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Plateau equation from Newtons law

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant 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


    3. 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?
     
  2. jcsd
  3. Nov 1, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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:
    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:
    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.
     
    Last edited: Nov 1, 2012
  4. Nov 2, 2012 #3
    Ah yes, I stupidly made an assumption that on [itex]x[/itex] and [itex]x + \mathrm{d}x[/itex] edges, where [itex]y = const.[/itex], the forces don't have the [itex]y[/itex] component. I then tailored my derivation around this assumption.

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

    Similarly as before I sum the forces over the edges and with [itex]\mathrm{d}_x[/itex] mark a small difference between [itex]x - \mathrm{d}x[/itex] (force of the left neighbouring element) and [itex]x + \mathrm{d}x[/itex] (force of the right neighbouring element). The forces are then,
    [itex]
    \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.
    [/itex]
    Divide this by [itex]2 \gamma \, \mathrm{d}x \, \mathrm{d}y[/itex] and you get the Plateau equation.


    Thank you for your input. It's been very helpful.
     
  5. Nov 2, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    That all looks correct to me. Good.
     
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