Deuterium density in seawater

Reading a discussion on this topic on the XKCD forums, it seems there is a whole book dedicated to this question. If you check e.g. page 51, the graph there indicates a much more detailed structure than an uniform one.

To be more precise, there are several mechanisms in play:

  • evaporation
  • ice formation
  • ice melting
  • influx of fresh water

Several quotes:

Of all natural waters, the ocean, which is a unique reservoir, remains most constant regarding its isotopic and other physicochemical properties. Several authors (Craig and Gordon 1965; Craig 1961b; Epstein and Mayeda 1953; Friedman et al. 1964) have shown that ocean water at depth of more than 500 m is homo- geneous in isotopic composition. This allowed Craig (1961b) to propose it as a standard for reporting concentrations of $D$ and $\phantom0^{18} O$ content in natural water.

The variation in the relative content of $D$ in deep ocean layer is about 4 per mil and that of $\phantom0^{18} O$ is of 0.3 per mil. In the surface ocean layer, the regional variations, depending upon water temperature, are 35 per mil for $D$ and about 3 per mil for $\phantom0^{18}O$. The lowered content of $D$ in the surface ocean layer occurs in those regions where ice-melting water affects isotopic composition. In high latitudes, where the surface layer of the ocean is freezing, isotopic fractionation factor is abt. 1.0180 for $D$ and abt. 1.0030 for $\phantom0^{18} O$. In the equatorial region of the ocean where intensive evaporation of water takes place, there is an enrichment of the surface layer with heavy isotopes.

Despite the considerable homogeneity of the isotopic composition of ocean water, the ranges of $D$ and $\phantom0^{18} O$ variations with latitude and depth are wide enough to be reliable indicators of processes occurring in the ocean. These ranges, for deep ocean waters, are about ten times greater than the accuracy of mass spectrometer measurements with which modern techniques and apparatus permit determination of $D$ (+-0.2 per mil) and $\phantom0^{18}O$ (+-0.02 per mil).

Previous answers make it clear that there are many factors that could lead to temporary non-uniform distributions of deuterium atoms in the majority of "normal" hydrogen atoms in sea-water. So, if one were to pour a beaker of $D_2O$ carefully and slowly into a beaker of normal water, the heavy water would be found concentrated at the bottom of the beaker.

However, the OP's question seems to me to be: If we mix the stratified beaker above, will the higher concentration of deuterium at the bottom of the beaker be restored?

In my opinion, the answer is yes, with a serious reservation. The concentration would depend on how large the difference in gravitational potential energy is, compared to the basic kinetic energy of the atoms in the solution. My thought is: not much. How often do you find a layer of alcohol at the top of an long-undisturbed bottle of beer?

In a similar vein, leaking chlorine (from, say a tank car) does concentrate in lower areas. However, a gymnasium full of thoroughly mixed chlorine-contaminated air will never have a centimeters-thick layer of poison at floor level. The gas molecules are travelling at speeds of the order of the speed of sound. Just compare $\frac{1}{2}v^2$ with $gh$.