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.

Visualization of decomposition of flow field

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!


1 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.

enter image description here

A derivation can be found here.

  • $\begingroup$ 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? $\endgroup$ May 16, 2021 at 21:37
  • $\begingroup$ Yes that's right, in your picture the dipole is aligned with the y-axis. $\endgroup$
    – Nick
    May 17, 2021 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.