# At the molecular level, how is the pressure at the bottom of a lake higher than at the top?

Surely the temperature of the molecules is the same throughout the water. Using $p = \rho g h$ seems to assume a constant density as well. But then how is it that the force per unit area on an object placed at the bottom of the lake will be higher than that on an object near the surface? My first thought is that the incompressible assumption is the approximation at fault, but it doesn't seem that a minute increase in the density of molecules at the bottom of the lake could account for many times more bombardments per square centimeter. What am I missing?

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Actually, the temperature of water isn't normally uniform throughout a lake. The water may be heated by sunlight or geothermal vents, or cooled by evaporation or conduction into the Earth, etc., and any of these processes will have a quicker and greater effect on the part of the lake near where they are working. But this question is still meaningful even if you're working with a body of water small enough that the temperature gradient is negligible. –  David Z Jan 20 '12 at 5:49
It is very simple. It is all about the mass of water above. That is all. Basically once water is in a liquid state, the molecule will not change much if at all, it is too simple and there are no voids to 'crush'. –  SkipBerne Jul 2 at 15:37

First of all, the temperature and pressure of a liquid are two independent intensive variables. Either of them may be lower or higher at the bottom of a lake, independently of the other. So let's focus on the pressure.

You are totally right that the incompressibility fails and this is the reason why the lake "knows" about the higher pressure: the density of atoms or molecules becomes somewhat higher. But it's still true that the liquid is "approximately incompressible" and this is exactly the reason why the changes of the pressure are so great even if the changes of the density are minuscule. In other words, $$\frac{\partial p}{\partial \rho}$$ is a very large number or, equivalently, $$\frac{\partial \rho}{\partial p}$$ is a very small number. That's what we mean by (approximate) incompressibility and that's why the changes of the density unavoidably linked to finite (or even huge) changes of the pressure are so tiny.

Liquids are nearly incompressible because of Pauli's exclusion principle; the electrons in the atoms are just not allowed to occupy the same state. One has to change the structure of the states but this, because of the repulsion of the charged nuclei and electrons from each other, leads to immense increases of energy. That's why the density of liquids (or, equivalently, the volume occupied by a single atom or molecule) is de facto calculable independently of the pressure.

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In a hard-gas model, if you make the atoms spheres, there are indeed lots more bombardments low down in the lake than high up, since hard sphere collisions are the only source of force, and the temperature (hence the mean velocity) is the same. The reason this is violating your intuition is because you are approaching the continuous collision limit, so that when the hard spheres are nearly touching, the number of collisions per-second diverges. The differences in pressure in the hard-sphere gas lead to many more collisions at the bottom then at the top.

But this model is stupid, because the force between atoms on the scale of their separation in a liquid is smoother than a hard sphere. In the real case, you just push the atoms together a tiny amount, and they push very much more against each other, but in a smooth way.

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