How Can Newton's Laws Derive the Plateau Equation for Soap Films?

[Sasha86]

Homework Statement

The Plateau equation (minimal surface of a soap film) can easily be derived from variational principle. We want to minimize the area of the soap film,
$S = \int \sqrt{1 + z_x^2 + z_y^2} \, \mathrm{d}x \, \mathrm{d}y$,
and through Euler-Lagrange equation we get the Plateau equation,
$\frac{\partial}{\partial x} \left( <br /> \frac{z_x}{\sqrt{1 + z_x^2 + z_y^2}}<br /> \right) +<br /> \frac{\partial}{\partial y} \left( <br /> \frac{z_y}{\sqrt{1 + z_x^2 + z_y^2}}<br /> \right) = 0.$
I'd like to derive this equation from Newtons law.

Homework Equations

$z_x = \frac{\partial z}{\partial x} \\<br /> z_y = \frac{\partial z}{\partial y} \\<br /> z_{xx} = \frac{\partial^2 z}{\partial x^2} \\<br /> z_{yy} = \frac{\partial^2 z}{\partial y^2}$
$\gamma$ - surface tension

The Attempt at a Solution

I'll write the forces for a small element of the film, whose projection to plane $z = 0$ is a square $\mathrm{d}x \, \mathrm{d}y$. Sum of the forces on each element must by Newton be 0. Area of the element is $\mathrm{d}S = \mathrm{d}x \, \mathrm{d}y \sqrt{1 + z_x^2 + z_y^2} = \sqrt{\mathrm{d}x^2 + \mathrm{d}z^2} \sqrt{\mathrm{d}y^2 + \mathrm{d}z^2} \frac{\sqrt{1 + z_x^2 + z_y^2}}{\sqrt{1 + z_x^2} \sqrt{1 + z_y^2}}$.

The work needed to increase a surface is $\mathrm{d}W = \gamma \mathrm{d}A$ ($F \, \mathrm{d}x = y \, \mathrm{d}x$ for a simple square). Imagine I want to stretch the element in the $x$ direction. Then the element stretches by $\mathrm{d} \left( \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} \right)$ and the force needed to overcome the tension is $\gamma \, \mathrm{d}y \, \sqrt{\frac{1 + z_x^2 + z_y^2}{1 + z_x^2}}$. I'll only be interested in $z$ component of the force so I need to multiply it by $\frac{\mathrm{d}z}{\sqrt{\mathrm{d}x^2 + \mathrm{d}z^2}}$.
Similarly for stretching in $y$ direction.

Now I'll mark with $\mathrm{d}_x$ a small diference between $x$ and $x + \mathrm{d}x$ and similarly for $y$. I then write the sum (over four sides of the small element) of the forces in $z$ direction on the small element,
$\mathrm{d}_x \left(<br /> \gamma \, \mathrm{d}y \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} z_x \right) +<br /> \mathrm{d}_y \left(<br /> \gamma \, \mathrm{d}x \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} z_y \right) = 0.$
Divide the expression bx $\gamma \, \mathrm{d}x \, \mathrm{d}y$ and get,
$\frac{\partial}{\partial x} \left( z_x \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} \right) +<br /> \frac{\partial}{\partial y} \left( z_y \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} \right) = 0,$
which isn't the Plateau equation.

Also, if I write the forces in the $x$ direction I get,
$\mathrm{d}_x \left(<br /> \gamma \, \mathrm{d}y \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} \right) = 0,$
which suggests that the expression between the braces depends only on $y$. Take this into account in the upper equation and get,
$\frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} z_{xx} +<br /> \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} z_{yy} = 0.$
This is even worse. What am I missing?

 

[TSny]

Hello, and welcome to PF!

A couple of things that I noticed that don't seem correct to me, but maybe I'm just not following your work.

First, you have the following expression for the force:

...the force needed to overcome the tension is $\gamma \, \mathrm{d}y \, \sqrt{\frac{1 + z_x^2 + z_y^2}{1 + z_x^2}}$.

If this is the force on one edge of the surface element, then I would think the magnitude of the force would be $\gamma$ times the length of the edge, which doesn't seem to be what you have.

Second, I don't think your expression for getting the z-component of the force is correct:

I'll only be interested in $z$ component of the force so I need to multiply it by $\frac{\mathrm{d}z}{\sqrt{\mathrm{d}x^2 + \mathrm{d}z^2}}$.

The force will point in a direction that is tangent to the surface and also perpendicular to the edge. In general, the force vector will have nonzero x, y, and z components. So, the factor for finding the z-component will be more complicated than your expression.

That's how I see it anyway.

 

[Sasha86]

Ah yes, I stupidly made an assumption that on $x$ and $x + \mathrm{d}x$ edges, where $y = const.$, the forces don't have the $y$ component. I then tailored my derivation around this assumption.

If I now try it again.
Force on $x$ and $x + \mathrm{d}x$ edges is $\gamma \sqrt{\mathrm{d}y^2 + \mathrm{d}z^2}$. This force is tangent to the surface and perpendicular to the edge. The normal to the surface is $\mathbf{n} = \left( -z_x, -z_y, 1 \right)$ (I'm not normalizing the vectors here) and the edge has the direction $\mathbf{s} = \left( 0, 1, z_y \right)$. The force then must be perpendicular to both this vectors and has the direction, $\mathbf{s} \times \mathbf{n} = \left( 1 + z_y^2, -z_x z_y, z_x \right)$. The $z$ component of the force is then proportional to $\frac{z_x}{\sqrt{\left( 1 + z_y^2 \right)^2 + z_x^2 z_y^2 + z_x^2}} = \frac{z_x}{\sqrt{1 + z_y^2} \sqrt{1 + z_x^2 + z_y^2}}$.
Similarly for the other two edges.

Similarly as before I sum the forces over the edges and with $\mathrm{d}_x$ mark a small difference between $x - \mathrm{d}x$ (force of the left neighbouring element) and $x + \mathrm{d}x$ (force of the right neighbouring element). The forces are then,
$\mathrm{d}_x \left( \gamma \mathrm\, {d}y \, \frac{z_x}{\sqrt{1 + z_x^2 + z_y^2}} \right) + \mathrm{d}_y \left( \gamma \mathrm\, {d}x \, \frac{z_y}{\sqrt{1 + z_x^2 + z_y^2}} \right) = 0.$
Divide this by $2 \gamma \, \mathrm{d}x \, \mathrm{d}y$ and you get the Plateau equation.


Thank you for your input. It's been very helpful.

 

[TSny]

That all looks correct to me. Good.