Say I have a series of tubes (not the internet) looking like this, where w represents water:

|    |      | |
|    |      | |
|wwww|      |w|
|wwww|      |w|

Why is it that if I put some water in either side, the water level changes on the other side until they equalize? Why does this work even if I put water on the thinner side? And why is it that when you tilt the whole contraption, the water level is still even, although slanted?

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    $\begingroup$ A basic principle of the internet is "net neutrality", meaning everything on the internet should be at the same level. The internet is a series of tubes, so water in a series of tubes must be at the same level. (PS - this is a joke! Real answer below). $\endgroup$ – Mark Eichenlaub Nov 15 '10 at 1:51

In a liquid like water, the pressure acts in an isotropic way.

That being said, imagine a slice of water in the middle tube; what are the forces acting on this slide ?

The force exerted by the pressure on the left side, and the one on the right side.

The one on the left depends and the height of the water column in the left pipe. The one on the right depends on the height in the right pipe.

If you want equilibrium, both have to be equal. Therefore the heights have to be equal.

About the pressure: pressure has dimension of force divided by surface, in common units: $N/m^2$.

The column of water on the left pipe exert a force, due to its weight (gravity) that is $g \rho S h$, where $\rho$ is the volumic mass, S is the cross section of the pipe and h is the height.

But the pressure is $g \rho h$, thus independent of the cross section of the pipe. This force (gravitational) acts downward, but the fluid make it acts in an isotropic way, thus being directed from left to right on the slice of water (see above).

A good schematic explanation is available in hyperphysics.


Altough the diameter of the left pipe is bigger, the force exerted on the "slice" of water is not higher because the pressure on a given infinitesimal volume depends only on the height of the column of water above it. Imagine two simple straight vertical pipes filled with water and with equal height, one with a large diameter and the other with a smaller. It is true that the force on the bottom of the big one is higher, but the pressure will be the same, because the force acts on a larger surface.

  • $\begingroup$ Hmm could you go more into how pressure is calculated? Doesn't the weight of the water have anything to do with it? Like there is more water on the left pushing on the right, so why doesn't the left side's height have to be smaller to have equilibrium? $\endgroup$ – Claudiu Nov 14 '10 at 22:59
  • $\begingroup$ @Claudiu: The weight has everything to do with the problem, it is "h" in the notation I used. Regarding your second question, pressure is exerted by one slice on another but it is not "additive". $\endgroup$ – Cedric H. Nov 14 '10 at 23:08
  • $\begingroup$ ah yeah that's the issue I haven't realized yet.. that only height matters for pressure, not diameter. it makes sense that the force is greater, but pressure is equal, since the force is on a greater area.. so what about if you have two tubes, one of a given width, and then another with equal height, but whose top is much wider, that tapers down to the bottom having the same width as the first. why is the pressure still equal then? isn't that more force on smaller area? $\endgroup$ – Claudiu Nov 15 '10 at 1:39
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    $\begingroup$ No because a fluid cannot sustain transverse constraints. You have to consider only the water just on top of the point (infinitesimal volume) where you want to have the pressure. The force exerted by the water is compensated by the wall of the pipe when they are not parallel. $\endgroup$ – Cedric H. Nov 15 '10 at 1:42

Another way to think of it is to take the rule that the water wants to minimize its potential energy. (This is a basic principle of statics.) The potential energy is just the average height of all the water, times its weight.

If the water level were higher on the right, we could skim a little water off the top there and dump it on the left. The water we moved would go down some while the rest of the water would stay at the same height, so the average level would go down. Hence, it wasn't at a minimum before. Only if the water level is the same everywhere is there no way to skim a little from one spot, dump it somewhere else, and reduce the potential energy.

Caveat: see comments on this answer for a little more explanation

  • $\begingroup$ This is not quite correct. The point about system "wanting" to attain a (local) equilibrium is completely omitted (it might not be clear to someone). If there was no friction, your argument would not work. On the other hand, once you require stability then thermodynamics already tells you that energy has to be minimized. So stability is an important point here and minimization of energy is only a corollary (although one that is very useful to keep in mind). $\endgroup$ – Marek Nov 15 '10 at 9:11
  • $\begingroup$ I don't completely understand you objection, but how about this: Suppose the equilibrium position of the water did not have the water level totally flat. Then I scoop some water from a high place and dump it on a low place, but as the water falls, I use the falling water to drive a water wheel. Now the water goes back to its equilibrium, and I repeat the process. This way I get infinite energy. Thus, if the equilibrium height profile of the water were not flat, the second law of thermodynamics would be violated. $\endgroup$ – Mark Eichenlaub Nov 15 '10 at 9:17
  • $\begingroup$ ...or the first law $\endgroup$ – Mark Eichenlaub Nov 15 '10 at 9:30
  • $\begingroup$ My objection is you didn't even mention equilibrium in your answer (perhaps dismissing it as obvious). But it's not obvious at all: without thermodynamics, there would be no equilibria. Equilibrium is attained precisely because of the second law (as you correctly state) and I'd like you to mention some of this in your answer. But that's up to you of course. $\endgroup$ – Marek Nov 15 '10 at 10:02
  • $\begingroup$ Okay, I see your point. Thanks for pointing it out. I think most people reading the answer will read these comments as well, but I'll just throw in a reference. $\endgroup$ – Mark Eichenlaub Nov 15 '10 at 10:06

The short answer is: the system (after long enough time passes) wants to attain a configuration which is stable (we say the system is in equilibrium).

Now would the system be stable if one side had higher column of water than the other? Of course not, because the pressure at the bottom would be different between the two sides, introducing some forces. It is these forces which lower the height of the higher column and vice versa. Now, if the system were without any friction this would result in harmonic oscillations with water going up left column, then right, then left, ad infinitum. But because of friction, energy is lost in the form of heat when water moves and it pretty quickly attains a stable configuration.

Still, I suppose you should be able to observe few oscillations if you make the height difference really big.

The slanted case is completely the same. Applying above intuition about stability, it should be clear that the surface of the water must always be perpendicular to the gradient of the gravitational potential and the surface must be a level surface of the gravitational potential (otherwise you introduce some nonequilibrium forces). In homogeneous potential this results in a plane perpendicular to the direction to the center of the Earth.


water pressure increases with depth, from the surface straight down in relation to gravity, so the pressure is the same at the same depth, whether the tube is 6 inches wide or 1 inch wide. raising the level in one will increase the water pressure at the bottom. so water will flow into the other tube until the pressure is equal, therefore the upper levels will be even horizontally. tilting it does the same thing water will flow through the connecting tube until the upper levels are even horizontally.


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