# Why sphere minimizes surface area for a given volume?

I was studying surface tension recently. Rain drops or bubbles of any kind which form are always of a spherical shape.

This is because the liquid tries to minimize the surface area as the molecules on the surface have higher potential energy. The shape which most efficiently manages to do this is a sphere. This is the part I didn't understand; how exactly do we reach the conclusion that a sphere would be the most efficient shape? Why couldn't it be any other shape?

• See e.g. math.stackexchange.com/q/1297870 Nov 29, 2015 at 8:40
• Nov 29, 2015 at 8:41
• Does the liquid really try to minimize the surface area? Is this a good physical answer? I mean, is there a principle of efficiency in physics, as with the alleged "simplicity" and "beauty" of certain explanations. Dec 8, 2017 at 1:26

The units of surface tension are $$[N/m]=[J/m^2]$$ which means surface tension can be interpreted as the energy cost of creating additional surface area. Imagine any shape in equilibrium; increasing its surface area will require an energy input to overcome surface tensile forces before it reaches a new equilibrium.

Now per volume the surface area of a cube of side $$s$$ is $$a_c = \frac{6s^2}{s^3} = \frac{6}{s}$$ whilst a sphere of radius $$r$$ is $$a_s = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}$$. So for equal volumes $$s^3=\frac{4}{3}\pi r^3\rightarrow \frac{s}{r} = \sqrt{\frac{4}{3}\pi}$$ we find: $$\frac{a_s}{a_c} = \frac{1}{2}\frac{s}{r} = \sqrt{\frac{\pi}{6}}<1$$

which mathematically shows that the specific surface area of a sphere is less than that of a cube. In fact this can be shown for any shape: As you can see the shape of a sphere has the lowest possible surface area to volume ratio and therefor requires the least energy to maintain its shape. The minimization of energy cost is usually what drives the physical world, hence natural objects like bubbles and raindrops tend to a spherical shape.

• Does this answer require exhaustive testing of all possible shapes to satisfy the OP's question? Nov 30, 2015 at 1:44
• @DJohnM With the details in this answer, Yes you would need to test all shapes. However, you can in fact write down a hamiltonian for the surface energy and solve that under the constraint of constant volume (technically mass), this will return a sphere. Dec 1, 2015 at 19:44
• Beautifully presented. Oct 2, 2016 at 3:44
• Could you add a bit of explanation to the math (why s/r is important and how you get the final relation 1/2 * s/r. Thanks. Nov 24, 2018 at 19:20
• @afuna s/r is the ratio between the sides of a cube and the radius of sphere which should equal that value for the cube and sphere to have the same volume. The final relation is simply the ratio of surface areas per volume of the cube and sphere stated earlier. Nov 25, 2018 at 12:05

$\mathbf{Geometrical \ approach}:$ Each point on the surface of a sphere is at an equal distance (equal to radius) from the center this results in the minimum surface area for a given volume. This can be proved analytically by comparing the surface area of a sphere with that of any other geometrical shape for a given volume.

For example, let's compare the surface areas of a sphere & a cube for a given volume say $V$ of the drop,

For the sphere of a radius $r$ $$\frac{4\pi}{3}r^3=V\implies r=\left(\frac{3V}{4\pi}\right)^{1/3}$$ surface area of the sphere $\color{red}{S_1}=4\pi r^2=4\pi\left(\frac{3V}{4\pi}\right)^{2/3}=\color{red}{\left(\frac{9}{2\sqrt \pi}\right)^{1/3}V^{2/3}\approx1.36 V^{2/3}}$

For a cube with edge length $a$ $$a^3=V\implies a=V^{1/3}$$ surface area of the cube $\color{blue}{S_2}=6 a^2=\color{blue}{6V^{2/3}}$

Comparing surface areas, $\color{red}{S_1}<\color{blue}{S_2}$ i.e. the surface area of a sphere is smaller than that of a cube for a given volume

Similarly, we can analytically compare surface area of a sphere with that of any other geometrical shape.

The minimum surface area of the sphere results in the minimum surface energy of the drop. that's why the drop takes spherical shape to minimize its potential energy