How to prove that a drop of water in the weightlessness of space is round in shape?

Question

How to prove that a drop of water in the weightlessness of space is round in shape theoretically? More specifically, how to prove that a drop of water in the weightlessness of space is round in shape with classical mechanics?

Answer 1

The root cause is in the forces between water molecules and air molecules. Water attracts water and air attracts air more strongly than air. This means that water next to water and air next to air is a lower energy state than water next to air.

Water drops flow under these forces to the lowest possible energy state. This is the state with the fewest possible molecules at the surface where water meets air. A Hamiltonian proof would be in terms of these energies.

Another way to solve it is to shift the way you think about it. Forces act to minimize the surface area. The drop behaves as if the surface was a physical membrane under tension, trying to pull itself into a smaller size.

Write a Hamiltonian that has a term proportional to surface area. There is a constraint that volume must be constant.

Answer 2

Here is a image of mercury drops on glass.

You will note that the smaller drops tend to be closer to being spherical.
Why is that?
It is because it takes energy to create a liquid-gas interface and the gravitational potential energy of a mercury droplet depends on the height of its centre of mass above some datum.
The mercury drop tries to minimize associated with the creation of liquid surface and the gravitational potential energy.
As the mercury drops get bigger it is energetically more favourable to create more surface whilst at the same time lowering the centre of mass of the droplet, thus the shape of the droplet moves further from being speherical.

In effective zero gravity conditions the contribution from gravitational potential energy is zero and so the drop tries to minimise its surface area.
It so happens that for a given volume the shape with the minimum surface area is a sphere.
This idea is discussed in the Wikipedia article Surface-area-to-volume ratio and two answers associated with the PSE question Why sphere minimizes surface area for a given volume?.

Answer 3

The shape of a liquid drop depends on basically two factors: surface energy and gravity. Surface energy has to do with the attractive forces that hold the molecules of the droplet together. Molecules on the surface have higher energy because they are only partially surrounded by other molecules. To minimize its energy the drop tries to decrease its surface area to the smallest possible area, which for a fixed volume is a sphere. Gravity distorts this shape because it tries to bringer the center of mass lower, so that the final shape is an ellipsoid rather than a sphere, as in the image posted by Farcher. In the absence of gravity surface energy is the only factor that plays a role and the equilibrium shape is exactly spherical.

If the drop is sitting on a surface (as in the posted image by Farcher) then there is an additional contribution due to the interaction between the drop and the material of the supporting surface. If the liquid wets the surface, i.e., the molecules of the liquid have high affinity for the molecules on the surface, the drop spreads out, even though this causes the area of the drop to increase. This effect would still be present in the absence of gravity.