Suppose we have a container of water with dissolved salt in it. Though it begins as a homogeneous mixture, gravity will have a tiny effect on the distribution of the ions. Another user found the result here:
The drift velocity of a particle of mass $m$ under gravity in a fluid of viscosity $\xi$ is $mg/\xi$, from which it follows that the relevant diffusion equation is
$$ \frac{\partial C}{\partial t} = D\frac{\partial^2 C}{\partial z^2}+\frac{mg}{\xi}\frac{\partial C}{\partial z} $$
The steady-state ($\partial C/\partial t=0$) solution is of the form
$$ C(z)=\alpha \frac{D\xi}{mg}e^{-mgz/\xi D}+\beta $$
where $\alpha,\beta$ can be solved by imposing appropriate boundary conditions.
where if you plug in numbers for salt ions, you get basically no variation in a modestly sized container of water.
However, I'm wondering if this solution is ignoring the other effect of gravity, buoyancy. Once water at the bottom becomes slightly saltier, I'm thinking it should become more dense and then prefer to sink and not mix as well with the less-salty water above it.
I found an answer on the web here, which says that the expected salt concentration by depth would be (using $C$ and $z$ for the concentration and depth instead):
$$C(z) = C(0)\exp\left(-\frac{mgz}{2kT}\right)$$
I don't know if this answer is right. He says it is analogous to the "law of atmospheres", but $kT$ shows up there because of the ideal gas law: pressure is related to temperature. The case of salt concentration in water seems different, although the diffusion constant above depends on temperature. Is there some kind of buoyant effect like this for salt dissolved in water or not? Which of these, if either, is the right answer? The exponential factors are about 8 orders of magnitude apart between them.
