Why dust drifts and cells settle instead of plummeting. In a slow, syrupy world the resistance on a sphere is not the squared, roaring drag of a falling stone — it is a gentle line: F = 6π·μ·r·v, resistance that grows in exact proportion to speed. Double the speed, double the drag. Balance it against the sinking weight and out falls a settling speed. Down the center the sphere goes in, the engine computes, the terminal velocity comes out. The blue team builds and defends; the red team tries to break it.
source Stokes, G. G., On the Effect of the Internal Friction of Fluids on the Motion of Pendulums (1851), Trans. Camb. Phil. Soc. 9, 8–106 — archive.org/details/b22464074. Rendered, not quoted.
Solve the creeping-flow equations (inertia dropped, μ∇²u = ∇p) around a sphere and the total force is F = 6π·μ·r·v — first order in every factor:
| double this… | …and F | because |
|---|---|---|
| speed v | ×2 | linear |
| radius r | ×2 | linear |
| viscosity μ | ×2 | linear |
Contrast the fast world (window 1): high-Reynolds drag goes as v². Stokes drag is the slow limit — and the drag always opposes the motion: F⃗ = −6π·μ·r·v⃗.
Stokes drag is what the-reynolds-number looks like when Re ≪ 1: viscosity dominates inertia, the flow is reversible, and resistance is linear in speed.
Balance the drag against the buoyant weight and the sphere stops accelerating — its settling speed, the neighbouring sphere the-terminal-velocity. Push Re past 1 and control passes to the-drag-equation's quadratic law. Each sphere is the next one's premise.
The blue team's live check: re-derive the linear ratio F(2v)/F(v) and the weight/drag balance at terminal velocity, and confirm them against the known laws. If red makes the drag quadratic, this badge is where it shows.
A single sphere drifting through a still fluid. Four numbers set the whole problem:
| symbol | quantity | units |
|---|---|---|
| r | sphere radius | m |
| μ | fluid viscosity | Pa·s |
| ρs, ρf | sphere / fluid density | kg·m⁻³ |
| v | speed through fluid | m·s⁻¹ |
Small r, high μ, slow v → the creeping regime where Stokes' line is exact. Feed them to the panel below.
The green line is the Stokes drag F = 6π·μ·r·v; the amber line is the net weight. They cross at the terminal velocity — every number is computed live from the laws, never looked up.
What the machine proves: drag linear in v, r and μ (F(2v)/F(v)=2 exactly), always opposing the motion, and — where it balances the buoyant weight — a Stokes terminal velocity vt = 2r²(ρs−ρf)g / 9μ. The current sphere's numbers are above; the law is the output.
The blue team's witness (left) confirms these live; the red team (right) tries to make the drag quadratic.
It also assumes a rigid sphere in an unbounded, continuous, steady fluid. Near a wall, drag rises; a bubble or drop obeys Hadamard–Rybczynski, not 6π; sub-micron particles feel slip (the Cunningham correction) because the air is no longer a continuum; and rapid changes add the unsteady Basset history force. Real, and each one bends the line.
"In air, everything falls at g." Cut. Only in vacuum. A 10 µm mote reaches a terminal velocity of a few mm/s in milliseconds — g never gets to act for long.
"Drag always grows with the square of speed." Kept, corrected. That is the high-Re law. At Re ≪ 1 drag is linear — the exact tamper the red team plays below.
"Stokes' law works for any sphere in any fluid." Cut. Only for Re ≪ 1, rigid, no walls, continuum fluid. Push past that and every term in window 1 wakes up.
The red team's move: swap the creeping-flow drag for a quadratic law (F ∝ v²) — Stokes' line wearing the high-Re form. The linear check must fail. The blue team's witness (window 7) is watching.
Force the drag quadratic and F(2v)/F(v) jumps to 4, not 2 — the witness recomputes, disagrees with the linear law, and turns red. Nothing is faked; the attack is real and it is caught.