Computing masses, spins and charges
You now have the tools (knots, invariants, Chern–Simons). Let us see how they produce a concrete particle spectrum — masses, spins, charges, and even the nucleon as a link.
1. The canonical mass formula
The central formula of the mass computation reads:
$$\boxed{m \;=\; \sqrt{\sigma_\chi}\;\times\;\mathcal{S}}$$A single dimensionful scale, $\sqrt{\sigma_\chi} = 418{,}47$ MeV — the tension of the $\chi$ string, read off the hadronic Regge slope — times a pure number $\mathcal{S}$ set by the particle's topological class. Each class has its sector law:
- BASE ($\mathcal{S} = 0$): the photon and neutrinos — free $\chi$ waves, no knot.
- LEPTONS: the Koide relation (amplitude $\sqrt{2}$ and phase $\pi/12$ derived) — $e$, $\mu$, $\tau$ reproduced to better than 0.005 %.
- BARYONS: the Regge base of the octet (the proton: 938.2 MeV, exact) + the ×3 generation step ($Z_3$) + the tensor correction.
- GOLDSTONE: the pion through the GMOR relation — $m_\pi$ to 0.06 %.
- HEAVY: stoikhion masses (the building-block knots $5_2$, $7_2$…) + the Cornell potential — D and B mesons, quarkonia.
- ELECTROWEAK: the asthenions (W, Z, H, top) at the VEV scale.
Global score: 89 states at 0.13 % median; the 226-state extension holds at 1.43 %.
2. One scale, sector laws
Why not a single formula applied to every knot? Because a particle's three attributes are not carried by the same structure. The “$\sigma_\chi$ prevails” principle:
$$\Delta M^2_{\rm Regge} = 2\pi\sigma_\chi\,, \qquad \Delta M^2_{V\!-\!P} = \pi\sigma_\chi = 0{,}5501\ \mathrm{GeV}^2$$the knot configuration carries the spin (braid arity) and the charge ($Q = \sum\chi + Lk$); the mass follows the law of its sector — the way the object grips the $\chi$ string. All the laws share the same scale: the $\mathcal{S}$ factors are pure numbers ($\sqrt{2}$, $\pi/12$, $3/4$, integer ratios…).
- a full turn of the ribbon costs $\Delta M^2 = 2\pi\sigma_\chi$ — the Regge slope, measured on hadrons for fifty years;
- a half-turn (flipping the ribbon) costs $\pi\sigma_\chi$ — the vector–pseudoscalar gap ($\rho$ vs $\pi$, $K^*$ vs $K$…);
- the theory re-reads these slopes as the twisting of the rings of the spacetime weave — zero new parameters.
It is the same move as in atomic physics: a single scale constant (the rydberg), and level laws specific to each situation.
3. What is calibrated, what is derived
The scale is calibrated — once, globally. And it is doubly constrained: the leptons (through Koide) and the octet base demand the same value to $7\times10^{-5}$:
$$\sqrt{\sigma_\chi} = 418{,}47\ \text{MeV}$$The full spectrum rests on 10 global parameters, 9 of them for the mass sector — never one parameter per particle. And many links are derived: Koide's $\sqrt{2}$ amplitude (quantum-dimension theorem), the $\pi\sigma_\chi$ half-turn, the ×3 generation step (the $Z_3$ triality of the trefoil), the electroweak scale $v$ (to 28 ppm).
No further parameter is adjusted afterwards: every mass below is an output of the same machinery.
4. Two concrete examples
The proton (octet base)
The nucleon is the base of the baryon sector: its factor $\mathcal{S}_{\rm octet}$ is built on the Regge base, with no proton-specific parameter:
$$m_p = \sqrt{\sigma_\chi}\times\mathcal{S}_{\rm octet} = 938{,}2\ \text{MeV}$$Measured: 938.27 MeV — a 0.0 % deviation.
The universal half-turn: $K^*$ vs $K$
The vector meson $K^*$ is the $K$ meson whose ribbon has flipped by half a turn. The cost is the quantum $\pi\sigma_\chi$:
$$m_{K^*}^2 - m_K^2 = 0{,}551\ \text{GeV}^2 \;\approx\; \pi\sigma_\chi = 0{,}5501\ \text{GeV}^2$$Deviation: 0.2 %. The same half-turn links every vector–pseudoscalar pair, from $\pi/\rho$ to $B_s/B_s^*$ (0.7 % median), with no parameter at all.
5. The spin rules
Spin is not computed: it is read off the object's braid structure, through three facts:
- the fermion/boson parity is the $4\pi$ parity of the ribbon (one turn does not bring it back, two turns do) — the $Z_2$ of spatial rotations;
- the spins of a link's components compose as in quantum mechanics;
- for excited states, the value of $j$ is read off the inverse Regge slope — 31 of the 32 spins tested on the 226-state extension.
Three typical readings:
| Structure | Spin $s$ | Example |
|---|---|---|
| single torus knot ($3_1$) | $1/2$ | electron, muon, tau |
| knot + mirror pair ($K + \bar{K}$: a meson) | $0$ or $1$ | $\pi$ (0), $\rho$ (1) — split by the half-turn |
| 3-component link ($\tfrac12\otimes1\otimes1$) | $1/2$ or $3/2$ | proton (octet), $\Delta$ (decuplet) |
Orbital excitations (P and D waves) add $L$: the heavy-spectrum bands follow closed rules, derived in the monograph.
6. The electric charge
For a link $L = K_1 \cup \cdots \cup K_n$ with several components, the total charge reads:
$$\boxed{Q(L) = \sum_{i=1}^{n} \chi(K_i) + Lk(L)}$$where:
- $\chi(K) = \sigma(K) \cdot b(K)$ — an integer attached to each component
- $Lk(L)$ — the total linking number between the components
$\sigma(K)$ is the curvature saturation (= $\pm 1$ for toric knots, 0 for amphicheiral ones), and $b(K)$ is the braid index. The result $Q \in \mathbb{Z}$ is a theorem — charge quantization is derived, not postulated.
7. The nucleon: a $3_1 + 7_2 + \bar{7}_2$ link
The proton and the neutron are, in TQNT, one and the same link: a trefoil ($3_1$) and a mirror pair of twist knots ($7_2 + \bar{7}_2$ — the mirror of $7_2$, of opposite chirality: their topological charges $+3$ and $-3$ compensate). The invariants:
| Component | $c_J$ | $U_{EM}$ | $b$ | $\chi$ |
|---|---|---|---|---|
| $3_1$ (trefoil) | 3.3 | 9.0 | 2 | $-2$ |
| $7_2$ (twist) | 8.1 | 33.8 | 3 | $+3$ |
| $\bar{7}_2$ (mirror twist) | 8.1 | 33.8 | 3 | $-3$ |
| Total | 19.5 | 76.7 | — | $-2$ |
Charge: $Q = \sum \chi + Lk = -2 + Lk$.
- Proton: $Lk = +3 \Rightarrow Q = +1$ ✓
- Neutron: $Lk = +2 \Rightarrow Q = 0$ ✓
8. The n–p mass difference = 1.293 MeV
The optimal linking is $Lk_{\rm opt} = +3$ (the proton, minimum energy). Any deviation from the optimum costs energy following a harmonic potential:
$$E_{\rm tension} = \beta\, (\Delta Lk)^2\, \langle\sqrt{c_J^{(i)}\, c_J^{(j)}}\rangle$$For the neutron, $\Delta Lk = -1$, hence:
$$E_{\rm tension}({\rm neutron}) = 0{,}210\, \text{MeV} \cdot 1 \cdot 6{,}15 = 1{,}293\, \text{MeV}$$This tension is exactly the measured mass difference:
$$m_n - m_p = 1{,}293\, \text{MeV} \quad \text{(exp.)}$$No parameter adjusted for this result (apart from $\beta = 0{,}210$ MeV, one of the 10 global parameters, calibrated once and for all — its mechanism, fluxoid trapping, is identified).
9. Historical note: the archived formulas
Early versions of the theory used a single formula applied to every knot: the empirical form $m = \alpha_0\, U_{EM}\, c_J^{3{,}55}$ (one calibrated scale, the knot's intrinsic EM energy, its topological complexity), then the $\Omega_{ts}$ operator derived from the Chern–Simons flow. They reproduced 118/122 masses (0.68 % median) — but that engine rested on the catalogue curves, 14 of 19 of which turned out to be topologically wrong, and on too much assignment freedom. Those chapters are archived in the monograph.
The canonical formula $m = \sqrt{\sigma_\chi}\times\mathcal{S}$ replaces them: fewer parameters, better accuracy, and sector laws several links of which are theorems (the $\pi\sigma_\chi$ half-turn, the Y-junction triality, Koide's $\sqrt{2}$ amplitude). It is the one presented throughout the rest of the site.
Mass reads $m = \sqrt{\sigma_\chi}\times\mathcal{S}$: a single scale (418.47 MeV, the tension of the $\chi$ string) and a pure number set by the topological class — 89 masses at 0.13 % median. Spin is read off the braid ($4\pi$ parity, inverse Regge); charge is $Q = \sum\chi + Lk$, integer by theorem. The nucleon is the $3_1 + 7_2 + \bar{7}_2$ link, and the $n - p = 1{,}293$ MeV mass difference comes out of the linking alone.
End-of-module quiz
Four questions on computing the particle spectrum.
1. What is the structure of the canonical mass formula?
2. Where does the proton's spin $1/2$ come from?
3. What is the electric-charge formula of a link in TQNT?
4. For the nucleon ($3_1 + 7_2 + \bar{7}_2$ link), why does the proton have $Q = +1$ and the neutron $Q = 0$?