Recently I've been working with the Young-Laplace equation and came across the following physical paradox that I couldn't wrap my head around. It goes like this:

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Imagine a spherical droplet (filled with water) in air. By the Young-Laplace equation, we know that the pressure in the spherical droplet is larger than the pressure in the air:

$$P_{\rm droplet}-P_{\rm air}=\frac{2\gamma}{R}$$

Where $\gamma$ is surface tension and R is the radius of the droplet.

Now, lets take a stationary fluid element that encompasses both side of the droplet. It is clear that the pressure on both sides of the fluid element are not equal, and by Newton's third law, all internal forces within the fluid element cancels out.

In this case, what keeps the fluid element stationary?

  • 1
    $\begingroup$ It seems to me you solve your own mystery when accounting for the surface tension. $\endgroup$ Dec 7, 2022 at 10:55
  • $\begingroup$ The surface of the drop is curved, so, at the edges of your fluid element, surface tension is exerting a force to the right. $\endgroup$ Dec 7, 2022 at 11:52

2 Answers 2


The Young-Laplace equation you give is exactly the force balance you are looking for.
Because it is curved (with total curvature $2H=2/R$), the surface area element inside the fluid element exerts a certain portion of its surface tension, namely $2H\gamma = \frac{2\gamma}{R}$, in the radial direction (add up the vectors from a curved portion of surface where all elements are pulling on each other: components parallel to the surface cancel, but small inward radial components add up).
The Laplace pressure $\Delta P$ that arises is defined as exactly the pressure difference needed to balance out the force from surface tension and keep the element stationary.


Pressure is uniform all around the droplet. Remember that stress is a vector quantity.

Equilibrium of an elementary part of the interface. The equilibrium of the elementary surface of the interface reads,

$\mathbf{t_n} dS = \Delta P \ \mathbf{\hat{r}} \ dS + d \boldsymbol{\Gamma}$,

being $\Delta P \ \mathbf{\hat{r}}$ the contributions of the pressure jump across the interface, and $d \boldsymbol{\Gamma}$ the elementary contribution of the surface tension.

Equilibrium of the bubble. Now, let's consider the equilibrium of the whole interface: the first contribution is the contribution of external actions, while the contribution of the surface tension is an internal contribution that cancels out (by action/reaction principle) when integrating over the whole system, to compute the resultant of the forces acting on the system.

If the bubble has a spherical shape with radius $R$, $\Delta P = \frac{2 \gamma}{R}$ is constant by Young-Laplace equation, and the normal vector points in the radial direction $\mathbf{\hat{n}} = \mathbf{\hat{r}}$.

Integration of this contribution to on the closed surface of the interface to get the resultant on it, gives us

$\mathbf{F}^{ext} = \displaystyle \oint_{\partial V}\mathbf{t_n} dS = \displaystyle \Delta P \oint_{\partial V} \ \mathbf{\hat{n}} \ dS = \mathbf{0} $,

where the last step comes from the gradient theorem (there is the divergence theorem, curl theorem, and also the always forgotten gradient theorem, poor gradient!),

$ \displaystyle \oint_{\partial V} f \mathbf{\hat{n}} \ dS = \displaystyle \int_{ V} \nabla f \ dV $,

with $f(\mathbf{r}) = 1$.


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