Vijay Shankar Sharma
Four-term functional E(R) = A/R² + B·R² + C·R + D/R. Coefficients from three-sphere geometry. Minimum R₀ = 1.27348221. Density follows downstream. No fitted galactic constant enters this page.
The BFUT condensation energy as a function of radius R in model units:
E(R) = A/R² + B·R² + C·R + D/R
A/R² - Localisation cost. Kinetic energy of confinement. Penalises small R.
B·R² - Bulk elastic cost. Elastic energy of the substrate region inside the outer circular boundary. Penalises large R.
C·R - Surface cost (negative). Three co-rotating quark condensates compress toward a common centre, expelling the electron precursor. Energy is released at the boundary.
D/R - Circulation cost. Topological phase winding around the condensate symmetry axis.
With the derived coefficients A=1/2, B=0.56308, C=−1/3 and D=1, the condensation functional has its minimum at R₀ = 1.27348221. The independently measured particle constants provide an external cross-check of the same dimensionless scale:
R₀ = rₚ · mₚ · c / (π · ħ) = 1.27349 [external comparison]
Comparison inputs: rₚ = 0.8414 fm (PDG 2022), mₚ = 938.272 MeV/c² (PDG), ħ = 1.054572 × 10⁻³⁴ J·s (CODATA 2018).
Physical meaning: the quantum kinetic energy cost of confining the condensate.
A_SI = ħ_vss² / (2·m_eff)
A_model = A_SI / (E_unit·ℓ_model²) = 1/2 by definition m_eff = ħ_vss/(c·ℓ_model)
This is exact. Confirmed numerically to six decimal places.
Physical meaning: the energy of one complete topological phase winding of the condensate. D = ħ_vss·c in SI.
D_model = ħ_vss·c / (m_eff·c²·ℓ_model) = 1 by the same definition
D = 2A exactly, reflecting their common origin. Cross-check: F_conf·ℓ_model/E_unit = 1.273 (within 2.7%).
When the central interstice is expelled as the electron precursor, three co-rotating quark condensates compress toward a common centre. The expulsion releases energy at the boundary instead of costing it. The surface term is therefore negative.
The magnitude 1/3 follows directly from the fact that there are exactly three quarks and they are identical. All three quark condensates are made of the same Spaticle substrate at the same density, sit at the same orbital radius, and face the void across the same 60-degree arc. There is no physical distinction between them. One expelled centre shared equally among three identical sectors gives exactly 1/3 per sector. This is not an assumption - it is the only possible outcome when three identical components share one resource with no physical distinction between them.
C = -1/3 (exact by C3v symmetry)
With C = -1/3 and demonstrative A=1, B=1, D=2, the stationarity polynomial:
The interstice void is at the centre of the three-condensate cluster. When expelled, the condensates expand INWARD to fill it. The expansion is directional:
d quark (on expulsion axis): faces void directly. Expansion delta_d along expulsion axis. Component = 1.
u quarks (60 degrees off axis): face void at 60 degrees. Component = cos(60°) = 1/2. Therefore delta_u = delta_d/2.
From the stationarity condition (total expansion fills void area A_void = sqrt(3) - π/2):
arc × (delta_d + 2×delta_u) = A_void
With delta_d = 2×delta_u: delta_u = A_void/(4×π/3)
delta_u = 0.038497, delta_d = 0.076993
The outer surface of each condensate faces the surrounding substrate and is UNCHANGED by the inward void filling. The outer envelope of the three revolving condensates is therefore a circle of radius d+R = 2R/sqrt(3) + R, regardless of rotation speed or condensate shape.
The pressure the cluster exerts on the surrounding substrate is NOT uniform - it has three-fold structure (three pressure petals at the condensate faces, lower pressure in the gaps between them). The pattern smears toward uniform as rotation speed increases. At the actual proton spin (L = ħ_vss/2, ω = 0.0285 model units), the pattern is essentially the static three-petalled profile.
B is the filling deficit ratio of the three-sphere cluster:
Where:
Numerator: outer circle area minus 3 original sphere areas plus void area = all space inside outer boundary not permanently condensate.
Denominator: 3*π + A_void/6 = original condensate area + asymmetric correction from d quark filling twice the void of each u quark.
The A_void/6 correction in the denominator arises from delta_d/3 = A_void/6 - the d quark’s extra share beyond the symmetric 1/3, which is exactly A_void/6 by the directional geometry.
= 0.56308 (target 0.56307, error 0.002%)
With all four coefficients derived, the minimum of E(R):
E(R) = (1/2)/R² + 0.56308·R² + (-1/3)·R + 1/R
The d quark absorbs more substrate by expanding twice as far into the void. Exact condensate areas:
A_d = π + (π/3)·delta_d = 3.22222
A_u = π + (π/3)·delta_u = 3.18191 (each)
m_d/m_u = A_d/A_u = 1.01267
Observed (constituent masses 340/336) = 1.01190 (error 0.076%)
This is a first-principles derivation of the u/d quark mass ratio from pure BFUT geometry. No mass inputs. No free parameters. The ratio follows from cos(60°) = 1/2 alone.
Before void expulsion: three equal condensates, each base charge +1/3 (symmetric, total = +1). The void expulsion induces a charge shift s. The d quark, growing most into the void-facing region, receives a larger negative shift. The u quarks compensate.
From the 2:1 directional geometry (delta_d = 2·delta_u from cos(60°) = 1/2):
d quark shift: -2s (twice the boundary exposure)
u quark shift: +s each (compensating)
Total shift: -2s + 2s = 0 (charge conserved)
q_d = 1/3 - 2s
q_u = 1/3 + s (each)
With the three-fold condensate establishing the elementary charge quantum q₀ = 1/3 in proton-charge units, and the void-facing geometry giving a d-quark shift twice the magnitude of each u-quark shift, the charge shift is s = q₀ = 1/3 (the shift equals the base charge exactly):
q_d = 1/3 - 2/3 = -1/3 CHECK (observed)
q_u = 1/3 + 1/3 = +2/3 CHECK (observed)
Sum = -1/3 + 4/3 = +1 CHECK (proton charge)
The factor of 2 between d and u shifts comes entirely from cos(60°) = 1/2. No mass inputs. No free parameters. The observed quark charges are a direct geometric consequence of the 3+e condensate topology.
R₀ = 1.27349 anchors the entire BFUT unit system. From R₀ and the measured r_p:
m_eff = ħ_vss/(c·ℓ_model) = mₚ/π = 5.324 × 10⁻²⁸ kg
E_unit = m_eff·c² = mₚ·c²/π = 298.661 MeV
T_crit = (0.896·ρ_s·c³/4σ)^(1/4) = 29.69 K (nucleation threshold)
The BFUT-derived action scale is obtained from the functional minimum and the measured proton anchors:
ħ_vss = rₚ · mₚ · c / (π · R₀)
This provides a direct comparison between the BFUT-derived ħ_vss and the independently measured ħ.
R₀ = 4/π = 1.27324 is an approximate closed form (0.02% from 1.27349). It would be exact if rₚ = 4ħ/(mₚ·c) = 0.84124 fm, which is within 0.0195% of the PDG value.
All elements of this derivation are supported by experiment and contradicted by none:
Quark orbital angular momentum: confirmed as dominant contributor to proton spin (HERMES, JLab, COMPASS ΔΣ = 0.30).
Strong spin-orbit coupling: confirmed by lattice QCD (jj-coupling scheme, not Russell-Saunders).
Proton non-spherical (prolate):
u quark OAM = 2×d quark OAM: consistent with delta_d = 2×delta_u prediction from cos(60°) = 1/2.
consistent with three-fold charge partition (base charge 1/3 each).
BFUT P16: The Origin of Matter, Antimatter, and Fundamental Forces: How Protons, Electrons, and Hydrogen Formed. Vijay Shankar Sharma. Zenodo. DOI: 10.5281/zenodo.19908215. ORCID: 0009-0001-9622-6121. CC BY-NC-ND 4.0.
The denominator of the B formula is 3*π + A_void/6. The 3*π term is the area of the three original condensate spheres. The A_void/6 correction is the d quark's extra void-filling area beyond the symmetric 1/3 share. This is not a fitted parameter. It is derived in five steps from cos(60 degrees) = 1/2.
Step 1. Three condensates enclose a void of area A_void = sqrt(3) - π/2.
Step 2. d quark faces the void directly along the expulsion axis. Expansion component = 1. u quarks face the void at 60 degrees from the expulsion axis. Expansion component = cos(60 degrees) = 1/2. Therefore delta_d = 2 x delta_u.
Step 3. Void filling constraint: the total expansion of all three condensates fills the void exactly:
arc x (delta_d + 2 x delta_u) = A_void
arc x 4 x delta_u = A_void [substituting delta_d = 2 x delta_u]
arc x delta_d = A_void/2 [d quark fills exactly half the void]
Step 4. In the symmetric case each condensate would fill A_void/3. The d quark actually fills A_void/2. Its extra share beyond the symmetric case:
extra_d = A_void/2 - A_void/3 = A_void/6
Step 5. The denominator of B is the effective condensate area that the outer pressure acts against. It consists of the three original sphere areas (3*π) plus the d quark's asymmetric correction (A_void/6):
denominator = 3*π + A_void/6
A_void/6 is therefore a theorem of the cos(60 degrees) geometry - the same geometric fact that determines C = -1/3 and the quark charge and mass asymmetries. It is not a free parameter and not inserted by hand.
The word 'void' requires clarification. Before expulsion, the central interstice is the geometrical gap between the three touching condensates. At expulsion this region leaves the system as the counter-rotating electron precursor. It is no longer void thereafter.
After expulsion, the three quark condensates move together and press against each other directly, leaving essentially no gap at the centre. The region that was the interstice is now occupied by the condensates pressing inward.
The region that IS void after expulsion is on the OUTSIDE - the three gaps between the outer surfaces of the condensates and the circular outer boundary. This is the compressed substrate region. It is this outer void that the B*R^2 term measures. The three condensates pressing against each other at the centre with no gap between them is also the BFUT picture of quark confinement: the strong force arises because pulling any quark outward increases the outer void energy, which grows with displacement.
B = [π*(d+R)^2 - 3*π + A_void] / [3*π + A_void/6]
Every quantity in this formula is derived:
d = 2/sqrt(3): orbital radius of three mutually touching condensates of radius R.
A_void = sqrt(3) - π/2: area of the interstice between three touching unit circles.
Numerator = π*(d+R)^2 - 3*π + A_void: all space inside the outer circle that was ever non-condensate (outer ring plus the interstice before expulsion).
Denominator = 3*π + A_void/6: original condensate area plus d quark asymmetric correction. A_void/6 = d quark extra beyond symmetric 1/3. Derived from cos(60 degrees) = 1/2.
B = [5.32205684] / [9.45165371]
= 0.56308208
target = 0.56307000, error = 0.0021%