OK this is the second of our videos about surface tension. In the first part we introduced surface tension and defined it in terms of
the energy or work associated with increasing the interfacial area between two fluids.
So here we're going to take that definition, start with it, and then apply it in a way that can let us look more deeply into the physics.
And obtain an equation, a relationship, called the Young-Laplace equation
which is useful for looking at surface tension or capillary effects of fluids.
So again the starting point is the definition of surface tension, which is the work or energy associated with
a change in interfacial area, an increase in interfacial area.
And we're going to apply it to an infinitesimal or a differential area element dA.
And so when we do that
the work or the energy associated with this change is equal to the surface tension coefficient times the differential area dA.
So we can look at the units of this to see that it makes sense. Work has units of Joules, or Newton-meters.
Surface tension, we said, has units of Newtons per meter. And area has units of meters squared. So this equation
makes sense dimensionally, which is always good to know.
So now we're going to apply this to a specific case. I'm going to look at a geometry here.
Let's consider a situation where we have
the work associated with increasing
a surface that we can describe in terms of two different radii of curvature, R1 and R2.
This is a generalized curved surface so let me draw what I'm talking about.
So I have two area elements on this curved surface. The green one is the initial area.
And the purple one is the expanded one. You can think of this as a balloon, for example.
If it's inflated to some amount, and then we draw square on the surface with a marker.
Then if we inflate the balloon a little bit more, the surface area is going to expand to a larger size, to this the purple element.
What's drawn here. And so I'm talking about how we can express this shape. And I can do that in terms of these two radii of curvature.
So R1 represents the curvature of the left and right hand sides of these curved area elements.
And R2 represents the curvature associated with the top and bottom sides. Okay, so these two
radii of curvature allow me to define this area element in space on a curved surface.
So now I know that I want to represent
this geometry. So since this is a differential element.
I can say that it's at some height x and some width y initially.
And these two elements are separated by some differential distance dz. So then in the deformed shape we go from x
to a height of x + dx. And the width increases from y to y + dy.
So this is the initial state and then this is the deformed state. So now with these definitions,
I can go in and describe the change of area associated with this deformation.
So this is the difference between the final and initial states. And so I know what those are. The final area is x + dx times y + dy.
and the initial area x times y. These are small shapes so I can approximate them as just length times width.
So then if I go through and multiply through the terms in this first equation,
and expand them out, I can see a couple of things. First is that this term, the product xy, cancels out.
So I can cross that out. And then I have this term dxdy. So I'm going to say that I can also neglect that term.
Why is that? Well it's what I'm calling a second order term.
So if we remember that we're talking about differential elements.
The length scales dx and dy are already small. They are infinitesimally small so those terms are already small. And the product
between two small numbers is something that's much smaller. So compared to all these
other terms in the equation, the product dxdy is much smaller.
So I can neglect it compared to these other two terms. So that allows me to simplify
the expression for the area. So I can plug it back in into our relationship up here.
So the work associated with this change in interfacial area is
the surface tension coefficient gamma times dA. Which is xdy + ydx.
OK so let me just rewrite that here on the next slide so that we can refer to it.
Now, because we're talking about a differential change, a differential displacement
the work associated with this change in interfacial area
can be associated with a corresponding change in pressure.
And so then I can
express the work done as force times displacement.
So I can map that to this pressure change. So here's what I mean. The work associated with
this change in interfacial area is equal to force times distance, force times displacement. Force is this pressure, which is
force per unit area times the area. So force times distance
associated with this displacement
from the initial to the final surface area.
So let me draw again, I'm just reproducing the drawing from the previous slide here so that we can refer to it and continue to refer to it.
As we continue. OK so now what we want to do is try to simplify our relationship because, specifically, we have
dx, dy, and dz. So we'd like to try to
reduce the number of variables that we have in the equation. We can do that
by comparing the geometry of these initial and deformed states and noting that they're similar, geometrically similar.
OK so let me show you what I'm talking about here.
So I can look at this
face of width x on a curved surface and I can see that it spans some interior angle here at a radius R1.
Then if I look at the corresponding face on the deformed surface of arc length x + dx,
I notice that it spans the same interior angle. So if you remember from your
trigonometry that the angle is equal to the ratio of the arc length to the radius. And so since they both span the same interior angle,
the ratio of these two arc lengths x and x+dx to the corresponding radii R1 and
on this side R1 + dZ, should be the same because they are geometrically equivalent.
So I can write that down here, x + dx
the width of this face over the radius
the ratio of those two quantities should be the same as x / R1.
Because, the reason being that they span the same interior angle.
So then with that I can solve for dx. I get dx is (x / R1) times (r1 + dz) - x. So now if I multiply through all these terms.
So I have x times R1 over R1 which is just one plus x times dz over R1 minus x. so the x's cancel.
And then I get finally an expression for dx in terms of dz.
So I did that with the face here that corresponds to the radius of curvature R1. So you can imagine that similarly
I can do the same analysis on the other faces that correspond to the radius of curvature R2.
And I arrive at a corresponding relationship that expresses dy in terms of dz.
So now I can return to the expression for the work
in terms of the surface tension
and the pressure times displacement
and I can now substitute in for
the corresponding terms in the area.
So the change in area we saw was x dy + y dx. Then I can substitute in directly from the expressions that I found from geometry
that expressed dy and dx in terms of dz. And this will allow me to simplify the equation further.
OK so let me again just rewrite that here at the top of this slide.
So if I solve for for delta P then I can substitute for dy and dx in terms of the relationship that we derived in the previous
slide in terms of dz divided by xy dz. Then if I multiply through, I have xy dz times 1 / R2
plus xy dz times 1 / R1
over xy dz. So notice that each of these terms has the quantity xy dz in common. So this will cancel out.
And so when I do this I get a much simpler equation. I get that the pressure drop
across the interface is related to the surface tension
plus the sum of the inverse of the radii of curvature. So this is the Young-Laplace equation.
And it comes basically just for a geometric argument associated with looking at the change in the surface area of this interface.
So far a sphere. We looked at before, a general curved shape
that could be expressed in terms of two independent radii of curvature.
For a spherical drop, if the shape is a sphere then there's one radius curvature. So R1 equals R2, and the equation
simplifies here. This is the the form of the Young-Laplace equation for a sphere.
So one of the things to notice here is that there's an inverse proportionality to the radius R. So as we get to small drops,
or small radii of curvature, this pressure drop across the interface becomes very large.
And so for that reason these surface tension forces, again like we said in the last slide,
become important, very important, at smaller and smaller length scales. So very small drops,
less then a micron or so, are extremely unstable for this reason.