# WikiJournal Preprints/Surface tension/Supplementary information
## Draft rewrite of lead to surface tension

Surface tension is an effect within the surface layer of a liquid that causes that layer to behave as an elastic sheet. It allows insects, such as the water strider, to walk on water. It allows small metal objects such as needles, razor blades, or foil fragments to float on the surface of water, and it is the cause of capillary action. Whenever a raindrop falls, or a child splashes in a swimming pool, or a cleaning agent is mixed with water, or an alcoholic beverage is stirred in a glass, the effects of surface tension are visible.
The physical and chemical behavior of liquids cannot be understood without taking surface tension into account. It governs the shape that small masses of liquid can assume and the degree of contact a liquid can make with another substance.
Applying Newtonian physics to the forces that arise owing to surface tension accurately predicts many liquid behaviors that are so commonplace that most people take them for granted. Applying thermodynamics to those same forces further predicts other more subtle liquid behaviors.

### end of draft rewrite of lead

The main surface tension article provides coverage of this physical phenomenon at a descriptive level. But a deeper understanding is available only through the application of mathematics and thermodynamics. This supplement provides a beginner's development of that understanding. Most of the material should be accessible to readers who have mastered high school algebra and basic physics. Some of it also requires first semester calculus.

## Basic Force Description

As described in the main article, due to surface tension, bodies of liquid behave as if their surfaces were covered with an elastic sheet. So it is useful to understand the basic nature of tension in a two-dimensional surface.
Tension in a surface is the two-dimensional analog to tension in a string (one dimension) or to stress in a solid body (three dimensions). Observe that stress in a solid has units of $\scriptstyle f\ \mathrm {per} \ x^{2}$ (where $\scriptstyle f$ is force and $\scriptstyle x$ is length) and tension in a string has units of $\scriptstyle f\ \mathrm {per} \ x^{0}$ (that is, simply a force). So it is reasonable to expect that tension in a surface should have units of $\scriptstyle f\ \mathrm {per} \ x^{1}$, or force per unit length.
This becomes clearer by imagining a surface under tension being cut in half and the halves stitched back together along a seam. If the tension in the surface is uniform, forces will be evenly distributed along the seam. So the seam feels a force per unit length of seam, and those forces are parallel to the surface but normal to the seam.
If a stressed rubber sheet has two such seams that cross at right angles to each other, it is easy to imagine a situation where the tension felt by one of the seams differs from that felt by the other. Hence, to fully describe the tension in a surface requires a vector of two elements, one for each of two tension axes. But for ordinary liquids, the surface tension in the two axes will always be identical. Indeed the surface tension will be uniform over the entire surface of the liquid provided that the liquid itself is uniform, and the material its surface contacts is uniform over the entire surface. Hence a single scalar can be used to describe the surface tension over the entire liquid surface.

### The Effect of Curveture on a Tensioned Surface

The Young-Laplace surface tension equation relates curveture and tension of a surface with the pressure differential across the surface. It applies not only to liquid surface tension, but to any curved surface under tension, a toy balloon for example.
The diagram shows a tiny rectangular patch cut from a curved surface. For discussion, we consider left-right to be the x axis, and front-back to be the y axis. Surface tension forces are shown at each of the four edges. Each is parallel to the surface and normal to its respective edge. These outward forces hold the patch to the surface it is cut from.
The curving in each axis is assumed to be circular, with a radius of curveture in the x axis of $\scriptstyle R_{x}$ and in the y axis of $\scriptstyle R_{y}$. By balancing the forces in this diagram we can deduce the Young-Laplace surface tension equation:
       $\Delta P\ =\ \gamma \left({\frac {1}{R_{x}}}+{\frac {1}{R_{y}}}\right)$
If the patch were perfectly flat, then $\scriptstyle F_{L}$ would exactly cancel $\scriptstyle F_{R}$, and $\scriptstyle F_{F}$ would exactly cancel $\scriptstyle F_{B}$. But because the surface curves, each of the four forces has a component toward the interior. We find those components by comparing each force arrow with the plane that is tangent to the patch at its center. The component toward the interior is that part of each force that is normal to that tangent plane.
In the left-right direction, the radian arc-angle spanned by the patch is $\scriptstyle \delta \theta _{x}$ and in the front-back direction the arc-angle spanned is $\scriptstyle \delta \theta _{y}$. Arc angle from the center of the patch to each edge is half of the respective arc-angle. The sines of each of the half-angles times each respective force give the components of the four forces normal to the tangent plane.
Adding up the normal components of all four forces on the patch of surface:
       $F_{normal}\ =\ \sin {\left({\frac {\delta \theta _{y}}{2}}\right)}F_{F}+\sin {\left({\frac {\delta \theta _{y}}{2}}\right)}F_{B}+\sin {\left({\frac {\delta \theta _{x}}{2}}\right)}F_{L}+\sin {\left({\frac {\delta \theta _{x}}{2}}\right)}F_{R}$
Since the $\scriptstyle \delta \theta$ values are infinitesimal, we can substitute the approximation, $\scriptstyle \sin {\left({\frac {\delta \theta }{2}}\right)}\ =\ {\frac {\delta \theta }{2}}$. The normal force total becomes:
       $F_{normal}\ =\ {\frac {\delta \theta _{y}}{2}}F_{F}+{\frac {\delta \theta _{y}}{2}}F_{B}+{\frac {\delta \theta _{x}}{2}}F_{L}+{\frac {\delta \theta _{x}}{2}}F_{R}$
Since each of the four forces is equal to the surface tension times the length of its respective edge, we have, for example, $\scriptstyle F_{L}\ =\ \gamma R_{y}\delta \theta _{y}$. So
       $F_{normal}\ =\ {\frac {\delta \theta _{y}}{2}}\gamma R_{x}\delta \theta _{x}+{\frac {\delta \theta _{y}}{2}}\gamma R_{x}\delta \theta _{x}+{\frac {\delta \theta _{x}}{2}}\gamma R_{y}\delta \theta _{y}+{\frac {\delta \theta _{x}}{2}}\gamma R_{y}\delta \theta _{y}$
or equivalently:
       $F_{normal}\ =\ \delta \theta _{y}\gamma R_{x}\delta \theta _{x}\ +\ \delta \theta _{x}\gamma R_{y}\delta \theta _{y}$
Observe that the area of the patch is $\scriptstyle A\ =\ R_{x}\delta \theta _{x}R_{y}\delta \theta _{y}$. When $\scriptstyle F_{normal}$ is divided by the area, the quotient yields the pressure differential required to balance the normal force total.
       ${\frac {F_{normal}}{A}}\ =\ \Delta P\ =\ \gamma \left({\frac {R_{x}+R_{y}}{R_{x}R_{y}}}\right)$
which is algebraically equivalent to the Young-Laplace equation given above.

## Balance of Forces at the Contact Point

Since no liquid can exist in a perfect vacuum, the surface of any liquid is an interface between that liquid and some other medium. The top surface of a pond, for example, is an interface between the pond water and the air. Surface tension, then, is not a property of the liquid alone, but a property of the liquid's interface with another medium. If a liquid is in a container, then besides the liquid/air interface at its top surface, there is also an interface between the liquid and the walls of the container. The surface tension between the liquid and air is usually different (greater than) its surface tension with the walls of a container. And where the two surfaces meet, their geometry must be such that all forces balance.
Where the two surfaces meet, they form a contact angle, $\scriptstyle \theta$. The diagram to the right shows two examples. The example on the left is where the liquid/solid surface tension, $\scriptstyle \gamma _{\mathrm {ls} }$, is less than the liquid/air surface tension, $\scriptstyle \gamma _{\mathrm {la} }$, but is nevertheless positive, that is
    $\gamma _{\mathrm {la} }\ >\ \gamma _{\mathrm {ls} }\ >\ 0$
In the diagram, both the vertical and horizontal forces must cancel exactly at the contact point. The horizontal component of $\scriptstyle f_{\mathrm {la} }$ is canceled by the adhesive force, $\scriptstyle f_{\mathrm {A} }$.
    $f_{\mathrm {A} }\ =\ f_{\mathrm {la} }\sin \theta$
The more telling balance of forces, though, is in the vertical direction. The vertical component of $\scriptstyle f_{\mathrm {la} }$ must exactly cancel the force, $\scriptstyle f_{\mathrm {ls} }$.
    $f_{\mathrm {ls} }\ =\ -f_{\mathrm {la} }\cos \theta$
Since the forces are in direct proportion to their respective surface tensions, we also have
    $\gamma _{\mathrm {ls} }\ =\ -\gamma _{\mathrm {la} }\cos \theta$
This means that although the liquid/solid surface tension, $\scriptstyle \gamma _{\mathrm {ls} }$, is difficult to measure directly, it can be inferred from the easily measured contact angle, $\scriptstyle \theta$, if the liquid/air surface tension, $\scriptstyle \gamma _{\mathrm {la} }$, is known.
This same relationship exists in the diagram on the right. But in this case we see that because the contact angle is less than 90°, the liquid/solid surface tension must be negative:
    $\gamma _{\mathrm {la} }\ >\ 0\ >\ \gamma _{\mathrm {ls} }$
Observe that in the special case of a water/silver interface where the contact angle is equal to 90°, the liquid/solid surface tension is exactly zero. Another special case is where the contact angle is exactly 180°. Water with specially prepared Teflon® is an example of this. Contact angle of 180° occurs when the liquid/solid surface tension is exactly equal to the liquid/air surface tension.
    $\gamma _{\mathrm {la} }\ =\ \gamma _{\mathrm {ls} }\ >\ 0\qquad \theta \ =\ 180^{\circ }$

## Bubble Physics

Using the analysis of surface tension forces as detailed in previous sections, it is possible to understand the behavior of soap bubbles. The diagram shows a soap bubble cut in half along its equator. By cutting along the equator, the crossectional plane is normal to the bubble's surface. Surface tension forces act parallel to the bubble's surface, as shown by the arrows in the diagram. The total surface tension force is equal to the surface tension, $\scriptstyle \gamma$, times the circumference:
    $f_{st}\ =\ \gamma 2\pi R$
where $\scriptstyle R$ is the radius of the bubble. The upward component of the force due to internal pressure of the bubble must exactly cancel $\scriptstyle f_{st}$. The upward pressure force is equal to the pressure difference (between interior and exerior), $\scriptstyle \Delta P$, times the crossectional area:
    $f_{P}\ =\ \Delta P\pi R^{2}$
Setting the two forces equal to each other and solving for $\scriptstyle \Delta P$, yields
    $\Delta P\ =\ {\frac {2\gamma }{R}}$
which is consistent with the Young-Laplace equation.
Anybody who has played soap bubble solution has seen the phenomenon of two bubbles fused together. The same method used to compute contact angle in the previous section also solves the for the angle where two bubbles are joined. The diagram shows three forces meeting at the joining point. Since all three surface tensions are equal, it follows that all three forces are equal. If $\scriptstyle f$ is that force and $\scriptstyle \theta$ is the angle between the downward force and either of the two diagonal forces, then, balancing the downward force with the vertical components of the two upward forces:
    $f+f\cos \theta +f\cos \theta \ =\ 0$
Dividing out $\scriptstyle f$, and dividing by two:
    ${\frac {1}{2}}\ =\ -\cos \theta$
or $\scriptstyle \theta \ =\ 120^{\circ }$. Even if the bubbles are of unequal size the fact remains that the three forces must be equal, so no matter what the bubble sizes are, the angle between the common surface and either of the individual surfaces must be 120°.
If the two bubbles are the same size, then the pressures on either side of the common surface are equal. In that case the Young-Laplace equation predicts that the radius of curveture of the common surface is infinite -- in other words, the common surface is flat. If the bubble sizes are unequal, then the pressures on either side of the common surface are also unequal, with the smaller bubble having the higher pressure. In this case, $\scriptstyle \Delta P$ is that pressure difference, and the Young-Laplace equation predicts that the common surface will curve outward from the smaller bubble and into the larger bubble.

## Minimization of Potential Energy

By analyzing the amount of work it takes to increase the size of a surface under tension, it is possible to see the equivalence between surface tension as force per unit length and as energy per unit area.
A rectangular film of soap solution with dimensions, $\scriptstyle L$ by $\scriptstyle W$ exerts an inward of force of $\scriptstyle \gamma L$ on each of the $\scriptstyle L$ length sides and $\scriptstyle \gamma W$ on each of the $\scriptstyle W$ length sides. So to increase the length of the rectangle by an amount, $\scriptstyle \Delta L$, it is necessary to pull against the force, $\scriptstyle \gamma W$ over a distance of $\scriptstyle \Delta L$. The total work done is the product of these two, $\scriptstyle \Delta U\ =\ \gamma W\Delta L$. The increase in area of the film is $\scriptstyle \Delta A\ =\ W\Delta L$. Hence the work done to increase the area of the film is $\scriptstyle \Delta U\ =\ \gamma \Delta A$. From this we can see that surface tension, $\scriptstyle \gamma$, is not only force per unit length, it is also work per unit area.
Note that the work done on the surface is reversible. That is, if you then shrink the area of the film by the same $\scriptstyle \Delta A$, the film gives back the exact amount of work it took to expand it in the first place. This means that when we stretch a surface, increasing its area, the work that it takes to do this is stored in the surface as potential energy.
Nature tries to find configurations that have minimal potential energy. An example of this is a weight hanging from a spring. Stretching the spring downward increases its potential energy. Lowering the weight decreases its potential energy. If left to its own devices, the weight-spring system will come to rest so that the sum of the potential energy of the weight and that of the spring is minimal. The same thing happens with surface tension. Clearly the potential energy of a liquid's surface is minimized when its surface area is minimized. But the liquid also has mass, and under the influence gravity, that mass finds its minimal potential energy when its center of mass is lowest. By finding a compromise between surface energy and other contributions to potential energy, it is possible to solve for the behavior of liquids in a container or on a horizontal surface. In the following sections we solve for minimal potential energy to discover some of the effects of surface tension.

## Behavior of Pools of Liquid
### Developement of the Thickness Formula

The Pool of Liquid on a Nonadhesive Surface section of the main article explains that a liquid on a non-adhesive horizontal surface will spread out only so far before its spread is limited by surface tension. We can find out just how far it can spread simply by applying the minimization of potential energy principle.
The simplest case is where the surface tension where the liquid is in contact with the surface is the same as the surface area elsewhere (contact angle = 180°). If the pool is big, we are justified in ignoring the area of the edges of the pool and take into account just the top and bottom side area. Experience also tells us that if the pool is big enough, its depth is more or less constant (except very near the edges).
If $\scriptstyle V$ is the volume of liquid in the pool and $\scriptstyle h$ is the nearly constant depth, then the top and bottom areas are each approximated very well by $\scriptstyle {\frac {V}{h}}$, for a total area of $\scriptstyle 2{\frac {V}{h}}$. We multiply that by the surface tension, $\scriptstyle \gamma$, to get the surface energy.
The center of mass of the pool will be at a height of $\scriptstyle {\frac {h}{2}}$ above the bottom of the pool. The total mass of the pool is $\scriptstyle \rho V$ (where $\scriptstyle \rho$ is the liquid's mass-density). So the gravitational potential energy is of the entire pool is $\scriptstyle {\frac {1}{2}}g\rho Vh$.
The total potential energy, $\scriptstyle U$, then is
    $U(h)\ =\ 2\gamma {\frac {V}{h}}\ +\ {\frac {1}{2}}g\rho Vh$
To minimize $\scriptstyle U(h)$, we solve for its derivative, $\scriptstyle {\frac {dU}{dh}}\ =\ 0$.
    ${\frac {dU}{dh}}\ =\ 0\ =\ -2\gamma {\frac {V}{h^{2}}}\ +\ {\frac {1}{2}}g\rho V$
Dividing out $\scriptstyle V$ and solving for $\scriptstyle h$ gives:
    $h\ =\ 2{\sqrt {\frac {\gamma }{g\rho }}}$
Doing this same calculation with differing surface tensions, $\scriptstyle \gamma _{\mathrm {la} }$ and $\scriptstyle \gamma _{\mathrm {ls} }$ (that is for contact angle < 180°), for the liquid/air and liquid/solid interfaces (that is for top surface and bottom surface respectively), you get:
    $h\ =\ {\sqrt {\frac {2\left(\gamma _{\mathrm {la} }+\gamma _{\mathrm {ls} }\right)}{g\rho }}}$
and substituting the contact angle formula into the above:
    $h\ =\ {\sqrt {\frac {2\gamma _{\mathrm {la} }\left(1-\cos {\theta }\right)}{g\rho }}}$

### Shape of the Edge of a Pool
#### Author's Notes

[The version below of this section is not even a draft -- it's just a copy of the author's notes]
* $\gamma {\frac {-{\frac {d^{2}x}{dh^{2}}}}{\left(1+\left({\frac {dx}{dh}}\right)^{2}\right)^{\frac {3}{2}}}}\ =\ g\rho \left(H-h\right)$$\mathrm {Integrating} \ -\gamma {\frac {\frac {dx}{dh}}{\sqrt {1+\left({\frac {dx}{dh}}\right)^{2}}}}\ =\ g\rho \left(Hh-{\frac {1}{2}}h^{2}\right)+C$ $\mathrm {Thickness\ equation} \ H\ =\ 2{\sqrt {\frac {\gamma }{g\rho }}}$
* $-\gamma {\frac {\frac {dx}{dh}}{\sqrt {1+\left({\frac {dx}{dh}}\right)^{2}}}}\ =\ g\rho \left(2h{\sqrt {\frac {\gamma }{g\rho }}}-{\frac {1}{2}}h^{2}\right)+C$
* $\mathrm {Solving\ boundary\ condition:}$
* $-\gamma {\frac {\frac {dx}{dh}}{\sqrt {1+\left({\frac {dx}{dh}}\right)^{2}}}}\ =\ g\rho \left(2h{\sqrt {\frac {\gamma }{g\rho }}}-{\frac {1}{2}}h^{2}\right)-\gamma$
* $\mathrm {Dimensionless} \ {\frac {-{\frac {dx}{dh}}}{\sqrt {1+\left({\frac {dx}{dh}}\right)^{2}}}}\ =\ 2h{\sqrt {\frac {g\rho }{\gamma }}}-{\frac {1}{2}}{\frac {g\rho h^{2}}{\gamma }}-1$
* $\mathrm {In} \ H:\ {\frac {-{\frac {dx}{dh}}}{\sqrt {1+\left({\frac {dx}{dh}}\right)^{2}}}}\ =\ 4{\frac {h}{H}}-2\left({\frac {h}{H}}\right)^{2}-1$
* $\mathrm {Normalized:} \ {\frac {\frac {dx}{dy}}{\sqrt {1+\left({\frac {dx}{dy}}\right)^{2}}}}\ =\ 2y^{2}-4y+1$
* $\mathrm {Isolate} \ {\frac {dx}{dy}}\ =\ {\frac {2y^{2}-4y+1}{\sqrt {1-\left(2y^{2}-4y+1\right)^{2}}}}$
* $\mathrm {Factoring} \ {\frac {dx}{dy}}\ =\ \pm {\frac {2y^{2}-4y+1}{2\left(y-1\right){\sqrt {-y(y-2)}}}}$
* $\mathrm {Subst} \ {\frac {dx}{du}}\ =\ \pm {\frac {2\left(u+1\right)^{2}-4\left(u+1\right)+1}{2u{\sqrt {1-u^{2}}}}}$
* $\mathrm {Simplify} \ {\frac {dx}{du}}\ =\ \pm {\frac {2u^{2}-1}{2u{\sqrt {1-u^{2}}}}}\ =\ \pm \left({\frac {\sqrt {1-u^{2}}}{u}}+{\frac {1}{2u{\sqrt {1-u^{2}}}}}\right)$
* $=\ \pm \left({\frac {u}{\sqrt {1-u^{2}}}}-{\frac {1}{2u{\sqrt {1-u^{2}}}}}\right)$
* $\mathrm {Integrating:} \$
* $x\ =\ \pm \left(-{\sqrt {1-u^{2}}}\ -\ {\frac {1}{2}}\ln {|u|}+{\frac {1}{2}}\ln {\left({\sqrt {1-u^{2}}}+1\right)}\right)+C$
* $\mathrm {Combining\ logs:} \$
* $x\ =\ \pm \left(-{\sqrt {1-u^{2}}}\ +{\frac {1}{2}}\ln {\left({\sqrt {{\frac {1}{u^{2}}}-1}}+{\frac {1}{u}}\right)}\right)+C=$
* $x\ =\ \pm \left(-{\sqrt {1-u^{2}}}\ +{\frac {1}{2}}\cosh ^{-1}{\left({\frac {1}{u}}\right)}\right)+C$
* Note: The last equation agrees with Pierre-Gilles de Gennes; Françoise Brochard-Wyart; David Quéré (2002). Capillary and Wetting Phenomena -- Drops, Bubbles, Pearls, Waves. Springer. ISBN 0-387-00592-7.  eq. 2.21.
$\scriptstyle h$ is the height, which is the independent variable. $\scriptstyle x$ is the horizontal position of the pool's edge, which is the dependent variable. $\scriptstyle H$ is the max thickness of the pool. First equation has curveture on the left and pressure on the right as a function of $\scriptstyle h$. We integrate curveture to get an equation for the slope of the curve. We replace $\scriptstyle H$ with the expression for the thickness. We know that at $\scriptstyle h\ =\ 0$ and at $\scriptstyle h\ =\ H$ that $\scriptstyle {\frac {dx}{dh}}\ =\ \pm \infty$ (respectively). In the limit, the left side is equal to $\scriptstyle -\gamma$ for $\scriptstyle h\ =\ 0$. We find that $\scriptstyle C\ =\ -\gamma$ solves this boundary condition. Near the other boundary of $\scriptstyle h\ =\ H$ we know that $\scriptstyle {\frac {dx}{dh}}\ \mathrm {goes\ to} \ -\infty$, and we find that $\scriptstyle C\ =\ -\gamma$ also solves this boundary condition.
Solving for $\scriptstyle {\frac {dx}{dh}}\ =\ 0$, we get $\scriptstyle h\ =\ H\left(1-{\frac {\sqrt {2}}{2}}\right)$
Normalizing $\scriptstyle x$ to units of $\scriptstyle H$ and letting $\scriptstyle y={\frac {h}{H}}$, we get a normalized equation from which it is easy to isolate $\scriptstyle {\frac {dx}{dy}}$ to one side. Amazingly it is possible to integrate the right-hand side without resorting to special functions. This is because the polynomial, $\scriptstyle P(y)\ =\ 2y^{2}-4y+1$ is equal to $\scriptstyle 1$ at $\scriptstyle y\ =\ 1$, where $\scriptstyle P$ also has a critical point. Hence its square also is equal to $\scriptstyle 1$ and has a critical point here too. This means that the expression under the radical has a double root at $\scriptstyle y\ =\ 1$, which we can divide out of the radical although whether we are dividing out $\scriptstyle y-1$ or $\scriptstyle 1-y$ remains unclear at this point). Making an elementary substitution of $\scriptstyle u\ =\ y-1$, and simplifying, it can be integrated using methods taught in 1st year calculus. Two equivalent forms of the final integrand are shown. Both have poles, as expected, at $\scriptstyle u\ =\ 0$ (equivalent to $\scriptstyle h\ =\ H$, which is the limiting value at the top surface) and at $\scriptstyle u\ =\ -1$ (equivalent to $\scriptstyle h\ =\ 0$, which is the limiting value at the bottom surface). The final integral of this is shown on the last line of the calculations, and has the correct characteristics expected of the pool-edge curve.

## The break up of streams of liquid into drops

In day to day life we all observe that a stream of water emerging from a faucet will break up into droplets, no matter how smoothly the stream is emitted from the faucet. This is due to a phenomenon called the Plateau-Rayleigh instability.
The explanation of this instability begins with the existence of tiny perturbations in the stream. These are always present, no matter how smooth the stream is. If the perturbations are resolved into sinusoidal components, we find that some components grow with time while others decay with time. Among those that grow with time, some grow at faster rates than others. Whether a component decays or grows, and how fast it grows is entirely a function of its wave number (how many peaks and troughs per centimeter) and the radius of the cylindrical stream. The diagram to the right shows an exaggeration of a single component.
By assuming that all possible components exist initially in roughly equal (but minuscule) amplitudes, the size of the final drops can be predicted by determining by wave number which component grows the fastest. As time progresses, it is the component whose growth rate is maximum that will come to dominate and will eventually be the one that pinches the stream into drops.
Although a thorough understanding of how this happens requires a mathematical development (see references), the diagram can provide a conceptual understanding. Observe the two bands shown girdling the stream -- one at a peak and the other at a trough of the wave. At the trough, the radius of the stream is smaller, hence according to the Young-Laplace equation (discussed above) the pressure due to surface tension is increased. Likewise at the peak the radius of the stream is greater and by the same reasoning pressure due to surface tension is reduced. If this were the only effect, we would expect that the higher pressure in the trough would squeeze liquid into the lower pressure region in the peak. In this way we see how the wave grows in amplitude over time.
But the Young-Laplace equation is influenced by two separate radius components. In this case one is the radius, already discussed of the stream itself. The other is the radius of curvature of the wave itself. The fitted arcs in the diagram show these at a peak and at a trough. Observe that the radius of curvature at the trough is, in fact, negative, meaning that, according to Young-Laplace, it actually decreases the pressure in the trough. Likewise the radius of curvature at the peak is positive and increase the pressure in that region. The effect of these components is exactly opposite the effects of the radius of the stream itself.
The two effects, in general, do not cancel. One of them will have greater magnitude than the other, depending upon wave number and the initial radius of the stream. When the wave number is such that the radius of curvature of the wave dominates the that of the radius of the stream, such components will decay over time. When the effect of the radius of the stream dominates that of the curvature of the wave, such components grow exponentially with time.
When all the math is done, it is found that unstable components are only those where the product of the wave number with the initial radius is less than unity ($\scriptstyle kR_{0}\ <\ 1$). The component that grows the fastest is the one whose wave number satisfies the equation:
       $kR_{0}\ \simeq \ 0.697$

## notes to be incorporated into article
### Surface Energy of Vaporization

Heat of vaporization of water is about 2260 J/gram, or equivalently 2.26×10$^{10}$ erg/gram. Dividing this by surface tension at 100°C of 75.9 erg/cm$^2$ gives an area of 2.98×10$^8$ cm$^2$ per gram. That is, if we increase the surface energy of one gram of water by the amount of energy it would take to boil that water, the surface area required for that energy becomes 2.98×10$^8$ cm$^2$. To have one gram of water divided into equal-sized spheres whose area adds up to this requires the radius of those spheres to be 1.01×10$^{–8}$ cm. The volume of each sphere is 4.27×10$^{–24}$ cm$^3$. Since the density of water is about 1 gram per cm$^3$, this is also the mass in grams of such a sphere when it is filled with liquid water. Dividing by 18 grams per mole of water and multiplying by Avagadro's number of 6.0×10$^{23}$ molecules per mole, this amounts to about 0.14 molecules per sphere. We see that the energy to boil a gram of water is within an order of magnitude of the surface energy of the individual molecules. Indeed the difference can easily be accounted for by observing that water molecules are not spherical, and therefore have a greater surface area to volume ratio than if they were. If you allow water molecules to have double the surface area to volume ratio of a sphere, then the same calculation gives 1.1 molecules per fragment of water.

## Surface tension and thermodynamics

[copied from User:Knights who say ni's sandbox]

#### Thermodynamic Definition

As stated above, the mechanical work needed to increase a surface is $\scriptstyle dW=\gamma dA$. For a reversible process, $\scriptstyle dG=VdP+\gamma dA-SdT$, therefore at constant temperature and pressure, surface tension equals Gibbs free energy per surface area:
$\gamma =\left({\frac {\partial G}{\partial A}}\right)_{P,T}$, where $\scriptstyle G$ is Gibbs free energy and $\scriptstyle A$ is the area.
From the definition it is easy to understand that decreasing the surface area of a substance is always spontaneous (ΔG<0), on the contrary, in order to increase its surface a certain amount of energy is needed, as the process is, per se, non spontaneous (ΔG>0). A measure of how spontaneous (or non-spontaneous) is the change in the surface area is precisely the surface tension.
The definition of Gibbs free energy can be arranged to $\scriptstyle H=G+TS$ so partial derivation yields $\scriptstyle \left({\frac {\partial H}{\partial A}}\right)_{P}=\left({\frac {\partial G}{\partial A}}\right)_{P}+T\left({\frac {\partial S}{\partial A}}\right)_{P}$ and applying the equations of state $\scriptstyle \left({\frac {\partial G}{\partial T}}\right)_{A,P}=-S$ we obtain $\scriptstyle \left({\frac {\partial \gamma }{\partial T}}\right)_{A,P}=-S^{A}$, where $\scriptstyle S^{A}$ means entropy per surface area.
Rearranging the previous expression Kelvin Equation I is obtained. It states that surface enthalpy or surface energy (different from surface free energy) depends both on surface tension and its derivative with temperature at constant pressure $H^{A}\ =\ \gamma -T\left({\frac {\partial \gamma }{\partial T}}\right)_{P}$

#### Influence of temperature on surface tension

Surface tension dependes on temperature, for that reason, when a value is given for the surface tension of an interface, temperature must be explicitely stated. The general trend is that surface tension decreases with temperature, reaching a value of 0 at the critical temperature. There are only empirical equations to relate surface tension and temperature:
* Eötvös:
       $\gamma V^{2/3}=k(T_{C}-T)\,\!$
$\scriptstyle V$ is the molar volume of that substance
$\scriptstyle T_{C}$ is the critical temperature
$\scriptstyle k$ is a constant for each substance.
For example for water k = 1.03 erg/°C, V = 18 ml/mol and T$_C$ = 374°C.
* Guggenheim-Katayama:
       $\gamma =\gamma ^{o}\left(1-{\frac {T}{T_{C}}}\right)^{n}$
$\scriptstyle \gamma ^{o}$ is a constant for each liquid and n is an empirical factor, whose value is 11/9 for organic liquids
Both take into account the fact that surface tension reaches 0 at the critical temperature.

#### Influence of solute concentration on surface tension

Solutes can have different effects on surface tension depending on their structure:
* No effect, for example sugar
* Increase of surface tension, inorganic salts
* Decrease surface tension progressively, alcohols
* Decrease surface tension and, once a minimum is reached, no more effect: surfactants
Gibbs isotherm states that $\Gamma \ =\ -{\frac {1}{RT}}\left({\frac {\partial \gamma }{\partial \ln C}}\right)_{T,P}$
$\scriptstyle \Gamma$ is known as surface concentration, it represents excess of solute per unit area of the surface over what would be present if the bulk concentration prevailed all the way to the surface. It has units of mol/m$^2$
$\scriptstyle C$ is the concentration of the substance in the bulk solution.
$\scriptstyle R$ is the gas constant and $\scriptstyle T$ the temperature
Certain assumptions are taken in its deduction, therefore Gibbs isotherm can only be applied to ideal (very dilute) solutions with two components.

#### Pressure jump across a curved surface

If viscous forces are absent, the pressure jump across a curved surface is given by the Young-Laplace Equation, which relates pressure inside a liquid with the pressure outside it, the surface tension and the geometry of the surface.
       $\Delta P=\gamma {\frac {dA}{dV}}$.
This equation can be applied to any surface:
* For a flat surface $\scriptstyle {\frac {dA}{dV}}\ =\ 0$ so the pressure inside is the same as the pressure outside.
* For a spherical surface $\scriptstyle P_{I}\ =\ P_{O}+{\frac {2\gamma }{R}}$
* For a toroidal surface $\scriptstyle P_{I}\ =\ P_{O}+\gamma \left({\frac {1}{R}}+{\frac {1}{r}}\right)$, where r and R are the radii of the toroid.

#### Influence of particle size on vapour pressure

Starting from Clausius-Clapeyron relation Kelvin Equation II can be obtained; it explains that because of surface tension, vapor pressure for small droplets of liquid in suspension is greater than standard vapor pressure of that same liquid when the interface is flat. That is to say that when a liquid is forming small droplets, the concentration of vapor of that liquid in the surroundings is greater, this is due to the fact that the pressure inside the droplet is greater than outside.
       $P_{v}^{fog}=P_{v}^{o}e^{\frac {V2\gamma }{RTr}}$
$\scriptstyle P_{v}^{o}$ is the standard vapor pressure for that liquid at that temperature and pressure.
$\scriptstyle V$ is the molar volume.
$\scriptstyle R$ is the gas constant
This equation is used in chemistry to estimate the pore size for solids.