Where does the equation of the velocity of a free vortex come from?

[I_laff]

I am aware of the equation of the velocity of a forced vortex, which is simply $v = r\omega$. However, the velocity for a free vortex is $v = \frac{c}{r}$ ($c$ is the 'circulation constant', $r$ is the radius, $v$ is the velocity and $\omega$ is the angular velocity). I understand why you should expect to see a difference in the equations for both the vortices since the free vortex doesn't have a radial velocity, but where does the equation come from?

 

[vanhees71]

I'm not so sure what you are after. First of all velocities are vectors (for you first equation) or vector fields.

Your first equation describes the velocity of a point particle rotating around an axis. The full equation is $$\vec{v}=\vec{\omega} \times \vec{r},$$
where $\vec{\omega}$ is the angular velocity and $\vec{r}$ the position vector of the particle (with the origin of the reference frame on the axis of rotation).

Proof: Take the rotation axis to be the $z$-axis of a Cartesian coordinate system and the particle being in the $xy$ plane. For a constant angular velocity you have
$$\vec{r}(t)=a \begin{pmatrix} \cos(\omega t) \\ \sin \omega t \\ 0 \end{pmatrix},$$
and the velocity is
$$\vec{v}(t)=\dot{\vec{r}}(t)=a \omega \begin{pmatrix} -\sin \omega t \\ \cos \omega \\ 0 \end{pmatrix}.$$
Now $\vec{\omega}=(0,0,\omega)^{\text{T}}$ and thus indeed
$$\vec{v}=\vec{\omega} \times \vec{r}.$$

The second equation describes the velocity field of a fluid. In Cartesian coordinates the correct equation is
$$\vec{v}(\vec{r})=\frac{c}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$
Note that this is a curl-free vector field everywhere except along the $z$-axis, where it is singular. Nevertheless the line integral along any closed curve going once around the $z$-axis gives the value $2 \pi c$. It's an example for a potential field in a non simply connected domain. It's thus called the "potential vortex".

 

[I_laff]

I'm not so sure what you are after. First of all velocities are vectors (for you first equation) or vector fields.

Your first equation describes the velocity of a point particle rotating around an axis. The full equation is $$\vec{v}=\vec{\omega} \times \vec{r},$$
where $\vec{\omega}$ is the angular velocity and $\vec{r}$ the position vector of the particle (with the origin of the reference frame on the axis of rotation).

Proof: Take the rotation axis to be the $z$-axis of a Cartesian coordinate system and the particle being in the $xy$ plane. For a constant angular velocity you have
$$\vec{r}(t)=a \begin{pmatrix} \cos(\omega t) \\ \sin \omega t \\ 0 \end{pmatrix},$$
and the velocity is
$$\vec{v}(t)=\dot{\vec{r}}(t)=a \omega \begin{pmatrix} -\sin \omega t \\ \cos \omega \\ 0 \end{pmatrix}.$$
Now $\vec{\omega}=(0,0,\omega)^{\text{T}}$ and thus indeed
$$\vec{v}=\vec{\omega} \times \vec{r}.$$

The second equation describes the velocity field of a fluid. In Cartesian coordinates the correct equation is
$$\vec{v}(\vec{r})=\frac{c}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$
Note that this is a curl-free vector field everywhere except along the $z$-axis, where it is singular. Nevertheless the line integral along any closed curve going once around the $z$-axis gives the value $2 \pi c$. It's an example for a potential field in a non simply connected domain. It's thus called the "potential vortex".

Thanks for the reply, my question better put is, where does the second equation,

$$\vec{v}(\vec{r})=\frac{c}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$​

come from. Is it simply that equation because the velocity field of that equation matches the description of a free vortex (i.e. irrotational).

 

[vanhees71]

Yes, it's simply used as an example for a curl- and source-free vector field (except along the $z$-axis, where it's singular!), for which
$$\int_C \mathrm{d} \vec{r} \cdot \vec{v}(\vec{r}) = 2 \pi c \neq 0$$
for any closed curve that encircles the $z$-axis.

It emphasizes the importance that form $\vec{\nabla} \times \vec{v}$ you can only conclude $\vec{v}=-\vec{\nabla} \phi$ locally, i.e., in simply-connected regions around a regular point.

Now globally, the Euclidean $\mathbb{R}^3$ with the $z$-axis taken out is not simply connected.

To see that $\vec{v}$ has a unique potential only in a region with an entire half-plane with the $z$-axis as boundary taken out, rewrite it in terms of the usual cylinder coordinates $(\rho,\varphi,z)$. It's easy to see that it can be written as
$$\vec{v}(\vec{r})=\frac{c}{\rho} \vec{e}_{\varphi}.$$
Obviously it has a local potential in every simply-connected neighborhood of any point not on the $z$-axis. Since it's only in direction of $\vec{e}_{\varphi}$, it's suggestive to assume that the potential is a function of $\varphi$ only. Indeed the gradient in cylinder coordinates gives
$$\vec{\nabla} \phi(\varphi)=\frac{1}{\rho} \vec{e}_{\varphi} \partial_\varphi \phi \stackrel{!}{=}-\vec{v}=-\frac{c}{\rho} \vec{e}_{\varphi} \; \rightarrow \; \phi(\varphi)=-c \varphi.$$
Now to get the potential unique, you have to take out some half-plane with the $z$-axis as boundary. You can, e.g., choose the $(x,z)$-half-plane with $x<0$. Then you can choose $\varphi \in ]-\pi,\pi]$. Then the potential makes a jump at any point on this half-plane by a value of $2 \pi c$, and indeed that's the value you get for the integral along a closed loop around the $z$-axis.