What would be the optimal diameter of gold grains if you want to fill a pool with them and swim in it? I was wondering, if I were to fill a pool with gold/gold dust, what would be the optimal diameter of the grains to minimalize the friction to a point at which you would be able to swim in it ((and if swimming is not possible, what size would pose the least resistance))?
This is assuming the grains are spherical and the normal gold-on-gold friction coefficient applies.
Thanks!
 A: It's not so much about a grainsize. As long as the grains are big enough to aloud air(gas) flow through them, you can make them liquid by"Fluidization"; 

"Fluidization (or fluidisation) is a process similar to liquefaction whereby a granular material is converted from a static solid-like state to a dynamic fluid-like state. This process occurs when a fluid (liquid or gas) is passed up through the granular material."

Basically You can easily swim in Gold, sand or any solid granular material if you fill a whirlpool with it, and blow air through it like shown in this video.
Edit: The given comment forces me to expand the answer. This is about Buoyanycy. It has nothing to do with density. (Explained in video at 6 min->) Another aspect is terminal velocity. If the gas speed is higher than the terminal velocity of the object floating in fluidized granulars, then the object will raise to the surface.
The terminal velocity of Human is 53 m/s, which means that it can be written; 
$$\frac{v^2}{2}=\frac{p}{\rho}$$
Where we know the $\rho=19.32$ (Gold, $kg/m^3$) and $v=53 m/s$, so we can solve; 
$$\frac{v^2\rho}{2}=p=27134 Pa$$
Which alouds us to solve the maximum granulat size. 
The area of the granulat is; 
$$A=\pi r^2$$
And the weight of the granulat is; 
$$m= \rho \frac{4}{3} \pi r^3$$
Which produces a simplifed model to calculate the maximum $r$, as $p=\frac{F}{A}$ and $F=ma$ and $a=g$, so
$$ p=\frac{\rho \frac{4}{3} \pi r^3 g}{\pi r^2}=\rho \frac{4}{3} \pi r g$$
So the r can be solved to be; 
$$ \frac{3p}{4 \pi g}=r$$
Which with $p=27134 Pa$ gives r= 660 m. 
(Somehow I don't believe this answer's magnitude, but I leave it here, and get back to it later. Good night)
