartesian. They're fairly simple to work out though.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.