# A Fluids and Vortices

1. May 2, 2017

### joshmccraney

Hi PF!

For an irrotational vortex, the circulation is zero along any closed contour that does not enclose the vortex axis; and has a fixed value, $\Gamma$, for any contour that does enclose the axis once. Why?

Secondly, why must vortex tubes close on themselves or begin/end at fluid boundaries?

Thanks so much!

2. May 3, 2017

3. May 4, 2017

### Joe Wolf

This is due to Helmholtz's Second Theorem

4. May 4, 2017

5. May 5, 2017

### vanhees71

If you have an irrotational vortex you have a flow filed $\vec{v}(\vec{x})$ (it can of course also be time dependent, but that's not important for my answer here), that means you have
$$\vec{\nabla} \times \vec{v}=0$$
everywhere except along some line. As an example take
$$\vec{v}=\frac{\Gamma}{2 \pi (x^2+y^2)} \begin{pmatrix} -y \\ x \\ 0 \end{pmatrix}.$$
This is defined everywhere except along the $z$ axis, and there you have
$$\vec{\nabla} \times \vec{v}=0.$$
To see, what's going on along the $z$ axis, we regularize this field by introducing a parameter $\epsilon>0$:
$$\vec{v}_{\epsilon}(\vec{x})=\frac{\Gamma}{2 \pi (x^2+y^2+\epsilon^2)} \begin{pmatrix} -y \\ x \\ 0 \end{pmatrix}.$$
Calculating now the curl leads to
$$\vec{\nabla} \times \vec{v}_{\epsilon}=\vec{e}_z \frac{\Gamma \epsilon^2}{\pi(\epsilon^2+x^2+y^2)^2}.$$
Now for $x^2+y^2 \neq 0$ you get of course $0$ in the limit $\epsilon \rightarrow 0$. For $x^2+y^2=0$ (i.e., along the $z$ axis) the limit $\epsilon \rightarrow 0$ diverges.

On the other hand you can integrate over any plane parallel to the $xy$ plane, leading to (using polar coordinates)
$$\int_{\mathbb{R}^2} \mathrm{d} x \mathrm{d} y \vec{e}_z \cdot \vec{\nabla} \times \vec{v}_{\epsilon}=2 \pi \int_0^{\infty} \mathrm{d} \rho \frac{ \rho \Gamma \epsilon^2}{\pi(\epsilon^2+x^2+y^2)^2}=\Gamma.$$
This means that
$$\vec{\nabla} \times \vec{v}=\lim_{\epsilon \rightarrow 0} \vec{\nabla} \times \vec{v}_{\epsilon}=\Gamma \delta(x)\delta(y).$$
Now calculate the circulation along any curve winding (once!) around the $z$ axis. Because $\vec{\nabla} \times \vec{v}=0$ everywhere except along the $z$ axis you can see using Stokes's theorem that for any such curve you get the same result, because you can cut the area encircelt by the two curves with two arbitrary lines and calculating the line integral over the corresponding two closed curves (leading finally to the integral along the one curve minus the one along the other), which don't contain the $z$ axis. Then you use Stokes theorem there, and since $\vec{\nabla} \times \vec{v}=0$ there you get $0$, i.e., the line integrals along the original two curves give the same.

Now use a circle parallel to the $xy$ plane, and you get
$$\int_C \mathrm{d} \vec{x} \cdot \vec{v}=\Gamma.$$

Last edited: May 5, 2017
6. May 5, 2017

### joshmccraney

So every irrotational vortex has zero curl everywhere except at some origin of the vortex, which is I guess singular?

7. May 6, 2017

### zwierz

Alexandre Chorin Jerrold E. Marsden A Mathematical Introduction
to Fluid Mechanics. Here $\boldsymbol \xi=\mathrm{rot}\,\boldsymbol v$

Last edited: May 6, 2017
8. May 6, 2017

### joshmccraney

What is the $\mathrm{rot}$ operator; some sort of rotation?

I'm trying to understand the last statement in the first paragraph, but I don't know what $\mathrm{rot} \boldsymbol v$ is.

Also, I'm unsure if a vortex could split in two; could you explain why or why not, or guide me to the answer?

9. May 6, 2017

rot=curl