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Imagine a soap bubble in water so that pressure outside the bubble is equal to pressure inside it ..... i.e excess pressure of soap bubble if it were in air, $P =\rho g h$ ($\rho$, density of water)

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  • $\begingroup$ Why is this different from a soap bubble in air with internal pressure of 1 atm? $\endgroup$ Sep 8, 2022 at 15:18

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The situation of a soap-film bubble is that the curvature of the surface IS the pressure inside (proportional to the pressure) because that curvature and the surface tension of the liquid are the only variables that matter. So, a bubble wall that has the same pressure on both sides, has zero curvature; this is the case for a twinned soap bubble, two identical-diameter bubbles in contact, the shared wall is... flat.

Soap Bubbles and the Forces Which Mould Them by C. V. Boys is recommended reading.

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My guess is that according to Young-Laplace equation, $$ \Delta p =-2\gamma H_{f} ,$$ where $H_{f}$ is mean curvature of a bubble surface, in case pressure difference $\Delta p = 0$, this gives per mean curvature definition, $$ H_f = \frac 12 \left(\frac 1R_1 + \frac 1R_2 \right) ,$$

that when mean curvature $H_f$ is zero, radii of principal curvatures $R_1,R_2 \to \infty$. Infinite radius corresponds to flat surface. So bubble must transform into flat soap film.

But the problem is that you can't get a plane from a sphere by performing a homeomorphic topological transformation, unless you remove 1 point from a sphere. "Making a hole" in a bubble breaks bubble integrity, so the final answer is that because it's impossible for bubble to transform into flat surface,- likely it will blew up (disintegrate) in such process.

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  • $\begingroup$ Let's compare w/ a soap bubble in air; internal pressure = $1 atm + \epsilon$ . If you spear the bubble with a pin, typically the bubble collapses into a shrunken skin, right? $\endgroup$ Sep 8, 2022 at 15:21
  • $\begingroup$ You don't need even a pin. After some time pressure will equalize by other means, due to expanding bubble, bubble becoming too dry, etc. So after pressure difference evaporates (no matter what reason, pin also included),- bubble collapses due to the fact that bubble curvature becomes unstable. That's what I said in the last paragraph. So what's your point then ? $\endgroup$ Sep 8, 2022 at 17:47
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    $\begingroup$ I was not going that far: in true "Spherical Cow" fashion, I was assuming a non-evaporating, non-spreading bubble made of unobtanium. Just trying to add a conceptual agreement. $\endgroup$ Sep 9, 2022 at 11:09
  • $\begingroup$ Well,... I always try to look deeper, even sometimes it seems there's nothing there. Sometimes it may be true, but if it's not - losses are bigger, so I dare to risk :-) $\endgroup$ Sep 9, 2022 at 12:12

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