Sorry if this is considered a duplicate, but I have further questions that are based on an old post. I was told that air pressure is strong enough to hold up water. Is this true? The vapour pressure of water at room temperature is 0.0313 atm which is far smaller than air pressure. But air pressure is 101325 N/m$^2$ and a tall column of liquid water with a surface area of 0.001m$^2$ and height of 1000m is 997 kg. This translates into a net gravitational force of 9771 N or 9771000 N/m$^2$ which is greater than that of air pressure in this contrived example.

Also, in the experiment of inverting a cup with water where there is a flat sheet cover on the opening of the cup, does the flat layer simply suppress amplitude oscillations of the water at the interface between the water and layer and thus stop Rayleigh-Taylor instabilities from growing? Does the sheet have to be perfectly flat? What is stopping the sheet from falling down? Are there important water-surface interactions that keep the surface from falling? Is this valid for any liquid (e.g. oil) and any surface (e.g. denser than air) that is not porous? Is it still consider the Rayleigh-Taylor instability once the evolution become nonlinear?


Yes, it is true, depending on the size of the vessel.

If you fill a vessel with water and place a massless, impermeable membrane across its mouth, then whether or not the water will fall out when you invert the vessel depends on the balance of fluid pressure pushing out and atmospheric pressure pushing in. The water will stay in the cup if $$P_{atm} \geq \rho_{water}gD$$

where $D$ is the depth of the cup. Roughly speaking, the trick works if $ D\leq 10 \text{ m}$, which is a pretty reasonable requirement.

Note that the pressure balance does not depend on the presence of the membrane. However, the presence of the membrane prevents the onset of the Rayleigh-Taylor instability by not allowing the air to "bubble up" through the water.

Because the membrane resists deformation, there is no instability; in the absence of such protection, the arrangement of water-above-air becomes unstable to perturbations in the air/water interface and the experimenter gets all soggy.

  • $\begingroup$ Thank you for the insightful response. Just a couple clarifications: what is keeping the membrane to the cup? Is it simply adhesive forces to the water? For example, I simply put a thin plastic cover on top of the cup (only contact is the edges of the glass) and the cover remained without falling. Why wouldn't the plastic cover simply fall down as it would normally without the cup present? Despite being solid, should not the plastic cover also fall under gravity because of an analog to the Rayleigh-Taylor instability? $\endgroup$ – Mathews24 Apr 27 '18 at 3:55
  • $\begingroup$ Also, perhaps this is a matter of interpretation, but is it the air "bubbling up" through the water or the water flumes forming downwards into the air that onsets the Rayleigh-Taylor instability? $\endgroup$ – Mathews24 Apr 27 '18 at 3:55
  • $\begingroup$ No, it's not adhesion - it's the fact that the net pressure is inwards. If the cup was evacuated instead of filled with water, the membrane would stay in place too. If, on the other hand, the vessel wass filled with air at atmospheric pressure, then the inward and outward pressures would cancel and the membrane would fall. As for the second question, those are the same thing - you can't have one without the other. $\endgroup$ – J. Murray Apr 27 '18 at 3:59
  • $\begingroup$ Thank you. If I have a column of 9m of water with 1m of air above the water, why would the water still be supported? Shouldn't the pressure of air above the water result in an additional force downwards on the membrane? Also, I realize RTI only applies to fluids, but since most solids have low surface pressures, shouldn't they be supported by air as well if they are of sufficiently low density? $\endgroup$ – Mathews24 May 1 '18 at 2:22
  • $\begingroup$ In that case the water would not be supported. That's why I said the vessel was filled with water. And yes, that's why the membrane is supported too - but if there is air above the membrane, it will fall. $\endgroup$ – J. Murray May 1 '18 at 13:11

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