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.