- #1
Sasha86
- 2
- 0
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,
[itex]
S = \int \sqrt{1 + z_x^2 + z_y^2} \, \mathrm{d}x \, \mathrm{d}y
[/itex],
and through Euler-Lagrange equation we get the Plateau equation,
[itex]
\frac{\partial}{\partial x} \left(
\frac{z_x}{\sqrt{1 + z_x^2 + z_y^2}}
\right) +
\frac{\partial}{\partial y} \left(
\frac{z_y}{\sqrt{1 + z_x^2 + z_y^2}}
\right) = 0.
[/itex]
I'd like to derive this equation from Newtons law.
Homework Equations
[itex]
z_x = \frac{\partial z}{\partial x} \\
z_y = \frac{\partial z}{\partial y} \\
z_{xx} = \frac{\partial^2 z}{\partial x^2} \\
z_{yy} = \frac{\partial^2 z}{\partial y^2}
[/itex]
[itex]\gamma[/itex] - surface tension
The Attempt at a Solution
I'll write the forces for a small element of the film, whose projection to plane [itex]z = 0[/itex] is a square [itex]\mathrm{d}x \, \mathrm{d}y[/itex]. Sum of the forces on each element must by Newton be 0. Area of the element is [itex]\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}}[/itex].
The work needed to increase a surface is [itex]\mathrm{d}W = \gamma \mathrm{d}A[/itex] ([itex]F \, \mathrm{d}x = y \, \mathrm{d}x[/itex] for a simple square). Imagine I want to stretch the element in the [itex]x[/itex] direction. Then the element stretches by [itex]\mathrm{d} \left( \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} \right)[/itex] and the force needed to overcome the tension is [itex]\gamma \, \mathrm{d}y \, \sqrt{\frac{1 + z_x^2 + z_y^2}{1 + z_x^2}}[/itex]. I'll only be interested in [itex]z[/itex] component of the force so I need to multiply it by [itex]\frac{\mathrm{d}z}{\sqrt{\mathrm{d}x^2 + \mathrm{d}z^2}}[/itex].
Similarly for stretching in [itex]y[/itex] direction.
Now I'll mark with [itex]\mathrm{d}_x[/itex] a small diference between [itex]x[/itex] and [itex]x + \mathrm{d}x[/itex] and similarly for [itex]y[/itex]. I then write the sum (over four sides of the small element) of the forces in [itex]z[/itex] direction on the small element,
[itex]
\mathrm{d}_x \left(
\gamma \, \mathrm{d}y \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} z_x \right) +
\mathrm{d}_y \left(
\gamma \, \mathrm{d}x \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} z_y \right) = 0.
[/itex]
Divide the expression bx [itex]\gamma \, \mathrm{d}x \, \mathrm{d}y[/itex] and get,
[itex]
\frac{\partial}{\partial x} \left( z_x \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} \right) +
\frac{\partial}{\partial y} \left( z_y \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} \right) = 0,
[/itex]
which isn't the Plateau equation.
Also, if I write the forces in the [itex]x[/itex] direction I get,
[itex]
\mathrm{d}_x \left(
\gamma \, \mathrm{d}y \, \frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} \right) = 0,
[/itex]
which suggests that the expression between the braces depends only on [itex]y[/itex]. Take this into account in the upper equation and get,
[itex]
\frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_x^2} z_{xx} +
\frac{\sqrt{1 + z_x^2 + z_y^2}}{1 + z_y^2} z_{yy} = 0.
[/itex]
This is even worse. What am I missing?