Isotropy of fluids vs. hydrostatic pressure (conflicting concepts to me)

Question

The formula for absolute hydrostatic pressure on Earth is

$$ p = ρgh + p_{atm}, $$

where $ρ$ is the density of the liquid, $g$ is the gravitational acceleration, $h$ is the depth of the liquid, and $p_{atm}$ is the atmospheric pressure just above the surface of the liquid.

What I feel is that this formula and the concept that pressure of fluids (gasses and liquids) are isotropic are conflicting.

In the derivation of the formula, you say that, since the fluids are static, $ΣF=0$. Then you analyze the forces on column in the liquid and say that the atmospheric pressure contributes with the downwards pointing force $F_{atm}$, and the weight of the liquid column, pointing downwards, is $F_{g,column}$. To counter-act this force, the pressure of the liquid acts at the depth with an upwards-pointing force $F_{pressure}$. Then you express the resulting forces and get $$ ΣF=F_{pressure}-F_{g,column}-F_{atm}=0 $$ $$ \Leftrightarrow F_{pressure}=F_{g,column}+F_{atm} $$ $$ \Leftrightarrow \frac{F_{pressure}}{A}=\frac{F_{g,column}}{A}+\frac{F_{atm}}{A} $$ $$ \Leftrightarrow p=\frac{ρgV}{A}+p_{atm} $$ $$ \Leftrightarrow p=\frac{ρghA}{A}+p_{atm} $$ $$ \Leftrightarrow p=ρgh+p_{atm}. $$

The way I understand the isotropy of the atmospheric pressure is that the forces within the air at each given point points in all directions, so that each force vector is canceled out. This is why we don't get compressed by the kilometres of airs above us, which is pulled down by gravity, or pressure imbalances in the air; the resulting force at every point in the air is zero, because fluids distribute forces evenly throughout themselves.

But if that is the case, why does the atmospheric pressure affect the pressure in the liquid? In the derivation, the force of the atmospheric pressure points downwards AND its magnitude is EXACTLY equal to the atmospheric pressure just above the surface of the liquid (at sea level, it would be about $101325 Pa$). It feels like it should be irrelevant, since it is cancelled out by the isotropic nature of fluids. But it is an added constant in the formula, being derived from seeing it as an externally downwards pointing force.

How can these two facts coexist — the cancellation of the 'atmospheric force' via isotropy and its significance as an added constant in the formula for hydrostatic pressure, being equal to the atmospheric pressure?

Answer 1

Here's a related question; let's say we extend an ideal, massless spring some distance beyond its rest length. As any introductory mechanics book will explain, the condition that the spring is massless requires tension to be uniform throughout the spring, so that the net tension cancels out everywhere within the spring. But the tension doesn't just disappear; the spring still carries it, because the ends of the spring don't experience isotropic tension, and thus the spring can exert a force through its ends.

Similarly, you can think of ocean particles roughly as vertical layers of little springs. Each layer has to support everything else above, so the lower layers are increasingly compressed, supporting increasing tension. If we isolate any spring in particular, it experiences no net force, as required by hydrostatic equilibrium. However, we can still compute the forces of pressure on little slices of water, since the springs at the edges of the slice can exert forces. The key idea is that, in this analogy, pressure isn't a measure of the net force on a given spring - that is always zero, as you point out. Rather, it's a measure of the compression of a particular layer of springs, which increases as you travel down the ocean. The effect of the atmospheric pressure is similar to the effect of adding a mass on top of the layers of springs, forcing the top springs to compress more to support the added weight, and that compression propagates downward to maintain mechanical equilibrium (this is Pascal's law).

It's worth noting that the above model has several limitations. For one, water doesn't actually compress in the same way that springs do. Also, there's the more nuanced point that whereas springs can only exert forces parallel to themselves, pressure acts "in all directions." If you're interested, the general treatment of pressure requires the stress tensor, which also accounts for anisotropic pressure. For fluids, however, the isotropicity of pressure means that all the messy internal contributions to pressure "cancel out," so you only have to consider the forces due to pressure at the edges of the parcel you're examining.

Answer 2

But if that is the case, why does the atmospheric pressure affect the pressure in the liquid?

The confusion appears to originate from a conflation between forces and pressure. In particular, the statement "the force of the atmospheric pressure points downwards AND its magnitude is EXACTLY equal to the atmospheric pressure just above the surface of the liquid (at sea level, it would be about 101325Pa)" is false; a force is not a pressure, and a force has units of newtons (in SI), not pascals.

At any point in a fluid, the forces act in all directions and sum to zero; thus, any small fluid region does not accelerate.

However, the pressure does not sum to zero; it itself arises from the forces acting in all directions.

If a liquid lies under an atmosphere, the top layer of the liquid has a pressure equal to that of the bottom layer of the atmosphere; we can show that from a free-body diagram and the equations of equilibrium.

The forces applied by the liquid container respond to this pressure accordingly. The pressure is isotropic at any point.

In addition, the pressure moving downward in the liquid increases with increasing depth (just as the atmospheric pressure does), with the reference value at the surface being the atmospheric pressure there. In other words, the pressure isn't isotropic over a finite region; it varies to balance out the weight of that region.