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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?

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    $\begingroup$ my guess is surface tension $\endgroup$
    – Razz
    Apr 16, 2023 at 4:57
  • $\begingroup$ I know that, but I hope proof of it. Some history about the question I just have found on the internet is here,. history.nasa.gov/SP-401/…. $\endgroup$ Apr 16, 2023 at 5:15
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    $\begingroup$ You would want to prove that a sphere has a smaller surface area than any other shape of that volume. $\endgroup$
    – mmesser314
    Apr 16, 2023 at 5:27
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    $\begingroup$ Observationally. In science we "prove" everything observationally. Theory is merely a description of our observations. $\endgroup$ Apr 16, 2023 at 5:44
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    $\begingroup$ As stated in comments above, surface tension is the tendency of a (volume of) liquid to minimize it's surface area. If you can prove that any volume's surface area is minimized when it is a sphere, you have essentially shown why water will form a sphere in space. Not sure of a proof using mechanics. It can definitely be done using pure mathematics or variational calculus. $\endgroup$
    – joseph h
    Apr 16, 2023 at 6:01

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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.

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  • $\begingroup$ Okay, I will try $\endgroup$ Apr 16, 2023 at 9:54
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Here is a image of mercury drops on glass.

enter image description here

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?.

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  • $\begingroup$ Thanks for answer $\endgroup$ Apr 16, 2023 at 14:43
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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.

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  • $\begingroup$ Thanks for answer $\endgroup$ Apr 16, 2023 at 14:44
  • $\begingroup$ minimizing $\mathcal {H}({\boldsymbol {p}},{\boldsymbol {q}},t)$? $\endgroup$ May 30, 2023 at 5:39
  • $\begingroup$ @XL_At_Here_There No, it minimizes the Gibbs energy. $\endgroup$
    – Themis
    May 30, 2023 at 15:37
  • $\begingroup$ Thanks, thermodynamic is Greek to me $\endgroup$ May 31, 2023 at 1:15
  • $\begingroup$ @XL_At_Here_There I am Greek :) $\endgroup$
    – Themis
    May 31, 2023 at 12:01

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