Visual walkthrough — Surface tension — origin, Young-Laplace equation

This is the visual companion to the parent note. Read that for the summary tables; read this to actually see why the numbers are what they are.

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Step 1 — A molecule with no top neighbour

WHAT. Look at two water molecules: one deep inside the liquid (the bulk), one sitting right at the top surface.

WHY. Everything about surface tension starts here. If we understand why the surface molecule feels a net inward pull, the rest is bookkeeping.

PICTURE.

• The bulk molecule (teal dot) is hugged by neighbours on every side. Each little arrow is one attractive tug. Add them all up and they cancel → net force zero.
• The surface molecule (orange dot) has neighbours below and beside it, but nothing above. The upward tugs are missing, so the leftover pull points straight down into the liquid.

That unbalanced inward pull is the seed of everything below.

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Step 2 — Fewer neighbours ⇒ the surface wants to be small

WHAT. Convert the "inward pull" idea into an energy idea.

WHY. Forces are hard to add over a whole curved sheet; energy is a single number we can minimise. Switching to energy is the trick that makes the derivation easy.

PICTURE.

To move a molecule from the crowded bulk up to the lonely surface, you must break some hand-holds (cohesive bonds). Breaking a bond costs energy. So:

We name the price of surface:

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Step 3 — Same $\gamma$ is "energy/area" AND "force/length"

WHAT. Show that "energy to make area" and "force pulling a line" are the identical number.

WHY. Later we balance a pressure force against surface energy. To do that we must be sure the two faces of $\gamma$ agree. This step earns that right.

PICTURE.

Picture a soap film on a U-shaped wire with a sliding bar of length $L$. Pull the bar out by a tiny distance $dx$.

$\underset{\text{pull on bar}}{\underbrace{F}}\underset{\text{distance moved}}{\underbrace{dx}}=\underset{\text{price per area}}{\underbrace{\gamma }}\underset{\text{new area made}}{\underbrace{2Ldx}}$

• $Fdx$ on the left is work — force times the distance it acts through.
• On the right, $2Ldx$ is the new area. The 2 is because a film has a front face and a back face — two surfaces, both grow (see Minimal surfaces & soap films).

Cancel $dx$ from both sides:

$F=2\gamma L⟹\gamma =\frac{F}{2L}$

Here $\frac{F}{2L}$ is exactly a force per length. Same $\gamma$, two disguises. (For a single surface — like a plain drop — drop the 2.)

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Step 4 — Why a curved skin needs extra pressure inside

WHAT. Argue, with arrows, that a flat skin makes no pressure difference but a curved skin does.

WHY. This is the whole reason the Young–Laplace equation exists. If you see this picture, the formula is almost inevitable.

PICTURE.

Tension always pulls along the skin (tangent to it), never outward. Watch what that means:

• Flat skin (left): the two tension arrows point in exactly opposite horizontal directions. They cancel — no leftover push in any direction → no pressure jump, $\Delta P=0$.
• Curved skin (right): the skin bends, so the two tension arrows tilt slightly toward each other. Their sum has a small component pointing inward (plum arrow). Something must resist that squeeze, or the surface caves in. That something is a higher pressure inside pushing back out.

We will now make "how much more pressure" exact.

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Step 5 — The energy balance for a drop (grow it by $dr$)

WHAT. Take a spherical drop of radius $r$ and imagine puffing it up by a hair's width $dr$. Balance the two energies.

WHY. At equilibrium the drop neither grows nor shrinks on its own. So the work the inside pressure does in pushing out must exactly equal the energy price of the new skin. Setting them equal solves for $\Delta P$.

PICTURE.

Two quantities change when $r\to r+dr$:

(a) Work done by the excess pressure as it pushes the whole surface outward: $d{W}_P=\underset{\text{extra pressure inside}}{\underbrace{\Delta P}}\times \underset{\text{surface area pushed}}{\underbrace{4\pi r^2}}\times \underset{\text{how far it moved}}{\underbrace{dr}}$ This is just (pressure)$\times$(area)$\times$(distance) = force $\times$ distance = work. The shaded shell in the figure is the swept volume $dV=4\pi r^{2}dr$.

(b) Energy price of the new skin. The area of a sphere is $A=4\pi r^2$. Grow $r$ a little and the area grows by $dA=8\pi rdr(\text{because }\frac{d}{dr}(4\pi r^2)=8\pi r)$

So the surface-energy cost is $d{W}_S=\underset{\text{price per area}}{\underbrace{\gamma }}\times \underset{\text{new area made}}{\underbrace{8\pi rdr}}$ A liquid drop has one interface (liquid–air), so no factor of 2 here.

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Step 6 — Solve it: $\Delta P=\frac{2\gamma }{r}$

WHAT. Set $d{W}_P=d{W}_S$ and cancel.

WHY. Equilibrium means the pressure's push pays exactly for the new skin — no surplus, no deficit.

PICTURE.

$\Delta P(4\pi r^{2}dr)=\gamma (8\pi rdr)$

Cancel the common pieces $4\pi rdr$ from both sides:

$\Delta P\cdot r=2\gamma ⟹\boxed{\Delta P=\frac{2\gamma }{r}}$

• $\gamma$ on top: stronger skin ⇒ more squeeze ⇒ more pressure. Makes sense.
• $r$ on the bottom: small drop ⇒ big $\Delta P$; huge drop ⇒ tiny $\Delta P$. The graph in the figure is a $1/r$ curve — it shoots up as $r\to 0$.

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Step 7 — Two skins: the soap bubble's factor 4

WHAT. Redo Step 5's energy count for a soap bubble instead of a drop.

WHY. A soap bubble is a thin liquid shell with air inside and outside — so it has two interfaces, not one. That single change doubles the answer, and it is the #1 exam trap.

PICTURE.

The pressure-work side is unchanged. But the area now grows on both faces: $d{A}_{\text{total}}=\underset{\text{inner face}}{\underbrace{8\pi rdr}}+\underset{\text{outer face}}{\underbrace{8\pi rdr}}=16\pi rdr$ So the balance becomes $\Delta P(4\pi r^2)=\gamma (16\pi r)⟹\boxed{\Delta P=\frac{4\gamma }{r}}$

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Step 8 — Every case: two radii, and the degenerate limits

WHAT. Generalise a sphere ($R_1=R_2=r$) to a patch that curves differently in two perpendicular directions, with principal radii $R_1,R_2$.

WHY. Real surfaces aren't all spheres — think of a cylinder, a saddle, or a flat pool. One formula must cover them all, including the degenerate cases where a radius becomes infinite (flat) or the sign flips (saddle).

PICTURE.

Running the same tilt-of-tension argument once for each direction and adding gives the full Young–Laplace equation:

$\boxed{\Delta P=\gamma \left(\frac{1}{R_1}+\frac{1}{R_2}\right)}$

• $\frac{1}{R}$ is called curvature — big for a tight bend (small $R$), small for a gentle bend (big $R$).
• Two directions, two curvatures, added up. That is the whole content.

Now walk every case — this is why the figure has four panels:

Case	$R_1,R_2$	$\Delta P$	What the picture shows
Sphere (drop)	$r,r$	$\frac{2\gamma }{r}$	both directions curve the same ⇒ $\frac{1}{r}+\frac{1}{r}$
Cylinder	$r,\infty$	$\frac{\gamma }{r}$	curves one way, straight the other ⇒ $\frac{1}{r}+0$
Flat	$\infty ,\infty$	$0$	no curve at all ⇒ $0+0$, matching Step 4's flat skin
Saddle	$+R,-R$	$0$	curves up one way, down the other ⇒ $\frac{1}{R}-\frac{1}{R}$

Recall Check the special cases fall out

Sphere: $\gamma (\frac{1}{r}+\frac{1}{r})=\frac{2\gamma }{r}$. ✓ Cylinder: $\gamma (\frac{1}{r}+\frac{1}{\infty })=\frac{\gamma }{r}$. ✓ Flat: $\gamma (0+0)=0$ — no pressure jump, exactly Step 4. ✓ (This flat case is why calm pools have no surface-tension pressure, and it connects to Pressure in fluids & Pascal's law.)

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The one-picture summary

Read it left to right: lonely surface molecule (Step 1) → skin wants least area, priced at $\gamma$ (Steps 2–3) → curving the skin tilts tension inward (Step 4) → balance push-work against new-skin energy (Steps 5–7) → the master formula (Step 8). Everything on this page is one of these arrows.

$\Delta P=\gamma \left(\frac{1}{R_1}+\frac{1}{R_2}\right)\stackrel{R_1=R_2=r}{\to }\frac{2\gamma }{r}(\text{drop})\stackrel{\text{two faces}}{\to }\frac{4\gamma }{r}(\text{bubble})$

Recall Feynman retelling — the whole walkthrough in plain words

A water molecule deep inside is hugged from all sides and feels nothing. A molecule on the top has no one above it, so it gets tugged down. Making more of these lonely surface molecules costs energy, so water tries to keep its surface as small as possible — that "cost per area" is $\gamma$, and it's the same number whether you measure it as energy-to-make-skin or as force-pulling-a-line.

Now the tension in that skin always pulls sideways along the surface. On a flat skin the sideways pulls cancel and nothing happens. But bend the skin into a ball and the pulls tilt slightly inward, squeezing the inside. To not collapse, the air inside must push back with extra pressure. How much extra? Puff the ball out by a hair and demand that the pressure's outward push pay exactly for the new skin you made — that balance gives $\Delta P=2\gamma /r$ for a drop. A soap bubble has two skins (inside face and outside face), so double it: $4\gamma /r$. And a general lumpy surface just adds up its curvature in the two directions it bends: $\gamma (1/R_1+1/R_2)$. Flat means infinite radius means zero — no squeeze, no extra pressure. That's the whole story.

Recall One-line self-tests

Why does a curved skin need higher inside pressure? ::: Tension pulls along the skin; curving tilts it inward, so the gas must push back harder. Where does the $8\pi r$ come from? ::: It's $\frac{d}{dr}(4\pi r^2)$ — the rate the sphere's area grows with radius. Drop vs bubble factor? ::: Drop $2\gamma /r$ (one skin), bubble $4\gamma /r$ (two skins). Cylinder $\Delta P$? ::: $\gamma /r$, because $R_1=r,R_2=\infty$ so $\frac{1}{r}+0$. A flat surface's $\Delta P$? ::: Zero — infinite radii mean zero curvature.