I'm quite confused with Laplace pressure. I know the formula is (at least considering a spherical surface) $\Delta P = P_{in}-P_{out} = \frac{4\sigma}{d}$. What exactly is the surface you have to consider? And I've heard and read that this is the pressure that occurs when two different phases meet, how do I know which pressure corresponds to which phase. Let's say water and air, is $P_{in}$ the one for water or air? Also, I know pressure doesn't really has a direction but I've seen many images that associate this extra pressure as being directed radially outside the liquid phase and actually some other which state that it is directed radially to the inside of a liquid. Which one is right? Hope someone can answer my questions.


2 Answers 2


Think of a balloon.

The surface of the balloon has some tension which wants to make the balloon smaller. This surface tension necessarily means that the pressure inside the balloon is larger than the pressure outside the balloon. The issue is not which one has which substance, the issue is which side of the balloon is concave. Concavity drives the direction.

This principle is similar to a guitar string: if I put my finger on it and pull it up, then the concave side points down and the guitar string wants to pull my finger down; if I put my finger on it and push it down, then the concave side points up and the guitar string wants to push my finger up. If you think of the surface as being made up of a mesh of springs, those springs always pull in the direction of the surface, and concavity is what directs that force on the average in one direction or another.

Another common physics trick: if something is static, you can determine the direction of the one force by removing the constraint that keeps it static. Just let the thing do what it “wants to” do. So if I tell you that the bottom hinge on a door actually pushes the door away from the wall, you can quickly visualize this by imagining what would happen if the hinge was not there, at which point the door would rotate into the wall. So this force must have been pushing it away from the wall. Similarly if we let the surface tension do what it “wants to do”, what does it do? It shrinks the bubble. The only reason it can't is because the air/water is incompressible and puts up a pressure to oppose it.

In fact the $4$ in your equation is, I think, twice as big as I remember. If I am right then in that case, what we are talking about is a soap bubble, we have air inside and outside, and a thin film of liquid between the two. So we have an air-water-air situation, and if the force were consistently from air to water or from water to air, then the pressure inside the bubble would have to be exactly the same as the pressure outside because you would be adding and subtracting the same number. But because it is not “which side is which substance,” it is only “which side is concave,” the two pressures stack rather than cancel.

  • $\begingroup$ Thank you for your time and answer. It's clearer to me now, and If I'm not mistaken the 4 is actually correct, the thing is that the equation I wrote has the diameter instead of the radius of the surface. $\endgroup$
    – Kevin
    Commented Sep 6, 2020 at 0:09
  • $\begingroup$ @CRDrost Sir , may you tell why we are able to give the terminilogy of surface tension in case of balloons too , as such what i know is that it arise due to surface energy changes happening in a fluid surface is balloons really a type of fluid you mean ? In this balloon how are we able to derive that laplace relation as such i know how to with respect to fluids. And do we take one or two surfaces with respect to balloon ? $\endgroup$
    – Orion_Pax
    Commented Apr 12, 2022 at 19:00
  • $\begingroup$ I would recommend asking your questions as their own question, I am unlikely to have free time this week and unlikely to remember your comment after this week $\endgroup$
    – CR Drost
    Commented Apr 12, 2022 at 22:56

The Laplace pressure reflects the energetic cost of creating additional interfacial area, as more surface area corresponds to additional unsatisfied bonds. (This is of course the origin of surface tension.) $P_\mathrm{in}$ is on the enclosed side of the curvature and is higher than $P_\mathrm{out}$ because Nature would essentially prefer the enclosed region to have a smaller volume to obtain a smaller surface area. It doesn't matter whether the enclosed material, i.e., the material on the concave or internal side of the curvature, is the water or air.

Annotating the associated schematic with arrows risks confusion, as is always the case when a proper free-body diagram hasn't been constructed. That's one reason why the images you saw might have implied contradictory results.

  • $\begingroup$ Thank you for your time and answer. It's clearer to me now $\endgroup$
    – Kevin
    Commented Sep 6, 2020 at 0:07

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