Part 4 of 6. Hardy, Pomeau and de Pazzis's square-grid lattice gas (1973) has a viscosity that depends on flow direction; Frisch, Hasslacher and Pomeau's hexagonal one (1986) obeys Navier–Stokes. One rank-4 tensor accounts for the difference.
Part 3 showed that a rule treating its neighbours alike keeps the symmetry of its lattice. A lattice gas asks whether that symmetry is enough for the gas to flow like an isotropic fluid. Particles sit on the sites of a lattice, each with one of a few discrete velocities, at most one particle per velocity per site. Each step, every particle hops one link along its velocity, then the particles at each site scatter by a local rule that conserves mass and momentum. A continuum fluid is symmetric under every rotation. The square lattice keeps four of them and the hexagonal lattice six; the question is whether the averaged flow can tell.
Both rules conserve mass and momentum exactly, and both can conserve more than a fluid does. No HPP velocity has both an x and a y part, and a collision swaps an x-pair for a y-pair on the same site, so the y-momentum of every column never changes: one extra conservation law per column and per row. FHP with head-on collisions alone keeps the momentum Da along each of its three lines separately, three numbers where momentum has two. Because c0 − c1 + c2 = 0, the combination Q = D0 − D1 + D2 carries no momentum at all, and nothing but the three-body rule can change it.
Figure 1. Two copies of one 64×74 periodic FHP gas, started with extra particles on the 0°, 120° and 240° links: large Q, net momentum only sampling noise. Rosettes: particles per direction at the chosen step, against the mean (hexagon). Without the three-body rule the triangular imbalance never relaxes; with it, Q is gone within about ten steps. Neither gas changes its momentum.
Averaged over many sites and steps, a Chapman–Enskog expansion turns the Boolean dynamics into equations for density and momentum. Both the convective and the viscous parts of the momentum flux come out built from one tensor, the fourth moment of the velocity set:
For the flow equations to be rotation-invariant this tensor must be isotropic. In two dimensions the only fully symmetric isotropic rank-4 tensor is
which is the same in every orthonormal frame. So the test is to rotate the frame: take a unit vector n at angle θ and its perpendicular t, and write T in components along them. An isotropic T gives the same numbers at every θ, with Tnnnn = 3Tntnt. Tntnt is the component that turns a shear along n into stress along t.
Figure 2. Left: each set's fourth moment in the rotated (n, t) frame, indexed by pairs of directions and scaled so the second moment is the identity. As θ turns, FHP's entries stay put and HPP's trade between nnnn and ntnt. Right: Tntnt over every direction of n (drag the handle, or use its arrow keys). HPP's clover touches zero on the axes; FHP's is a circle. Eight directions with equal weights err the other way; slide their axis : diagonal weight to 4 : 1, D2Q9's 1/9 to 1/36, and the curve is a circle.
Adding diagonals does not raise the point group: the eight-direction stencil is still D4. It adds a second speed, and with it a weight that can be tuned. That is the route Part 5 takes.
For a regular polygon of n unit velocities, the rank-k moment along n is f(θ) = Σa cosk(θ − 2πa/n). Expanded in harmonics cos mθ (with m ≤ k and of the same parity), the sum over the polygon kills every harmonic unless n divides m. For rank 4 on the square, m = 4 survives: the second D4-invariant quartic, x4 + y4 beside (x2 + y2)2. Under D6 only (x2 + y2)2 is left.
Figure 3. Angular variation of the rank-k moment of a regular n-gon of velocities: max − min of f over 720 angles, as a percentage of the continuum value; 0 is isotropic. Select a cell (click or arrow keys) to plot f(θ). The square fails at rank 4; the hexagon holds through rank 5 and fails at rank 6. Pentagons and octagons pass rank 4, but no lattice has five- or eight-fold rotations (Part 1).
A shear wave is a velocity along t that varies as a sine along n. It has no advection, so it only decays, as exp(−νk2t) for wavenumber k, and fitting that decay measures the shear viscosity ν for that orientation.
Figure 4. Four periodic lattice gases started from a sinusoidal shear flow. Left: transverse momentum in 4×4 blocks, averaged over four seeded copies. Right: amplitude on a log axis, with ν fitted per panel. HPP's wave with k along x is made of column momenta and cannot decay (the readout checks every column each frame); tilted 45° it is gone in a few hundred steps. FHP's two waves decay together near the Chapman–Enskog value ν = 1/(12d(1−d)3) − 1/8 (grey). Below a link density of about 0.2 they separate, because a particle then travels several sites between collisions.
Two orientations per lattice leave room for luck. Figure 5 measures ν at five angles, on lattices twice as wide so that the tilted waves have about the same wavenumber as the aligned ones.
Figure 5. Viscosity fitted to the decay of one wave per angle, 300 steps each, on 128×128 HPP and 128×148 FHP lattices at link density 0.25; the angle is the wave vector's. The run starts when the figure scrolls into view. Add seeds to see the spread (earlier seeds hollow).
HPP's viscosity is zero on both axes and largest at 45°, and no collision table on these four velocities can change that: their fourth moment has the wrong shape whatever happens at the sites. FHP's points stay within about 10% of the Chapman–Enskog line but are not quite flat. Seed after seed, waves near a lattice direction (0°, 63.4°) decay a little faster than the one 30° off (90°). That sixfold pattern is the rank-6 row of Figure 3. It enters the decay rate at order k4, so it fades at longer wavelengths; the Chapman–Enskog value is the long-wavelength limit.
All runs are periodic and seeded, and start with each link occupied with probability d(1 + 2ca·u) for a shear field u of amplitude 0.3.
Part 5 replaces the Boolean occupations with real-valued populations on nine velocities and chooses their weights so that the square lattice's fourth moment comes out isotropic.