Categories: Fluid mechanics, Physics, Surface tension.

Young-Laplace law

In liquids, the Young-Laplace law governs surface tension: it describes the tension forces on a surface as a pressure difference between the two sides of the liquid.

Consider a small rectangle on the surface with sides d1\dd{\ell_1} and d2\dd{\ell_2}, orientated such that the sides are parallel to the (orthogonal) principal directions of the surface’ curvature.

Surface tension then pulls at the sides with a force of magnitude αd2\alpha \dd{\ell_2} and αd2\alpha \dd{\ell_2}, where α\alpha is the energy cost per unit of area, which is the same as the force per unit of distance. However, due to the surface’ curvature, those forces are not quite in the same plane as the rectangle.

Along both principal directions, if we treat this portion of the surface as a small arc of a circle with a radius equal to the principal radius of curvature R1R_1 or R2R_2, then the tension forces are at angles θ1\theta_1 and θ2\theta_2 calculated from the arc length:

θ1R1=12d2θ2R2=12d1\begin{aligned} \theta_1 R_1 = \frac{1}{2} \dd{\ell_2} \qquad \qquad \theta_2 R_2 = \frac{1}{2} \dd{\ell_1} \end{aligned}

Pay attention to the indices 11 and 22: to get the angle of the force pulling at d1\dd{\ell_1}, we need to treat d2/2\dd{\ell_2} / 2 as an arc, and vice versa.

Since the forces are not quite in the plane, they have a small component acting perpendicular to the surface, with the following magnitudes dF1\dd{F_1} and dF2\dd{F_2} along the principal axes:

dF1=2αd1sinθ12αd1θ1=αd1d2R1=αR1dAdF2=2αd2sinθ22αd2θ2=αd2d1R2=αR2dA\begin{aligned} \dd{F_1} &= 2 \alpha \dd{\ell_1} \sin\theta_1 \approx 2 \alpha \dd{\ell_1} \theta_1 = \alpha \dd{\ell_1} \frac{\dd{\ell_2}}{R_1} = \frac{\alpha}{R_1} \dd{A} \\ \dd{F_2} &= 2 \alpha \dd{\ell_2} \sin\theta_2 \approx 2 \alpha \dd{\ell_2} \theta_2 = \alpha \dd{\ell_2} \frac{\dd{\ell_1}}{R_2} = \frac{\alpha}{R_2} \dd{A} \end{aligned}

The initial factor of 22 is there since the same force is pulling at opposide sides of the rectangle. We end up with α/R1,2\alpha / R_{1,2} multiplied by the surface area dA=d1d2\dd{A} = \dd{\ell_1} \dd{\ell_2} of the rectangle.

Adding together dF1\dd{F_1} and dF2\dd{F_2} and dividing out dA\dd{A} gives us the force-per-area (i.e. the pressure) added by surface tension, which is given by the Young-Laplace law:

Δp=α(1R1+1R2)\begin{aligned} \boxed{ \Delta p = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) } \end{aligned}

The total excess pressure Δp\Delta p is called the Laplace pressure, and fully determines the effects of surface tension: a certain interface shape leads to a certain Δp\Delta p, and the liquid will flow (i.e. the surface will move) to try to reach an equilibrium.


  1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.
  2. T. Bohr, Surface tension and Laplace pressure, 2021, unpublished.