# On a liquid-air surface, does the pressure gradient always have the same direction as the normal vector of the surface?

I am reading a CFD paper http://www.cs.columbia.edu/cg/surfaceliquids/droplets.pdf. In page 5 paragraph "Tangential velocity", it says that in a liquid-air surface $\Gamma$, the pressure gradient $\nabla p$ always has the same direction as the normal vector $\bf n$ of surface $\Gamma$, where $p$ is the pressure term in the incompressible Euler equation. They argue that

For the scenarios we consider, air density is negligible compared to liquid density. Imagine that the pressure gradient $\nabla p$ has a tangential component: it would effect an infinite tangential air velocity, in turn instantly restoring equilibrium.

I am not convinced by this argument. I think it does not hold when we consider the external body force (such as gravity) and surface tension. Here is my counter-example: Imagine we have a cup of water. Due to the gravity, the pressure gradient on the surface is vertically downward. However, due to the surface tension, the direction normal vector of the liquid-air surface is not always vertical.

Am I wrong?

• I don't understand your counterexample. Why should surface tension make 'the direction normal vector of the liquid-air surface ... not always vertical'? – lemon Oct 14 '16 at 9:20
• @lemon Here I mean the normal vector of liquid-air surface of the water in a cup is not in the direction of gravity. This is because the liquid surface has curvature due to the surface tension. – Yichao Zhou Oct 14 '16 at 9:28

If surface tension is neglected, then the free surface is a constant pressure surface (matching the air pressure), and the pressure gradient must be normal a surface of constant pressure. If surface tension is included and the curvature is changing along the surface, then immediately below the surface, the pressure is changing in the tangential direction. Therefore, the pressure gradient is not normal to the surface.

• Yes, but surface tension wouldn't really come in the middle portions but only at the edges, which the article doesn't seems to discuss – Pranshu Malik Oct 14 '16 at 13:38
• Thus it would be normal always (for the case under discussion) – Pranshu Malik Oct 14 '16 at 13:38

At the edges it won't be vertical, so just ignore the edge effects (since the pressure gradient you are talking about is for an infinite area). At the surface, the average force on all particles is normal (downwards) to the surface because of symmetry in the attraction forces. All extend like a bunch only in the downwards direction (semi circular laterally), thus leaving just one normal force which is also the reason for surface energy and is the surface tension. Your counter argument is supportive of the idea conveyed in the passage.

P.S. The edge effects come in due to the absence of particles or bulk phase on one half and the adhesive forces of the cup.

• I don't fully understand your point. What does the "infinity area" for pressure gradient mean? You mean for every point on the liquid surface, the pressure gradient is in the normal direction of surface? So the pressure gradient is not continuous? – Yichao Zhou Oct 14 '16 at 10:04
• What I mean by infinite area is symmetry in all directions. So that you have the entire bulk phase below the surface attracting the molecules on the surface. Due to symmetry the horizontal components cancel out leaving behind only the net downward force. This wouldn't be possible if you were at the edge of the cup, since the coffee bulk wouldn't be on one side and thus cover only a half of the bulk available to the surface at the exact center. – Pranshu Malik Oct 14 '16 at 13:37

"...$\nabla p$ always has the same direction as the normal vector $\textbf{n}$ of surface $\Gamma$..." which is correct. Nowhere do they say that $\nabla p$ is vertical. If fluid surface is curved, then normal vector changes and $\nabla p$ will be along the normal vector at that point.

Where the interface is curved, there will be jump in pressure across the interface, but that change in pressure is again along the normal to the surface. As they have said $\nabla p$ tangential to the surface cannot be sustained because there is no force present in that direction that would balance it.