Surface tension — origin, Young-Laplace equation: Worked examples

Before we begin, let us pin down the tools we keep reusing, so we never use a symbol before it is earned.

What "two principal radii" means (before we use $R_{1}, R_{2}$ )

Stand on any point of a curved skin. Slice the surface with a plane and you get a curved line; that line hugs a circle of some radius. Now rotate your slicing plane around — the radius of the hugging circle changes as you turn. It turns out there are two special, perpendicular slicing directions where the radius is smallest and largest. Call those two radii the principal radii of curvature, written $R_{1}$ and $R_{2}$ .

Where the general formula comes from (a careful sketch)

We keep leaning on $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ , so let us actually build it, using the $R_{1}, R_{2}$ we just defined. Look at Figure 1 while you read.

Take a tiny rectangular patch of skin with sides $d x$ (running along the $R_{1}$ direction) and $d y$ (running along the $R_{2}$ direction), curved gently outward.

The tension pulls along every edge. On the two $d y$ -length edges (the edges facing each other across the $d x$ span), the surface tension $γ$ pulls with force $γ d y$ each — force per length times the edge length. These two pulls point outward along the surface, tangent to it.
- Why $γ d y$ ? Because $γ$ is force per unit length of the line it acts along, and that line has length $d y$ .
Because the patch is curved, those two pulls are not parallel. Along the $R_{1}$ direction the surface bends through a small angle. If the patch spans arc-length $d x$ on a circle of radius $R_{1}$ , the two tangent directions differ by the angle $d θ_{1} = \frac{d x}{R _{1}}$ (arc = radius × angle).
- Why $d x / R_{1}$ ? That is the very definition of the angle subtended by an arc: split the arc $d x$ by the radius $R_{1}$ .
Each tilted pull therefore has a tiny inward component. A force $γ d y$ tilted inward by the half-angle $d θ_{1} /2$ contributes $γ d y \cdot sin (d θ_{1} /2) \approx γ d y \cdot \frac{d θ _{1}}{2}$ inward. The pair (both edges) gives $2 \times γ d y \cdot \frac{d θ _{1}}{2} = γ d y d θ_{1} = γ d y \cdot \frac{d x}{R _{1}} = \frac{γ d x d y}{R _{1}}$ .
- Why the small-angle step $sin x \approx x$ ? Because $d θ_{1}$ is infinitesimal; that is the whole reason we take a tiny patch — the geometry becomes exact in the limit.
Do the same in the perpendicular direction (the $R_{2}$ direction, edges of length $d x$ ): they contribute $\frac{γ d x d y}{R _{2}}$ inward by the identical argument.
Balance against the pressure push. The excess pressure pushes the patch outward with force $Δ P \cdot (d x d y)$ . At equilibrium the outward push equals the total inward pull: $Δ P (d x d y) = \frac{γ d x d y}{R _{1}} + \frac{γ d x d y}{R _{2}} .$
Cancel the area $d x d y$ : $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) .$

For a sphere both radii equal $r$ , so $\frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{2}{r}$ — that is where the "2" in $2 γ / r$ is born. Notice the key link the sketch makes concrete: the inward component of tension is exactly $γ d x d y / R$ because the edge tilts by the arc-angle $d x / R$ .

The scenario matrix

Every problem in this topic lives in one of these cells. The Example column tells you which worked example nails that cell.

Cell	What makes it distinct	Example
A. One interface (drop)	Single liquid–air skin ⇒ factor 2	Ex 1
B. Two interfaces (soap bubble)	Film has inside + outside skin ⇒ factor 4	Ex 2
C. Gas cavity in liquid	A bubble of gas inside water — still one skin	Ex 3
D. Cylinder (one radius $\infty$ )	$R_{1} = r, R_{2} = \infty$ ⇒ factor 1	Ex 4
E. Flat / degenerate limit	$r \to \infty$ ⇒ $Δ P \to 0$	Ex 4 (aside)
F. Direction of flow (sign of $1/ r$ )	Two connected bubbles; which is at higher $P$ ?	Ex 5
G. Limiting behaviour $r \to 0$	$Δ P \to \infty$ : why tiny drops "resist"	Ex 6
H. Energy method / word problem	Splitting one drop into many; energy released	Ex 7
I. Exam twist — coalescence	Conserve air (volume) + surfaces to find new radius	Ex 8
J. Negative curvature (saddle)	One radius negative ⇒ terms cancel, $Δ P$ can be 0	Ex 9

Every numeric answer below is machine-checked at the bottom.

Cell A — one interface

Worked example Ex 1 · Excess pressure inside a raindrop

A spherical water raindrop has radius $r = 2 mm$ . Surface tension of water $γ = 0.072 N/m$ . Find the pressure inside compared to the surrounding air.

Forecast: guess first — is $Δ P$ closer to $1 Pa$ , $70 Pa$ , or $7000 Pa$ ? (Atmospheric pressure is $\sim 1 0^{5} Pa$ , so it should be tiny.)

Identify the number of interfaces. A liquid drop has one liquid–air skin.
- Why this step? The factor in the formula is decided entirely by counting skins. One skin ⇒ use $2 γ / r$ , not $4 γ / r$ .
Convert the radius to metres: $r = 2 mm = 2 \times 1 0^{- 3} m$ .
- Why this step? $γ$ is in N/m; mixing millimetres would give a wrong power of ten.
Substitute: $Δ P = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 72 Pa .$
- Why this step? For a sphere both principal radii equal $r$ , so the general formula gives $γ (1/ r + 1/ r) = 2 γ / r$ — the outward pressure push balances the inward pull of the single skin.

Verify: Units: $\frac{N/m}{m} = \frac{N}{m ^{2}} = Pa$ ✓. Magnitude $72 Pa$ is a whisper next to $1 0^{5} Pa$ atmospheric — sensible for a soft-skinned drop.

Cell B — two interfaces

The figure below shows the crucial difference side by side: the drop (left) has a single lavender boundary, while the soap bubble (right) shows two coral rings — the inner and outer skins that together earn the factor 4. Look at how the bubble is a shell of liquid with air on both faces.

Figure — Surface tension — origin, Young-Laplace equation

Figure 1 — One skin (drop) versus two skins (soap bubble): counting boundaries decides the factor 2 vs 4.

Cell C — gas cavity in liquid

Worked example Ex 3 · An air bubble underwater

A tiny air bubble of radius $r = 0.5 mm$ sits inside water, $γ = 0.072 N/m$ . Find the excess pressure of the air inside over the surrounding water.

Forecast: is this a "drop" case (one skin, factor 2) or a "bubble" case (two skins, factor 4)? Look at the picture before deciding.

Count interfaces. There is water on the outside and air on the inside, separated by one liquid–air surface only. There is no thin film — the water is a bulk ocean.
- Why this step? "Soap bubble" has two skins because it is a thin film. A gas cavity in bulk liquid has just one skin, exactly like a drop.
Apply the one-skin formula: $Δ P = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{0.5 \times 1 0 ^{- 3}} = 288 Pa .$
- Why this step? Same physics as the raindrop — the curved skin squeezes the enclosed gas.

Verify: Units give Pa ✓. Smaller radius than Ex 1 (0.5 mm vs 2 mm), so pressure should be larger: $288 > 72$ ✓. The word "bubble" did not mean factor 4 here — geometry beats vocabulary.

In the figure, the air cavity sits in a sea of mint-green bulk water with a single lavender skin. The coral arrows point inward — that is the tension squeezing the trapped air. There is no second skin anywhere, so the factor is 2, exactly like the raindrop.

Figure 2 — An air cavity in bulk water: one interface (factor 2), not a two-skinned film.

Cell D — cylinder (and Cell E — flat limit)

Worked example Ex 4 · A liquid jet (cylinder) and its flat limit

A cylindrical column of water (a thin jet) has radius $r = 1 mm$ , $γ = 0.072 N/m$ . Find $Δ P$ across its curved side. Then ask: what happens as $r \to \infty$ ?

Forecast: a cylinder curves like a circle across its width but is straight along its length. Should its pressure be $2 γ / r$ , $γ / r$ , or something between?

Find the two principal radii. Across the circular cross-section the radius of curvature is $R_{1} = r$ (positive — it bulges toward the inside). Along the axis the surface is a straight line, so it does not curve: $R_{2} = \infty$ .
- Why this step? The general Young–Laplace formula needs both curvatures, and a cylinder is the case where one of them vanishes to a straight line ( $1/\infty = 0$ ).
Substitute into the general form: $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = γ (\frac{1}{r} + \frac{1}{\infty}) = \frac{γ}{r} .$
- Why this step? $1/\infty = 0$ kills the axial term, leaving a single factor of 1 — half the drop's factor 2.
Numbers: $Δ P = \frac{0.072}{1 \times 1 0 ^{- 3}} = 72 Pa .$

Cell E — the flat limit: now let the radius blow up, $r \to \infty$ . Both terms become $1/\infty = 0$ , so $Δ P \to 0$ .

Why this step? A perfectly flat skin has no inward-pointing component of the tension, so it needs no pressure difference to hold it up. This is the degenerate case that anchors the whole formula.

Verify: Units Pa ✓. Cylinder factor 1 is exactly half the drop's factor 2, and the flat surface gives $0$ as the parent's table demands ✓.

The figure makes the two curvatures visible: the coral arrow is the finite radius $R_{1} = r$ across the round cross-section, while the mint double-arrow runs straight along the axis — that is the direction with $R_{2} = \infty$ . Only the curved direction contributes, so the factor is 1.

Figure 3 — A cylinder has one finite radius ( $R_{1} = r$ ) and one infinite one ( $R_{2} = \infty$ ), giving $Δ P = γ / r$ .

Cell F — direction of flow (the sign of $1/ r$ )

Worked example Ex 5 · Two connected soap bubbles — which way does the air rush?

Bubble A has radius $r_{A} = 3 cm$ , bubble B has radius $r_{B} = 1 cm$ . They are joined by a short tube with a valve. $γ = 0.030 N/m$ . Open the valve — which way does air flow, and what are the two pressures?

Forecast: intuition screams "big bubble = more air = more pressure, so air leaves the big one." Trust the formula, not the gut.

Write each pressure. Both are soap bubbles (two skins), so $Δ P = \frac{4 γ}{r}$ . $Δ P_{A} = \frac{4 ( 0.030 )}{0.03} = 4 Pa, Δ P_{B} = \frac{4 ( 0.030 )}{0.01} = 12 Pa .$
- Why this step? Each bubble sets its own internal excess pressure from its own radius; the tube lets them compare.
Compare. $Δ P \propto 1/ r$ , so the smaller bubble (B) has the higher pressure.
- Why this step? A tighter curve squeezes the gas harder — the whole "Petite = Pressurised" mnemonic.
Conclude direction. Air flows from high pressure to low pressure ⇒ from B (small) to A (big). B shrinks and A grows.
- Why this step? Gas always drifts down a pressure gradient; here the gradient points from the small bubble to the big one.

Verify: $Δ P_{B} /Δ P_{A} = 12/4 = 3 = r_{A} / r_{B} = 3/1$ ✓ — the ratio must equal the inverse radius ratio. The counter-intuitive small-to-large flow is confirmed.

The figure shows both bubbles joined by a tube, each labelled with its radius and pressure. The lavender flow arrow points from the small, high-pressure bubble toward the big, low-pressure one — trace it and see how the air deserts the tightly curved bubble.

Figure 4 — Small bubble (12 Pa) drains into the large one (4 Pa): flow follows the pressure, which follows $1/ r$ .

Cell G — limiting behaviour $r \to 0$

Worked example Ex 6 · Halving the radius of a droplet

A drop of radius $r$ has excess pressure $Δ P$ . It is squeezed to half the radius, $r /2$ . By what factor does $Δ P$ change? Illustrate with $r = 1 μ m = 1 0^{- 6} m$ and $γ = 0.072 N/m$ .

Forecast: does the pressure double, halve, or quadruple?

Use $Δ P = 2 γ / r$ . Replacing $r$ by $r /2$ : $Δ P^{'} = \frac{2 γ}{r /2} = \frac{4 γ}{r} = 2 Δ P .$
- Why this step? Pressure scales as $1/ r$ , so halving the radius doubles the pressure.
Push to the tiny limit. For $r = 1 0^{- 6} m$ : $Δ P = \frac{2 ( 0.072 )}{1 0 ^{- 6}} = 1.44 \times 1 0^{5} Pa .$
- Why this step? At micrometre scale the excess pressure already rivals a full atmosphere — this is why mist droplets and nucleating bubbles resist forming.
Take the formal limit $r \to 0$ : $Δ P = 2 γ / r \to \infty$ .
- Why this step? The degenerate small-radius end of the matrix: infinitely tight curvature demands infinite pressure. It is hard to make a very small drop or bubble — the birth of "nucleation barriers."

Verify: $Δ P^{'} /Δ P = 2$ exactly ✓, and $1.44 \times 1 0^{5} Pa \approx 1.4$ atm — the promised "rivals atmosphere" ✓.

Cell H — energy method / word problem

Worked example Ex 7 · Splitting one drop into a thousand

A single water drop of radius $R = 1 mm$ is sprayed into $n = 1000$ identical tiny droplets. $γ = 0.072 N/m$ . How much surface energy must be supplied?

Forecast: does spraying release energy or cost energy? (Which shape has more total surface — one big drop or a thousand small ones?)

Conserve volume to find the small radius $r$ . Volume is fixed: $\frac{4}{3} π R^{3} = n \cdot \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{n ^{1/3}} = \frac{1 0 ^{- 3}}{100 0 ^{1/3}} = \frac{1 0 ^{- 3}}{10} = 1 0^{- 4} m .$
- Why this step? No water is created or destroyed; only its surface changes. We need $r$ before we can total up the new area.
Compute total surface area before and after. $A_{before} = 4 π R^{2}, A_{after} = n \cdot 4 π r^{2} = 1000 \cdot 4 π (1 0^{- 4})^{2} .$ Numerically $A_{before} = 4 π (1 0^{- 3})^{2} = 4 π \times 1 0^{- 6} m^{2}$ and $A_{after} = 4 π \times 1 0^{- 5} m^{2}$ .
- Why this step? Surface energy is $γ$ times surface area (parent's energy definition of $γ$ ). Area is the thing that changed.
Energy required = $γ Δ A$ : $Δ E = γ (A_{after} - A_{before}) = 0.072 (4 π \times 1 0^{- 5} - 4 π \times 1 0^{- 6}) .$ $= 0.072 \cdot 4 π \cdot 1 0^{- 6} (10 - 1) = 0.072 \cdot 4 π \cdot 9 \times 1 0^{- 6} \approx 8.14 \times 1 0^{- 6} J .$
- Why this step? Positive means energy must be put in — splitting increases area, so you cannot spray for free.

Verify: $A_{after} = 10 A_{before}$ (area grows by $n^{1/3} = 10$ ) ✓. Sign is positive ⇒ energy supplied, matching "many small drops have more surface" ✓. Numeric $Δ E \approx 8.14 μ J$ . (See Energy methods in mechanics for the general "energy = $γ \times$ area change" bookkeeping.)

Cell I — the exam twist: coalescence

Worked example Ex 8 · Two soap bubbles merge — find the new radius

Two soap bubbles of radii $r_{1} = 3 cm$ and $r_{2} = 4 cm$ coalesce. Solve it properly (conserve the air) in the limit where surface energy dominates, so the $P_{0}$ terms are negligible. Find the new radius $r$ .

Forecast: will the new radius be closer to $5 cm$ , $7 cm$ , or exactly $r_{1} + r_{2} = 7 cm$ ?

Start from the honest conservation of air. At constant temperature, $\sum P_{i} V_{i}$ is conserved. With $P_{i} = P_{0} + \frac{4 γ}{r _{i}}$ and $V_{i} = \frac{4}{3} π r_{i}^{3}$ , this is the parent cubic.
- Why this step? Air cannot vanish; only the surface rearranges. Volume (through $P V$ ) is the physically conserved quantity, not area.
Take the surface-dominated limit $P_{0} \to 0$ . Then $P_{i} \to \frac{4 γ}{r _{i}}$ , and $\sum P_{i} V_{i}$ becomes proportional to $\sum r_{i}^{2}$ : $\frac{4 γ}{r _{i}} \cdot \frac{4}{3} π r_{i}^{3} = \frac{16}{3} π γ r_{i}^{2} \Rightarrow r^{2} = r_{1}^{2} + r_{2}^{2} .$
- Why this step? The $r_{i}^{3}$ from volume and the $1/ r_{i}$ from pressure combine to leave $r_{i}^{2}$ — which is why the neglect- $P_{0}$ model looks like "area conservation." It is really the $P V$ law in disguise.
Substitute: $r = r_{1}^{2} + r_{2}^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = 5 cm .$
- Why this step? This is the neat 3–4–5 giveaway examiners love — the radii add "in quadrature," not linearly.

Verify: $r = 5 cm$ , and $r < r_{1} + r_{2} = 7 cm$ as it must (one bubble is more area-efficient than two) ✓. Also $r > max (r_{1}, r_{2}) = 4 cm$ since the merged bubble holds both volumes-worth of air ✓.

Cell J — negative curvature: the saddle where pressure cancels

Worked example Ex 9 · The soap-film neck (catenoid) — why

$Δ P = 0$ A soap film spans the gap between two coaxial rings, pulling into a waisted "neck" shape (a catenoid). At the narrowest point of the neck the film curves inward across the circular waist with radius $R_{1} = 1 cm$ , but curves outward along the axis (a saddle) with radius of curvature $R_{2} = 1 cm$ pointing the other way. $γ = 0.030 N/m$ . Find $Δ P$ across the film.

Forecast: a curved soap surface — surely there is a pressure jump like every other example? Watch the signs.

Assign signs with our convention. Across the waist the surface bulges toward the inside air, so $R_{1} = + 1 cm$ . Along the axis the surface is a saddle — it curves the opposite way (centre of the hugging circle on the outside), so $R_{2} = - 1 cm$ is negative.
- Why this step? This is the case the sign convention was built for. A saddle curves toward the inside in one direction and away from it in the perpendicular direction — opposite signs.
Apply the general formula, keeping the signs: $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = γ (\frac{1}{+ 0.01} + \frac{1}{- 0.01}) = γ (100 - 100) = 0.$
- Why this step? The two curvatures are equal and opposite, so their reciprocals cancel exactly. The inward pull from one direction is undone by the outward "unbending" of the other.
Interpret. $Δ P = 0$ means the air pressure is the same on both sides of the film — the film is not squeezing anything. Such an equal-pressure film is a minimal surface (locally area-minimising); the catenoid is the classic example (see Minimal surfaces & soap films).
- Why this step? When there is no pressure jump to support, the only thing surface tension can do is shrink area, so the film settles into the least-area shape it can.

Verify: $\frac{1}{+ 0.01} + \frac{1}{- 0.01} = 100 - 100 = 0$ , so $Δ P = 0$ exactly ✓. This is the promised edge case: with one radius negative, the terms cancel and a genuinely curved surface can still have zero pressure jump — impossible if you forget the sign.

The figure shows the neck: the coral circle is the inward waist curvature $R_{1} > 0$ , while the mint saddle-arc along the axis is the outward curvature $R_{2} < 0$ . Trace how one bends "in" and the other bends "out" — that is why they cancel.

Figure 5 — A saddle-shaped soap-film neck: $R_{1} > 0$ (waist) and $R_{2} < 0$ (axis) cancel, giving $Δ P = 0$ .

Recall

Recall Cover the answers and test yourself

What are the two principal radii $R_{1}, R_{2}$ ? ::: The radii of the two best-hugging circles in the perpendicular directions of most and least curving. Which cases give factor 2, 4, and 1? ::: Drop and gas cavity ⇒ 2 (one skin); soap bubble ⇒ 4 (two skins); cylinder ⇒ 1 (one radius infinite). An air bubble in bulk water uses which factor? ::: 2 — it is one skin, not a thin film. As $r \to \infty$ what is $Δ P$ ? ::: Zero — a flat surface needs no pressure jump. As $r \to 0$ what is $Δ P$ ? ::: Infinity — tiny curves squeeze infinitely hard. When is a principal radius counted negative, and what can happen? ::: When the surface curves the other way (a saddle/dimple), centre on the outside — then the two terms can cancel to $Δ P = 0$ (a minimal surface). Two connected bubbles: air flows which way? ::: From the small (high-pressure) bubble to the large one. Coalescence: what is physically conserved, volume or area? ::: The air (volume / $P V$ ). "Area conservation" is only the neglect- $P_{0}$ shortcut.