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,
<br /> S = \int \sqrt{1 + z_x^2 + z_y^2} \, \mathrm{d}x \, \mathrm{d}y<br />,
and through Euler-Lagrange equation we get the Plateau equation,
<br /> \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.<br />
I'd like to derive this equation from Newtons law.

Homework Equations

<br /> 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}<br />
\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,
<br /> \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.<br />
Divide the expression bx \gamma \, \mathrm{d}x \, \mathrm{d}y and get,
<br /> \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,<br />
which isn't the Plateau equation.

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