In short

Surface tension \gamma is the force per unit length along a liquid's free surface (equivalently, the energy per unit area of that surface), measured in N/m. Molecules at the surface are pulled inward by their bulk neighbours — with none above to pull them back — and this imbalance makes the surface behave like a stretched elastic film. For a spherical drop, surface tension creates an inward-pointing pressure, giving an excess pressure \boxed{\Delta P = 2\gamma/r}; for a soap bubble (two surfaces — inner and outer — sandwiching a thin water film), \boxed{\Delta P = 4\gamma/r}. In a narrow tube dipped in a wetting liquid, surface tension pulls the liquid up to a height \boxed{h = \dfrac{2\gamma \cos\theta}{\rho g r}}, where \theta is the contact angle. The same formula predicts capillary depression (mercury in glass, \theta > 90°, \cos\theta < 0, so h < 0).

Set a needle gently on the surface of a still glass of water. If your hand is steady enough, the needle floats. It has no business floating — steel is 7.8 times denser than water, so Archimedes says it should sink. And if you push the needle down a millimetre, it does sink; the floating is not buoyancy. Look closely and you will see the water surface dipping slightly around the needle, as if the surface were a thin rubber sheet being pressed by a small load.

It is, in a sense, a thin rubber sheet. Every liquid has a surface that behaves like a membrane under tension — pulling inward, storing energy, and trying to minimise its area. That behaviour, called surface tension, is not a new force invented to explain the floating needle; it is a macroscopic consequence of how liquid molecules attract each other and how that attraction is different at the surface than in the bulk. The same surface tension is why a drop of mustard oil beads into a bright amber sphere before it spreads, why a soap bubble is perfectly spherical, why water rises in the narrow capillaries of a cotton wick, and why mercury sinks in a glass tube instead of rising.

This article builds surface tension from its molecular origin, derives the three quantitative results you need for JEE — excess pressure in a drop, excess pressure in a bubble, and the capillary rise formula — with every algebraic step shown, and ends with the energy-minimisation argument that underlies all three. By the end, you will know why a soap bubble's \Delta P has a factor of 4 and a raindrop's has a factor of 2, and why the difference is a lesson in counting surfaces.

The molecular origin of surface tension

Molecular forces in bulk vs at the surface Two water molecules are highlighted. The first, deep in the bulk, is surrounded on all sides by neighbours and feels attractive forces from every direction that cancel out. The second, at the free surface, has neighbours only below and to the sides, so the net attractive force on it points downward (into the bulk). This imbalance is why the surface behaves like a membrane under tension. free surface (air above) bulk surface net force (downward)
A bulk molecule (lower) is surrounded by neighbours on every side; the attractive forces from them cancel. A surface molecule (upper) has neighbours only below — no water above, just air — so the net attractive force on it points into the bulk. This inward pull is the molecular basis of surface tension.

Water molecules attract each other — that is what holds liquid water together rather than letting it disperse into a gas. Every molecule in the bulk is surrounded on all sides by neighbours, each tugging with a short-range attractive force (hydrogen bonds, for water). Because the neighbours are symmetric all around, the forces on a bulk molecule cancel out: no net pull in any direction.

A molecule at the free surface is different. Above it is air, whose molecules are thousands of times further apart and exert negligible attraction. Below it is water, pulling down with the usual short-range force. Around it, other surface molecules pull sideways. The net result is that a surface molecule feels a net inward-pointing force that a bulk molecule does not.

Two consequences

(a) The surface stores energy. To increase the area of the surface, you must pull molecules out of the bulk — against the inward force — and place them on the new surface. That work must come from somewhere. It becomes stored as surface energy. Every new square metre of surface you create costs a fixed amount of energy, called the surface energy per unit area.

(b) The surface behaves like a stretched membrane. Because stretching the surface costs energy, the surface "resists" being stretched — exactly like a rubber sheet. If you draw a line across the surface and consider the two sides, each side pulls on the other with a force per unit length along that line. This force per unit length is the surface tension.

The two pictures — energy per unit area and force per unit length — are the same quantity, expressed differently. We call it \gamma (sometimes T or \sigma):

\gamma = \frac{\text{force}}{\text{length}} = \frac{\text{energy}}{\text{area}}

with SI units of N/m, which is the same as J/m² — as it must be, since 1 \text{ J/m}^2 = 1 \text{ N} \cdot \text{m/m}^2 = 1 \text{ N/m}.

Typical values

Liquid \gamma (N/m) at 20 °C
Water 0.0728
Mercury 0.465
Mustard oil ~0.033
Soap solution 0.025–0.030
Kerosene 0.025
Milk ~0.050
Molten iron (1550 °C) ~1.7

Mercury's surface tension is huge compared to water's — six times larger — which is part of why mercury beads up into perfect spheres when you spill it (those tiny quicksilver balls on a broken thermometer are a demonstration of surface tension). The "soap solution" number is lower than pure water, not higher — soap is a surfactant (surface-active agent) that disrupts the hydrogen bonding at the water surface and reduces \gamma. This is central to how soap bubbles form and how detergents wet dirty cloth.

Surface tension as force per unit length

Imagine a rectangular wire frame with one side free to slide. Dip it in a soap solution so a thin soap film spans the frame. The free slider will be pulled toward the film — the film is trying to contract. To hold the slider in place, you must apply an outward force F.

Soap film on a wire frame with a sliding bar A wire frame shaped like a U, with a sliding bar on the right. A soap film spans the rectangle. The film pulls the sliding bar leftward with force F_film; an external force F pulls the bar rightward to hold it in place. Because the soap film has two surfaces (top and bottom), the film's inward pull is 2 γ L where L is the bar length. soap film sliding bar $F_{\text{film}}$ $F_{\text{ext}}$ $L$ film has TWO surfaces (top + bottom)
A soap film spans a U-shaped wire frame. The right bar, of length $L$, is free to slide. The film pulls it inward with total force $2\gamma L$ — a factor of 2 because a soap film has two free surfaces (front and back).

If the film has length L along the slider, what is the force the film exerts on the slider? Each surface of the soap film (the one you can see on one side, and the one on the back) pulls with a force per unit length equal to \gamma, so each contributes \gamma L. A soap film has two surfaces — a front one touching air and a back one touching air — so the total inward pull is

F_{\text{film}} = 2 \gamma L.

This is the operational definition: surface tension is the force, per unit length, that one side of a free liquid surface exerts on the other along any line drawn in the surface. It acts parallel to the surface and perpendicular to the line.

(If you had only a single surface — say a regular water surface open to the air, with no film — you would measure \gamma L instead of 2 \gamma L. The factor 2 is specific to soap films, which have two surfaces sandwiching the thin water layer.)

Surface tension as energy per unit area — same thing

Now push the slider outward by a distance dx. The external force F_{\text{ext}} = 2\gamma L does work dW = 2\gamma L \, dx, and the area of the film increases by dA = 2 L \, dx (top surface and bottom surface each grow by L \, dx). The work done per unit increase of surface area is

\frac{dW}{dA} = \frac{2 \gamma L \, dx}{2 L \, dx} = \gamma.

So \gamma is also the work done per unit area of new surface created, i.e., the surface energy per unit area. The units match: \text{N/m} = \text{J/m}^2. This is the energetic definition of surface tension, and it is the one you use in energy-minimisation arguments.

Excess pressure inside a liquid drop

Here is the payoff argument. Why is a raindrop spherical? Because the sphere is the shape with the smallest surface area for a given volume — and surface tension drives liquids to minimise their surface area, which means minimising surface energy.

Force balance on half a liquid drop A spherical liquid drop of radius r is cut by an imaginary plane through its centre. The excess pressure inside the drop (ΔP) pushes the two halves apart across the cut plane; surface tension around the rim of the cut pulls them together. Force balance on one hemisphere gives ΔP times the cross-sectional area = γ times the perimeter. That is ΔP × π r² = γ × 2π r, giving ΔP = 2γ/r. cut plane $\Delta P$ pushes out $\gamma$ (along rim) $\gamma$ (along rim) $r$ Force balance on the right half: $\Delta P \cdot \pi r^2 = \gamma \cdot 2\pi r$ $\Rightarrow \Delta P = 2\gamma / r$
Slice a spherical liquid drop through its equator. Inside pressure pushes the two halves apart; surface tension around the perimeter of the cut holds them together. The balance gives the famous excess-pressure formula.

But a spherical drop has a subtle consequence: the inside of the drop must be at a higher pressure than the outside. Surface tension is squeezing the drop inward, and if the inside were not at a higher pressure, the drop would collapse.

Derive the excess pressure as follows.

Step 1. Take a spherical liquid drop of radius r. Mentally cut it into two hemispheres by a plane through the centre. Consider the right hemisphere as a free body.

Step 2. Find the pressure force on the flat (cut) face. The excess pressure inside is \Delta P = P_{\text{inside}} - P_{\text{outside}}. This pressure, acting on the circular cross-section of area \pi r^2, produces a net horizontal force:

F_{\text{pressure}} = \Delta P \cdot \pi r^2.

Why: pressure outside the cut (which pushes right on the right hemisphere) is cancelled by pressure from the fluid outside the drop pushing left on the curved hemisphere surface; the net is only the excess pressure times the cross-sectional area.

Step 3. Find the surface-tension force around the rim of the cut. The cut exposes a circular rim of length 2\pi r. Surface tension pulls the right hemisphere leftward along this rim with a force

F_{\text{tension}} = \gamma \cdot 2\pi r.

Why: the rim is where one surface meets the imaginary cut plane. Each metre of rim pulls with \gamma newtons. Total rim length is the circumference 2\pi r.

Step 4. Balance. The hemisphere is in equilibrium:

\Delta P \cdot \pi r^2 = \gamma \cdot 2\pi r.

Step 5. Solve for \Delta P.

\boxed{\; \Delta P = \frac{2\gamma}{r}. \;}

Why: dividing both sides by \pi r leaves the standard result. Notice \Delta P \propto 1/r — the smaller the drop, the higher the inside pressure. A mist-droplet of radius 1 μm has a hundred times the excess pressure of a 100 μm raindrop.

Why \Delta P grows as the drop shrinks

This 1/r law has striking consequences. A very small water droplet (say r = 1 μm = 10^{-6} m) has

\Delta P = \frac{2 \times 0.073}{10^{-6}} = 1.46 \times 10^5 \text{ Pa} \approx 1.5 \text{ atm}.

A micrometre-sized water droplet has 1.5 atmospheres of excess pressure inside it — more than ambient — all because of surface tension. This extra pressure is why water spontaneously evaporates from tiny droplets (the higher internal pressure raises the vapour pressure in equilibrium with the droplet), and why clouds consisting of tiny droplets are thermodynamically unstable — small droplets tend to evaporate and their mass migrates to bigger droplets via gas-phase diffusion. This is the Ostwald ripening mechanism that grows raindrops in clouds.

Excess pressure inside a soap bubble

A soap bubble looks like a drop, but the physics is different in one crucial way: a soap bubble is a thin shell of liquid, not a solid ball. It has two surfaces — the outer surface (liquid meeting outside air) and the inner surface (liquid meeting the air trapped inside the bubble) — separated by a thin layer of liquid (of thickness nano- to micrometres, much smaller than the bubble radius).

Cross-section of a soap bubble showing inner and outer surfaces A soap bubble cross-section. Two concentric circles bound a thin liquid film: the inner surface at radius r (where air inside the bubble meets the soap solution) and the outer surface at radius r plus small thickness (where the soap solution meets the outside air). Both surfaces contribute surface tension γ, so the excess pressure formula has a factor of 4 instead of 2. $P_{\text{in}}$ (inside) outer surface inner surface $P_{\text{out}}$ (outside) $r$ $\Delta P = \dfrac{4\gamma}{r}$
Two air–liquid interfaces — one on the outside and one on the inside — each contribute a surface tension $\gamma$. The excess pressure is therefore twice that of a simple drop.

Repeat the hemisphere analysis for a soap bubble. The pressure difference between inside and outside is still \Delta P, and it still pushes outward on the cut cross-section with force \Delta P \cdot \pi r^2. The rim of the cut now passes through both surfaces — outer and inner — so the total rim length contributing to surface tension is 2 \times 2\pi r = 4\pi r:

\Delta P \cdot \pi r^2 = \gamma \cdot 4\pi r.

Solving:

\boxed{\; \Delta P = \frac{4\gamma}{r}. \;}

The factor of 4 (compared to 2 for a simple drop) is the signature of a soap bubble. Any time a liquid has two free surfaces around an enclosed volume, you get a factor of 4; any time it has one surface around a solid ball of liquid, you get a factor of 2. Counting surfaces is the whole difference.

Numerical comparison

For a soap bubble of 1 cm radius (r = 10^{-2} m) with \gamma = 0.025 N/m (soap solution):

\Delta P = \frac{4 \times 0.025}{10^{-2}} = 10 \text{ Pa}.

Just 10 pascals — one hundred thousandth of atmospheric pressure. Very small bubbles (smaller than 1 mm) have larger \Delta P, which is why tiny bubbles (below ~0.1 mm) pop quickly — the internal pressure drives rapid evaporation and film thinning.

A pure water drop of 1 cm radius would have \Delta P = 2 \times 0.073 / 10^{-2} = 14.6 Pa — a bit more than the bubble, mostly because water's surface tension is higher than soap solution's.

Counting surfaces — the general rule

If you have n free surfaces between inside and outside, each contributing surface tension \gamma, the excess pressure is

\Delta P = \frac{2n\gamma}{r}.

The last case matters for boiling: a tiny steam bubble inside a pot of boiling water has an excess pressure of 2\gamma/r, which delays its formation — water has to be slightly superheated before bubbles can nucleate.

Capillarity — the rise (and depression) in a narrow tube

Dip a thin glass tube into a glass of water, and water climbs up inside the tube — defying gravity, apparently — to a height of a few centimetres. Dip the same tube into a dish of mercury, and mercury does the opposite: it sinks below the outside level. Both effects are capillarity, and both are explained by the same formula.

Capillary rise in water vs capillary depression in mercury Two narrow tubes side by side, dipped in two different liquids. In the water tube (left), water climbs up the tube with a concave meniscus — the contact angle is less than 90 degrees. In the mercury tube (right), mercury is pushed down below the outside level and forms a convex meniscus — the contact angle is greater than 90 degrees. water (concave meniscus) $\theta < 90°$ $h > 0$ water rises mercury (convex meniscus) $\theta > 90°$ $|h| > 0$ mercury depresses
Water rises in a glass capillary (it wets the glass, $\theta < 90°$); mercury depresses (it does not wet the glass, $\theta > 90°$). The height is proportional to $\cos\theta$, which is positive for wetting liquids and negative for non-wetting ones.

The contact angle

When a liquid meets a solid surface, the surface of the liquid near the solid is not flat — it curves. The angle between the solid and the tangent to the liquid surface, measured through the liquid, is the contact angle \theta.

The contact angle is a property of the pair of substances (liquid and solid) plus any contamination of the interfaces; it is not a property of the liquid alone.

Deriving the capillary rise formula

Take a narrow cylindrical tube of inner radius r dipped vertically into a pool of liquid of density \rho and surface tension \gamma. The contact angle with the tube wall is \theta. Find the height h to which the liquid rises (or depresses) inside the tube.

Capillary rise derivation — force balance on the liquid column Inside a narrow vertical tube of radius r, a column of liquid of height h is held up against gravity. At the top, the liquid meets the tube wall at a contact angle θ. Surface tension γ acts along the line where liquid, tube, and air all meet — a circle of circumference 2 π r. The vertical component of the surface tension pull is γ cos θ per unit length, so the total upward force on the column is 2 π r γ cos θ. This must equal the weight of the column, ρ g π r squared h. surface tension pulls up and out $\theta$ $h$ $2r$ weight $\rho g \pi r^2 h$
The liquid column in a capillary tube is held up by surface tension acting along the meniscus-tube contact line. The vertical component of that tension supports the weight of the column.

Step 1. Identify the force balance on the liquid column inside the tube. The column is a cylinder of radius r and height h, with a curved meniscus on top. It is in equilibrium under two vertical forces: the upward pull of surface tension along the meniscus, and the downward weight of the liquid column.

Step 2. Compute the upward force from surface tension.

Along the line where the liquid, the tube wall, and the air all meet — a horizontal circle at the top of the column with circumference 2\pi r — surface tension pulls on the liquid with magnitude \gamma per unit length, directed along the liquid surface. The liquid surface makes the contact angle \theta with the vertical tube wall, so the surface tension vector has a vertical component of magnitude \gamma \cos\theta per unit length.

F_{\text{up}} = (\gamma \cos\theta)(2\pi r) = 2\pi r \gamma \cos\theta.

Why: surface tension points along the liquid–air interface, not vertically. The vertical component is \gamma \cos\theta — maximum when \theta = 0 (the liquid surface is horizontal, tangent to the vertical wall means all of \gamma pulls up), zero when \theta = 90° (no vertical component), negative when \theta > 90° (the tension pulls the liquid down, producing capillary depression).

Step 3. Compute the downward weight of the column.

The column has volume V = \pi r^2 h (neglecting the small correction from the curved meniscus) and density \rho, so its weight is

W = \rho V g = \rho g \pi r^2 h.

Step 4. Balance.

2 \pi r \gamma \cos\theta = \rho g \pi r^2 h.

Step 5. Solve for h.

\boxed{\; h = \frac{2 \gamma \cos\theta}{\rho g r}. \;}

Why: cancelling \pi r from both sides leaves the standard result. Notice h \propto 1/r — thinner tubes produce higher rises. If \cos\theta > 0 (wetting, \theta < 90°), the rise is positive. If \cos\theta < 0 (non-wetting, \theta > 90°), the rise is negative — the liquid is pushed below the outside level, as mercury does in glass.

Numerical check

Water in a clean glass capillary of radius 0.5 mm, with \gamma = 0.073 N/m, \rho = 1000 kg/m³, g = 9.8 m/s², \theta = 0° (so \cos\theta = 1):

h = \frac{2 \times 0.073 \times 1}{1000 \times 9.8 \times 5 \times 10^{-4}} = \frac{0.146}{4.9} = 0.0298 \text{ m} \approx 3 \text{ cm}.

A millimetre-diameter glass capillary lifts water about 3 cm. Halve the radius and the rise doubles.

Mercury in a glass capillary of the same radius, with \gamma = 0.465 N/m, \rho = 13{,}600 kg/m³, \theta = 140° (so \cos\theta = -0.766):

h = \frac{2 \times 0.465 \times (-0.766)}{13{,}600 \times 9.8 \times 5 \times 10^{-4}} = \frac{-0.712}{66.64} = -0.0107 \text{ m} \approx -11 \text{ mm}.

Mercury is depressed by about 11 mm in the same tube — less in magnitude than water rises, because mercury's density is much higher than its surface tension can overcome.

Worked examples

Example 1: How high does a cotton wick draw mustard oil?

A traditional diya uses a cotton wick soaked in mustard oil. The wick can be modelled as a bundle of capillaries, each of average radius 0.05 mm. Mustard oil has \gamma = 0.033 N/m, \rho = 920 kg/m³, and the contact angle with cotton is about 25°. Find the height to which mustard oil rises in a single capillary of the wick. Take g = 9.8 m/s².

Mustard oil rising by capillarity up a cotton wick A diya with mustard oil below, a cotton wick sticking up. The wick is made of many fine fibres; each fibre acts as a capillary tube. Oil rises up the fibres until the flame at the top burns it. The capillary rise formula predicts roughly 12 cm for a 0.05 mm radius capillary. flame cotton wick mustard oil $h \approx 12$ cm
Capillary action draws mustard oil up the cotton fibres of a diya wick. Thinner fibres lift oil higher — the rise scales as $1/r$.

Step 1. Identify the known quantities.

\gamma = 0.033 N/m, \rho = 920 kg/m³, \theta = 25° (so \cos\theta = 0.906), r = 5 \times 10^{-5} m, g = 9.8 m/s².

Step 2. Apply the capillary rise formula.

h = \frac{2 \gamma \cos\theta}{\rho g r}.

Why: cotton wets well with oil, so the contact angle is small and \cos\theta is close to 1 — the capillary effect is strong. The narrow fibre radius makes the rise substantial.

Step 3. Plug in.

h = \frac{2 \times 0.033 \times 0.906}{920 \times 9.8 \times 5 \times 10^{-5}}.

Numerator: 2 \times 0.033 \times 0.906 = 0.0598. Denominator: 920 \times 9.8 \times 5 \times 10^{-5} = 920 \times 0.00049 = 0.451.

h = \frac{0.0598}{0.451} \approx 0.133 \text{ m} \approx 13 \text{ cm}.

Why: the answer is of the right order — cotton wicks typically need to pull oil several centimetres to feed the flame at the top. The 13 cm is a single-capillary estimate; real wicks have a distribution of fibre diameters, but the formula captures the main effect.

Result: Mustard oil climbs about 13 cm in a single 0.05 mm radius cotton fibre. This is why a modest length of wick can keep a diya burning all evening without anyone needing to lift oil manually.

What this shows: Capillarity is not a curiosity — it is the mechanism by which plants pull water up from roots to leaves, by which blotting paper absorbs ink, and by which a traditional diya draws oil from the pot to the flame. In every case, the liquid is climbing against gravity, and the force driving the climb is the vertical component of surface tension along a narrow channel.

Example 2: Pressure inside a pressure-sensitive soap bubble

A child in Chennai blows a soap bubble of radius 3 cm using soap solution with \gamma = 0.025 N/m. (a) Find the excess pressure inside the bubble. (b) If the bubble shrinks to 1 cm radius as it floats (due to evaporation), by what factor has the excess pressure changed? (c) Compare with a water drop of the same radius.

Soap bubbles of different sizes and their excess pressures Two soap bubbles are drawn side by side. The larger bubble (3 cm radius) has a small excess pressure inside; the smaller bubble (1 cm radius) has a larger excess pressure because ΔP scales as 1/r. The ratio of pressures is 3 to 1. $r = 3$ cm $\Delta P = 3.3$ Pa $r = 1$ cm $\Delta P = 10$ Pa smaller bubble, 3× the excess pressure
Smaller soap bubbles have larger excess pressures. The factor of 3 difference in $\Delta P$ between the two bubbles above follows directly from $\Delta P = 4\gamma/r$.

Step 1. Use the soap-bubble formula \Delta P = 4\gamma/r.

For r = 3 cm = 0.03 m:

\Delta P_1 = \frac{4 \times 0.025}{0.03} = \frac{0.1}{0.03} \approx 3.33 \text{ Pa}.

For r = 1 cm = 0.01 m:

\Delta P_2 = \frac{4 \times 0.025}{0.01} = \frac{0.1}{0.01} = 10 \text{ Pa}.

Why: the factor of 4 (not 2) is because a soap bubble has two free surfaces — inner and outer.

Step 2. Ratio of excess pressures.

\frac{\Delta P_2}{\Delta P_1} = \frac{10}{3.33} = 3.

This is the same as the ratio of radii inverted: 3/1 = 3. The excess pressure scales as 1/r, so shrinking the radius by a factor of 3 raises \Delta P by a factor of 3.

Why: \Delta P = 4\gamma/r is inversely proportional to r, so any ratio of pressures equals the inverse ratio of radii.

Step 3. Compare with a water drop of the same 1 cm radius.

Water, with \gamma = 0.073 N/m, has \Delta P = 2\gamma/r (only one surface):

\Delta P_{\text{drop}} = \frac{2 \times 0.073}{0.01} = 14.6 \text{ Pa}.

Why: the formula is 2\gamma/r (not 4\gamma/r) because a water drop has only one surface — the outer surface separating liquid from air. The inside of the drop is continuous liquid, not a separate surface.

Result: (a) \Delta P = 3.3 Pa for the 3 cm bubble; (b) \Delta P triples to 10 Pa when the bubble shrinks to 1 cm — a factor-of-3 change because \Delta P \propto 1/r; (c) a water drop of 1 cm radius has \Delta P = 14.6 Pa — higher than the 1 cm soap bubble, because water's surface tension (0.073 N/m) is larger than soap solution's (0.025 N/m) — the factor of 4 vs 2 does not fully compensate for the factor of ~3 drop in \gamma.

What this shows: The factor of 4 for a bubble and 2 for a drop comes from counting surfaces, not from any intrinsic difference between bubbles and drops. The 1/r scaling means smaller things always have more excess pressure — which is why small soap bubbles are harder to make (they want to collapse) and small water drops evaporate faster than big ones.

Common confusions

If you have the three formulas — \Delta P = 2\gamma/r for a drop, 4\gamma/r for a bubble, h = 2\gamma\cos\theta/(\rho g r) for capillary rise — and the intuition for what each means, you have everything JEE Mains and Advanced will test on surface tension. What follows is the energy-minimisation derivation of excess pressure (much cleaner than the force-balance version), the Jurin height relation as the balance between surface energy and gravitational PE, and the Young equation that pins down the contact angle from three pair-specific surface tensions.

Excess pressure from energy minimisation

Take a spherical drop of radius r. Its volume is V = \tfrac{4}{3}\pi r^3 and its surface area is A = 4\pi r^2. The total surface energy is

E = \gamma A = 4\pi \gamma r^2.

Suppose the drop expands slightly to radius r + dr. The surface area changes by

dA = 8\pi r \, dr,

so the surface energy changes by

dE = \gamma \, dA = 8\pi \gamma r \, dr.

This energy has to come from somewhere. It comes from the pressure difference \Delta P doing work on the expanding drop. The volume change is

dV = 4\pi r^2 \, dr,

and the work done by the pressure difference is

dW = \Delta P \, dV = 4\pi r^2 \, \Delta P \, dr.

In equilibrium, the drop has minimum surface energy consistent with its volume. Expanding costs energy dE (new surface) but gains energy dW (pressure × volume) — and for equilibrium these balance:

dE = dW
8\pi \gamma r \, dr = 4\pi r^2 \, \Delta P \, dr
\Delta P = \frac{2\gamma}{r}.

Why: this argument is physically cleaner than the force-balance one. The excess pressure is exactly what is needed to supply the surface energy when the drop expands, which is why Laplace's formula applies to drops of any liquid regardless of the details of molecular interactions — it is a consequence of energy conservation and the surface energy per unit area being \gamma.

For a soap bubble, A = 2 \times 4\pi r^2 = 8\pi r^2 (two surfaces), so dA = 16\pi r \, dr and the balance gives \Delta P = 4\gamma/r. The factor of 4 drops out of the surface-area count directly. Same physics, different surface accounting.

This derivation generalises to non-spherical surfaces via the Young–Laplace equation:

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

where R_1, R_2 are the two principal radii of curvature. For a sphere, R_1 = R_2 = r, recovering \Delta P = 2\gamma/r. For a cylindrical film (like a soap film bridging two rings), one radius is infinite, so \Delta P = \gamma/r. For a saddle shape with R_1 = -R_2, \Delta P = 0 — which is why soap-film minimal surfaces (saddles) can exist without a pressure difference across them.

Capillary rise as energy minimisation — Jurin's law

The force-balance derivation of h = 2\gamma\cos\theta/(\rho g r) is quick, but an energy argument is more general. Consider a liquid rising from height 0 to height h inside a capillary tube of radius r. Two energy terms matter.

(a) Change in surface energy. When the liquid rises, some of the tube's dry inner wall (area 2\pi r h) becomes wet. Let \gamma_{SL} be the solid–liquid interfacial energy per unit area and \gamma_{SV} the solid–vapour interfacial energy per unit area. Wetting replaces \gamma_{SV} area with \gamma_{SL} area, an energy change

\Delta E_{\text{surface}} = 2\pi r h \, (\gamma_{SL} - \gamma_{SV}).

By the Young equation (below), this is related to \gamma \cos\theta, so \Delta E_{\text{surface}} = -2\pi r h \gamma \cos\theta — negative because wetting is energetically favourable.

(b) Gain in gravitational PE. The liquid column of mass \rho \pi r^2 h has its centre of mass at height h/2:

\Delta E_{\text{PE}} = \rho \pi r^2 h \cdot g \cdot \frac{h}{2} = \frac{\pi r^2 \rho g h^2}{2}.

In equilibrium, the total energy is minimised with respect to h:

\frac{d}{dh}\left[ -2\pi r h \gamma \cos\theta + \frac{\pi r^2 \rho g h^2}{2} \right] = 0
-2\pi r \gamma \cos\theta + \pi r^2 \rho g h = 0
h = \frac{2\gamma\cos\theta}{\rho g r}.

Same answer as the force balance, different route. This is Jurin's law, and the derivation makes it clear why the rise happens: the system lowers its total energy by wetting more of the tube (gaining negative surface energy) at the cost of lifting liquid against gravity (gaining positive gravitational energy). The equilibrium is where the marginal trade is zero.

The Young equation for the contact angle

At the rim where liquid, solid, and air all meet, three surface tensions pull on every point of the contact line: liquid–vapour (\gamma_{LV}, usually just called \gamma), solid–liquid (\gamma_{SL}), and solid–vapour (\gamma_{SV}). Horizontal force balance along the solid surface gives

\gamma_{SV} = \gamma_{SL} + \gamma_{LV} \cos\theta.

Rearranging:

\boxed{\; \cos\theta = \frac{\gamma_{SV} - \gamma_{SL}}{\gamma_{LV}}. \;}

This is the Young equation. It fixes the contact angle in terms of the three interfacial tensions. If \gamma_{SV} > \gamma_{SL}, the right side is positive and \theta < 90° (wetting). If \gamma_{SV} < \gamma_{SL}, the right side is negative and \theta > 90° (non-wetting). If |\gamma_{SV} - \gamma_{SL}| > \gamma_{LV}, the equation has no solution — the liquid either spreads completely (\theta = 0) or forms a sessile bead that does not even have a well-defined contact angle.

The Young equation explains two stubborn puzzles.

(i) Why water wets clean glass but not greasy glass. Clean glass has a high \gamma_{SV} (glass–air interfacial energy is large because glass is ionic-polar). When water touches it, \gamma_{SL} is low (water's polarity matches glass's), so \gamma_{SV} - \gamma_{SL} is large and \cos\theta comes close to 1 — almost complete wetting. A greasy layer (grease is non-polar) has low \gamma_{SV} and high \gamma_{SL} with water, so \cos\theta becomes small or negative — water beads up.

(ii) Why mercury does not wet glass. Mercury has enormous cohesive forces (hence \gamma_{LV} = 0.465 N/m, six times water's). Its adhesion to glass is weak because mercury is a liquid metal and glass is an electrical insulator — the electron sharing that binds mercury atoms to each other does not extend to glass. So \gamma_{SL} (mercury–glass) is large, \gamma_{SV} - \gamma_{SL} is negative, and \cos\theta is negative — \theta > 90°, giving capillary depression.

The Young equation is the microscopic justification for everything you saw macroscopically in this article: surface tension, the contact angle, the wetting/non-wetting dichotomy, and the capillary rise/depression formula all descend from a single balance of interfacial forces at the three-phase contact line.

Where this leads next