Surface Tension and Capillary Action

Surface tension and capillary action

What is capillary action?

Capillary action, the result of a pressure differential at the surface of a liquid in a slender tube, is the rise or fall of the liquid in the tube above or below the original liquid line of the container into which the tube is inserted.

Preliminaries:

1. Density = Mass/Volume; Force = Mass*Acceleration (N); Pressure = Force/Area (Pa);

    Work = Force*distance (J)

1. Newton (N) = kg-m/sec²; Pascal (Pa) = N/m²; 1 atm= 101.3 kPa = 760 mmHg = 760 torr; 1 liter (L) = 0.001 m³ = 1 (dm)³ = 1000 cc; Joule (J) = Nm; the joule is also the unit of Energy (the capacity to do work)
2. In general, if a liquid has density D and the pressure at the surface is P, the pressure at depth H is P+ gDH where g is the acceleration of gravity. Let’s see why. If we consider a cylinder of the liquid of height H, radius R and top at the surface then the mass of the column is M = D*(π*R²*H) and the downward force due to this mass is, Mg =  D*(π* R²*H)*g. Therefore, the additional pressure at the bottom of the cylinder due to this column of liquid is the force per area or g* D*(π*R²*H)/π*R² = gDH. Thus the total pressure at depth H is P + gDH. For example, water has density 1.0 kg/L, the “deep water” additional pressure at depth H is gH where g is 9.8 m/sec² and H is measured in meters. That is, 9.8 m/sec²*1.0 kg/L*H m = 9.8*H kPa. So at 10m below the surface of a pool the pressure is nearly twice the standard atmospheric pressure (1 atm). If a tube is inserted 5 cm into a container of water, and the container is in a room with 101.3 kPa pressure, the total pressure at the base of the tube is 101.3 + 9.8(.05) kPa = 101.8 kPa.
3. When we look at a pressure difference on either side of an interface between two fluids we measure it in the direction opposite to the outward normal (the normal pointing away from the center of curvature). That is, we measure it in the direction towards the center of curvature. In this way we will always have ∆P > 0.

Given a sufficiently thin tube, whether the liquid in the tube rises above or falls below the original liquid line of the container depends on the relative strengths of two forces: cohesion between the molecules of the liquid and adhesion of the liquid molecules to the walls of the tube. For example,  water in a glass tube is a case in which adhesion (water to glass) is stronger than cohesion (water molecules bonded by hydrogen bonds), whereas liquid mercury is a case where cohesion is stronger than adhesion. We will get back to this after explaining surface tension.

Surface tension is a phenomenon that takes place at the interface of two substances. 

For our discussion we will assume an interface between a liquid and a vapor although the arguments can apply to any two immiscible liquids.

There are several ways to explain where surface tension comes from and what it is. Because of this there can be confusion about how surface tension acts. In fact, the way it acts can be described in different ways. There is a very good explanation of surface tension at: doc.utwente.nl/79082/1/why_is_surface.pdf

In particular, this paper fills in the gap that most explanations leave as to why the surface tension acts parallel to the surface of the interface.

The difficulty arises because most descriptions of surface tension point to one phenomenon, the difference in the number of bonds that exist for bulk molecules versus interface molecules due to the asymmetry at the surface, as the origin of surface tension. The difference in molecular density on the two sides of the interface and the subsequent reduction in bonds established with molecules above the surface by the surface molecules, leads to increased free energy in the interface molecules which means that the system tends to minimize the number of such higher energy surface molecules. That is, the system tends to minimize its surface area. The work needed to create new surface area is then defined as surface tension. The units of work (energy) per unit area are equivalent to force per unit length and so this is also used to define surface tension. But from this description it is not obvious why the force that arises from surface tension is parallel to the surface. 

In the paper cited above, the direction of the force that arises from surface tension is explained in terms of two forces acting on the molecules near the surface (within a few molecular diameters). Not only the attractions used to explain the difference in free energy in the surface molecules, but also an isotropic close-range repulsive force that acts as a counterbalance to the atrractive forces. Although best described in the detail of the paper cited, here we will summarize the argument.

The repulsive force is a very close-range force that arises only when molecules are near enough to each other that their electron clouds repel each other (according to the Pauli exclusion principle which says that no two electrons can occupy the same quantum state, that is, have the same four electronic quantum numbers, in the same atom). This repulsion is isotropic - the same in all directions.

The attractive force between molecules are relatively longer range and are anisotropic - not necessarily the same in all directions. In particular, at the surface, where there is an imbalance in molecules above and below the surface, the net attraction on a surface molecule will be inward towards the liquid.

A thought experiment can explain the interaction and the resultant net force from these forces. Consider imaginary surfaces of length w, at depths d1, just below the surface, and d2 a few molecules deeper. Now consider the two subsystems, ad1, the liquid above d1, and bd1, the liquid below d1. Similarly we can consider the subsystems ad2 and bd2, the bodies of liquid above and below d2 respectively. Now consider the overall attraction and repulsion of ad1 on bd1 and ad2 on bd2 across the width w. Since our system is in equilibrium we must have attraction equal to repulsion at each depth. But the attraction of ad1 is certainly less than the attraction of ad2 because there are fewer attracting molecules in ad1. Since attraction equals repulsion at each level we must have that the repulsion is also less at d1 than at d2. Let a1, r1, a2, r2 be the attraction and repulsion forces of subsystems ad1 on bd1 and ad2 on bd2 respectively. We have a1=r1 < a2=r2. Note the overall forces are proportional to w. As we go deeper into the liquid we soon arrive at the state where all forces are perfectly balanced as they are in the bulk of the liquid.

Next we consider two subsystems called left-hand-side (lhs) and right-hand-side (rhs) where the liquid is divided into two parts by a plane perpendicular to the surface (interface) and again of width w. Now let us consider the attraction and repulsion of lhs on rhs. Since repulsion is isotropic, the repulsive force is still r1 at d1 and r2 at d2 in this horizontal direction. But unlike the vertical case, the horizontal attractive forces are balanced and the same at every depth. (Note: since the horizontal forces are balanced for both repulsion and attraction at every depth (net zero) there is no requirement that they balance each other (as was the case in the vertical case). Thus at d1 we have a net attractive force which has diminished at d2 and will go to zero in the bulk. This is the key result which is missing from most discussions of surface tension and I am indebted to the team in the cited paper for this insight.

Taking the thought experiment a few steps further, we see that the line drawn on the surface to establish the lhs/rhs subsystems was arbitrary so that for any point P on the surface it is attracting in all directions. Therefore, for interior points on the surface the net attraction is zero. However, at the boundary of the surface there is a net attraction parallel (tangent) the surface and normal to the boundary curve (the non-normal components cancel and the normal components sum to the net). Again the overall attraction is proportional to the length of the boundary curve. Finally, if the surface is curved we make the same argument in the limit by considering the tangent plane at P as the approximation to the case discussed.

So we have the following result: surface tension is a force per unit length along the boundary of the interface which acts in a direction tangent to the surface and normal to the boundary. Thus, if we think of the interface as a sheet, the surface tension acts as a pulling on the sheet around the edges to make the sheet taut (i.e. introduces tension in the sheet). If an object is placed on the sheet the sheet can hold it as long as the upward force (vertical component of surface tension of sheet times length of edge) is greater than or equal to the weight of the object.

The surface tension acts parallel to the surface. Thus, where the surface is flat the surface tension is all horizontal and where the surface is curved the surface tension acts along the tangent to the curve so it has both a horizontal and a vertical component.

Example of surface tension at work: When the circular pod of radius r of a water-walking insect rests on the water, it presses down a bit on the surface creating a tangential angle z around the circumference where the water curves back to the surface. Since all other surface tension components are horizontal and mutually cancelling, the vertical component, T sin z, where T is the surface tension of the water, is the net force which is applied around the circumference for a total net force upwards of (T*sin z)*(2πr) = (vertical component of surface tension) * (distance around circumference). If this force is at least equal to the weight w of the insect then the insect will not break through the water surface. That is, we need the force (2πrTsin z) ≥ w for the insect to rest atop the water....not floating, but held atop by the surface tension of the water.

Another example of surface tension at work: A small quantity of water wants to be a sphere because that is the shape with the minimal surface area for a given volume (in this case the volume of the drop of water). That is, any deformation of the sphere would increase the surface area thus increasing the number of molecules in a higher energy state which doesn’t happen any more than a marble resting anywhere but at the bottom of a U-shaped trough as systems always try to go to the least energy state. [Note: When the volume increases gravity comes more into play in shaping the water, for example, the teardrop shape or such a blob.]

Another example of surface tension at work: The liquid/air interface surface in a tube is called the meniscus. In a sufficiently narrow tube the meniscus will be curved either in a convex or concave shape as seen from above. The surface tension in the liquid will try to minimize the surface area of the meniscus thereby shaping it into a portion of a sphere (that is a shape of constant curvature) just as in the last example.

Suppose we had a meniscus of radius of curvature R in a tube of radius r. Suppose that z is the angle the (tangent to the) curve makes with the downward tube at the air/liquid interface. Then we can show that cos z = ± r/R depending on z being less than or greater than 90º. To see this, let Q be a point on the meniscus/tube intersection and let O be the center of the sphere of which the meniscus is a part. Then the length of OQ is R. Let B be the intersection of a vertical line through O and the perpendicular to this line through Q. Let L be the vertical line through Q (the line of the tube). Let E be the intersection of the tangent line and the vertical line through O. We know that the tangent line is perpendicular to the radial line through Q. The the length of BQ is r and triangle ∆OBQ is a right triangle with hypotnuse R. First consider the case z < 90º. Angle z is the complement of angle BQE which in turn is the complement of angle BQO so angle BQO = z and cos z = r/R. Now consider the case z >90º. Let x be the angle between the radial line through Q and the line L. Then x = z-90º and x equals angle BOQ as opposite angles of a transversal through parallel lines L and the vertical through O. Thus sin x = r/R = sin (z-90º) = -sin(90º-z) = -cos z. So cos z = -r/R.

Cohesion and adhesion in a capillary system.

Case 1: Let’s first look at the case of water in the narrow tube (sufficiently small radius so adhesion/cohesion dominate gravitational effects on the water surface). When we submerge one end of the tube into the water container to a depth H we notice that the water in the tube rises above the container water level by a height of h and we notice that when we look down the tube from above the surface of the water in the tube is concave. If we use a photograph we could measure the angle z that the tangent to the curve makes with the vertical tube at the air/water interface (z < 90º). If we think about the water surface being higher around the tube than at the center it makes sense when adhesion is stronger than cohesion because what happens in the water is that molecules are attracted more to the outside, towards the tube, away from the center, so with less density near the center there is a pressure differential, ∆P, at the center between the air pressure above the surface, p1, and the pressure just below the surface, p2. Here ∆P = p1-p2 as in preliminaries 4. This difference in pressure pushes the water down while the adhesion acts to hold the water to the walls and together they cause the concave water surface as seen from the top. This differs from the situation where a horizontal interface exists where the pressure above and just below the surface are the same (∆P=0) until at deeper depths where the added pressure due to the weight of a column of water increases the pressure in the water as discussed in preliminaries 3.

  Suppose p1 is the ambient air pressure in the room. Then the pressure on the container surface water is p1 as is the pressure at the top of the water in the tube. Suppose we observe the water in the tube at height h above the container level. As in preliminaries 3, the pressure at the bottom of the tube is p1 + gdH pushing up on the water in the tube. But since the pressure, p2, in the water is less than p1, the stronger pressure p1 + gdH at the bottom of the tube pushes the column of water upwards until it reaches a height h above the original water level where now the downward pressure of the water in the tube, p2 + dgh + dgH matches the upward pressure. The increase of downward pressure in the tube comes from the pressure from the newly created h-high “capillary-effect” column of water in the tube. Thus, equilibrium is reached when p1 + gdH = p2 + dgh + dgH, or p1-p2 = ∆P = dgh, where ∆P is the change in pressure at the water/air interface in the tube in the direction towards the center of curvature of the surface.

Case 2: Now suppose cohesion dominates, so cohesion is greater than adhesion. Let’s use mercury as such a liquid. Molecules in the mercury are pulled together more than they are attracted to the tube wall so the density increases and the pressure in the mercury, p2, is greater than the atmospheric pressure, p1 at the interface. We note that the surface of the mercury is now lower around the tube than it is in the middle as the difference in pressure pushes up on the middle against the more limited adhesive attraction to the wall. Again, we could measure the angle z that the tangent to the surface curve makes with the vertical tube at the air/water interface (z > 90º). As in case 1, let H be the depth of the tube in the container of mercury, d the density of mercury and h the height needed for equilibrium. Just as before we must have p1 + gdH = p2 + gdH + gdh, or ∆P = p2-p1= -gdh. But now, since ∆P is positive, we must have h < 0. This means that the mercury column in the tube must go h units below the original mercury level in the container.

The analysis so far has been based on pressure and from this we derived the height h (positive or negative) that a liquid attains in a tube relative to the original liquid level in a container in terms of ∆P. We saw that h was positive when adhesion dominated cohesion and that in this case the meniscus was concave so the contact angle z satisfied z < 90º. We also saw that h was negative when cohesion dominated adhesion and that in this case the meniscus was convex so the contact angle z satisfied z > 90º. There is more to the story.

What is ∆P?  It turns out that we can actually compute ∆P as a function of the surface tension of the liquid and the shape of the interface. This is the Young-Laplace equation which says ∆P = γ (∇･n) where n is the outward unit normal (away from center of curvature), γ is the surface tension, and ∆P is the pressure difference in the direction towards the center of curvature (so ∆P ≥ 0). In the case of a spherical shape, this reduces to ∆P = 2γ/R where R is the radius of curvature. This result is explained in detail with some examples in another blog article called Young-Laplace Equation. Using this result we can write (for a spherical interface)

∆P = dgh = 2γ/R.

Now, using the example we worked out at the end of the section on surface tension, we know that cos z = ±r/R where r is the radius of the tube and z is the angle between the curve and the downward verticle. Thus we get

∆P = dgh = 2γ/R = ± (2γcos z)/r where + is for z < 90° and - for z > 90°.

This result, 

γ = ± dghr/2cos z 

allows you to determine the surface tension from experimental measurements. 

Another approach for the special case of a spherical interface.

Finally, we can view the equilibrium condition as one in which the vertical component of surface tension at the interface is counterbalancing the weight (gain/loss ) of the liquid in the tube compared with the container level weight. 

In case 1, where z < 90°, the vertical component of the surface tension is γcos z. We see this by noting that the surface tension has magnitude γ in the direction tangent to the curve and this tangent makes the angle z with the downward verticle. So resolving this into verticle and horizontal components, the horizontals cancel around the boundary and the verticals give us a total force of 2πrγcos z upward. As before, the increased weight of the water being held up by the surface tension is πr²hdg. Thus, at equilibrium we get πr²hdg = 2πrγcos z or

rhdg = 2γcos z , as before.

In case 2, where z > 90°, the vertical component of the surface tension is 

γcos(π-z) = -γcos z so the net vertical force due to surface tension is -2πrγcos z (a downward force) which is balanced by the decreased weight of the water -πr²hdg. So again, 

rhdg = -2γcos z

Thus,

γ = ± dghr/2cos z  where + is for z < 90° and - for z > 90°, like before.

This second derivation did not rely on the Young-Laplace equation which provides a much more general framework than the spherical case which can be handled with a less sophisticated tool.

Young-Laplace Equation

Young-Laplace Equation:  To show that ∆P=γ(∇･n) where n=unit normal to surface, γ is the surface tension of the liquid, ∆P=pressure difference P_l - P_v, in moving across a vapor-liquid interface from a vapor with pressure P_v to a liquid with pressure P_l. In the case of a spherical surface of radius R as the interface, show ∆P=2γ/R.

Preliminaries:

1. The gradient operator, ∇ = (δ/δx₁, δ/δx₂, δ/δx₃) can act on a scalar function producing a vector, or on a vector function producing a tensor.  If f(x₁,x₂,x₃) is a scalar function and g(x₁,x₂,x₃) is a vector function, g=(g₁,g₂,g₃) where the g_i are scalar functions, then ∇f=(δf/δx₁, δf/δx₂, δf/δx₃), a vector, and ∇g= (∇g₁,∇g₂,∇g₃), a tensor. It is sometimes convenient to think of ∇g as a 3x3 matrix with each gradient as a column vector. Also recall that (∇･g) is called the divergence of g,
   div g = (δ/δx₁)g₁ + (δ/δx₂)g₂ + (δ/δx₃)g₃, a scalar, and (∇ x g) is called the curl of g,
   curl g = ((δ/δx₂)g₃ - (δ/δx₃)g_2, (δ/δx₃)g₁ - (δ/δx₁)g_3, (δ/δx₁)g₂ - (δ/δx₂)g₁), a vector.
   
2. The dot product, a･b, of two vectors a and b, is the scalar Σ a_ib_i i=1,2,3. As a matrix multiply this is (1x3) times (3x1). For a constant vector a, (a･∇) is a differential operator, (a₁ δ/δx_1, a₂ δ/δx_2, a₃ δ/δx₃ ), and is not the same as the scalar value (∇･a), or div a.  Also, (a･∇)f = a･∇f. The dot product of a vector a and a tensor (b,c,d), a･(b,c,d), is the vector (a･b,a･c,a･d). As a matrix multiplication this is (1x3) times (3x3) yielding (1x3) where b, c, d are the columns of the matrix. The dot product of a tensor (b,c,d) and a vector e , (b,c,d)･e, is the vector (e･b,e･c,e･d). As a matrix multiplication this is (3x3) times (3x1) yielding (3x1) yielding (3x1) where b, c, d are the rows of the matrix. Using these definitions it is easy to show that a･((b,c,d)･e) =  (a･(b,c,d))･e so we can write a･(b,c,d)･e to mean one or the other.
   In particular, considering the tensor ∇f, we have (a･∇)f = a･∇f = a･∇(f₁,f₂,f₃)
   = (a･∇ f₁, a･∇ f₂, a･∇ f₃), and  a･∇f･n = (a･∇f)･n = a･(∇f･n) =  Σ a_i δf_j/δx_i n_j i,j=1,2,3.
   
3. The total force, f_p, due to constant pressure P on a surface S is the product PA where A is the surface area and where the force acts normal to the surface at all points. Taking ∆S as an approximation of a surface area element by an element of the tangent plane at some point on that surface area element and the force on this element as P∆Sn where n is the unit normal to this tangent plane, then f_p = ∫_S PndS, where ∫_S  indicates a surface integral. Recall, if the surface is parameterized as r(u,v)=(x(u,v),y(u,v),z(u,v)) over the region R_uv, then n = r_u x r_v/| r_u x r_v |.
   
4. Similarly, the total force, f_t, due to surface tension γ along a closed curve C which forms the boundary of the surface S is the product γL where L is the length of C and where the force acts normal to C and parallel (tangent) to the surface at all points. The direction of f_t is given by t x n where t is the unit tangent to C and n is the unit normal to S (thus their cross product is normal to C and parallel to S). We can compute f_t as a line integral, f_p = γ ∫_C t x n dr, where the total force due to the surface tension is represented as a limiting sum of the approximation forces  (γ ∆r t x n) taken along C.
   
5. Stokes’ Theorem states that  ∫_C F･dr = ∫_C F･t dr = ∫_S (∇  x F)･ndS.
   
6. In general, curl(a x b) = ∇ x (a x b) = (b･∇)a-(a･∇)b + a (∇･b) - b (∇･a). If a is a vector function and b is a constant (vector function), then ∇ x (a x b) = (b･∇)a  - b (∇･a), since any differential operation on a constant vector function leaves the zero vector.
   
7. For any vectors a, b, c, the triple product a･b x c = b･c x a = c･a x b = - a･c x b where a･b x c =  determinant of the matrix with rows a, b, c.
   
8. a. For scalar function f and vector function v we have
   ∇(fv)･v = (δ(fv)/δx₁, δ(fv)/δx₂, δ(fv)/δx₃)･v
   =  (δ(fv)/δx₁･v, δ(fv)/δx₂･v, δ(fv)/δx₃･v)
   = ((v₁[ v₁δf/δx₁+f δv₁/δx₁] + v₂[v₂δf/δx₁+f δv₂/δx₁ ] + v₃[v₃δf/δx₁+f δv₃/δx₁]), ___, ___)  So by symmetry we can write,
   = ( (v･v)δf/δx₁ + f δv/δx₁ ･ v, (v･v)δf/δx₂ + f δv/δx₂ ･ v, (v･v)δf/δx₃ + f δv/δx₃ ･ v)
   = (v･v)∇f + f (∇v･v).
   b. In particular, if v is a unit vector, v･v = 1 , so the first term reduces to ∇f and
   ∇v･v = ( δv/δx₁ ･ v, δv/δx₂ ･ v, δv/δx₃ ･ v) where
    2 δv/δx_i ･ v = δ(v･v)/δx_i = 0. So when v is a unit vector, ∇(fv)･v = ∇f.
   
9. Since for scalar function f and vector function v, div(fv) = ∇f･ v + f div v, when ∇f is perpendicular to v, div v = f div v.
   
10. The surface integral  ∫_S f(x,y,z) dσ is a limiting sum of products f(x_i,y_i,z_i) dσ_i taken over area elements dσ_i of the surface S where (x_i,y_i,z_i) is a point on dσ_i. When we have a vector function F instead of the scalar function f we define  ∫_S F ･ dσ  = ∫_S F ･ n dσ where n is a unit normal to the surface. In this paper we will also consider the case where we are evaluating an integral of the form  ∫_S F dσ ( a sum of vectors ) which we define as  ∫_S F dσ = (  ∫_S F₁ dσ, ∫_S F₂ dσ, ∫_S F₃ dσ). In particular we will be integrating to get the total force obtained by the limiting sum of products (pressure times area) or (surface tension time length) over elements of surface area on a surface or arc length along a curve. See (3) and (4) below.

Main argument:

Let S be the liquid-vapor interface surface, C  the closed curve boundary of the interface, γ the surface tension of the liquid and ∆P the difference in pressure
 P_l - P_v between the liquid and its vapor. If the system is at equilibrium, the total force on the surface is zero. By (3) and (4) in the preliminaries this gives us the equilibrium equation:
0 = f_p + f_t  or   ∫_S ∆P ndS = - γ ∫_C t x n dr . 

From (5) we have  ∫_C F･dr = ∫_C F･t dr = ∫_S (∇  x F)･ndS , so letting F = g x b where g is a vector function and b is a constant (vector funtion),
 ∫_C (g x b)･t dr = ∫_S (∇  x (g x b))･ndS. Using (6) and (7) we can write
b･ ∫_C t x g dr = ∫_S ((b･∇)g - b(∇･g))･ndS. Since (∇･g) is a scalar we can use (2) to write
b･ ∫_C t x g dr = b･ ∫_S ( ∇g･n - (∇･g)n ) dS. Since this is true for an arbitrary b, we have
∫_C t x g dr =  ∫_S ( ∇g･n - (∇･g)n ) dS.  Now substitute g = γn and using the fact (4) that ∇γ is tangent to the surface (normal to n) and the result in (9) we have 

 ∫_C t x γn dr = γ ∫_C t x n dr =  ∫_S ( ∇(γn)･n - (∇･ γn)n ) dS = ∫_S ( ∇(γn)･n -  γ(∇･n)n ) dS. Using (8b) we get  γ ∫_C t x n dr  =  ∫_S ( ∇γ -  γ(∇･n)n ) dS. Now if we assume γ is constant on the surface then ∇γ = 0, and we have  γ ∫_C t x n dr  = -  ∫_S γ(∇･n)ndS. From the equilibrium equation we then have ∫_S ∆P ndS =  ∫_S γ(∇･n)ndS or ∫_S (∆P - γ(∇･n)) ndS = 0. Since this is true for any S, we must have ∆P = γ(∇･n). QED

For constant γ we have P_l - P_{v =} γ(∇･n) where n is the unit normal to the surface away from the liquid. When the surface is a plane we have ∇･n = 0, since n is constant. In general, it can be shown that  (∇･n) measures the local mean curvature of the interface given by (κ₁ + κ₂)/2. In the case of a spherical surface, (∇･n) = 2/R  where R is the radius of curvature (radius) of the sphere (κ₁ = κ₂ = 1/R). We show this in the examples below.

Note: P_l - P_v is the pressure difference going from the vapor to the liquid which is the opposite direction we are taking for the (outward) normal n as positive orientation. In considering a droplet or bubble this means the pressure inside is greater than the pressure outside. Thus the bubble tries to expand but this expansion is countered by the surface tension. Equilibrium occurs when these forces are balanced. In particular, the Young-Laplace equation says that the difference in pressure (at equilibrium) is proportional to the surface tension and inversely proportional to the radius. Thus smaller bubbles have greater pressure differences, etc.

Note: Soap bubbles have two surfaces with the air (with a thin film of liquid in between). Therefore the pressure difference on the two sides of a soap bubble is twice that of a single surface.

Example 1: Use ∆P = γ(∇･n) to show that on a spherical surface, ∆P = 2γ/R.

We need to show ∇･n = 2/R.

For the sphere x² + y² + z² = R², the unit (outward) normal is given by n = (x,y,z)/R where R = (x² + y² + z²)^1/2 . Consider δ/δx(x/R) = (R - x²/R)/R². The same type of expression is obtained for δ/δy(y/R) and δ/δz(z/R) so we have
∇･n = (R - x²/R)/R² + (R - y²/R)/R² + (R - z²/R)/R²  = (3R - (x² + y² + z²)/R)/R²
= 2R/R² = 2/R.

Example 2: Show directly that on a half-sphere ∆P = 2γ/R. (Not using Young-Laplace)

Step 1. Show that the total force on the half-sphere x² + y² + z² = R² , z ≥ 0, where constant pressure P is acting normal at every point is (0, 0, PπR²).

∫_S P ndS = P ∫_S ndS = ( P ∫_S n₁dS, P ∫_S n₂dS, P ∫_S n₃dS). Parameterize to spherical coordinates x = Rsinφcosθ, y = Rsinφsinθ, z = Rcosφ for 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2. Then r_θ = (-Rsinφsinθ, Rsinφcosθ, 0) and r_φ = (Rcosφcosθ, Rcosφsinθ, -Rsinφ). Thus
r_θ  x r_φ = (-R²sin²φcosθ, -R²sin²φsinθ, -R²sinφcosφ). Since 0 ≤ φ ≤ π/2, the final component is negative so we take the unit normal n = - r_θ x r_φ/|r_θ  x r_φ| as the outward normal to the surface. We have P ∫_S ndS = P ∫_Rθφ n |r_θ  x r_φ| dφdθ
= P ∫∫ (R²sin²φcosθ, R²sin²φsinθ, R²sinφcosφ) dφdθ
= PR² ( ∫∫ sin²φcosθ dφdθ, ∫∫ sin²φsinθ dφdθ, ∫∫ sinφcosφ dφdθ) for 0 ≤ θ ≤ 2π,
0 ≤ φ ≤ π/2. We integrate these components to get PR² (0, 0, π) and so the total force due to pressure is (0, 0, PπR²).

Step 2. Show that the total force due to surface tension is given by (0,0,-γ2πR) around the equator of the half-sphere.

Here C is the equator x² + y² + z² = R² , z = 0, so the unit tangent at (x,y,0) on the curve is t = (-y,x,0)/R and the unit normal to the surface at (x,y,0) is n = (x,y,0)/R so the direction of the surface tension force is t x n = (0,0,-1). The curve (equator) has length 2πR, so the total force due to surface tension is (0,0,-γ2πR).

Step 3, By the equilibrium equation, PπR² = γ2πR so P = γ2/R.

Note: Although the Young-Laplace equation for the spherical shaped surface can be established without the higher mathematics used in the general result, the general result is a much more powerful tool and saves the work of establishing the result for specific cases like we did here.