# Vorticity Stretching in an Axisymmetric Flow Without Swirl

For an axisymmetric flow with zero swirl the velocity field is $$u=u_r \hat e_r+u_z\hat e_z$$ which results in the following vorticity field $$\omega=\omega_\theta\hat e_\theta$$

Here, $\omega_\theta=\partial_zu_r-\partial_ru_z$ from taking the curl.

When I look at the incompressible inviscid vorticity equation $$D_t\omega=u\cdot\nabla\omega=\omega\cdot\nabla u$$ I run into the following problem.

Normally in 2D flows the vortex stretching term is zero, that is $\omega\cdot\nabla u=0$. This is because $\omega\cdot\nabla u=\omega_\theta\frac{1}{r}\partial_\theta u$ and since $u=u(r,z)$ only. However, $\hat e_r=\begin{pmatrix}\cos(\theta)\\\sin(\theta)\\0\end{pmatrix}$ so $\partial_\theta\hat e_r=\begin{pmatrix}-\sin(\theta)\\\cos(\theta)\\0\end{pmatrix}=\hat e_\theta$, expanding $\omega\cdot\nabla u$ with product rule gives (in it's fully simplified form) $\omega_\theta \frac{u_r}{r}\hat e_\theta$ not zero.

How can $u\cdot\nabla\omega=0$ and $\omega_\theta \frac{u_r}{r}\hat e_\theta$? I'm not sure if because the flow is axisymmetric if that means the $\frac{1}{r}\partial_\theta$ in $\nabla$ vanishes or not. If it does, then obviously the only remaining derivatives render the vortex stretching term zero.

Two-dimensional flow is defined with reference to a Cartesian coordinate system as a velocity field that (1) depends on only two spatial coordinates (say $x$ and $y$) and (2) has a zero component in the third direction ($u_z$ = 0 in this case). Even without swirl, axisymmetric flow is not two-dimensional. There are non-zero velocity components along all three Cartesian directions.

In two-dimensional flow the vorticity must be perpendicular to the plane of the flow and there is no vortex-line stretching. For a velocity field $(u_x(x,y), u_y(x,y)$) the vorticity is aligned in the $z-$direction and the vortex stretching term is

$$\mathbb{\omega} \cdot \nabla\mathbb{u}= \omega_z \frac{\partial u_x}{\partial z}\hat e_x + \omega_z\frac{\partial u_y}{\partial z}\hat e_y = 0.$$

However, the vortex stretching term need not be zero in axisymmetric flow. In the case of axisymmetric flow without swirl, we have

$$\mathbb{\omega} \cdot \nabla\mathbb{u}= \frac{u_r \omega_\theta}{r} \hat e_\theta.$$

There must be some non-zero radial velocity component for the vortex-line stretching to be present. In fully-developed laminar flow in a pipe, there is only a non-zero axial velocity component and the stretching is absent.

• thank you for the clarification on 2D flows. That is what I thought regarding the axisymmetric vortex stretching. Commented Mar 4, 2016 at 22:53
• @NWernerC456: You're welcome. I can se where the confusion might arise as axisymmetric flow has only two non-zero components.
– RRL
Commented Mar 4, 2016 at 23:02
• Right -- but even though they are orthogonal there is some curvature effect making the vortex stretching term survive.
– RRL
Commented Mar 4, 2016 at 23:12