How does a velocity field (in a fluid) give rise to the gradient of a potential?

Question

I'm trying to understand Ch. 3.2 of the paper On Bubble Rings and Ink Chandeliers by Padilla et al.: I don't understand how $u(\gamma', 0)$ gives rise to the gradient of a potential (last sentence on p. 129:5 left column).

$\gamma$ is a vortex filament, $\gamma'$ its deformation. $M_0$ denotes the exterior of the filament, $M_1$ the interior. Considering the exterior $M_0$ in a plane orthogonal to the tangent vector of $\gamma$, the velocity field due to the filaments deformation only (Circulation C is assumed to be zero) gives rise to the gradient of a potential.

Figure from that paper, the case in question is the middle one.

I'm struggling to understand what the gradient of a potential would be here. In other words, how would the underlying (implicit) potential look like? I fail to see a scalar field that would yield such a vector field as its gradients - or is this the wrong way to look at it?

Also, if I'm missing out on some basic theory, I would appreciate some pointers to appropriate reading material!

Answer 1

The middle picture above shows a velocity field due to a 2D dipole, the superposition of the flow due to a line source and line sink of equal and opposite strength in the limit as the distance between them goes to zero.

The potential is

$$ \phi = -\frac{\bf{\mu}\cdot\bf{r}}{2\pi r^2}, $$

where $\bf{\mu}$ is the strength and direction of the dipole and $\bf{r}$ is the position vector. If we have a dipole aligned with the positive $x$-axis then

$$ \phi = -\frac{\mu x}{2\pi r^2}=-\frac{\mu \cos{\theta}}{2\pi r}, $$

plotted below.

A derivation can be found here.