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?


2 Answers 2


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.

  • $\begingroup$ @Motl does that mean that when we say the liquids are really incompressible, we come to the conclusion that liquid pressure does not change with depth and we do this by violating the nature of forces? $\endgroup$ Oct 2, 2018 at 18:13
  • $\begingroup$ No, I didn't write anything of the sort and what you wrote is clearly wrong. The pressure depends on the depth as $h\rho g$, it's easy to calculate, if we talk about the same thing. Incompressibility really means the exact opposite of what you think. It means that it is easy to change the pressure by a big amount without changing the volume. $\endgroup$ Oct 3, 2018 at 5:59
  • $\begingroup$ @I'd really appreciate if you eloborate on how the pressure changes with depth in molecular level. Thank you👍 $\endgroup$ Oct 3, 2018 at 6:19
  • $\begingroup$ @Motl, I mean if incompressibility did not fail and the density did not change, how would the number of bombardments increase with depth in a liquid. I just can't understand this. $\endgroup$ Oct 3, 2018 at 6:37
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    $\begingroup$ @Motl, thank you. This was actually exactly what I had in mind after reading your original answer. I guess I just coulnt express myself well. Perfect 👍 $\endgroup$ Oct 6, 2018 at 6:06

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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