Aether Science

The Physics of the Aether — Research by Harold Aspden

Aspden’s Aether from the Bottom Up, Part 1: The Smallest Units

A three-light-year-tall pillar of gas and dust in the Carina Nebula, photographed by the Hubble Space Telescope for its 20th anniversary
Three-light-year-tall star-forming pillar in the Carina Nebula. Credit: NASA, ESA and M. Livio — Hubble 20th Anniversary Team (STScI) · NASA Image Library ↗

Companion to Testing Harold Aspden’s Aether Physics Against Modern Measurements. Every test here is also listed in the test register. Downloads: Word · Markdown · analysis scripts.

Quons, the charge continuum, the photon unit and the cell energy: how each number is obtained, what holds together, and what measurements since the 1990s say

By Doug Westmoreland (project lead: direction and review) and Claude, Anthropic (research and analysis) · Companion to Testing Harold Aspden’s Aether Physics Against Modern Measurements (working paper v2.1) · Version 1.5, 31 October 2026 (1.1 adds section 6 on the 0.6 ppm gap, U20–U26; 1.2 adds section 7, the photon unit in an endless lattice, U27–U31; 1.3 adds section 8, the self-consistent model of how the lattice responds to light, U32–U36, and corrects U11–U15, with every derivation set out in the supplement Part 1A; 1.4 adds four objections considered, U37–U40; 1.5 splits input I1 into its two assumptions, adds U41 on the postulate that the two frames move at c relative to each other, and revises U40 to test the frame Aspden actually proposed, carried with the Earth)

Abstract

Harold Aspden built his physics on a few elementary units. Space holds a uniform continuum of charge. Embedded in it is a simple cubic lattice of "quons", each carrying the electron’s charge. The two systems circle each other in orbits fixed by the electron’s Compton frequency. This paper rebuilds those units from his 2003 account (The Physics of Creation, chapter 7), traces them back through Physics Unified (1980) and Physics without Einstein (1969), and tests them one at a time. There are 41 numbered tests, U01–U41. The supplement, Part 1A, derives the model behind them step by step. Each states Aspden’s claim and source, the calculation step by step, the inputs, the script output, the verdict and the assumptions open to challenge.

The findings:

  1. The dimensionless chain is internally consistent. His force balance, quon mass, quon radius, photon-unit inertia and cell energy agree with one another exactly once his own fine-structure constant is used. The only numerical gap anywhere in the chain is that constant itself, 0.6 ppm from the measured value (U20). Once the electron fixes the scale, the chain has no further freedom. Its most consequential assumption is that a quon’s inertia is half its Thomson mass, by a buoyancy analogy; with the full mass, 1/α comes out 0.56% low (U05). The orbit radius rests on a postulate that the two frames move at the speed of light relative to each other, each at c/2, which Aspden calls an assumption. Worked through his equations, it cancels out of α, N and the lattice spacing, and only decides which odd N is chosen (U41).
  2. His scale matches two real quantum scales. His orbit radius r = ħ/2mₑc is the Dirac zitterbewegung amplitude and the range of quantum electrodynamics’ vacuum polarization. His orbital frequency is the electron’s Compton frequency.
  3. The lattice cannot carry light. Aspden’s restoring force on a quon (4πσqD) is the field of the displaced lattice itself, as his own derivation of it says (Part 1A, section 4). Treated with Coulomb’s and Maxwell’s laws, a lattice held only by its own field behaves like a plasma whose frequency is exactly the electron’s Compton frequency: his orbit is that plasma oscillation, which is a credit. But the same plasma blocks every wave below 511 keV, including radio, visible light and X-rays, and cancels static electric fields beyond about 10⁻¹⁰ m. It is the same as giving the photon the electron’s mass, 5 × 10²³ times the measured upper limit (U32–U36).
    • Objections tested (U37–U40).
      • Removing restoring forces does not help, because a plasma needs none: the ionosphere blocks radio below about 9 MHz for the same reason.
      • Letting the lattice flex elastically gives shear waves at 0.2–0.3c, which are unstable along the cube axes.
      • Background (CMB) motion of the quons is far too small to matter, and adds linearly anyway.
      • Aspden’s own frame for light travels with the Earth but does not turn with it. A laboratory then crosses it at the Earth’s surface speed, 283 m/s at Berlin, which rotating cavities exclude by a factor of about 1,500. An aether turning with the Earth is excluded by ring lasers instead (U40).
  4. His 1982 resonance picture fails too. In 1982 he described the vacuum as resonating at the Compton frequency, as if each quon had its own spring. Version 1.2 of this paper tested that picture (U11–U15), and it fails as well: gamma rays of 662 keV, 847 and 1238 keV (from a supernova ten million light-years away) and 6 GeV all propagate normally, and photons from about 10 keV to 31 GeV from gamma-ray burst GRB 090510 arrived together after 7.5 billion years. His "dual displacement" remedy gives no equations, and by his own statement stops at the pair-creation frequency.
  5. A lattice of spacing d = 6.37 × 10⁻¹³ m cannot carry waves shorter than 2d (about 1 MeV). Photons are now detected at 1.4 PeV, 1.4 × 10⁹ times shorter. Shrinking the lattice to fix this would break the tie to the electron that sets every scale in the theory.
  6. Three smaller claims fail.
    • The cell energy, which he calls a virtual muon pair, is 0.21% below two muon masses.
    • The aether energy density is 2.5 × 10³⁵ times the observed dark-energy density. Mainstream physics has the same kind of cancellation problem, and Aspden acknowledged his.
    • The semiconductor light-hole masses he cited in support of the quon mass vary tenfold between crystals.
  7. The 0.6 ppm gap cannot be closed by correcting a slip.
    • His printed figures all reproduce.
    • The final α depends only on N = 1843 and two relations. Closing the gap needs one of them changed by 0.41 ppm (photon unit) or 3.7 ppm (size ratio), right to 1 part in 4,000.
    • His restoring-force coefficient is justified by Maxwell’s equations.
    • The postulate behind the c/2 orbital speed cancels out of α for a given N, so it cannot be the source either (U41).
    • The photon unit’s outer quons spin at 0.41c and 0.58c. With the speed-dependence of mass that Aspden accepted, its angular momentum rises 19% and 1/α becomes 163. Effects his formula leaves out are 1,000 to 450,000 times larger than the gap, so the 0.6 ppm agreement cannot reflect the precision of the calculation.
    • One step can be improved: his free constant k (2003) can be calculated from his own description (k = 11.7). This leaves α unchanged, and without the integer rule it brings the pure lattice from 140 ppm to 37 ppm low (U20–U26).
  8. Treated as an endless crystal, the lattice confirms the orbit but not the photon unit.
    • The uniform orbital motion is exact in an endless lattice: the quons’ forces on one another cancel, and the orbital current is balanced by the displacement current.
    • Electrostatically, the simple cubic lattice is unstable: 18% of its vibration modes have imaginary frequencies. A body-centred cubic lattice, which Aspden himself names as the minimum-energy structure, is stable.
    • Magnetic coupling between the moving quons of a spinning photon unit adds 9–29% to its inertia, and carried through his chain gives 1/α of 149–177.
    • Treating the photon as a wave does not lower the speeds while 27 quons carry ħ at a quarter of the Compton frequency: their kinetic energy is 11% of their rest energy whatever the pattern of motion. Spreading the same angular momentum over about 7.5 million quons, a region about 1.3 × 10⁻¹⁰ m across, would make the speed corrections small enough. But then the number 36 in 36m₀d², from which Aspden obtains Planck’s constant, is no longer fixed by the geometry and has to be put in by hand (U27–U31).

Version history of the units: the 1969 lattice gave 1/α = 137.038, using a graviton-linked correction that reproduces exactly (0.000455 against his printed 0.000456). The 1980 version replaced it with the odd-integer rule. The 2003 version added a constant k whose stated value, 8.34, also reproduces (8.347).

1. How to read and challenge this paper

Each test has an identifier (U01, U02, …). The same identifier labels:

  • its block in the script aether_units.py;
  • the lines of the script’s output quoted here;
  • its entry in the combined test register, which also lists the tests in the main paper (M01, M02, …).

Every entry uses the same headings, so any step can be checked or disputed on its own:

  • Aspden’s claim, with work, year and equation;
  • Steps, each with the formula and the number it produced;
  • Inputs, the measured values used, with sources;
  • Output, the exact lines printed by the script;
  • Verdict: Credited (his step is sound or justified), Consistent, Disagrees, Excluded or Open;
  • Challenge points, the choices made in this analysis rather than by Aspden, on which the verdict depends.

Aspden wrote in Gaussian units. Here his q² is converted to SI as e²/(4πε₀). Constants are CODATA 2022.

2. The units

Unit What it is in his theory His value Recomputed here How it is obtained
Charge continuum (part of the "G frame") Uniform charge density σ filling space, opposite in sign to the quons; "the theory assumes the continuum proper has zero energy" σ = q/d³ 6.19 × 10¹⁷ C/m³ Assumed; "I have no idea how to explain or justify the existence of the continuum" (POC p. 113)
Quon ("lattice particle" before about 1997) Charge e at each site of a simple cubic lattice (the "E frame") Charge e Charge e Assumed
Lattice spacing d Side of the cubic cell holding one quon 6.374 × 10⁻¹¹ cm 6.374 × 10⁻¹³ m From r and r/d (U03)
Orbit radius r Both frames circle with radius r at speed c/2, relative speed c, displaced 2r from each other r = h/4πmₑc (7.1) 1.931 × 10⁻¹³ m Inputs I1a and I1b: the electron’s frequency and the postulated relative speed c (U01, U41)
Orbital frequency Ω Ω = (c/2)/r the Compton electron frequency ħΩ = 511.0 keV Input I1a (U02)
Lattice ratio r/d Fixed by a zero-energy lattice condition, then an odd-integer rule (3/4)(8/1843)^(1/6) 0.3029159 Main paper §5; U19
Quon effective mass m₀ Half the Thomson mass (hydrodynamic buoyancy analogy) m₀ = q²/3bc² (7.4) 0.04078 mₑ = 20.84 keV U04, U05
Quon charge radius b From the mass and the force balance b/d = (d/r)²/96π (7.15) 2.303 × 10⁻¹⁴ m (b/d = 0.036) U05
N Quon charge volume over electron charge volume, (b/a)³; must be odd 1843 1844.54 before the integer rule U05; main paper §5.2
Photon unit A 3×3×3 block of 27 quons spinning about a lattice axis; its spin energy sets Planck’s constant I = 36m₀d² = 4hr/cπ (7.21) size 3d = 1.91 × 10⁻¹² m U06
Cell energy E₀, the "virtual muon pair" Energy per cell, taken equal to the quon’s internal energy density times the cell volume 412.6658 mₑc² (7.29) 210.87 MeV U08
Gravitons, dimuon, taon, proton Built from the units above; covered in the main paper, §§6–8; Part 2 of this series

3. How the numbers follow from one another

Eight inputs carry the chain. Everything else follows by algebra, and every identity in steps 3–7 holds in the script to 0.6 ppm, the gap between his α and the measured one.

Inputs (the places to challenge Aspden):

  • I1a. The two frames orbit at the electron’s Compton frequency: ħΩ = mₑc² (POC 6.14).
  • I1b. The two frames move at the speed of light relative to each other, Ω·2r = c, so each moves at c/2 (POC 6.6, 6.13). Aspden calls this an assumption (The Theory of Gravitation, 1998 note; Part 1A, section 1.2).
  • I2. The quon charge equals the electron charge, and the continuum holds one opposite charge per cell: σ = q/d³ (7.7).
  • I3. The quon’s inertia is half its Thomson mass-energy over c², by analogy with a sphere accelerating through a fluid of equal density (7.4).
  • I4. The lattice’s electrostatic energy, including the orbital displacement and a finite-size term, is zero (7.18). The potential is referred to a block of whole cells (main paper §5.3).
  • I5. The photon unit is a 3×3×3 cube. Its energy is Ω times its angular momentum, and it radiates at four times its spin rate (7.19–7.24).
  • I6. N is odd, and is rounded down from the zero-energy value (7.26).
  • I7. The energy density of the aether equals the energy density inside a quon’s charge sphere, by pressure balance (7.3).

Chain:

  1. I1a + I1b → r = ħ/2mₑc = 1.9308 × 10⁻¹³ m (7.1) and Ω = mₑc²/ħ.
  2. I4 + I6 → r/d = (3/4)(8/1843)^(1/6) = 0.30291588, so d = 6.3740 × 10⁻¹³ m.
  3. Force balance m₀Ω²r = (4πσq)(2r) with I2 gives m₀c² = 32πσqr² (7.5–7.6). This gives m₀ = 0.0407813 mₑ.
  4. Combining step 3 with I3 gives b = d³/(96πr²), so b/d = 0.036136.
  5. With the electron’s Thomson radius a = 2e²/3mₑc²: N = (b/a)³ = (9/8)³(d/r)⁶ = 1843. Equivalently, m₀/mₑ = (4/9)(r/d)² (7.25).
  6. I5 → 36m₀d² = 4hr/cπ, so ħc/e² = 144π(r/d) = 137.0359148 (7.24).
  7. I7 → E₀ = (1/2π)q²d³/b⁴ = (3/4π)(108π)³N^(−4/3) mₑc² = 412.6658 mₑc² (7.29).

Inputs I3, I4, I5 and I6 are where his physical assumptions enter. I1a is where the electron enters, and with it every absolute size in the theory. I1b fixes only the orbit radius: it cancels out of α, N, d, m₀, b and E₀, and reaches them only through I4, which decides which odd N is used (U41).

4. How the units changed, 1960–2003

Year Work What the units were What changed
1960 The Theory of Gravitation (48-page booklet) Lattice of charges in a continuum; quon inertia known from its motion at the Compton frequency; size not yet known First printed version (POC p. 97; Fifth Question)
1966 The Theory of Gravitation, 2nd ed. Quon radius 2.3 × 10⁻¹² cm from the pressure balance and the two-muon energy per cell; α obtained P. N. McNeall later found a small numerical error in this edition (POC p. 105)
1969 Physics without Einstein, ch. 6 "Lattice particles" of mass m₀ = e²/3bc² (6.39); r = 1.93 × 10⁻¹¹ cm; r/d = 0.30292 (6.58) after a "priming energy" correction y = 0.000456 e²/d tied to one electron per graviton (6.57, 6.74–6.75); 1/α = 137.038 (6.61); lattice density 144 g/cc linked to Mercury’s perihelion U18; main paper §9.2
1972 Aspden and Eagles, Phys. Lett. 41A, 423 Zero-energy r/d computed by D. M. Eagles and C. H. Burton: 0.302874 Hand value 0.30288 confirmed by computer
1980 Physics Unified, ch. 6, eqs (128)–(160) Finite-size term (d/r)⁴/4608π (144); odd integer N = 1843; r/d = (3/4)(8/1843)^(1/6) (156); 1/α = 137.035915 (157); E₀ = 412.665816 mₑc², "tempting to suggest … a pair of virtual muons" (160) The 1969 priming-energy correction is replaced by the integer rule
1982 Wireless World 88, 37 (reprinted as Lecture 24, 1998) The vacuum’s natural frequency is the Compton electron frequency; "dual displacement" proposed to remove dispersion U10, U12–U14
by 1997 Research Notes, Tutorial Notes 6–8 The name "quon" (also "sub-electron") Name only
2003 The Physics of Creation, ch. 7 ("an edited and slightly extended version of pages 103–115 of … Physics Unified") A factor (1 − 2/3k) on the finite-size term (7.18), with k ≥ 8.34 needed to keep N = 1843; cell energy "deemed to be that of a pair of virtual muons" (7.3); gravity reinterpreted as an "inverted" electrostatic force, the unit values unchanged U19

5. Tests

U01. The orbit radius equals the zitterbewegung amplitude

  • Aspden’s claim: "the continuum and q charge systems, the G and E frames … are displaced 2r one from the other owing to their relative motion at speed c in circular orbits of radius r given by r = h/4πmc" (POC 7.1).
  • What was tested: whether r is derived or put in, and how it compares with known quantum scales.
  • Steps:
    1. r = ħ/(2mₑc) = 1.9307963 × 10⁻¹³ m.
    2. The amplitude of the electron’s zitterbewegung (trembling motion) in Dirac’s theory is ħ/(2mₑc), the same expression.
    3. The vacuum-polarization (Uehling) potential of quantum electrodynamics falls off as exp(−2mₑc x/ħ). Its range is ħ/(2mₑc), again the same.
  • Inputs: ħ, mₑ, c (CODATA 2022).
  • Output: 1 r = hbar/(2 m_e c) 1.9307963e-13 m
  • Result: r is put in, not derived. It combines the electron’s Compton frequency (I1a) with the postulate that the two frames move at c relative to each other (I1b, U41). It coincides exactly with two scales of established physics.
  • Verdict: Credited (correspondence). The correspondence is not evidence for the lattice: both known scales follow from the electron’s mass alone, so any theory built on ħ/mₑc lands on them.
  • Challenge points: whether Aspden derives r elsewhere. Physics without Einstein ch. 4 introduces it through a photon unit compensating an electron’s motion (the Schrödinger-equation argument), which still takes mₑ as given.

U02. The orbital frequency is the Compton frequency

  • Aspden’s claim: the frames move at c/2 in circles of radius r (POC 7.1); the quon’s "natural rhythmic motion at the Compton electron frequency" (Fifth Question, 2001).
  • Steps:
    1. Ω = (c/2)/r = 7.7634407 × 10²⁰ rad/s.
    2. ħΩ/mₑc² = 1.0 exactly.
    3. Dirac’s zitterbewegung frequency 2mₑc²/ħ is 2Ω.
  • Output: 2 hbar Omega / (m_e c^2) (= 1: the electron Compton angular frequency) 1.0
  • Verdict: Consistent. Aspden’s frequency is half Dirac’s, because his speed is c/2 where Dirac’s trembling speed is c.
  • Challenge points: none in the arithmetic. The step that matters is U10, which shows this is also the vacuum’s natural frequency.

U03. Lattice spacing

  • Aspden’s claim: d = 6.374 × 10⁻¹¹ cm (Fifth Question, 2001). d/a = 108π, where a is the electron’s Thomson radius (POC ch. 2, 7.24–7.26).
  • Steps:
    1. r/d = (3/4)(8/1843)^(1/6) = 0.3029158782.
    2. d = r/(r/d) = 6.3740348 × 10⁻¹³ m.
    3. d/a = 339.29222, against 108π = 339.29201.
    4. d/r_e = 226.195 = 72π.
  • Output: 2 d = r/(r/d) 6.3740348e-13 m
  • Verdict: Consistent. The 0.6 ppm difference in step 3 is the gap between his α and the measured one (U06).
  • Challenge points: d depends on the integer rule through r/d. With the pure lattice value r/d = 0.3028738, d would be 0.014% larger.

U04. The quon mass from the force balance

  • Aspden’s claim: m₀(c/2r)²r = 4πσq(2r) (7.5), giving m₀c² = 32πσqr² (7.6) and m₀/mₑ = (4/9)(r/d)² (7.25).
  • What was tested: whether the three expressions agree when computed independently.
  • Steps:
    1. Restoring force per unit displacement, SI form: k = σe/ε₀ with σ = e/d³.
    2. m₀ from (7.5) is 0.04078132126 mₑ.
    3. m₀ from (7.25) is 0.04078134634 mₑ.
    4. Ratio 0.999999385.
    5. m₀c² = 20.839 keV.
  • Output: 3 ratio (identity: should be 1) 0.999999385229
  • Verdict: Credited. The 0.6 ppm gap is the α gap again. The main paper (§5.2) found and corrected a typo in the printed form of (7.5) in Tutorial Note 8. The POC form is correct.
  • Challenge points: the force 4πσq per unit displacement is the field inside a uniformly charged medium (plasma-like). It treats the displacement as collective, so that all quons move together. That is the natural reading of a lattice moving as a frame.

U05. The quon radius, the Thomson mass and N

  • Aspden’s claim: E = 2q²/3b (7.2); the effective mass is half of E/c² (7.4); b/d = (d/r)²/96π (7.15); N = (9/8)³(d/r)⁶ (7.26), "approximately 1844.53" with r/d = 0.302874.

  • Steps:

    1. b = d(d/r)²/96π = 2.3032933 × 10⁻¹⁴ m (b/d = 0.036136; POC rounds to 0.035).

    2. m₀ from q²/3bc², divided by m₀ from (7.25): 0.999999385.

    3. Quon Thomson energy: 41.678 keV, which is mₑc²/N^(1/3).

    4. (b/a)³ with his r/d is 1843.0034, equal to 1843 by construction.

    5. With the pure-lattice r/d: 1844.5379.

    6. The mass factor f, where m₀c² = f × 2q²/3b, was varied, re-solving the zero-energy lattice and applying his integer rule each time:

      Mass factor f N before rule Odd N 1/α vs measured
      ½ (Aspden: buoyancy) 1844.54 1843 137.0359148 −0.6 ppm
      1 (full Thomson mass) 15,254 15,253 136.26294 −5,641 ppm
      5/4 (own mass plus a fluid added mass of half the displaced ½) 30,583 30,583 135.66837 −9,980 ppm

  • Output: 6 (9/8)^3 (d/r)^6 with the pure-lattice r/d (POC: 1844.53) 1844.5379; 7 f = 1 (full Thomson mass): … 15254.2 15253 136.262937 -5641.3 ppm

  • Verdict: Credited (consistent). The halving of the mass (input I3) is an analogy, not a derivation, and it carries the result.

  • Challenge points:

    • The hydrodynamic factor ½. In ideal-fluid hydrodynamics, buoyancy offsets weight, not inertia. A sphere accelerating through a fluid behaves as if heavier, its own mass plus half the mass of the fluid it displaces. Aspden’s text notes this "apparent increase in mass", but it uses buoyancy to halve the mass. The direct hydrodynamic reading is the f = 5/4 row, which moves 1/α by 1%. Only f = ½ gives N near 1843. So I3 is the most consequential single assumption in the chain, and the same 1843 enters his proton and G derivations.
    • The fluid’s density. The 5/4 row assumes the displaced "fluid" has half the quon’s energy density, as his pressure-balance argument states (POC pp. 112–113).

U06. The photon unit and Planck’s constant

  • Aspden’s claim: the 3×3×3 photon unit has moment of inertia 36m₀d² = 4hr/cπ (7.21, PWE 6.59), giving ħc/e² = 144π(r/d) (7.24).
  • Steps:
    1. 36m₀d²/(4hr/cπ) = 1.0.
    2. 144π × 0.30291588 = 137.035914847.
    3. Measured 137.035999177(21).
    4. Difference −0.615 ppm.
  • Output: 4 difference, ppm -0.6154 ppm
  • Verdict: Credited for the algebra. The comparison with α is the main paper’s test (§5; register M-entries), not repeated here.
  • Challenge points: the four-fold radiation rule and "energy = Ω × angular momentum" (I5) are assumptions. The main paper checks the algebra, not the physics of these rules.

U07. Number and mass density of the lattice

  • Aspden’s claim: "3.87 × 10³⁰ lattice particles per cc … equivalent to … 144 gm/cc" (PWE p. 117).
  • Steps:
    1. 1/d³ = 3.862 × 10³⁶ m⁻³ = 3.862 × 10³⁰ cm⁻³.
    2. m₀/d³ = 143.5 g/cm³.
  • Output: 3 mass density m0/d^3 143.5 g/cm^3
  • Verdict: Consistent. Aspden used this density in 1969 to explain Mercury’s perihelion. That mechanism is excluded (main paper §9.2).
  • Challenge points: none in the arithmetic.

U08. The cell energy and the "virtual muon pair"

  • Aspden’s claim: E₀ = 412.6658 mₑc², "the energy quantum that is regarded as the virtual muon pair, the occupants of each cubic cell of aether" (POC 7.29). In 1980: "very close to the energy of two muons … it is tempting to suggest" (PU 160).
  • Steps:
    1. E₀ from (7.29) = 412.6658162 mₑc² (210.87 MeV).
    2. E₀ from (7.3), using b and d from U03 and U05, divided by step 1: 0.99999939.
    3. Two muon masses: 413.5365654 mₑ.
    4. E₀/2m_μ − 1 = −2106 ppm.
  • Inputs: m_μ/mₑ = 206.7682827(46) (CODATA 2022).
  • Output: 4 E0 / (2 m_mu) - 1, ppm -2105.6 ppm
  • Verdict: Disagrees, if read as two real muons: 0.21% low, against a muon mass known to 22 parts per billion. Aspden presents it as an approximate identification. In 2003 he flags the "approximation" himself and answers "or rather hoping" that the muons’ interaction energy makes up the difference (POC p. 112).
  • Challenge points: the proton-mass derivation (main paper §6) uses E₀ itself, not the muon mass. So the 0.19 ppm proton result does not depend on this identification.

U09. The energy density of the aether

  • Aspden’s claim: E₀ per cell is "the energy density of the aether, on average" (POC 7.3). It is balanced by "a negative interaction energy that exactly matches the positive energy of the quon and its equal measure of associated graviton energy" (POC p. 112).
  • Steps:
    1. E₀/d³ = 1.305 × 10²⁶ J/m³.
    2. Observed dark-energy density Ω_Λ·3H₀²c²/8πG = 5.25 × 10⁻¹⁰ J/m³.
    3. Ratio 2.5 × 10³⁵.
  • Inputs: H₀ = 67.4 km/s/Mpc, Ω_Λ = 0.6847 (Planck 2018); G (CODATA 2022).
  • Output: 3 ratio 2.48e+35
  • Verdict: Open. Mainstream physics has the same kind of problem. A vacuum energy at the strong-interaction scale is about 3 × 10³⁴ J/m³, comparable to Aspden’s, and the full quantum estimate is vastly larger; no accepted mechanism cancels it. Aspden’s cancellation is stated as a hope, not derived.
  • Challenge points: whether this energy gravitates in his theory at all. His gravity comes from gravitons displacing the continuum (main paper §7), so he could argue the cell energy is not a source of gravity. That argument is not made in the texts examined.

U10. The natural frequency of the quon system is the Compton frequency

  • Aspden’s claim: "the ether sets a critical frequency threshold (the Compton electron frequency) and so may have an electrical structure conforming with this resonant frequency … in proposing a kind of crystal structure for the vacuum medium and establishing … that it has a natural frequency … one is led directly into the question of frequency dispersion" (Wireless World, 1982; Lecture 24).
  • What was tested: whether his own units give that natural frequency without further assumptions.
  • Steps:
    1. Both frames orbit with radius r about their common centre, so their inertias per cell are equal (M_G = m₀).
    2. The relative displacement X obeys X” = −k(1/m₀ + 1/M_G)X, with k = σe/ε₀ (U04).
    3. A circular orbit under a linear restoring force is itself the free oscillation, so its frequency is the natural frequency: √(2k/m₀)/Ω = 0.9999997.
    4. ħ × natural frequency = 510.999 keV.
    5. If the continuum did not move (infinite M_G), it would be 361.3 keV.
  • Output: 2 hbar x natural frequency 510.99879 keV
  • Verdict: Consistent. Aspden says so himself, and his units give exactly mₑc². This is the plasma frequency of the two frames (U33).
  • Challenge points: the equal-inertia step. It follows from his description of two frames circling a common centre with equal radii. If the G frame were much heavier, the resonance would move to 361 keV, which does not change any conclusion below.

U11. The static polarization of the lattice equals the vacuum’s

  • Aspden’s claim: displacement currents in the vacuum are displacements of the quon lattice (POC ch. 6). Electromagnetic waves are "oscillations of the quon lattice system".
  • Steps:
    1. A static field E displaces the frames by X = eE/k.
    2. The polarization is P = neX = ne²E/k, and with k = ne²/ε₀ this is P = ε₀E exactly.
    3. The same holds for any ratio of the two inertias (script: M_G = 10m₀ gives 1.0).
  • Output: 1 static P / (eps0 E) (=1: quon displacement alone gives exactly the vacuum D = eps0 E) 1.0
  • Verdict: Withdrawn as a credit (version 1.3). The restoring force k used here is the field of the displaced lattice itself (Part 1A, section 4). So "P = ε₀E" means the lattice’s own field cancels the applied field inside it. That is complete screening, as in a metal, not a vacuum (U36).
  • Challenge points: this is a general property of any charge lattice in a neutralizing continuum, whatever its spacing. It shows consistency, not evidence for these particular values.

U12. Aspden’s resonance picture: dispersion below 511 keV and no light above it

  • Aspden’s claim: none. This is the consequence of U10 and U11 that his 1982 article set out to avoid. In his words, "the major problem confronting aether theory, namely the issue of its non-dispersion property" (1982a commentary).
  • What was tested: how light would travel if the quons alone carried the displacement current, in two readings.
    • Reading B: the quons are the whole displacement field, so ε_r(ω) = Ω²/(Ω² − ω²).
    • Reading A: a bare vacuum plus the quons, ε_r = 1 + Ω²/(Ω² − ω²), with the bare speed set so that radio waves still travel at c.
  • Steps:

    1. Reading B, group velocity v_g = c(1 − ω²/Ω²)^(3/2):

      Photon energy 1 − v_g/c
      2 eV (visible light) 2.3 × 10⁻¹¹
      1 keV 5.7 × 10⁻⁶
      10 keV 5.7 × 10⁻⁴
      100 keV 0.057
      300 keV 0.47

    2. Reading B: above 511.0 keV, ε_r < 0, so no wave propagates. A 661.7 keV gamma ray would decay within 2.5 × 10⁻¹³ m.

    3. Reading A: no wave propagates between 511.0 and 722.7 keV (6.1 × 10⁻¹³ m decay length at 661.7 keV). Above that band light travels faster than c, reaching √2 c = 1.414c at GeV energies.

  • Output: B reading B: eps_r < 0 (no propagating wave) for every photon energy above 510.999 keV; A reading A, photon 6.0e+6 keV: v_group/c 1.414214
  • Verdict: this is the prediction of Aspden’s 1982 resonance picture, in which each quon has its own spring of frequency Ω. U13–U14 compare it with data. Version 1.3 note: the lattice he describes has no such spring (Part 1A, section 4). Its self-consistent response is a plasma (U34), which fails at all energies below 511 keV instead.
  • Challenge points:
    • The wave’s magnetic force. The quons move at c/2, so the wave’s magnetic force on them is comparable to the electric force. It does not enter here: the quon and its continuum charge have opposite charges and opposite velocities, so the magnetic forces on them are equal. They move the pair together, not apart, and so do not polarize the lattice.
    • Resonance damping. Damping is ignored. It would broaden the resonance without removing it.

U13. Gamma rays above 511 keV travel, and at the speed c

  • Aspden’s claim: above "the frequency at which electron-positron pair creation occurs … there is no counter-displacement" (Lecture 24, 1982 article). So above that frequency his remedy is absent by his own account, and the single-displacement response of U12 applies.
  • Steps:
    1. Propagation.
      • Caesium-137 (661.7 keV) and cobalt-60 (1173 and 1332 keV) gamma rays cross metres of air and vacuum in every gamma-spectroscopy laboratory.
      • INTEGRAL detected the 847 and 1238 keV cobalt-56 lines from supernova SN 2014J in M82, just over 10 million light-years away.
      • Reading B forbids all of these; reading A forbids the 662 keV line, which lies inside its 511–723 keV stop band.
    2. Speed. Alväger et al. (1964) timed gamma rays above 6 GeV from decaying neutral pions at c′ = (2.9977 ± 0.0004) × 10⁸ m/s, that is 0.999925c. Reading A predicts 1.414c, excluded by 3,105σ.
  • Inputs: Churazov et al., Nature 512, 406 (2014); MPA press release (distance); Alväger, Farley, Kjellman and Wallin, Phys. Lett. 12, 260 (1964).
  • Output: 3 reading A excluded by (sigma) 3105.0 sigma
  • Verdict: Excluded, for both readings of the resonance picture. (The self-consistent plasma model of U34 lets these gamma rays through, slowly, but blocks everything below 511 keV.)
  • Challenge points:
    • Pair-creation threshold. Aspden’s "pair creation frequency" may mean mₑc² or 2mₑc² (1.022 MeV). Either way, the cobalt and GeV results lie above it.
    • The 511 keV line is not used. The Galactic annihilation line (Siegert et al., A&A 586, A84, 2016) sits at the resonance, but light just below it is slowed rather than stopped. A steady source would still be seen, so the line does not discriminate.

U14. GRB 090510: the same arrival time from keV to 31 GeV after 7.5 billion years

  • Aspden’s claim: below the limit frequency the vacuum "adapts to any frequency and exhibits the properties of a tuned LC system … the local vacuum resonates at the optimum frequency set by these displacements. In this way the signal frequency is preserved over vast distances" (Lecture 24).
  • What was tested: first, how far the simple lattice (U12) misses this burst; second, whether his tuned-vacuum remedy can be tested at all.
  • Steps:
    1. The light-travel time to z = 0.903 (Planck 2018 flat ΛCDM) is 7.535 billion years.
    2. Fermi detected a 31 GeV photon 0.829 s after the burst began. Abdo et al. limit the delay between it and the low-energy (keV–MeV) emission to |Δt| < 0.859 s.
    3. So |Δv|/c ≲ 0.859 s ÷ 7.535 Gyr = 3.6 × 10⁻¹⁸. This is approximate; dedicated analyses weight the distance by (1 + z).
    4. Reading B: v(10 keV) − v(300 keV) = 0.47c. The arrivals would spread over 3.5 billion years, about 10¹⁷ times the limit, and the 31 GeV photon could not travel at all.
    5. Reading A: v(10 keV) − v(31 GeV) = 0.41c, about 2 × 10¹⁷ times the limit.
    6. Aspden’s tuned vacuum gives no equation from which a delay can be computed. His description makes the vacuum at each point tune to the frequency of the signal passing through it. The burst sent photons from about 10 keV to 31 GeV through the same space at the same time, as ordinary starlight sends all its colours, so a single local tuning cannot serve them all. A response that depends on the signal is also a nonlinear one, under which crossing beams would interact.
  • Inputs: Abdo et al., Nature 462, 331 (2009); Planck 2018 cosmology (H₀ = 67.4, Ω_m = 0.315).
  • Output: 5 excess over the limit 1.3e+17; 7 excess over the limit 2.05e+17
  • Verdict: Excluded for the simple lattice. Open but unsupported for the 1982 remedy: no equations are given, and its central idea conflicts with light of many frequencies sharing the same space.
  • Challenge points:
    • The distance measure; using the (1 + z)-weighted distance changes the limit by less than a factor of 2.
    • Whether a full mathematical version of the dual-displacement idea could avoid superposition problems. None exists in the texts examined. This is the place where a defender of the theory would need to supply one.

U15. What the scale would have to be

  • What was tested: how high the vacuum resonance would have to sit for an oscillator-type vacuum to pass GRB 090510, and what that would do to his units.
  • Steps:
    1. Well below resonance, 1 − v_g/c ≈ (3/2)(E/E_res)². Requiring this to be under 3.6 × 10⁻¹⁸ at 31 GeV gives E_res > 2.0 × 10¹⁰ GeV.
    2. That is 3.9 × 10¹³ times mₑc².
    3. Since Ω = c/2r, the orbit radius would have to be below 4.9 × 10⁻²⁷ m.
    4. But r = ħ/2mₑc is what ties the lattice to the electron (I1). It also fixes m₀, b, N and E₀ in electron units.
  • Output: 3 in electron rest energies (Aspden: exactly 1) 3.91e+13
  • Verdict: Excluded as a repair. Because α and h depend only on r/d, a smaller lattice could keep them. But the electron would then have to be about 4 × 10¹³ times heavier than it is. For the self-consistent plasma model, the requirement is far stronger: ħΩ below the photon-mass limit of 10⁻¹⁸ eV (U35).
  • Challenge points: the quadratic form assumes an oscillator-like response. A non-oscillator response (U14 step 6) is not covered by this bound.

U16. Lattice discreteness

  • Aspden’s claim: the photon "will emerge as the largest symmetrical unit of that lattice structure", and electromagnetic waves are lattice oscillations (POC 7; ch. 6).

  • Steps:

    1. A lattice of spacing d carries no wave shorter than 2d = 1.275 × 10⁻¹² m. That is a photon of 0.97 MeV.

    2. The highest nearest-neighbour lattice frequency is ħ·2c/d = 0.62 MeV.

    3. LHAASO detects photons up to 1.4 PeV (wavelength 8.9 × 10⁻²² m), 1.4 × 10⁹ times shorter than 2d.

    4. Below the cutoff, a cubic lattice slows short waves by about (kd)²/24:

      Light Speed deficit (kd)²/24 Direction difference, [100] vs [111], (kd)²/36
      500 nm (visible) 2.7 × 10⁻¹² 1.8 × 10⁻¹²
      10 keV X-ray 4.4 × 10⁻⁵ 2.9 × 10⁻⁵
      100 keV gamma 4.4 × 10⁻³ 2.9 × 10⁻³

  • Inputs: Cao et al. (LHAASO), Nature 594, 33 (2021).

  • Output: 5 2d / that wavelength 1.44e+9

  • Verdict: Excluded, unless high-energy photons are something other than lattice waves. The one passage found on this point asks whether "those high energy gamma rays are really nothing other than references to energy transactions involving the aether" (Fifth Question, 2001), which does not say how they travel.

  • Challenge points:

    • The (kd)²/24 form assumes nearest-neighbour coupling; long-range Coulomb coupling changes the coefficient but not the cutoff.
    • The visible-light direction effect (about 10⁻¹² at 500 nm, with four-fold symmetry under rotation) has not been compared here with the rotating-cavity experiments. Those are usually analysed for two-fold signals, and checking this is listed as an open lead.

U17. Semiconductor light holes as evidence for the quon mass

  • Aspden’s claim: the quon "has an effective mass 0.04078 times that of the electron. It is debatable whether the effective mass of positive holes in semi-conductors may in some way reflect the transitional states of electrons and sub-electrons … Experiments on p-type germanium crystals have given effective masses m* of positive holes as m*/m = 0.04 and m*/m = 0.3" (Lecture 1 appendix, citing Ehrenberg).

  • Steps: light-hole masses compared with his 0.04078:

    Crystal Light hole (mₑ) Ratio to 0.04078 Source
    InSb 0.015 0.37 Ioffe
    Ge 0.043 1.05 Ioffe
    Ge 0.044 1.08 Van Zeghbroeck
    GaAs 0.082 2.01 Van Zeghbroeck
    Si 0.16 3.92 Van Zeghbroeck

    The spread is a factor of 10.7.

  • Inputs: Ioffe Institute semiconductor database (Ge, InSb); B. Van Zeghbroeck, Principles of Semiconductor Devices (Ge, Si, GaAs).

  • Output: 2 spread of light-hole masses across these crystals (max/min) 10.7

  • Verdict: Not supported. Light-hole masses follow each crystal’s band structure (k·p theory ties them to the band gap at the zone centre), so a universal aether particle cannot set them. Germanium’s match is one point inside a tenfold spread. Aspden himself called the link "debatable".

  • Challenge points: the comparison uses room-temperature band-edge masses. Light-hole masses also vary with temperature, strain and direction (band warping), which only widens the spread.

U18. The 1969 version of the lattice

  • Aspden’s claim: r/d = 0.30292 after a correction y = 0.000456 e²/d, "introduce[d] … at this stage without proof" (PWE 6.57–6.58). It is derived later in the chapter from one electron paired with each graviton, 5,063 lattice particles per electron, and energy ½m₀(c/2)² doubled for the G frame (6.74–6.75). 1/α = 137.038, "exactly the value measured" (6.61).
  • Steps:
    1. y = 2(m₀c²/8)/5063 in units of e²/d = 0.0004555 (printed 0.000456).
    2. 144π × 0.30292 = 137.03778.
    3. Rounding r/d to five digits allows ±16.5 ppm.
    4. The July 1969 adjustment gave 1/α = 137.03602(21). Aspden’s value is 12.8 ppm above it, inside his rounding.
  • Inputs: CODATA 1969 value as tabulated in the Wikipedia article "Fine-structure constant" (historical values table).
  • Output: 1 y = 2 (m0 c^2/8)/5063 in units of e^2/d (PWE prints 0.000456) 0.0004555
  • Verdict: Consistent at the precision he printed. The correction is reproducible from his description. It ties α to the graviton mass (5,063 mₑ, the same number that sets his G). The 1980 integer rule replaced it, so his α derivation was changed after 1969 by swapping one small correction for another.
  • Challenge points:
    • The sixth digit of 137.038 is not reproducible from the five printed digits of r/d.
    • The measured value he compared with in 1969 has not been identified. The pre-1969 accepted value is not checked here.

U19. The 2003 k term

  • Aspden’s claim: the finite-size term carries a factor (1 − 2/3k), where kb is the mean distance to which the continuum charge displaced by a quon moves (7.17–7.18). "This is when r/d has the value 0.302916 and k is 8.34" (POC p. 109).
  • Steps:
    1. Let F = 1 − 2/3k be the factor on the finite-size term. The F at which the zero-energy r/d equals the N = 1843 value is 0.920134 with his printed lattice sum 2.31368, giving k = 8.347 (his 8.34). With the exact sum 2.3136987 it is 0.922413, giving k = 8.593.
    2. If the displaced charge stays within its cell (kb ≤ d/2), then k ≤ d/2b = 13.84 (F ≤ 0.952).
    3. So any k from about 8.4–8.6 up to 13.8 keeps N = 1843 and leaves 1/α at 137.035915.
  • Output: 2 k at that point (Aspden's 2.31368; POC: 8.34) 8.347
  • Verdict: Credited (reproduced and consistent). The term was absent in 1980. It adds a parameter, but over a range wide enough not to affect the result. It ties in with the main paper’s finding that the finite-size coefficient must lie in a narrow window (§5.3): k values near 8.5 sit at that window’s lower edge.
  • Challenge points: the k term is a physical correction added in 2003. It is needed only if the continuum yields to compression, as Aspden notes, offering a third alternative (POC p. 113).

6. The 0.6 ppm: where it comes from and whether it can be fixed

This section answers one question in order. Why is the chain 0.6 ppm off, and is there anything in Aspden’s figures or reasoning that can be corrected to close it?

U20. The 0.6 ppm is one number

  • Aspden’s claim: 1/α = 108π(8/N)^(1/6) = 137.035915 with N = 1843 (PU 157; POC 7.24–7.26).
  • What was tested: whether the small mismatches seen in U03–U08 are separate errors or one.
  • Steps:
    1. His 1/α divided by the measured 1/α is 0.999999384616, that is −0.615 ppm.
    2. The quon mass from the force balance, using the measured e², divided by the quon mass from (7.25), is 0.999999385229: the same −0.615 ppm (U04).
    3. Repeating step 2 with the e² that his own α implies (q² = αħc with his α) gives exactly 1.
    4. The same holds for b from (7.4) against (7.15) and for E₀ from (7.3) against (7.29): exactly 1 with his α.
  • Output: 3 m0 (force balance, his alpha) / m0 (7.25) 1.0
  • Result: every relation among his units is exact. The 0.6 ppm appears wherever the measured electron charge meets his lattice ratio r/d. It is a single number: the difference between his α and the measured α.
  • Verdict: Credited (internally exact). The question becomes why his α is 0.615 ppm low.
  • Challenge points: none in the arithmetic. The measured value used is CODATA 2022, 1/α = 137.035999177(21). The atom-recoil values (Rb 2020, Cs 2018) move the gap by less than 0.002 ppm.

U21. His printed figures

  • What was tested: whether any arithmetic or transcription error in his published numbers explains the gap.

  • Steps:

    Quantity Aspden printed Recomputed
    N with r/d = 0.302874 "approximately 1844.53" (POC p. 108) 1844.53
    1/α = 108π(8/1843)^(1/6) 137.035915 (PU 157) 137.0359148
    E₀ = (3/4π)(108π)³ 1843^(−4/3) mₑc² 412.665816 (PU 160) 412.6658162
    Lattice term in (7.18) 2.31368 (hand sum) 2.3136987 (exact Ewald sum; main paper §5.2)
    b/d 0.035 (POC p. 109) 0.036136
    kb for k = 8.34 "more than 0.29d" 0.301d

  • Output: 5 1/alpha with his hand sum and with the exact sum, after the integer rule 137.0359148

  • Result: every figure reproduces. His hand lattice sum is 1.9 × 10⁻⁵ below the exact value, and his b/d is rounded, but neither changes the final α, because the integer rule absorbs them.

  • Verdict: Credited. No slip in his figures causes the gap.

U22. What a correction would have to do

  • What was tested: which parts of the derivation can move the final α, and by how much they would have to move.
  • Steps:
    1. Under the integer rule the final value is 1/α = 108π(8/N)^(1/6). It depends only on N = 1843 and on two relations:
      • the photon-unit relation, ħc/e² = 144π(r/d);
      • the size relation, N = (9/8)³(d/r)⁶.
    2. The lattice sum, the finite-size term and the k factor only decide which odd integer is chosen. Any finite-size factor F from 0.93 to 1.02 gives exactly the same 137.035914847.
    3. A fractional change ε in the photon-unit relation moves 1/α by 1.5ε. Closing the gap needs ε = +0.410 ppm.
    4. A fractional change δ in the size relation moves 1/α by δ/6. Closing the gap needs δ = +3.69 ppm, which is the same as replacing N = 1843 by 1842.9932.
    5. A change η in the orbit radius r = ħ/2mₑc enters N as its cube. Closing the gap needs η = +1.23 ppm.
    6. Today’s measurement fixes 1/α to 0.00015 ppm. So a real correction must predict the 0.615 ppm gap to within about 1 part in 4,000.
  • Output: 1 required change to the photon-unit relation, eps (1/alpha moves by 1.5 eps) 0.41026 ppm
  • Verdict: this defines the target for U23–U26.
  • Challenge points: step 2 assumes the integer rule. Without the rule, the lattice details set α directly (U25).

U23. The restoring-force coefficient is justified

  • Aspden’s claim: the force on a quon displaced by D from the continuum is 4πσqD (POC 7.5, quoting his chapter 6 eq. 6.1). His basis is "the Maxwell displacement field theory governing that motion" (Fifth Question, 2001).
  • What was tested: whether the coefficient 4π is forced, since electrostatics alone gives different values for different shapes.
  • Steps:
    1. Shape dependence in electrostatics. For a finite body displaced uniformly against its background, the electrostatic restoring field depends on shape: 4πσD for a flat slab, 2πσD for a cylinder, (4π/3)σD for a sphere.
    2. No boundaries. In an unbounded medium there is no surface charge. A uniform displacement still makes a uniform current, and Maxwell’s equations with uniform fields give dE/dt = −4πJ, so E = −4πP = −4πσD in every direction. This is the long-wavelength plasma mode, and it gives Aspden’s coefficient.
    3. Alternatives. With the sphere’s coefficient the chain gives N = 67 and 1/α = 79.4. With the cylinder’s it gives N = 229 and 1/α = 97.0.
  • Output: 2 electrostatic sphere coefficient 4 pi/3 instead: odd N, 1/alpha 67 79.363639
  • Verdict: Credited. His coefficient is the one Maxwell’s equations require for an unbounded lattice. It is better justified than his text shows, and no other value comes close.
  • Challenge points: step 2 treats the aether as unbounded on the scale of the motion. Near matter, or at the edge of an aether region, the coefficient would change, and with it α. That is not considered by Aspden.

U24. The photon unit’s inertia leaves out effects far larger than the gap

  • Aspden’s claim: the 3×3×3 photon unit has moment of inertia 36m₀d², from twelve quons at distance d from the axis and twelve at √2 d (POC 7.21). It spins at a quarter of the orbital rate, since it "radiates at four times its spin rate" (main paper §5.1).
  • What was tested: whether the effects this sum leaves out are small enough for a 0.6 ppm result to be meaningful.
  • Steps:
    1. Spin rate Ω/4 = c/8r. The quons at distance d move at 0.413c, and those at √2 d at 0.584c.
    2. Speed-dependent mass. Aspden accepted that mass rises with speed; Physics Unified ch. 4 relies on "the relativistic formula for increase of mass with speed", which he held could be justified without Einstein. Applying it to the spin speeds alone raises the angular momentum by 18.7%. Carried through the chain, N becomes 3083 and 1/α becomes 162.6.
    3. Quons turning with the unit. If each quon also turns with the unit, its own inertia adds 1.26% when its Thomson field mass is spread out to d/2 (1/α = 138.76), or 0.039% as a uniform sphere of radius b (N jumps to 1845, 1/α = 137.092).
    4. Continuum turning with the unit. If the continuum’s equal inertia (U10) were carried round with the unit, the inertia would rise 112% (1/α = 291).
    5. The change actually needed is 0.000041% (U22). The omitted effects are 1,000 to 450,000 times larger.
  • Output: 3 angular momentum increase if mass rises with speed (Aspden accepts this, PU ch. 4) 18.69 %
  • Result: the formula 36m₀d² treats the photon unit as 27 rigid point masses moving slowly. Their actual speeds are 41–58% of the speed of light.
  • Verdict: Disagrees, internally. Taken as a mechanical calculation, the photon-unit step cannot be accurate to better than about 10%. The 0.6 ppm agreement therefore cannot come from the calculation’s precision. Either some reason, not given, makes these effects cancel exactly, or the closeness of 137.036 is a coincidence. The chance of that is estimated in the main paper, §11.
  • Challenge points:
    • The speed estimate. Step 2 counts only the spin speeds. Adding the orbital c/2 motion of every quon would change the size of the correction but not its order.
    • What "spin" means. A defender could argue that the spin is a travelling pattern of phase in the lattice rather than a rotation of mass. But then I = 36m₀d² has no basis, because a moment of inertia describes moving mass.
    • The continuum. Whether the continuum turns with the unit is not stated by Aspden. Step 4 is included to show the size of that question, not as his claim.

U25. The 2003 constant k can be calculated from his own description

  • Aspden’s claim: the continuum charge pushed aside by a quon’s volume moves "to a mean distance kb, where kb can have a value approaching d/2 if the displaced charge is spread over much of the aether cell" (POC 7.17). He requires only k ≥ 8.34.
  • What was tested: whether his description fixes k, turning a free constant into a calculated one.
  • Steps:
    1. In (7.17), kb is the distance at which the displaced charge’s interaction energy equals that of the spread charge, so 1/(kb) is the mean of 1/R over where the charge goes.
    2. The mean of d/R over a cube of side d about its centre is 2.380077. This is his own constant I₀ in (7.12), 2.380077382. It has the closed form 2[3 ln((1 + √3)/√2) − π/4].
    3. Excluding the quon’s own void of radius b changes it to 2.3723.
    4. So k = d/(b × 2.3723) = 11.665, inside his allowed range of 8.35–13.8 (U19), and F = 1 − 2/3k = 0.94285.
    5. With his integer rule, N stays 1843 and 1/α stays 137.0359148.
    6. Without the integer rule, the pure lattice gives 1/α = 137.0308995. That is −37.2 ppm, against −139.6 ppm with his F = 1.
  • Output: 2 k = d/(b x mean of d/R outside the void) 11.665
  • Verdict: Credited, as an improvement. His free constant can be calculated from his own words, and the value fits his constraints. It does not touch the 0.6 ppm gap, which the integer rule fixes, but it removes one adjustable number from the 2003 version.
  • Challenge points:
    • Uniform spreading. Spreading the displaced charge uniformly through the cell is one reading of "spread over much of the aether cell". Spreading it to a thinner shell near the cell wall would raise k, up to 13.8.
    • Neighbouring voids. The voids in neighbouring cells are ignored. They occupy 0.02% of each cell.

U26. Small numbers of the right size are not fixes

  • What was tested: whether simple small quantities of about the needed size close the gap, as a check against reading a match as a mechanism.

  • Steps: each candidate value of ε was put into the chain:

    Candidate ε Size (ppm) Resulting 1/α Remaining gap In today’s σ
    α³ 0.389 137.035994724 −0.033 ppm −212σ
    1/N² 0.294 137.035975364 −0.174 ppm −1,134σ
    (α/2π)² 1.349 137.036192113 +1.41 ppm +9,187σ

  • Output: 1 eps = alpha^3 = 0.3886 ppm: 1/alpha, residual ppm, residual in today's sigma 137.035994724 -0.0325 ppm -212.0 sigma

  • Verdict: none closes the gap. α³ comes nearest and still leaves a 212σ miss. With enough candidates one will always fall near the target, so a correction counts only if a mechanism produces it, derived before comparing with the measurement, and it lands within 1 part in 4,000 (U22). No such mechanism was found.

  • Challenge points: the main paper (§6.5) notes that the proton/electron residual is close to α³/2. One rule fixing both residuals would be worth a look. Applying α³ here leaves 212σ, so a single α³ rule does not do it.

Summary of section 6

  • Why the chain is 0.6 ppm off. Every relation among Aspden’s units is exact. The only gap is his fine-structure constant, which is 0.615 ppm low.
  • His figures are not the cause. They all reproduce.
  • His force law is sound. The restoring-force coefficient he uses is the one Maxwell’s equations require.
  • What is off is the photon-unit step. It treats 27 quons spinning at 41–58% of the speed of light as slow rigid point masses. The effects this leaves out are 10³ to 10⁵ times larger than the gap, so the near-agreement with α cannot be the product of an accurate calculation as the model stands.
  • The one improvement found. His 2003 constant k can be calculated from his own description rather than left free.
  • What would be needed to make the result meaningful. Someone would have to supply a treatment of the photon unit at those speeds, and show that it still gives 144π(r/d) with a correction of +0.41 ppm.

7. The photon unit in an endless lattice

The tests in section 6 treated the photon unit as an isolated 3×3×3 block. The block is in fact part of an endless lattice, a crystal with lattice vectors. Moving charges couple to one another, so the simple formulas for orbital speed, radius and angular momentum must be checked in that setting. The particles may also be better described as waves. This section does those checks. The script is lattice_dynamics.py. It treats the lattice as a crystal of point charges in a uniform neutralizing continuum, the arrangement Aspden describes, and sums the infinite lattice by the Ewald method.

U27. The orbit in endless space is exact

  • Aspden’s claim: both frames orbit with radius r at speed c/2, and the restoring force is 4πσq per unit displacement (POC 7.1, 7.5).
  • What was tested: whether the coupling between all quons in an endless lattice changes the orbit.
  • Steps:
    1. No net force from the other quons. In the orbit every quon moves by the same vector at the same time. Their relative positions never change, so the force on any quon from all the others, summed over the endless lattice, is zero at every instant.
    2. The Ewald check. The lattice’s dynamical matrix at long wavelength gives exactly the plasma frequency for motion along the wave and zero across it: 1.000000 -1.31e-10. The only restoring force on the uniform motion is the continuum’s (U23).
    3. Magnetic coupling. The two frames carry opposite charges moving in opposite directions, so their currents add: σc = 1.86 × 10²⁶ A/m². In an endless medium a uniform current is exactly cancelled by the displacement current of the same motion (Maxwell: dE/dt = −4πJ). So no magnetic field arises, and there is no magnetic coupling in the orbit.
  • Output: 1 D(k -> 0 along x) / w_p^2: longitudinal xx (expect 1), transverse yy (expect 0) 1.000000 -1.31e-10
  • Verdict: Credited. The orbit radius, speed and frequency survive the endless-lattice treatment unchanged, given his inputs.
  • Challenge points: the speed c/2 is relativistic. Aspden absorbs this into his "effective mass in orbit" m₀ (POC 7.4); that choice is his, not tested here.

U28. The endless simple cubic lattice is unstable

  • Aspden’s claim: the quons form a simple cubic lattice. He also says that if the lattice were allowed its minimum energy "the aether lattice particles would all be at rest … in a structure that was body-centred cubic, as in iron" (POC ch. 7), and that the zero-energy condition and the motion prevent this.
  • What was tested: whether the simple cubic arrangement is mechanically stable. In crystal terms this means checking that every lattice wave (every wavevector k in the Brillouin zone, every polarization) has a real frequency.
  • Steps:
    1. The method. The dynamical matrix D(k) of point charges q, mass m₀, spacing d in a rigid uniform continuum was built by Ewald summation. Checks:
      • the trace of D equals the plasma frequency squared at every k, as it must for Coulomb forces;
      • the results do not depend on the Ewald splitting parameter.
    2. Along the symmetry lines Γ–X–M–Γ–R–X, the two transverse branches have ω² < 0 along Γ–X, X–M and M–Γ. The worst point is M = (π/d, π/d, 0), with ω² = −0.093 ω_p².
    3. Over a 16³ grid of the Brillouin zone, 18.4% of all modes are unstable.
    4. The same code for a body-centred cubic lattice finds every mode stable, with lowest ω² = +0.003 ω_p². This also checks the method.
    5. At the zone face X, the transverse frequencies are imaginary (82.5i keV) and the longitudinal one is 379.7 keV, using ħω_p = 361.3 keV for the quons against a still continuum.
  • Output: 5 fraction of grid modes with w^2 < 0 (unstable) 0.184
  • Verdict: Disagrees, on electrostatics alone. A simple cubic lattice of like charges in a uniform background shears spontaneously toward another structure. Aspden’s text accepts that the static minimum would be body-centred cubic and relies on the orbital motion to hold the simple cubic form, but he gives no calculation showing that the motion stabilizes these shear modes.
  • Challenge points:
    • The orbital motion. The quons move at c/2, so velocity-dependent (magnetic) forces between disturbed quons are about (v/c)² = 25% of the electric ones. That is the same order as the instability (9% of ω_p²), so the motion could in principle stabilize the lattice. Whether it does is untested here and would need a relativistic lattice-dynamics calculation.
    • A rigid continuum. The continuum is held rigid for these internal waves, as a frame that moves only as a whole. A continuum that can compress would change the longitudinal branch, not the transverse instability.
    • If the lattice were body-centred cubic. His lattice sums (the Ewald value 2.8373 and the π/6 cell term) and with them r/d, N and α would all change.

U29. Coupling between moving quons changes the photon unit’s inertia by 9–29%

  • What was tested: how much the magnetic coupling between the 27 moving quons of a spinning photon unit changes its moment of inertia, and so Planck’s constant in his formula.
  • Steps:
    1. The coupling. To first order in (v/c)², the coupling between moving charges is the Darwin interaction, L = (q²/2c²r)[vᵢ·vⱼ + (vᵢ·n)(vⱼ·n)]. Summed over the 351 pairs of a 3×3×3 block spinning about a lattice axis, the sum is 97.44 (in units of ω²d).
    2. The added inertia. This acts as extra inertia, I_D/I = (3b/d) × 97.44/36 = 29.3%, using q²/c² = 3bm₀.
    3. Screening by the surrounding lattice. The rest of the endless lattice screens the magnetic field over the skin depth c/ω_p = 0.857d. Including that cuts the addition to 9.1%.
    4. Through his chain. Carried through his chain, the added inertia gives 1/α = 177.2 (unscreened) or 149.4 (screened), against 137.04.
    5. The change actually needed is 0.000041% (U22).
  • Output: 4 added inertia with the surrounding lattice screening the field (Yukawa, length c/w_p) 9.059 %
  • Verdict: Disagrees. Coupling between the moving charges does change the simple formula, as the challenge said, but by about 10–30%. That moves the result further from α, not closer, and is far beyond the 0.41 ppm a fix needs.
  • Challenge points:
    • Leading order only. The Darwin form is the leading order in (v/c)². At 0.4–0.6c, higher orders are not small.
    • The screening model. Screening is modelled as a simple exponential over the skin depth. A full calculation would use the lattice’s actual magnetic response.
    • The continuum. The continuum inside the spinning block is taken not to turn with it (U24).

U30. The speeds do not depend on the picture

  • What was tested: whether describing the photon unit as a wave rather than a rigid body lowers the quons’ speeds.
  • Steps:
    1. Energy of any circular motion. For any motion in which particles circle at angular frequency ω (a rigid spin, or any circularly polarized wave), the kinetic energy is exactly half the angular momentum times ω. With angular momentum ħ and ω = Ω/4 this is ħΩ/8 = 63.9 keV.
    2. Compared with rest energy. The 27 quons’ rest energy is 27 × 20.84 = 562.7 keV, so the kinetic energy is 11.4% of it.
    3. The speed. The rms speed is therefore √(2 × 0.1135) c = 0.476c, whatever the pattern of motion, as long as these 27 quons carry the angular momentum.
  • Output: 4 rms speed sqrt(2 KE / 27 m0), any pattern of circular motion at that frequency 0.4765 c
  • Verdict: Disagrees. A wave description of the same 27 quons gives the same speeds. The speeds follow from the energy, not from the rigid-body picture.

U31. Spreading the wave over more quons

  • What was tested: whether a wave spread over many quons could carry the same angular momentum slowly while keeping Aspden’s inertia, and what that costs.
  • Steps:

    1. Spreading keeps the inertia. The inertia I = Σm₀aᵢ² stays 36m₀d² if the same sum of squared radii is shared among more quons. Their speeds then fall as 1/√(number of quons).

    2. Speeds and mass corrections:

      Quons sharing the angular momentum rms speed mass correction v²/2c²
      27 0.476c 114,000 ppm
      1,000 0.078c 3,070 ppm
      100,000 0.0078c 31 ppm
      10,000,000 0.00078c 0.31 ppm

    3. How many are needed. The correction falls below the 0.41 ppm a fix needs at about 7.5 million quons. That is a cube about 195 lattice spacings, or 1.25 × 10⁻¹⁰ m, on a side, roughly twice the Bohr radius.

  • Output: 2 quons needed for the mass correction (v^2/2c^2) to fall below 0.41 ppm 7.47e+06
  • Verdict: Open, and it carries a cost. A wave spread over roughly atomic dimensions could carry ħ slowly. But Aspden’s value of h comes from the number 36, the moment of inertia of a rigid 3×3×3 cube in units of m₀d². In a spread-out wave, the sum of squared amplitudes is fixed by requiring the angular momentum to be ħ, so 36m₀d² would have to be imposed rather than derived. The derivation of h, and with it the formula for α, would become an input.
  • Challenge points: the spread-out wave would also need magnetic coupling (U29) and the lattice’s own wave modes (U28) included before its inertia could be stated. No such calculation exists in Aspden’s work or here.

Summary of section 7

  • What the endless lattice confirms. The orbit is exact, and the restoring force is right.
  • What it does not support. The photon unit as Aspden treats it fails in this setting in three ways:
    • the simple cubic lattice is electrostatically unstable;
    • coupling between the moving quons changes the photon unit’s inertia by tens of percent;
    • a wave picture of the same quons has the same speeds.
  • The one way to lower the speeds is to spread the photon over millions of quons, and that removes the geometric origin of the 36 in his formula for h.
  • What a working version would need: a full relativistic lattice-dynamics treatment (and, if waves are taken seriously, a quantum one) that keeps a stable lattice and still produces 144π(r/d) to within 0.4 ppm. That is a substantial research problem, and it is where the theory would have to be rebuilt.

8. The self-consistent model: the vacuum as a plasma

Each step below is derived in full in the supplement, Part 1A. The numbers are printed by quon_model.py (blocks W3–W7 for U32–U36, W9–W12 for U37–U40, W14 for U41).

U32. A quon feels only the average field

  • Aspden’s claim: under an applied field the quons move together, and "there will be no change in the Coulomb force on any particle due to the action of its neighbours". The restoring force 4πσqD comes from the uncovered "planar slice" of continuum charge (POC ch. 6, eq. 6.1).
  • What was tested: whether a quon has any restoring force of its own when the whole lattice shifts, or only the average field of the displaced system.
  • Steps:
    1. Its own cell’s continuum (a sphere) pulls the quon back with −(4π/3)σqu.
    2. All other cells form a polarized medium with P = σu. At the quon, they give the average field plus the Lorentz cavity field +(4π/3)P.
    3. The two local terms cancel exactly. The quon feels only qE, the average field.
    4. Exact Ewald lattice sums confirm it:
      • for long-wavelength motion along the wave, the restoring force is exactly the plasma value;
      • across the wave, it is zero;
      • for one quon moved alone, it is one-third of the plasma value.
  • Output: 6 Ewald check, k -> 0: transverse / w_p^2 (no local restoring force) = -1.31e-10
  • Verdict: Credited. Aspden’s statement is exact. His restoring force 4πσqD is therefore the field of the displaced lattice, not a separate spring.
  • Challenge points:
    • The cancellation holds for cubic lattices of point charges. A quon of finite size changes it by about (b/d)³ ≈ 5 × 10⁻⁵.

U33. The orbit is the plasma oscillation of the two frames

  • Steps:
    1. Equations of motion. m₀ü_E = qE and Mü_G = −qE, with E = −4πP and P = nq(u_E − u_G).
    2. Combined, P̈ = −4πnq²(1/m₀ + 1/M)P. With M = m₀ this gives Ω_pl² = 8πnq²/m₀, Aspden’s 7.5.
    3. Numbers. ħΩ_pl = 510.999 keV; Ω_pl/Ω = 1.0.
  • Output: 3 Omega_pl / Omega (Aspden's orbital frequency) = 1.0
  • Verdict: Credited. His orbit is the plasma oscillation of the quon frame against the continuum, consistent with Coulomb’s and Newton’s laws, at exactly the Compton frequency.

U34. Light below 511 keV cannot travel

  • Steps:
    1. Driven by a light wave, P(ω) = −(Ω²/4πω²)E, the response of a plasma.
    2. With Maxwell’s equations, c²k² = ω² − Ω². No wave exists below Ω. A blocked wave dies away within c/Ω = 2r = 3.9 × 10⁻¹³ m.
    3. Predicted:
      • radio, visible light, 10 keV X-rays and 100 keV gamma rays are all blocked;
      • 661.7 keV gamma rays travel at 0.635c;
      • between 35 MeV and 31 GeV the speed difference is 1.1 × 10⁻⁴ of c, 3 × 10¹³ times the GRB 090510 limit.
  • Output: visible, 500 nm -4.2466e+10 no
  • Verdict: Excluded, by the fact that light reaches us at all.
  • Challenge points: this assumes Maxwell’s equations with their usual vacuum term. If the quons carry all the displacement current with no separate vacuum term, as Aspden says, then c²k² = −Ω² and nothing propagates at any frequency (Part 1A, 6.6).

U35. The same thing as a photon mass

  • Steps:
    1. The equivalence. c²k² = ω² − Ω² is the wave equation of a photon with rest energy ħΩ = 511 keV.
    2. The limits:
      • Particle Data Group, from the solar wind’s magnetic field (Ryutov 2007): below 10⁻¹⁸ eV;
      • laboratory (Lakes 1998): below 10⁻¹⁷ eV;
      • test of Coulomb’s law (Williams, Faller and Hill 1971): below 10⁻¹⁴ eV.
  • Output: PDG limit (Ryutov 2007, solar wind): limit (eV) and ratio = 1.0e-18 x 5.1e+23
  • Verdict: Excluded by a factor of 5 × 10²³. A lattice tied to the electron (ħΩ = mₑc²) cannot pass. Passing would need ħΩ < 10⁻¹⁸ eV, which would mean r = ħc/2E > 10¹¹ m.

U36. Static charges would be screened

  • Steps:
    1. Exact lattice sums. An applied field that varies over λ is reduced to the fraction 1 − ω_p²/ω_L(k)² of itself.
    2. Fraction left:
      • λ = 200 lattice spacings (1.3 × 10⁻¹⁰ m): 0.003%;
      • λ = 20 spacings: 0.26%;
      • λ = 2 spacings: 9%.
    3. His own equation says the same. Aspden’s 6.1 says the quon stops where the restoring force equals Vq. In Coulomb’s law that is where the total field at the quon is zero, for every other charge there as well.
  • Output: k = 0.01 pi/d w_L^2/w_p^2 = 1.00003 field left = +0.00003 of the applied field
  • Verdict: Excluded. A charge’s field would vanish beyond about 10⁻¹⁰ m, so there would be no electrostatics at laboratory scale.
  • Challenge points: avoiding this needs a force law in which the lattice’s own field acts on quons but not on ordinary matter. Aspden does not state one, and it would contradict Coulomb’s law (Part 1A, 6.7).

U37. No restoring force is needed for a plasma

  • Objection tested: the quons need not have any restoring force.
  • Steps:
    1. Agreement. Section 4 of Part 1A already finds no local restoring force (U32).
    2. What sets the plasma frequency. ω_p² = nq²/ε₀m contains only density, charge and mass, with no spring constant.
    3. A real example. The Earth’s ionosphere has about 10¹² free electrons per m³ with nothing holding them. Its cutoff is f_p = 8.98 MHz, and it reflects radio below that.
    4. Aspden’s quons. The same formula gives 361 keV, or 511 keV with the continuum moving.
  • Output: ionosphere F layer, electrons n = 1e12 m^-3: f_p = 8.97866e+6 Hz
  • Verdict: Consistent with U34. The plasma result does not depend on a restoring force; removing one makes the lattice more plasma-like, not less.

U38. The lattice flexing as an elastic solid

  • Objection tested: the medium flexes to carry the waves.
  • Steps:
    1. The shear waves. A crystal’s elastic flexing (neighbours shifting relative to one another) carries shear waves. The exact lattice sums give their speeds at long wavelength:
      • along [111], 0.215c;
      • along [110], 0.279c, plus one unstable mode;
      • along [100], both modes unstable (0.134c × i).
    2. What light would need. To carry light at c the lattice would have to be 13–22 times stiffer, stable along the cube axes, and the same in every direction to 10⁻¹⁷.
  • Output: shear-wave speeds along [111] body diagonal = 0.215c, 0.215c
  • Verdict: Excluded as the carrier of light. Flexing as a polarizable medium is U34; flexing as a solid is this test. Both fail with Aspden’s charges, masses and spacing.

U39. Background (CMB) driven motion of the quons

  • Objection tested: the quons may be in periodic motion driven by the cosmic background.
  • Steps:
    1. Energy available. kT at 2.725 K is 2.3 × 10⁻⁴ eV.
    2. How far and how fast the quons move. A free quon would move at 55 km/s. In the lattice the thermal amplitude is 5.8 × 10⁻¹⁷ m, 0.0001 of the spacing.
    3. Still a solid. The crystal’s coupling parameter is 1.6 × 10⁷ (melting near 175).
    4. Already larger. Aspden already gives every quon an orbital motion at 149,896 km/s.
    5. No effect on the response. The response to a wave is linear and adds to any existing motion without being changed by it.
  • Output: thermal amplitude in the lattice, sqrt(kT/(m0 w_p^2)), and as a fraction of d = 5.79761e-17 m 9.09566e-5
  • Verdict: Does not change U34. Background motion is far too small to matter, and would not alter the response even if larger, while it stays linear.

U40. Light’s reference frame: the background’s frame and Aspden’s Earth-carried frame

  • Objection tested: the medium is the reference frame against which light travels, and it moves with the planet.
  • Aspden’s claim: the Earth’s electromagnetic reference frame "is a non-rotating frame of reference which moves linearly with the space lattice" (Physics Unified, ch. 9). If a laboratory’s West–East speed, "some 350 m/s at 40oN", showed up in light-speed measurements confined to the laboratory, "this would clearly refute Einstein’s theory" (1983b). His statements are collected in Part 1A, section 1.3.
  • Steps:
    1. The background’s frame (version 1.4 of this test; not Aspden’s). The Earth moves through it at 369.8 km/s (Planck 2018). Without length contraction, light’s speed in the laboratory would depend on direction by ½(v/c)² = 7.6 × 10⁻⁷, against cavity limits of about 10⁻¹⁷ to 10⁻¹⁸: a factor of 7.6 × 10¹⁰.
    2. Aspden’s frame. The 370 km/s disappears, because the frame travels with the Earth. A laboratory turning with the Earth crosses the frame at Ω_E R cos(latitude): 283 m/s at Berlin (52.5°N), 356 m/s at 40°N.
    3. The prediction. A laboratory-fixed signal at twice the turntable rate, ½(v/c)² = 4.4 × 10⁻¹³ at Berlin.
    4. The measurement. Herrmann et al. (2009), Berlin: laboratory-fixed signals below 3 × 10⁻¹⁶. The ratio is about 1,500. The largest steady flow of the frame past the laboratory that this allows is 7.3 m/s.
    5. A frame turning with the Earth would hide the signal, but a ring laser would then sense no rotation. The Wettzell ring laser senses the full rotation to 48 parts per billion. By Stokes’s theorem, if ring lasers sense the full rotation, the frame’s mean flow past the ground round Berlin’s latitude circle is 283 m/s, 38 times the cavity limit.
  • Inputs: the Earth’s rotation rate 7.2921150 × 10⁻⁵ rad/s and mean radius 6,371 km; Planck 2018 dipole 369.82 km/s; Herrmann et al. 2009 and Nagel et al. 2015 (refs 22–23); Wettzell ring laser (main paper §3.3).
  • Output: ratio = 1481.93 (Aspden’s frame, W12 line 9); ratio = 7.6e+10 (background’s frame, W12 line 4)
  • Verdict: Excluded: by a factor of about 1,500 in Aspden’s own frame, and 7.6 × 10¹⁰ in the background’s frame, unless moving rulers and clocks change exactly as relativity says. That is Lorentz’s aether, which cannot be detected (main paper, section 3.3).
  • Challenge points:
    • The coefficient ½(v/c)² assumes no length contraction. With full contraction the signal is zero.
    • The limit used is the laboratory-fixed one (3 × 10⁻¹⁶). The overall limit, about 10⁻¹⁷, would strengthen the result.
    • Version 1.4 of this test used the background’s frame, which was this analysis’s choice, not Aspden’s. Corrected in version 1.5.

U41. The postulate that the two frames move at c relative to each other

  • Aspden’s claim: "the relative velocity between the system of aether particles and the medium constituting the background continuum is assumed to be the speed of light" (The Theory of Gravitation, 1998 note); "a likely candidate is the relative velocity between the E-frame and the G-frame, meaning the quantity Ω(2r)" (POC 6.6). Each frame then moves at c/2 (POC 6.13–6.15). His other statements, from 1960 to 2003, are collected in Part 1A, section 1.2.
  • What was tested: whether the dynamics fixes the relative speed, and which of his results depend on it.
  • Steps:
    1. The force balance contains no radius. m₀Ω² = 8πq²/d³ (7.5 with σ = q/d³). The orbit’s frequency is fixed, and its size is free. In the self-consistent model the orbit is a plasma oscillation, whose frequency does not depend on its amplitude (U33).
    2. His chain with a general relative speed. Write Ω·2r = s·c. Aspden converts (7.19), I = 2h/πΩ, into (7.20) "because Ω is c/2r". Keeping s, (7.19) and the force balance give ħc/q² = 144π(r/d)/s in place of (7.24), m₀/mₑ = (4/9)(r/d)²/s² in place of (7.25), and N = (9/8)³s⁶(d/r)⁶ in place of (7.26).
    3. For a given N, s cancels. 1/α = 108π(8/N)^(1/6). The lattice spacing d = cN^(1/6)/(2Ω(9/8)^½), the quon mass, the quon radius and the cell energy are unchanged. Only r = s·ħ/2mₑc and r/d scale with s.
    4. Where s still enters. In the zero-energy condition (7.18), through r/d and the finite-size term, which goes as s⁴. With the exact lattice sum, N = 1843 is chosen for any s from 0.999862 to 1.0000414 (−138 to +41 ppm). Outside that range N becomes 1841 or 1845, and 1/α jumps by 181 ppm.
  • Inputs: Aspden’s equations 7.18–7.26; the lattice sum 2.3136987 (U21).
  • Output: range of s that keeps N = 1843 (the zero-energy N between 1843 and 1845) = 0.99986198 to 1.0000414 (W14 line 6)
  • Result: The postulate is not derived, and Aspden calls it an assumption. In his own chain it fixes only the orbit radius. It reaches α only through the choice of N, which tolerates a change of −138 to +41 ppm, so it cannot be the source of the 0.6 ppm gap.
  • Verdict: Open: an assumption, not derived. α, N and the lattice spacing do not depend on it.
  • Challenge points:
    • The finite-size term is taken to scale as s⁴, from b/d = s²/96π(r/d)² (7.15 with s). If that term were fixed independently of the force balance, the range would shift slightly.
    • Whether the frames’ relative speed could set light’s speed at all. In the self-consistent model, light’s speed comes from the field equations (Part 1A, section 6). The relative velocity turns with the orbit, so there is no steady stream for a wave to ride on (Part 1A, section 1.2).

Summary of section 8

  • Coulomb’s and Maxwell’s laws make his lattice a plasma. Applied to the lattice exactly as Aspden describes it, they show that his orbit is a plasma oscillation, and that his restoring force is the lattice’s own field.
  • A plasma at that frequency cannot be the vacuum. It blocks light below 511 keV, screens charges, and amounts to a photon of electron mass.
  • What would be needed. Keeping his picture requires a new rule for how ordinary matter feels the lattice’s field, which none of his texts supplies.
  • Light’s frame. Aspden’s own frame, carried with the Earth but not turning with it, is excluded through the Earth’s rotation by a factor of about 1,500 (U40). The c/2 orbital speed rests on a postulate that his chain needs only for the orbit radius (U41).

9. What has merit

  • A consistent chain. Given his inputs, every identity among the units holds to 0.6 ppm (U03–U08). His published numbers for N, k, y, E₀ and the lattice density reproduce. The postulate behind the c/2 orbital speed cancels out of α, N and the lattice spacing (U41).
  • Real scales. His orbit radius is the zitterbewegung amplitude and the range of QED vacuum polarization. His frequency is the Compton frequency. His statement that the vacuum has a critical frequency at the electron scale has a counterpart in standard physics: pair creation and vacuum polarization both turn on at that scale.
  • The orbit as a plasma oscillation. Treated with Coulomb’s and Newton’s laws, his orbit is the plasma oscillation of the quon frame against the continuum, at exactly the Compton frequency (U33). His statement that the neighbours exert no net force under a collective displacement is exact (U32).
  • The force law. His restoring coefficient 4πσq is what Maxwell’s equations require for an unbounded lattice (U23).
  • A free constant removed. His 2003 constant k follows from his own description of the displaced charge (U25).
  • The orbit in an endless lattice. The uniform orbital motion is exact when the lattice is treated as an endless crystal, with no net force from the other quons and no magnetic field (U27).
  • Candour. He identified dispersion as "the major problem confronting aether theory". He admitted he could not explain the continuum, and called his treatment of the cell energy a hope. Those admissions locate the weak points.

10. What a repair would have to do

  1. Let light through at all energies. In the self-consistent model the lattice is a plasma that blocks everything below 511 keV (U34–U36). Any repair must stop the lattice from responding to transverse fields, or must find a rule by which its field does not act on matter. Then it must propagate light without dispersion or a cutoff from radio to at least 1.4 PeV, with arrival times matching to about 10⁻¹⁸ (U13–U16), and without direction dependence at the 10⁻¹⁷ level of modern cavity tests (main paper §3). In standard physics the vacuum does polarize at exactly Aspden’s scale, ħ/2mₑc, yet light passes undisturbed, because that polarization is Lorentz invariant. It leaves freely travelling photons alone and shows up only where charges are close: the running of α, and the −27 MHz vacuum-polarization part of hydrogen’s Lamb shift. A repaired quon lattice would have to behave the same way. That means giving up a lattice with a fixed spacing at rest in a preferred frame as the carrier of light.
  2. Keep the electron tie. Shrinking the lattice to push the cutoff above observed energies breaks r = ħ/2mₑc (U15).
  3. What could survive. The dimensionless numbers (α from r/d, N, the proton and G chains) do not depend on the lattice’s absolute size or on how light travels through it. They could survive only as properties of some structure that is not a medium carrying light through space. That is a lead, not a result. Part 2 will test whether the particle-mass chain built on these units (dimuon, proton, taon, graviton) holds up out of sample, as the main paper (§14, item 2) proposes.

11. Summary of tests

ID Test Verdict
U01 Orbit radius = zitterbewegung amplitude and QED vacuum-polarization range Credited (correspondence; input, not derived)
U02 Orbital frequency = Compton frequency Consistent
U03 Lattice spacing d = 6.374 × 10⁻¹³ m = 108πa Consistent
U04 Quon mass from the force balance (7.5) = (7.25) Credited
U05 Quon radius, Thomson mass, N Credited (the halved-mass analogy carries the result: f = 1 gives 1/α 0.56% low)
U06 Photon-unit inertia gives 144π(r/d) Credited (algebra)
U07 Lattice density 143.5 g/cm³ (1969: 144) Consistent
U08 Cell energy vs two muon masses Disagrees (−0.21%)
U09 Aether energy density vs dark energy Open (2.5 × 10³⁵; same kind of problem as mainstream)
U10 Natural frequency = Compton frequency Consistent (Aspden 1982 agrees)
U11 Static lattice polarization = ε₀E Withdrawn as a credit (it is complete screening; U36)
U12 Aspden’s resonance picture: dispersion below 511 keV, no light above Prediction (of his 1982 picture)
U13 Gamma rays above 511 keV propagate at c Excluded (3,105σ for reading A; reading B forbids propagation)
U14 GRB 090510 arrival times; the 1982 tuned-vacuum remedy Excluded (10¹⁷); remedy open but unsupported
U15 Resonance scale needed to pass GRB 090510 Excluded as a repair (electron 4 × 10¹³ too heavy)
U16 Lattice cutoff vs 1.4 PeV photons Excluded (factor 1.4 × 10⁹)
U17 Light holes as evidence for the quon mass Not supported
U18 1969 version (y = 0.000456; 1/α = 137.038) Consistent at printed precision; reproduced
U19 2003 k term (k = 8.34) Credited (reproduced, 8.347)
U20 The 0.6 ppm is one number (all unit relations exact with his α) Credited
U21 His printed figures Credited (all reproduce)
U22 What a correction must do (0.41 ppm photon unit, 3.69 ppm size ratio, to 1 part in 4,000) Target defined
U23 Restoring coefficient 4πσq Credited (Maxwell, unbounded medium)
U24 Photon-unit inertia at spin speeds 0.41–0.58c Disagrees internally (omitted effects 10³–10⁵ × the gap)
U25 k calculated from his description (11.67) Credited (improvement; α unchanged)
U26 Small numbers of the right size (α³, 1/N², (α/2π)²) None closes the gap
U27 The orbit in an endless lattice Credited (exact; no magnetic field)
U28 Stability of the endless simple cubic lattice Disagrees (18% of modes unstable electrostatically; bcc stable)
U29 Magnetic coupling between the photon unit’s moving quons Disagrees (adds 9–29% to its inertia; 1/α 149–177)
U30 Speeds in a wave picture of the same 27 quons Disagrees (0.48c whatever the pattern)
U31 Spreading the wave over more quons Open (7.5 million quons needed; the 36 then has to be put in)
U32 A quon feels only the average field (no local spring) Credited (Aspden’s statement is exact)
U33 The orbit is the plasma oscillation of the two frames Credited
U34 Light below 511 keV cannot travel Excluded
U35 The same as a photon of electron mass Excluded (5 × 10²³ above the limit)
U36 Static charges screened beyond about 10⁻¹⁰ m Excluded
U37 No restoring force is needed for a plasma (ionosphere example) Consistent with U34
U38 The lattice flexing as an elastic solid (shear waves 0.2–0.3c, unstable on cube axes) Excluded as the carrier of light
U39 Background (CMB) driven motion of the quons Does not change U34
U40 Light’s reference frame: the background’s frame and Aspden’s Earth-carried frame Excluded (about 1,500 in Aspden’s frame; 7.6 × 10¹⁰ in the background’s)
U41 The postulate that the frames move at c relative to each other Open (assumed, not derived; cancels out of α, N and d)

References

  1. H. Aspden, The Physics of Creation, 2nd ed. (2003), chapter 7, eqs (7.1)–(7.29). https://aetherscience.org/book-physics-of-creation-2nd-ed/book-physics-of-creation-2nd-ed-2edpocch7/ (also https://aetherscience.org/2edpocch7/); chapter 6: https://aetherscience.org/book-physics-of-creation-2nd-ed/book-physics-of-creation-2nd-ed-2edpocch6/
  2. H. Aspden, Physics Unified (1980), chapter 6, eqs (128)–(160). https://aetherscience.org/book-physics-unified/book-physics-unified-puchapter6/
  3. H. Aspden, Physics without Einstein (1969), chapter 6, eqs (6.39)–(6.75). https://aetherscience.org/book-physics-without-einstein/book-physics-without-einstein-pwech6/
  4. H. Aspden, "The Ether – An Assessment", Wireless World 88, 37 (October 1982); reprinted as Lecture No. 24 (1998). https://aetherscience.org/www-energyscience-org-uk-le-le24/ ; commentary: https://aetherscience.org/aspden-www-aspden-org-papers-bib-1982a/
  5. H. Aspden, "The Fifth Question: What is a Neutrino?" (2001). https://aetherscience.org/www-energyscience-org-uk-question-es0105/
  6. H. Aspden, Appendix to Lecture No. 1. https://aetherscience.org/www-energyscience-org-uk-le-le01app/
  7. H. Aspden and D. M. Eagles, Phys. Lett. 41A, 423 (1972). https://aetherscience.org/aspden-www-aspden-org-papers-bib-1972a/
  8. H. Aspden, Tutorial Notes 7 and 8. https://aetherscience.org/www-energyscience-org-uk-tu-tu07/ ; https://aetherscience.org/www-energyscience-org-uk-tu-tu08/
  9. E. A. Uehling, "Polarization effects in the positron theory", Phys. Rev. 48, 55 (1935).
  10. T. Alväger, F. J. M. Farley, J. Kjellman and I. Wallin, "Test of the second postulate of special relativity in the GeV region", Phys. Lett. 12, 260 (1964). https://docente.unife.it/guido.zavattini/allegati/alvaeger-pl1964.pdf
  11. E. Churazov et al., "Cobalt-56 γ-ray emission lines from the type Ia supernova 2014J", Nature 512, 406 (2014). https://arxiv.org/abs/1405.3332 ; distance: Max Planck Institute for Astrophysics press release, https://wwwmpa.mpa-garching.mpg.de/mpa/institute/news_archives/news1408_ddd/news1408_ddd-en.html
  12. T. Siegert et al., "Gamma-ray spectroscopy of positron annihilation in the Milky Way", A&A 586, A84 (2016). https://www.aanda.org/10.1051/0004-6361/201527510/pdf
  13. A. A. Abdo et al., "A limit on the variation of the speed of light arising from quantum gravity effects", Nature 462, 331 (2009). https://www.openu.ac.il/personal_sites/yoni-granot/papers/GRB090510_Nature.pdf
  14. Z. Cao et al. (LHAASO), "Ultrahigh-energy photons up to 1.4 petaelectronvolts from 12 γ-ray Galactic sources", Nature 594, 33 (2021).
  15. Planck Collaboration, "Planck 2018 results. VI. Cosmological parameters", A&A 641, A6 (2020).
  16. Ioffe Institute, semiconductor database (NSM): Ge https://www.ioffe.ru/SVA/NSM/Semicond/Ge/bandstr.html ; InSb https://ioffe.ru/SVA/NSM/Semicond/InSb/bandstr.html ; Si (mirror) https://www-f9.ijs.si/~matevz/silikon/Semicond/Si/bandstr.html
  17. B. Van Zeghbroeck, Principles of Semiconductor Devices, "Effective mass in semiconductors" (1997). https://apachepersonal.miun.se/~gorthu/halvledare/Effective%20mass%20in%20semiconductors.htm
  18. Historical values of 1/α (CODATA 1969: 137.03602(21)): Wikipedia, "Fine-structure constant". https://en.wikipedia.org/wiki/Fine-structure_constant
  19. CODATA 2022 recommended values. https://physics.nist.gov/cuu/pdf/all.pdf
  20. Particle Data Group, photon mass limit m < 1 × 10⁻¹⁸ eV (Ryutov 2007); laboratory limit 1 × 10⁻¹⁷ eV (Lakes 1998); Coulomb’s-law test 1 × 10⁻¹⁴ eV (Williams, Faller and Hill 1971). https://pdg.lbl.gov/2021/listings/rpp2021-list-photon.pdf
  21. Aspden’s Aether from the Bottom Up, Part 1A: The Quon–Continuum Model, Derived Step by Step (supplement). https://aetherscience.org/aspden-quon-model/
  22. S. Herrmann et al., "Rotating optical cavity experiment testing Lorentz invariance at the 10⁻¹⁷ level," Phys. Rev. D 80, 105011 (2009). https://arxiv.org/pdf/1002.1284
  23. M. Nagel et al., "Direct terrestrial test of Lorentz symmetry in electrodynamics to 10⁻¹⁸," Nat. Commun. 6, 8174 (2015). https://pmc.ncbi.nlm.nih.gov/articles/PMC4569797
  24. H. Aspden, The Theory of Gravitation (1960), web reprint with 1998 notes. https://aetherscience.org/www-energyscience-org-uk-tg-pr1/
  25. H. Aspden, Physics Unified (1980), ch. 5 and 9. https://aetherscience.org/book-physics-unified/book-physics-unified-puchapter5/ ; https://aetherscience.org/book-physics-unified/book-physics-unified-puchapter9/
  26. H. Aspden, 1983b: optical sensing of a laboratory’s West–East speed. https://aetherscience.org/www-energyscience-org-uk-papers-bib-1983b/

Appendix. Program listings and output

The scripts aether_units.py (U01–U26) and lattice_dynamics.py (U27–U31), with their full output, are in the Word version of this paper and, as runnable files, in analysis-scripts.zip. Each output block heading matches a test ID above.