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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


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!

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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.

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