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In the frame of a course on instabilities, I am trying to prove that the soap film between two rings has the form of a catenoid. Since pressure is equal on either side of the film, we expect to have a surface that has zero mean curvature everywhere, i.e.:

$$\frac 1 {r_1}+\frac 1 {r_2}=0 \tag1 $$

where $r_1$ and $r_2$ are the two principal radii. (Assume we have two perpendicular planes, the $(x,r)$ plane and the $(r,\theta)$ plane). In the $(x,r)$ plane, the radius is:

$$r_1=\frac{r''(x)}{\left(1+r'(x)^2\right)^{3/2}} \tag2$$

while the radius is "constant" at each section in $(r,\theta)$ with $r_2=-r(x) \tag3 $

so that the first relation becomes: $$r(x)r''(x)=\left(1+r'(x)^2\right)^{3/2} \tag4 $$

which is not the relation given by finding the minimal surface

$$ rr''=1+r'^2 \tag 5$$ that leads to the correct catenoid form.

Where did I go wrong? I feel I'm using the wrong formula for $r_1$ but can't seem to find an alternative. Can someone throw a hint my way?

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    $\begingroup$ The two principal radii are to be taken along perpendicular directions in the tangent plane at any point. The $(r,\theta)$ plane is not perpendicular to this tangent plane. So $-r(x)$ is not equal to $r_2$. However, by the fact that the surface is one of revolution, your $r_1$ is correct. $\endgroup$ – Chrystomath Nov 28 '17 at 18:04
  • $\begingroup$ @MMS Agreed, the way I wrote it only applies in the middle. I should've taken an inclined plane making an angle $\alpha$ with respect to the $r$ axis. And since the curvature $k=||dT/ds||$ we should replace $dT$ with $$dT_2=dT cos\alpha=\frac{dT}{\sqrt{1+\tan^2\alpha}}=\frac{dT}{\sqrt{1+r'^2}}$$ so that $$r_{corrected}=r(x)\sqrt{1+r'^2}$$ $\endgroup$ – György von Fesz Nov 28 '17 at 18:58
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For any curve $r(\theta)$, the curvature is given from the formula $$r(\theta)=r(0)+\dot{r}(0)\theta+\frac{1}{2}\ddot{r}(0)\theta^2+o(3)=r(0)+\dot{s}t+\frac{1}{2}\dot{s}^2\kappa n+o(3)$$ that is, $\kappa=\frac{\ddot{r}(0)\cdot n}{\dot{s}^2}$.

In the problem of the soap film, the surface is given by $$\mathbf{r}(x,\theta)=\begin{pmatrix}r(x)\cos\theta\\r(x)\sin\theta\\x\end{pmatrix}$$ There are two relevant perpendicular curves, those at constant $x$ and at constant $\theta$. The tangents in the different directions are $$t_x=\frac{d}{dx}\mathbf{r}=\begin{pmatrix}r'(x)\cos\theta\\r'(x)\sin\theta\\1\end{pmatrix}$$ $$t_\theta=\frac{d}{d\theta}\mathbf{r}=\begin{pmatrix}-r(x)\sin\theta\\r(x)\cos\theta\\0\end{pmatrix}$$ The normal vector is then $$n=\frac{t_x\times t_\theta}{|t_x\times t_\theta|}=\frac{1}{\sqrt{1+r'(x)^2}}\begin{pmatrix}-\cos\theta\\-\sin\theta\\r'(x)\end{pmatrix}$$

$\dot{s}=|t|$ in each case; so for the constant $\theta$ curve, $\dot{s}=|t_x|=\sqrt{1+r'(x)^2}$, while for the constant $x$ curve, $\dot{s}=|t_\theta|=r$.

Thus the two curvatures are given by $$\kappa_x=\frac{t_x'\cdot n}{\dot{s}^2}=\frac{-r''(x)}{\sqrt{1+r'(x)^2}^{3/2}}$$ $$\kappa_\theta=\frac{t_\theta'\cdot n}{\dot{s}^2}=\frac{r(x)}{r(x)^2\sqrt{1+r'(x)^2}}$$

For a minimal surface $\kappa_x+\kappa_\theta=0$, so $r''r=1+(r')^2$.

As you can see, the principal curvature in the $\theta$ direction is not $1/r$ as you might think.

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  • $\begingroup$ Agreed. I tried to fix it in the comment, but your way is much more rigorous. Thank you $\endgroup$ – György von Fesz Nov 28 '17 at 22:11
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$$\frac 1 {r_1}+\frac 1 {r_2}=0 \rightarrow \frac {r_2}{r1}=-1 \tag{11} $$

where $r_1$ and $r_2$ are the same two principal radii of principal curvatures. $r_2$ should be taken correctly normal to meridian as:

$$ r_2= {r(x)}{\left(1+r'(x)^2\right)^{1/2}} \tag{22}$$

$$r_1=\frac{-\left(1+r'(x)^2\right)^{3/2}}{r''(x)} \tag{33}$$

Divide above two equations

$$ \frac{r(x) \, r''(x)}{1+r'(x)^2}=1 \tag{44/5}$$

Also you have written curvature incorrecty for radius of curvature at equation 2). Also your equation 3) incorrect. What you wrote is true only for a cylinder.

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