Try to separate two ends of a knotted vortex and the force between them never weakens — it stays constant, then the string snaps and makes new particles. Spatial Displacement Theory says this isn't a new force at all. It's the pressure that's already there, squeezing a tube of throughput so it can't spread.
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Pull two magnets apart and the pull fades fast — by the square of the distance. Same for gravity, same for electric charge. Spread the field lines over a bigger and bigger sphere and each patch of space feels less. That is the famous 1/r² falloff, and almost everything in physics obeys it.
The strong force inside a proton does the exact opposite. Try to drag one strand out and the energy keeps climbing in a straight line — the harder you pull, the more the tube fights back, forever, until something gives. You never get a lone strand. That is confinement, and the standard story buries it inside quarks and gluons and colour charge.
SDT refuses all of those. There are no quarks here, no new charge, no extra field. A proton is a (2,3) trefoil knot — a single throughput vortex tied so it crosses itself three times. The question of PPT05 is blunt: can you get a straight-line, never-weakening potential out of the pressure field that already exists, with no new force bolted on? The answer is yes, and it comes in three moves.
Before we can talk about pulling on the knot, we need to know the knot is even a stable object. Throw a ripple onto the surface of the vortex — a little corrugation — and ask whether it grows or fades. SDT's answer is clean: the finer the ripple, the faster it dies.
Here's the mechanism in one breath. The vortex spins at a fixed budget angle, and that spin pushes outward exactly enough to balance the convergent pressure squeezing in. Put a bump on the surface and the spin there is now at the wrong radius — the balance breaks and a restoring push appears. Crucially, a ripple with harmonic number n (n full wiggles around the loop) oscillates back with a stiffness that grows like n²:
So a coarse wobble (small n) limps back slowly; a fine corrugation (large n) snaps back violently and radiates its energy away in a flash. After the dust settles, every wrinkle is gone and what's left is the smooth torus. The shape forgets its bumps — but it cannot forget how the flow is threaded through itself. The winding numbers (p,q) = (2,3) and the chirality live in the flow phase, not the surface, so they survive every smoothing. That is why a proton is a durable, identical object and not a blob that slowly melts.
Now pull. Take one of the three crossovers and drag its two strands apart by a distance L. The throughput that used to swirl around both strands at once now has to bridge the gap — it stretches into a tube running from one end to the other.
Here is the whole trick, and it's purely mechanical. The tube is sitting inside the same convergent pressure that fills all of space, pressing in equally from every side with strength P_conv/3. That isotropic squeeze does two things at once:
So the tube is a cylinder of constant width, no matter how long you stretch it. (This is exactly how a superconductor pins a magnetic flux tube — the expelled field squeezes it to a fixed thickness.) A cylinder of fixed cross-section has a volume that grows straight in proportion to its length, so the stored energy does too:
No spreading means no 1/r² dilution — the energy density inside the tube never drops, so every extra femtometre costs the same fixed amount. That constant is the string tension σ. When the engine grinds the proton's torus geometry through this, it lands at:
That number is derived from the proton's knot geometry alone — nothing from quantum field theory is fed in. It is a falsifiable prediction, to be tested against the string tension measured from hadron spectra. The real win is the shape of the law (linear, not 1/r²); pinning the exact magnitude is the one open item, flagged honestly in the scoreboard.
So the energy climbs and climbs as you pull. Where does it end? Not with a free strand — that path costs infinite energy, since the tube would have to run to infinity. Instead, the tube hits a ceiling: the moment its stored energy is enough to conjure a brand-new vortex pair out of the medium, it's cheaper to do that than to keep stretching.
The cheapest closed vortex you can make is a (1,1) ring — a pion. Making a particle and its anti-partner costs 2 m_π c². Set the tube energy equal to that and solve for the breaking length:
So at about a fifth of a femtometre, snap — the tube breaks in the middle and each raw end immediately re-closes into a new loop. You started trying to extract one strand; you ended up with two pions and the original proton still intact. The total number of crossings is conserved through the whole event. You can never isolate a strand, because the universe would rather build new particles than leave an open end. That is confinement, stated mechanically.
Confinement and string-breaking come out of convergent-pressure geometry acting on the trefoil — no new fundamental force was added. The structure of the result is the prize; here is the honest accounting, including the one number that lands wide.
| Theorem | What it shows | Result | Status |
|---|---|---|---|
| A | ripples damp ∝ n² → smooth torus survives, (p,q) topology is durable | PROVEN | DERIVED |
| B | collimated tube → E(L) = σL is linear, not 1/r² | PROVEN | DERIVED |
| B′ | numerical σ_SDT = 1.230 GeV/fm — SDT prediction (magnitude pending the pressure profile) | prediction | COMPUTED |
| C | L_c = 2m_πc²/σ ≈ 0.23 fm → string snaps to pions, topology conserved | PROVEN | DERIVED |
| C′ | free strand costs infinite energy → forbidden | PROVEN | DERIVED |
E(L)=σL with constant cross-section, straight from the isotropic P_conv/3 squeeze. No quarks, no colour, no new force. This is the result.
L_c falls out of σ and the measured pion mass, and the snap conserves crossing number. Free strands are energetically forbidden. Survives.
σ_SDT = 1.23 GeV/fm is SDT's prediction, derived from knot geometry with no quantum-field-theory input. The linear shape is locked; pinning the exact magnitude still needs the internal pressure profile of the crossover. Open problem.
Ripples damp like n², so the knot is a durable object and its topology is an honest invariant. Survives.
That is the shape of the answer to why a strand can never come loose. Not because of a mysterious new charge, but because the pressure that already fills space won't let a throughput tube spread, so its energy climbs forever, and long before it frees a strand it would rather spend that energy building new particles.