Surface curvature

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Surface Curvature

If no force acts normal to a tensioned surface, the surface must remain flat. But if the pressure on one side of the surface differs from pressure on the other side, the pressure difference times surface area results in a normal force. In order for the surface tension forces to cancel the force due to pressure, the surface must be curved. When all the forces are balanced, the resulting equation is known as the Young–Laplace equation.

<math>\Delta p\ =\ \gamma \left( \frac{1}{R_x} + \frac{1}{R_y} \right)</math>

The variables are:

  • Δp is the pressure difference.
  • γ is surface tension.
  • Rx and Ry are radii of curvature in each of the axes that are parallel to the surface.

Mathematically speaking, the Young-Laplace equation is

<math>\Delta p\ =\ \gamma H_f</math>

Where γ is the surface tension, and Hf is the mean curvature. Mean curvature is defined in general as follows: consider the set of all curves through some point p on a surface. Each of these curves, if parametrized by <math>c(t) = (x(t),y(t))</math>, has a curvature

<math>k = \frac{x'y-y'x}{(x'^2+y'^2)^{3/2}}.</math>

We can then find the minimum <math>k_1</math> and maximum <math>k_2</math> curvature of all of these curves. The mean curvature <math>H_f</math> is then defined as <math>H_f=(k_1+k_2)</math>.

Pierre Simon, Marquis de Laplace

Curvature of a liquid surface implies a pressure drop across that surface, the larger the curvature, the greater the pressure drop. The fundamental work was by Pierre Simon, Marquis de Laplace. An interesting side note is that much of career was in astronomy.

www.swlearning.com

Laplace equation

deGennes et al. give a simple (but not general) derivation for the relation between the surface tension, surface curvature, and pressure drop:

Overpressure inside a drop of oil “o” in water “w”. de Gennes (2004) Fig. 1.5
If the surface is perturbed from a sphere: <math>\delta W=-p_{o}dV_{o}-p_{w}dV_{w}+\sigma _{o/w}dA</math>
For a sphere: <math>\begin{align} & dV_{o}=4\pi R^{2}dR=-dV_{w} \\ & dA=8\pi RdR \\ \end{align}</math>
At equilibrium: <math>\begin{align} & \delta W=0 \\ & p_{o}-p_{w}=\Delta p=\frac{2\sigma _{o/w}}{R} \\ \end{align}</math>
For oil/water

<math>\Delta p=\frac{2\sigma _{o/w}}{R}\approx \frac{2\cdot 30\cdot 10^{-3}{N}/{m}\;}{R}</math>

For a 1mm drop: <math>\Delta p\approx \frac{6\cdot 10^{-2}{N}/{m}\;}{10^{-6}m}\approx 6\cdot 10^{4}Pa\approx 0.6\text{ atm}</math>






Ostwald ripening

For drops (or bubbles) of different sizes in the same emulsion (or foam), the internal pressures are different. The smaller is at higher pressure than the larger so that molecules diffuse from small drops to large ones. A process called Ostwald ripening.

BinanryBubbles.png

Ostwald ripening was first discovered in 1896 by Wilhem Ostwald. This process is driven by the fact that larger droplets are more energetically favorable. This is because they are more thermordynamically favored whereas small droplets are more kinetically favored. In an effort to reduce energy, states with the molecules well ordered and packed are more favorable. The large volume to surface area ratio of large droplets allows for a lower energy state. This will cause the smaller droplets to transform into large crystals. (http://xray.bmc.uu.se/~terese/crystallization/tutorials/tutorial6.html)

In this course, Ostwald ripening is significant because it is a key mechanism in the destabilization of emulsions (for example, by creaming and sedimentation). Relating back to the previous posts about the importance of crystal formation in ice cream, Ostwald ripening can play a big role in making ice cream taste old and less creamy. The water in the ice cream recrystallizes causing the larger crystals to grow competitively with the small crystals. This destabilizes the homogeneity of the emulsion and leads to the gritty taste of older ice cream. http://www.answers.com/topic/ostwald-ripening

The term Ostwald ripening is also used in geology to describe the formation of large crystals in a fine grained crystal matix as igneous rocks cool.

Ostwald ripening in solutions is a process by which quantum dots may be formed. Initially crystals of a variety of sizes will form in a solution but over time the small crystals disappear since they are more soluble due to their small size. The larger crystals will continue to grow at the expense of the smaller ones.

http://www.answers.com/topic/ostwald-ripening


General form of Laplace equation

A more general form of the Laplace equation is <math>\Delta p=\sigma \left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)</math> . Where R1 and R2 are the principle radii of curvature.

(This section needs (a) a derivation of the Laplace equation and (b) a description of the radii of curvature.)







Minimial surfaces

A minimal surface is defined as one having mean curvature zero (see Surface Curvature above.) Given some boundaries (say a wire frame), a soap bubble meeting those boundaries forms a minimal surface. Thus given a wire frame, one can analytically (or numerically) find the soap bubble that will form by minimizing the mean curvature.

Because the mean curvature on any point of a soap bubble is the same as at any other point (see Plateau's Laws), the Laplace-Young equation implies that the energy of a soap bubble is proportional to its surface area. Since the volume of air inside of soap bubble cannot change as it deforms, one can use soap bubbles to find the surfaces that enclose maximal volume. For example, the fact that a single bubble is a sphere proves that spheres have the optimum surface area to volume ratio, though this was proven by Schwarz in 1884. However, it was not proven until 2000 that the surface formed by two connected bubbles is the surface area to volume optimum for two disconnected volumes.

Bubble structures often imitate natural structures possible indicating that these shapes are controlled by the same forces that control foams:

(From lantern slide.)

It may be possible to use a version of the ideal gas law to determine the total surface area of a foam composed of soap bubbles. The gas law has been shown for single bubbles and attached pairs of bubbles to be:

<math>P V + \frac{2}{3} \sigma A = n k T</math>

where P is the pressure, V is the volume, <math>\sigma</math> is the surface tension, A is the total surface area, n is the number of atoms, k is Boltzmann's constant, and T is the temperature. An open question is whether this same equation apples to a bulk collection of bubbles. The temperature, pressure, volume, and mass of the gas can all be controlled experimentally, leaving the surface area as the only unknown in the equation.

Plateau's Laws

Plateau's Laws describe soap films in foams, or soap films with fixed (i.e. wire frame) boundaries. These laws were formulated in the 19th century by Joseph Plateau from his experimental observations, though they were proved mathematically by Jean Taylor.

  1. Soap films are made of smooth surfaces.
  2. The mean curvature of a portion of a soap film is always constant on any point on the same piece of soap film.
  3. Soap films always meet in threes, and they do so at an angle of cos−1(−1/2) = 120 degrees forming an edge called a Plateau Border.
  4. These Plateau Borders meet in fours at an angle of cos−1(−1/3) ≈ 109.47 degrees to form a vertex.

Configurations other than those of Plateau's Rules are unstable and the foam will quickly tend to rearrange itself to conform to these rules.

As an example, say we wanted to find the shortest set of lines that connects four points on a square

Soap bubble - crossroads.gif

It might tempting to draw the above figure as a solution with length <math>2\sqrt{2}=2.83</math>, however we know that it cannot be the minimum by Plateau's Laws. After more experimentation, we may notice the following solution:

Soap bubble - shortest.gif

which has length <math>1+\sqrt{3}=2.73</math>. In the above figure, the films meet at an angle of 120 degrees, as required.


Lord Kelvin

William Thomson, 1st Baron Kelvin, (26 June 1824 – 17 December 1907):

Weaire, 1996 Front Cover (in reverse?)

The Kelvin Problem

The Kelvin problem is that of partitioning 3D space into cells of equal volume and minimum surface area. Kelvin proposed the solution as a curved tetrakaidecahedron, six squares and eight hexagons. All edges are curved; all faces with no curvature, but none flat.


Weaire, p. 56

However, structure with even smaller surface areas have been found recently and are discussed in Weaire, 1996.



Bubble rise

The motion of bubbles is an active field of research. (A description of Prof. Stone's research should be added here.)


A model for the effect of viscosity on bubble formation has been given by Krotov (p. 203f).


Without viscous effects the pressure is determined by the Laplace equation: <math>p^{\alpha }\left( t \right)=p+\frac{2\sigma \left( t \right)}{R\left( t \right)}</math>

Adding the viscous effects gives: <math>p^{\alpha }\left( t \right)=p+\frac{2}{R\left( t \right)}\left( \sigma \left( t \right)+2\eta \frac{dR\left( t \right)}{dt} \right)</math>

The viscous term is negligible for aqueous systems but might be important for the maximum bubble pressure experiment or ink jetting. e.g.

<math>\begin{align}

 & \text{For glycerol:} \\ 
& \eta =1.49\text{ }Pa-s\text{  and with }\frac{dR\left( t \right)}{dt}\approx 10^{-3}\text{ }m-s \\ 
& 2\eta \frac{dR\left( t \right)}{dt}=3\text{ }{mN}/{m}\; \\ 

\end{align}</math>



Capillary adhesion

A liquid drop between two flat plates has curved surfaces, but with curvatures in opposite directions:

de Gennes, 2004, Fig. 1.8

The large radius of curvature, R, is in the plane of the plates. The smaller one is perpendicular and opposite in sign: <math>-{}^{H}\!\!\diagup\!\!{}_{\left( 2\cos \theta _{e} \right)}\;</math>

The Laplace pressure is: <math>\Delta p=\sigma \left( \frac{1}{R}-\frac{2\cos \theta _{e}}{H} \right)\sim -\frac{2\sigma \cos \theta _{e}}{H}</math>

The force pushing the plates together is the drop area times Laplace pressure: <math>F=\pi R^{2}\frac{2\sigma \cos \theta _{e}}{H}</math>

<math>\text{For }R=1\text{ cm, }H=5\mu \text{m, and }\theta =0</math> the pressure is about 1/3 atm and the force 10 N.


(What is the stable state? What are the dynamics?)

Drops on a fiber

For small drops, gravity is not significant so the pressure is constant throughout the drop, hence the curvature is constant everywhere:

de Gennes, 2004 Fig.1-10


<math>\begin{align}

 & \Delta p=\sigma \left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)\text{ }\Rightarrow \text{ }\frac{\Delta p}{\sigma }=\left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)=\text{constant} \\ 
& \text{ or  }\left( \frac{{\ddot{z}}}{\left[ 1+\dot{z}^{2} \right]^{{3}/{2}\;}}+\frac{{\dot{z}}}{x\left[ 1+\dot{z} \right]^{{1}/{2}\;}} \right)=\text{constant} \\ 

\end{align}</math>

de Gennes (2004, p. 11) gives the solution by considering the forces on the vertical dotted line. For any position on the x-axis, where the thickness of the drop is z, the sum of the component of surface tension pulling to the right and the force of pressure to the left and the force of wetting the fiber is zero: <math>2\pi z\sigma \cos \theta -\Delta p\pi \left( z^{2}-b^{2} \right)-2\pi b\sigma =0</math>

<math>\text{From p}\text{. 12 }\cos \theta ={1}/{\sqrt{1+\dot{z}^{2}}}\;</math>

The equation applies for all z, hence when <math>\dot{z}=0\text{ and }z=L</math>

Hence: <math>\Delta p=\frac{2\sigma }{L+b}</math>





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