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I suspect this has something to do with thermodynamics and the isoperimetric inequality and I'm interested in a mathematical derivation of this result.

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    $\begingroup$ See "minimum energy configuration" ? $\endgroup$ – Carl Witthoft Sep 30 '14 at 11:39
  • $\begingroup$ Do you mean this physics.stackexchange.com/q/83295 ? $\endgroup$ – user29305 Sep 30 '14 at 11:42
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    $\begingroup$ I think so :-) . Take a look at the stress on a section of the bubble and how that stress will increase if there's nonuniform curvature. I'm too old and creaky to do the math. $\endgroup$ – Carl Witthoft Sep 30 '14 at 11:44
  • $\begingroup$ Cool. I'll look into that. :) $\endgroup$ – user29305 Sep 30 '14 at 11:45
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    $\begingroup$ Related: physics.stackexchange.com/q/113092 $\endgroup$ – David Hammen Sep 30 '14 at 18:00
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For the sake of the explanation I will assume you mean a gas bubble in a liquid*.

David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length).

If surface tension is indeed dominant than the pressure in the gas bubble is meanly determined by the Laplace pressure jump across the bubble interface, $\Delta P_c$. This pressure consists of two terms: the surface tension between the gas and the liquid, $\gamma$ and the local curvature of the interface, $\kappa$, which is usually written as the sum of the inverse of two principle radii of curvature like this: $$\Delta P_c = \gamma \left(\frac{1}{R_1}+\frac{1}{R_2}\right) $$

When we look for an equilibrium situation, still in the case where surface tension is dominant, we are looking for a situation in which the pressure in the bubble is equal everywhere. If that is not the case we get a flow from one place to the other. This means that we are looking for a situation where $\Delta P_c$ is constant on the entire interface of the bubble. The only way this can happen is if $\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$ is constant all over the interface, i.e. constant mean curvature. A sphere is the only closed shape for which this is the case, hence explaining why a gas bubble evolves towards a spherical shape in the absence of other forces.


*The logic applies to the soap bubbles in the other answers as well, but soap bubbles are in fact a bit more complicated

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Rising bubbles of air in a liquid oftentimes are anything but spherical.

These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape.

Here's a rather non-spherical giant-sized soap bubble:

The extreme size of the bubble and gravity get in the way of the bubble taking on a nice, simple spherical shape.

You're probably asking about bubbles like this one:

This bubble is isolated, is more or less stationary, is more or less the same density as the surrounding air, isn't that large, and lives for a long enough time to relax to that nearly spherical shape.

Why a sphere? A sphere is the unique geometrical shape that minimizes the surface area given an object of a fixed volume. The reason this is the preferred shape is entropy. Minimizing the surface area minimizes the potential energy of the bubble, and that in turn maximizes entropy. See Why does a system try to minimize potential energy? for details.

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The thing you'll notice about a sphere is that it's symmetrical. very symmetrical. No matter how you rotate it, it looks the same. the surface tension pulls the surface of the bubble into a shape that has even surface tension over the entire bubble. The shape with even surface tension is a sphere. a sphere has the smallest possible surface area for an enclosed shape. Thus the bubble will be the thickest and toughest when the shape it forms is a sphere.

artificial cube http://www.bubbles.org/images/cube.gif

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    $\begingroup$ I think you are confusing the terms surface tension and curvature. Surface tension, $\gamma$, is a physical property of the fluids at the interface and is not dependent on shape. Curvature, $\kappa$, is the term that multiplies with the surface tension to give the laplace pressure ($P=\gamma \kappa$). It is this curvature (crudely put the second derivative of the function describing the shape) that is shape dependent. $\endgroup$ – Michiel Sep 30 '14 at 18:50

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