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

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!

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.

• Thanks, it makes more sense to me now! Do I get this right, so in above middle picture of the question, source and sink would be on top and bottom of the pipe (dipole in y-axis direction), right? Commented May 16, 2021 at 21:37
• Yes that's right, in your picture the dipole is aligned with the y-axis.
– Nick
Commented May 17, 2021 at 13:02