From hydrostatics, it is said that when you have a liquid with no moving parts, the pressure does not vary on any horizontal slice of the liquid. The pressure only varies with depth.

Thinking about this led me to consider the following scenario that might contradict this.

Suppose I have water in a fully sealed and fully rigid container. We say it is located on Earth, so gravity is present as usual. Now suppose there is a small air bubble at the top of the container, as shown in the sketch below.

enter image description here

To a good approximation, the air bubble has a uniform pressure $P_{\text{Air}} = \text{const}$ everywhere inside the air bubble. The water presumably has a linear pressure gradient downward: $$ P(z) = \gamma z $$ where $z$ is the distance from the top of the container to the given level and $\gamma$ is the specific weight of the water. Maybe more accurately, the air bubble causes the water at the top to have some pressure as well, so maybe we should write $$ P(z) = P_{0} + \gamma z. $$ Let's say for the sake of the questions, the air bubble is $h = 3\text{ mm}$ in height.

enter image description here

Now the problem is that there seems to be a contradiction between the following three ideas:

  1. Pressure must be continuous with respect to position, including across the air-water interface.
  2. The pressure of the water must be the same at every horizontal slice, and it only linearly increases with depth.
  3. The pressure is approximately uniform/constant in the air bubble.

All three of these can't be correct simultaneously. Which one(s) is/are wrong?

What I really wonder is, what is the pressure distribution $P = P(x, y, z)$ throughout the whole container, and how can you derive this from first principles? I am really curious to know if there is anything out there that deals with these kinds of scenarios.

  • 1
    $\begingroup$ Remember to include surface tension. $\endgroup$
    – Jon Custer
    Aug 25 at 17:26
  • 1
    $\begingroup$ The first concept is incorrect, as interfaces cost energy. See Laplace pressure. $\endgroup$ Aug 25 at 17:58

1 Answer 1


There is a contradiction between those three ideas mainly because point 1. is wrong: pressure is not continuous across the air-water interface because of the surface tension.

Point 2. and point 3. are ok. In a continuous domain (water and air inspected independently) of a fluid at rest, Stevino's law holds, i.e.

$P(z) = P(z_0) + \rho g ( z - z_0 )$

with $z$ pointing downward. This formula readily describes the linearly increasing pressure in water (remember to use water density $\rho_{H_{2}O}$). For the air bubble, you can assume that the pressure is approximately uniform, if $\rho_{air}$ is small if compared with $\rho_{H_{2}O}$ (usually it is about $1/1000$), and anyway the small size of the bubble leads to a small difference of the $z$-coordinate in the air bubble.

The interface surface behaves as a membrane, and the pressure jump across it, at each point of the surface $\mathbf{r}_s$, can be described with Young-Laplace formula

$P_{air}(\mathbf{r}_s) - P_{H_2O}(\mathbf{r}_s) = 2 \gamma H(\mathbf{r}_s) = \gamma \left( \dfrac{1}{R_1(\mathbf{r}_s)} + \dfrac{1}{R_2(\mathbf{r}_s)}\right)$,

where $H = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$ is the average curvature of the surface, and $\gamma$ is the surface tension. The angle of contact between air, water and the solid walls of the container is typically a property of the combinations of the three materials, independent from the pressure.

Approximate solution

We can find an approximate solution, assuming:

  • that the bubble is really small, that the pressure in the water surrounding it can be treated as constant;
  • spherical symmetry (or better, cylindrical symmetry): since the pressure jump is constant, the mean curvature of the surface is constant and there is no reason (or at least, it seems so to me) to have different principal radii of curvature, and thus the bubble has the shape of a spherical cup with the flat surface corresponding to the upper wall of the container.

Since the contact angle is determined by the materials, the shape of the spherical cup is determined. The only thing that remains to be determined is the dimension of the bubble, i.e. its radius. The dimension of the bubble, and thus the pressure jump, the "offset" of the pressure profile in water and the pressure in the air bubble and all derived quantities, can be derived using

  • the conservation of the mass of the air trapped in the container
  • the equation of state of the air if we have some details about the filling process of the container, where air has been trapped.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.