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I've been puzzling over the following thought experiment: Consider two versions of a perfectly rigid, sealable box. Suppose first we fill the first box with water at sea level, so the hydrostatic pressure inside the box is basically 1 atm, plus the pressure of the water (but let's say the box is very shallow). Now seal the box and push down into a deep body of water and attach it there.

Second, suppose that we leave the second box open, attach it under water, and then seal it there. In that case, the pressure inside the box is the same as the (high) ambient water pressure.

How should I imagine the difference between the two boxes? The first box experiences strong inward pressure (driving it to implode), which we assume is absorbed by the box, whereas the second one does not. (Only the material of the box itself is under pressure.) But the water inside and outside is incompressible, and there's nowhere for the force to act, no energy to spend (?).

Is there any theoretical mechanism by which one state could be transformed into the other, in either direction? It seems that displacing content is not an option. What if I slowly open the seal of the first box -- does the interior pressure immediately go up? Is there any net force on the lid?

(Aspects of this question appear in this related question about valves.)

Box 1:

  +-------+
  | Water |   ---->
--+-------+--
             \~~~~~~~          ~~~~~~~~~~~~~~~~~~~~~~~~~~
                                         .
                                         .
                                         .




                                         .
                                     +-------+
                                     | Water |
                                     +-------+
                               ##########################

Box 2:

     /
    /
   /
  +
  |  Air  |   ---->
--+-------+--
             \~~~~~~~     ~~~~~~~~~~~~~~~~~~~         ~~~~~~~~~~~~~~~
                                    .                        .
                                    .                        .
                                    .                        .


                                                ---->

                                    .                        .
                                    .                        .
                                    .                        .

                                   /
                                  /
                                 /
                                +                        +-------+
                                |       |                | Water |
                                +-------+                +--------
                          #####################      #################
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  • $\begingroup$ Please consider providing a diagram. What you are asking is not very clear. $\endgroup$ – sammy gerbil Jan 24 '17 at 23:26
  • $\begingroup$ @sammygerbil: OK, I tried something -- does that help? $\endgroup$ – Kerrek SB Jan 24 '17 at 23:33
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If the box is sealed and perfectly rigid, the fluid pressure inside it will not change even if it is submerged to the bottom of the sea.

A real box is compressible to a variable extent, which depends on its structure and the material it is made from. The fluid pressure inside the box will only change in response to either a change in the volume of the box, or a change in the temperature of the fluid, or a change in the amount of fluid inside. If the box is sealed, not compressible, and held at constant temperature, the fluid pressure inside it will not change.

No material is really incompressible. Incompressible simply means that the decrease in volume is very small for the typical forces you are considering.

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  • $\begingroup$ To be clear, both boxes are meant to be full of water. Box one is filled and shut at sea level, Box 2 is only shut deep underwater. So it's the water pressure inside the boxes I am thinking about. $\endgroup$ – Kerrek SB Jan 24 '17 at 23:34
  • $\begingroup$ Whether the boxes are full of air or water makes no difference. Any change in the internal fluid pressure in a sealed box depends on the box being compressed. If there is no change in volume of the box, there is no change in internal fluid pressure. $\endgroup$ – sammy gerbil Jan 24 '17 at 23:42
  • $\begingroup$ Yes, right, but now I want to open the lid of the low-pressure box. If there's air inside, this will obviously cause movement. But if there's already water inside, then when happens? $\endgroup$ – Kerrek SB Jan 24 '17 at 23:49
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    $\begingroup$ If you open the box, or make a hole in it, the internal pressure will equalise with the pressure in the water outside. The water inside will be compressed very sightly, and a very small amount of water will flow in from outside until the internal and external pressures are equal. $\endgroup$ – sammy gerbil Jan 24 '17 at 23:54
  • $\begingroup$ And is there any mechanism to go the other way -- i.e. take the sealed box 2 and reduce its internal pressure? Without changing the contained (idealized) volume, so that it attains a state like that of box 1? $\endgroup$ – Kerrek SB Jan 24 '17 at 23:55
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I believe that the problem as stated is indeterminate, in that you have combined two absolute properties: an absolutely rigid box, and absolutely incompressible fluid in the box.

Assume that at the bottom of the sea, the box resists a fraction $x$ of the applied pressure (without changing its volume) and transmits $(1-x)$ of the pressure to the liquid in the box (without any change in the fluid's volume) So the box and liquid have the same volume as before.

And this is true for any values of x, such that $0<x<1$.

If the box is rigid, the fluid could be as compressible as you like, even a gas. And if the liquid is incompressible, the container could be as flimsy as you like...

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  • $\begingroup$ If the container were flimsy, then the water in both boxes would have the same pressure (equal to the ambient pressure) at the bottom of the sea, right? The rigidity of the box makes a big difference? $\endgroup$ – Kerrek SB Jan 25 '17 at 0:58
  • $\begingroup$ A flimsy box is the case where x=0 in my answer... $\endgroup$ – DJohnM Jan 25 '17 at 4:00

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