Why can’t you simulate isotropic fluid flow on a square lattice?

There are easy methods for discrete simulations of gas dispersion in two dimensions. If you take a large square lattice, each cell of which is assumed to contain at most one gas molecule, and you move the molecules from cell to adjacent cell at random, the large-scale results are in many ways a good simulation of gas behavior. In particular, even though molecules are individually moving in only the four cardinal directions, at a large scale the simulation does not show any bias to these four directions. If you start with a large square lump of material, it will rapidly diffuse into a circle. This property of the simulation is called isotropy.

Long ago when I was involved in research in this area I was told that this worked fine for gas flow simulations, but not for incompressible fluids. If one wanted to make an isotropic two-dimensional simulation of incompressible fluid flow, I was told, one had to use a hexagonal lattice instead of a square lattice. I was told that one could deduce from the Navier-Stokes equations that that any simulation of incompressible fluid on a square lattice would necessarily be anisotropic.

Is this correct? If so, how does the proof go? If the argument is complex, is it possible to get an intuitive idea of why compressible and incompressible fluids are different in this regard? What is a reference that would include the full proof?

Addendum: Despite the answer I posted below, I still don’t understand any of the details. I would be glad to award the bounty to someone who could explain it.

Answer

It seems that this might be the original source: Lattice gas automata for the Navier-Stokes equations, by Frisch et al. (PDF)

The [square lattice] HPP automaton is invariant under π/2 rotations. Such a lattice symmetry is insufficient to ensure the isotropy of the fourth degree tensor relating momentum flux to quadratic terms in the velocity. …
Observe that when the underlying microworld is two-dimensional and invariant under the hexagonal rotation group (multiples of π/3) the tensor T is isotropic…

Unfortunately it will be a long time before I am able to understand this properly; I would still prefer to see an answer that explains what is going on.

Attribution
Source : Link , Question Author : MJD , Answer Author : MJD

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