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I was watching a YouTube video of Steve Mould about water ripples in streams of water. At some moment in the video he shows this experience of soaps surfaces, where you take two loops with soap, you put one over the other (same axis of symmetry) and then you rise one up.

The image is shown here: Minute 5:24

The surface created seems like an hyperboloid (it is actually a catenoid). But I'm not completely sure, so I wanted to derive it from standard thermodynamics. So I propose that at fixed temperature, I have to minimize:

$$dG=\gamma\text{d}S-p\text{d}V$$ where $\gamma$ is the surface tension, $S$ is the surface, $p$ is the pressure difference and $V$ is the volume.

I tried to calculate a relation for $p$ and $\gamma$ by saying $$\text{d}S= 2\pi r\: \text{d}h$$ $$ \text{d}V= \pi r^2\: \text{d}h $$ where $h$ is a differential of height in the direction of the axis of symmetry (up) and the radius is a function of $h$, $r=r(h)$.

Of course by inserting this into $\text{d}G$, and minimizing with respect to $h$, I get the usual Young-Laplace formula for a spherical surface: $$p=\frac{2\gamma}{r}$$ but I'll with expect to get: $$p=\gamma\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$$ which is the general result, $r_1$ and $r_2$ being the two curvature radius.

IMPORTANT: I'm assuming that inside the liquid fills the volume and not that is a two sided surface. What can I do to get a better result? without assuming that I should get an hyperboloid?

Edit:

Following some corrections from the comments I now have:

$$p=\frac{2\gamma}{r}\sqrt{1+\left(\frac{\text{d}r}{\text{d}h}\right)^2}$$

How should I interpret this equation? How should I proceed? in order to prove it is an hyperboloid?

Edit 2.2:

I think @MichaelSeifert is totally on the right track. If I suppose it is just a surface as in the original video, I may forget the volume-pressure term and get the surface parametrization through variational calculus. This soap two-side surface problem has as solution, a catenoid.

The problem I defined above is different, as I am considering an internal volume and only one surface. I may keep the problem open though as it is not clear what shape it should have as the soap two-sided surface problem (the convexity depends on the sign of the pressure too).

New question: is this problem (with volume filled) experimentally plausible?

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  • $\begingroup$ I think your surface element should be $dS = 2\pi r \sqrt{1+(\frac{dr}{dh})^2}dh$. $\endgroup$ – jacob1729 Jun 10 '18 at 14:58
  • $\begingroup$ @jacob1729 why is that? $\endgroup$ – Mauricio Jun 10 '18 at 15:51
  • $\begingroup$ rewrite what @jacob1729 wrote $dS=2\pi r \sqrt{(dh)^2+(dr)^2}$ $\endgroup$ – hyportnex Jun 10 '18 at 16:36
  • $\begingroup$ Ok @hyportnex, but how should I proceed from that? $\endgroup$ – Mauricio Jun 10 '18 at 18:31
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    $\begingroup$ Just to be clear: the video asks about a soap film spanning two loops. But you're asking about a water droplet (or a volume of some other fluid) that has a similar shape, spanning two surfaces. Is that correct? The two problems aren't obviously physically equivalent, which is why I ask. Also: note that for the soap film, the surface is not a hyperboloid (a rotated hyperbola) but a rotated catenary. $\endgroup$ – Michael Seifert Jun 10 '18 at 18:55

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