# Is the pressure-gradient force that causes hydrostatic equilibrium entropic, contact, or both?

This question considers classical fluid mechanics, particularly the kinetic theory of gasses.

In the simple one-dimensional gravitational setting, a fluid is in hydrostatic equilibrium if $$\frac{dP}{dh} = - \rho(h) g(h)$$, where $$h$$ is the height, $$P$$ is the pressure, $$\rho$$ is the density, and $$g$$ is the gravitational acceleration field. More generally, a fluid is in hydrostatic equilibrium if the internal pressure-gradient forces balance out any external forces acting on the fluid. Note that this equation makes no reference to entropy, temperature, or any other statistical mechanics concepts.

The standard derivation of this equation starts out by assuming that the inward internal force per unit area on a fluid parcel at each side is given by by the pressure at that side. But I don't actually see why this would be true for an ideal (i.e. noninteracting) gas.

The paradigmatic explanation of buoyancy imagines immersing an impenetrable solid within a fluid. In this situation, it's very clear how the buoyancy works: at the molecular level, air molecules are continuously colliding elastically off the surface of the solid object and thereby transferring tiny amounts of momentum to it via contact forces. If one could track this process with microscopic resolution, then the net force on the solid object could be easily calculated by just tallying up the momenta transferred from each particle to each face of the object.

But the molecules in an ideal gas are noninteracting by definition, so it isn't clear to me why an ideal gas would "feel its own pressure" and redistribute accordingly. One possible explanation is that the pressure-gradient force is purely entropic and only emerges at the macro level. This question raises that possibility, but the one answer seems to disagree that the internal forces are indeed entropic. It seems odd to me that an entropic force would not contain any reference to temperature at all (although maybe the temperature is "hidden" in the relationship with the number density given by the ideal gas law $$P = n T$$).

To make things even more complicated, different fluids self-interact in qualitatively different ways. Ideal gas are non-interacting, real gasses are weakly (but not always neglibly) interacting, and liquids are quite strongly interacting (which leads to phenomena like incompressibility). So for liquids, there clearly are physical contact (as opposed to entropic) forces at the microscopic level. And yet, despite the pressure-gradient forces' very different origin, they apparently follow the exact same simple equation.

Anyway, I'll stop rambling and get to my two main questions:

A. Which of the following claims is the most accurate?

1. Even an ideal noninteracting gas under generic initial conditions would eventually reach hydrostatic equilibrium solely because of entropic pressure-gradient forces, as suggested by the other SE question. (An interesting corollary to this answer would be that the weight of the column of air above you doesn't really determine the pressure at ground level, like it would for a tall stack of bricks. Instead, entropy maximization leads to an interesting nonlocal perfect correlation between the (nonlocal) weight of the column of air and the (local) pressure at ground level.)

2. Ideal gasses aren't ergodic and would never reach hydrostatic equilibrium. The weak intermolecular interactions in a real gas are the sole cause of hydrostatic equilibration. The process of hydrostatic equilibration is conceptually separate from the process of thermal equilibration, and you don't need stat mech to understand it (e.g. it might occur on a completely different time scale than it takes for the fluid to reach a uniform temperature).

3. Real gasses reach hydrostatic equilibrium because of both entropic forces and microscopic-scale physical forces, which are of comparable conceptual importance. You can't understand this process without fundamentally incorporating stat mech concepts.

B. Is the answer to A the same for incompressible liquids like water, which are necessarily strongly interacting?

It seems strange to me that the same equation for hydrostatic equilibrium holds for both gasses and incompressible liquids, despite their qualitatively different microscopic self-interaction behavior.

This question about water is somewhat similar to mine, but mine focuses more on the gas case, which I find more confusing.

Two thoughts occur to me.

First, we can model an ideal gas by allowing that there are short-range repulsive forces between molecules (the 'hard spheres' model), and the molecules themselves take up a negligible volume compared to that of the chamber holding the gas. In this model we will find that both Boyle's law holds and the internal energy is a function of temperature alone, so such a gas is correctly called ideal. In this case the standard argument about pressure gradient from forces on a thin parcel of gas will hold since each parcel of gas does indeed exert a force on surrounding parcels.

If, on the other hand, you want to think of the gas particles as never even hitting each other, then the thermalization takes place via collisions with the walls. In this case I think the standard treatment of chemical potential still holds and you can use an entropy argument to show that the density (and hence the pressure) is uniform in the absence of gravity, and there is a density gradient (and hence a pressure gradient) in the presence of gravity.

(I note that in the question you say an ideal gas is not ergodic and would never reach equilibrium. In order to maintain that you would have to treat collisions with walls as either not happening at all or else acting like simple reflections, which is not the case. I have pointed out above that in order for a gas to be ideal is not necessary for inter-particle collisions to be negligible. But even if you insist on reserving the term 'ideal' for a model with no inter-particle collisions at all (which I think would be a non-standard terminology), then collisions with the walls would still suffice to produce ergodic behaviour.)

• Thanks for the thoughtful answer. To clarify, I'm not claiming that any of claims #1, #2, or #3 are true - in fact, they're mutually exclusive. I'm asking which one of them is true. Why don't elastic collisions of classical point particles with an impenetrable flat wall act "like simple reflections"? I would think that they would. Jun 25, 2022 at 23:08
• So the punch line is that in real gasses as well as liquids, the interactions between fluid parcels are mediated via very short-range physical (not entropic) forces? Jun 25, 2022 at 23:12
• But if the pressure-gradient force is caused by physical interactions between molecules, then why doesn’t it depend on the nature of those interactions (e.g. the radius of the spheres in the hard-sphere model)? Are you saying that the macroscopic equilibration time is discontinuous and is infinite for zero interactions but then goes to a constant for infinitesimal interactions? Jun 26, 2022 at 13:31
• @tparker My answer is mainly focused on the desire to model a real gas using a model which is complete enough to capture the behaviour under study, but simple enough to be easy to calculate. The hard spheres model is such a model. I think you may be asking about another issue, namely what can be discovered about pressure gradient in a gas with no internal interactions at all. A case which may be useful to consider is a small flat solid object in such a gas. There would be forces on each side, and I think its buoyancy would go to zero if its density matched that of the local gas. ... Jun 26, 2022 at 17:01
• @tparker When a molecule arrives at a wall it lands and sticks to the surface. It remains for a short time called the dwell time, before being knocked off by another atom or shaken off by thermal motion. On departing it leaves in a random direction. Jun 27, 2022 at 20:12

Upon reflection, I think that the pressure-gradient forces within fluids result from a combination of entropic and physical forces.

I think the right microscopic way to think about these pressure-gradient is in terms of momentum transfer between small volume cells. (I'm taking an Eulerian rather than a Lagrangian perspective of the fluid flow, where we aren't tracking individual particles, but the net flow through fixed positions in space.) If a molecule with momentum $$2{\bf p}$$ freely passes from volume cell $$A$$ to an adjacent volume cell $$B$$, then cell $$B$$ gains momentum $$2{\bf p}$$ due to cell A. You could say that this momentum transfer results from an "entropic" force that cell $$A$$ exerts on cell $$B$$, since it simply results from the probabilistic diffusion of free particles.

On the other hand, consider a situation where a molecule with momentum $$-{\bf p}$$ in cell $$B$$ is initially on track to pass into cell $$A$$, but at the interface it collides elastically with a molecule in cell $$A$$ (due to a short-range repulsive intermolecular interaction), reverses its momentum, and stays within cell $$B$$. In this case, the net change of momentum in cell $$B$$ due to cell $$A$$ will be $$2{\bf p}$$, just as the in the previous situation. But in this case, the force causing the change in momentum is clearly a physical contact force, not an entropic force.

But from a macroscopic perspective, these two interactions have equivalent effects, which is that cell $$A$$ transfers momentum $$2{\bf p}$$ to cell $$B$$ at the interface. In a macroscopic system with many such momentum micro-tranfers happening continuously, how many of them are "entropic" vs. "contact" will depend on the microscopic nature of the interactions: it will be entirely entropic for a completely non-interacting gas, and entirely contact within a solid, and a combination of both within a real gas or liquid. But the pressure concept conveniently combines both types of effects together in a way that lets us abstract out the microscopic interaction details.