Why is liquid pressure the same in all directions?

Question

I have a question which I have been unable to find a satisfying answer to: how come in liquids, given a depth, the pressure exerted by that liquid is the same in all directions? I have read this thread and this thread already, but they did not make any wiser.

I have seen many derivations of the classic formula $p = \rho g h$ but these all feel a bit dubious since the only force being considered is the weight of the liquid "column" being considered, which acts downward. Nothing about this seems to indicate that pressure would be the same in all directions, only from above. What causes the pressure to act upward? Or from the sides?

Intuitively, I thought this may have something to do with how energy gets distributed in the water; the water molecules get pulled downward by gravity and the increase in kinetic energy due to gravity gets distributed to nearby water molecules through random collisions. Since the collisions are random, the movement of the affected water molecules would also be random, but the average kinetic energy of the water molecules at a certain depth would never change. This would mean the pressure caused by these random collisions would be the same from all sides, and thus only depend on depth.

This is something which has always bothered me so I hope you have some insights!

Answer 1

By definition, ideal fluids have no shear strength; any shear stress causes them to rearrange freely, as the molecules aren't all that well bonded.

There is only one stress state that involves no shear in any orientation: the equitriaxial (same in all three axes) or dilatational stress state. The compressive version is known as hydrostatic stress or uniform pressure. So this is what we find along any horizontal level of a fluid at rest (or any point at all, if gravity is ignored). If we try to introduce a deviatoric stress, the fluid tends to rearrange to eliminate it through flow.

Answer 2

This is actually one of the things that is easier thought of using forces rather than energies. And some of the answer is that "fluid pressure is the same in all directions" is a simplification. It's a good simplification, but I think the steps they take might be causing you issues.

First off, let's consider pressure without gravity. Perhaps its fluid inside a container on the International Space Station. Fluid pressure is the statistical measure of the effect of myriad collisions between the fluid molecules and the environment (such as the tank holding the fluid). The reason why pressure is the same in all directions is because the molecules are all bouncing off each other unpredictably. Were there to be a difference in pressure on one side versus the other, then that environment (e.g the wall) would be applying an unbalanced force to the fluid, and statistically, the fluid would accelerate away from the wall. In practice, unless you're dealing with impulses near the speed of sound in the fluid, we ignore the statistical bit and instead just treat the fluid as one body which gets accelerated. So there's the first piece of the puzzle: if pressure were to be different from different directions, the fluid would accelerate. The rule that pressure is the same in all directions assumes no acceleration.

But what if there is a force? Take the same fluid container, and now put it on Earth. Now gravity is affecting every molecule. In this case, we find that the pressures must not be equal on all sides of the container. The bottom must have more pressure because it must apply a force to counteract the force of gravity -- else the fluid would need to accelerate. This force is computed exactly as you say: $P=\rho g h$, multiplied by area to yield $F=\rho g h A$ where $A$ is the area of the bottom surface.

So how do we get from "the forces on top and bottom are different" to "the forces are the same in all directions?" First we consider a small slug of fluid. There's no obligation that our "container" be the outside walls of something solid. It could just be a small volume of fluid, where the pressure pushing in from the nearby volume of fluid is exactly equal to the pressure pushing out from our own volume of fluid. The net behavior is the same as in the example enclosed by a physical tank: the fluid does not accelerate because there is no net force on the fluid. Were there to be a net force, it would quickly be distributed from the molecules closest to the source of the force to the rest of the body (the rate this happens is actually the speed of sound in that material).

What if we make this volume shallow? What if we take a thin slice parallel to the ground (perpendicular to the force of gravity?) Let's consider the difference between the pressures at the top and the pressures at the bottom. With a simple subtraction we see that $\Delta P = \rho g \Delta h$. So the smaller the difference in height between the top and bottom of your volume, the less the difference between the pressures.

When we say "fluid pressure is the same in all directions," we typically mean this to be true at a point. A point's height is 0, so we trivially get $\Delta P = 0$. The forces top and bottom are the same. And since the only asymmetry in the problem is the top-vs bottom difference due to gravity, it's not a big stretch to see that pressure should be the same in all directions (handwaving math a bit).

The weirdness is when you try to figure out how, if fluid pressure is the same in all directions, you somehow get a difference in pressure across top and bottom. It feels like adding 0 a bunch of times cannot possibly result in a positive value. To resolve this, one needs calculus. Calculus deals which such infinitesimal values rigorously, yielding the intuitive physics we expect. In calculus, we might write $dP=\rho g dh$, where the lower case "d" has the same meaning as $\Delta$ did in the earlier paragraphs, except it's an infintessimally small amount (or, even more rigorously, we'd write $\frac{dP}{dh}=\rho g$, but I chose to write the form that better mirrors the other equation). We can see that the "limit" of the difference in pressure as $dh$ goes to 0 is, in fact 0. This lines up with the idea that fluid pressure is the same in all directions. However, if we sum them all up across some real height, using what calculus calls an integral, we end up calculating $\Delta P=\int \rho g dh$, which accounts for the fact that we're summing an unlimited number of infinitely small slices.

If you haven't taken calculus, you may just have to take my word for it. It quite literally took millennia to invent (I like to measure it from Zeno's paradoxes being posed to when they finally were resolved by Newton and Leibnitz). Calculus does bridge these two apparently paradoxical definitions (pressure on a volume of fluid is greater on the bottom than the top vs. pressure is the same in all directions). That bridge just happens right near the $\Delta h = 0$ mark where math got really squirrely until calculus straightened it out.

Answer 3

Imagine we hold a thin tube so that its lower end is at depth $h$ in the container of liquid. The tube is bent so that its lower, open end faces in a direction of our choice. The tube has a moveable piston, area $A$, near its lower end.

We push the piston a little way, $\Delta x$, outwards, so a small volume, $A\Delta x$ of liquid is displaced from depth $h$. This is equivalent to moving $A\Delta x$ from depth $h$ to the surface. There is a gain in potential energy of $\rho A\Delta x g h$. Assuming no viscous forces or friction, the gain in PE must equal the work done pushing the piston against the liquid pressure, $p$. So $$pA\Delta x= \rho A\Delta x\ g h.$$ So we have $$p=\rho g h.$$ Note that we've shown this to hold whatever direction the piston is facing.

This answer might appear not to make use off what we all know about fluids: absence of shear strength. But the energy equation above DOES rely on this property, as it insists on there being no (dissipative) viscous forces (achieved to as good an approximation as we like by very slow fluid movement).

Answer 4

If a small hole is made at a depth $h$ of a column of water, the velocity of the horizontal water jet outside and close to the column is $$v = \sqrt{2 g h} \implies \frac{1}{2} \rho v^2 = \rho g h$$ The kinetic energy inside the column is practically zero except very close to the hole (if that is small compared with the volume of the column). So this relation can be viewed as the work energy theorem. Some horizontal force (per unit of volume) accelerates the water horizontally from zero to the final velocity, and the work done is the change of the kinetic energy. $$F_v\Delta x = \rho g h \implies \frac{p\Delta A}{\Delta V}\Delta x = \rho g h \implies p = \rho g h$$ But that can be shown to be the vertical pressure. So, the horizontal pressure is equal to the vertical pressure.

Answer 5

Liquids have equal pressure in all directions because if you push down on them they try to spread out sideways where they meet resistance for example from the bottom of a container. But the sideways spreading force gets deflected off the walls of the container and back to where the push was initiated.This can only happen because liquid molecules can move freely - a solid wouldn't be able to do this because it's molecules are fixed in place.Hence water can have equal pressure in all directions but not ice.