What Happens When You Fire an Electron at a Proton?
Three Energy Regimes, Three Answers, One Algebra

Three Energy Regimes, Three Answers, One Algebra

Martin Scholl — Independent Researcher  ·  It Is All One — Working Paper Series  ·  July 2026 (working draft)

April 2026

We take one physical system — an electron approaching a proton — and ask the simplest possible question: what comes out? The answer depends entirely on how hard the electron hits. At 13.6 eV, nothing flies out at all: the electron is captured, and you get a hydrogen atom. At 100 MeV, the electron bounces off and the proton recoils, but both remain intact. At 4 GeV, the proton gets excited and spits out a pion. At 20 GeV, the proton shatters, and you discover that it was built from quarks all along. Every number in this document is computed from first principles. No magic, no secret constants — just masses, α = 1/137, and ħc = 197.3 MeV·fm. The formulas are numbered for reference; all numerical results follow directly from them.

1The Setup: One Electron, One Proton, One Question

Imagine you have a proton sitting on a table. You pick up an electron and throw it at the proton. What happens? The answer depends on one thing: how fast you throw. And “how fast” is really “how much energy.” In physics, we measure this energy in electron-volts (eV). One electron-volt is the energy an electron gains when it falls through a one-volt battery — about 1.6 × 10⁻¹⁹ joules — the natural unit for atomic and particle physics. The beautiful thing about this experiment is that as you increase the energy, you peel back layers of reality like an onion. At low energy, you see the proton as a point. At medium energy, you see it as a ball. At high energy, you see inside the ball and discover it is made of smaller things. The mathematics changes at each layer — from quaternions to octonions — and this document shows you exactly where and why.

11 The two numbers that govern everything

Two numbers control all that follows.

Proton radius: rₚ = 0.8414 fm (1 femtometre = 10⁻¹⁵ m, a million times smaller than an atom)

Electron de Broglie wavelength: λ = h/p = 2πħc/p (the “resolution” of the electron as a probe)

Think of the electron as a flashlight: its wavelength is the size of the smallest detail it can illuminate. The fundamental rule of wave physics is that you cannot resolve anything smaller than your wavelength.

(1)λ = 2πħc / p = h / p

When λ ≫ rₚ: the electron sees a featureless dot. When λ ≈ rₚ: it starts to see shape. When λ ≪ rₚ: it sees inside. The electron does not change. The proton does not change. What changes is the resolution. The resolution criterion λ ≫ rₚ (= 0.84 fm): proton looks like a point → Regime 1 and 2 λ ≈ rₚ: proton’s surface visible → Regime 3 λ ≪ rₚ: interior visible → Regime 4 The transition energy (where λ = rₚ) is p = 2πħc/rₚ = 2π × 197.3/0.8414 = 1474 MeV/c.

12 Fundamental constants used throughout

α = 1/137.036 Fine structure constant (strength of the electromagnetic force)

ħc = 197.3 MeV·fm The natural unit conversion: energy times length

mᵉ = 0.511 MeV/c² Electron rest mass

Mₚ = 938.272 MeV/c² Proton rest mass

Mπ⁰ = 134.977 MeV/c² Neutral pion mass (lightest hadron)

Mπ⁺ = 139.570 MeV/c² Charged pion mass

MΔ = 1232 MeV/c² Delta baryon mass (proton first excited state)

2Regime 1: The Gentle Touch (13.6 eV)

Nothing flies out — you get a hydrogen atom

At room temperature, an electron has about 0.025 eV of kinetic energy. Even at 13.6 eV, it is barely crawling by particle physics standards. Let us compute its wavelength from equation (1). At 13.6 eV, the electron momentum (non-relativistic, since 13.6 eV ≪ mᵉc² = 511,000 eV) is p = √(2mᵉT) where T = 13.6 eV. Converting: T = 13.6 / 10⁶ MeV = 1.36 × 10⁻⁵ MeV. Then: λ = 2πħc / √(2mᵉc²·T) = 2π × 197.3 / √(2 × 0.511 × 1.36×10⁻⁵) = 332,559 fm = 3.32 Å Compare to the proton radius rₚ = 0.8414 fm. The ratio λ/rₚ = 395,300. The electron’s flashlight is nearly four hundred thousand times wider than the proton. It cannot see the proton at all — it sees a point charge, nothing more. Something remarkable happens: instead of bouncing off, the electron is captured. It falls into the lowest orbit around the proton. The system radiates a photon carrying exactly 13.6 eV, and you have a hydrogen atom.

21 The hydrogen energy levels

The energy levels of hydrogen are one of physics’ most beautiful results. In quaternion language, the electron’s state is described by:

Q = E + Lₓ·ι1 + Lᵧ·ι2 + Sᵣ·ι3

where E is the energy, Lₓ, Lᵧ are angular momentum components, Sᵣ is spin, and ι1, ι2, ι3 are the three quaternion imaginary units. The energy eigenvalues come from the requirement that the quaternion norm be preserved under rotation: Eₙ = −mᵉc²α² / (2n²) = −13.606 eV / n² n = 1, 2, 3, ... For n = 1: E₁ = −13.606 eV (ground state). For n = 2: E₂ = −3.401 eV. The photon emitted in the 2→1 transition carries E₂ − E₁ = 10.205 eV, wavelength 121.6 nm — the Lyman alpha line. Regime 1 summary: 13.6 eV Wavelength: λ = 332,559 fm (λ/rₚ = 395,300) Electron speed: v = 0.00730 c (non-relativistic: γ = 1.000027) What comes out: one photon (13.6 eV, ultraviolet) Process: electron capture → hydrogen atom Algebra: ℍ × ℍ (quaternion × quaternion)

3Regime 2: The Hard Bounce (100 MeV)

Elastic scattering — the proton stays intact

Now throw the electron much harder: 100 MeV, about 7.4 million times the energy of the hydrogen case. The electron is now ultra-relativistic. We compute the Lorentz factor: γ = E/mᵉc² = 100/0.511 = 195.7. The electron travels at β = √(1−1/γ²) = 0.999987c. Its de Broglie wavelength from equation (1):

λ = 2πħc / E = 2π × 197.3 / 100 = 12.39 fm (λ/rₚ = 14.7)

The electron’s flashlight is 15 times wider than the proton — still much too coarse to see inside. The proton stays intact. The electron bounces off the proton’s Coulomb field.

31 Elastic scattering kinematics

In elastic scattering, both particles are the same before and after. Only the momenta change direction. The scattered electron energy E′ at angle θ is given exactly by:

(5)E' = E / ( 1 + (2E/Mₚ) sin²(θ/2) ) [elastic, proton at rest]

Let us compute E′ and the kinetic energy given to the proton Tₚ = E − E′ for several angles: Even at backscatter (180°), the proton receives only 17.7 MeV and moves at 0.19c. The 1836:1 mass ratio means the electron is like a marble bouncing off a bowling ball.

32 The Mott cross section

Not all angles are equally likely. The differential cross section for relativistic electron scattering from a point charge is the Mott formula:

(2)dσ/dΩ = (αħc / 2E)² × cos²(θ/2) / sin⁴(θ/2)

The sin⁴(θ/2) factor in the denominator makes forward scattering overwhelmingly more probable. Let us compute dσ/dΩ in units of fm²/sr at 100 MeV:

theoremForward scattering at 10° is ten thousand times more likely than 90° scattering. Most electrons sail through with barely a nudge. Note that at exactly 180°, the cos²(θ/2) = cos²(90°) = 0 term kills the cross section — a purely relativistic effect absent in the non-relativistic Rutherford formula.

33 Centre-of-mass energy and the pion threshold

Can we break the proton at 100 MeV? We need to compute the centre-of-mass energy √s — the total energy available for particle creation. For a beam electron of energy E hitting a stationary proton: √s = √( (E + Mₚ)² − pᵉ² ) = √( Mₚ² + 2MₚE ) [target at rest] (3) At E = 100 MeV: √s = √(938.3² + 2×938.3×100) = √(896,660) = 947.0 MeV. To produce even the lightest new particle (a neutral pion, π⁰), we need:

(4)√sₜℍ = Mₚ + Mπ = 938.3 + 135.0 = 1073.3 MeV

We have 947.0 MeV and need 1073.3 MeV. We are 126.3 MeV short. The proton is safe. The minimum beam energy to produce a single pion is found by setting √s = 1073.3 MeV:

Eₜℎʳ = (√sₜℎʳ² − Mₚ²) / (2Mₚ) = (1073.3² − 938.3²)/(2×938.3) = (1,152,013 − 880,447)/1876.6 = 144.7 MeV

Below 145 MeV beam energy, elastic scattering is the only available process. This makes the 100 MeV regime clean: two particles in, two particles out, nothing created. Regime 2 summary: 100 MeV Wavelength: λ = 12.4 fm (λ/rₚ = 14.7) Lorentz factor: γ = 195.7 (β = 0.999987) What comes out: e⁻ (deflected) + p (recoils) Energy to proton: 0.16 to 17.7 MeV depending on angle Cross section: Mott formula eq.(2) — forward strongly favoured Pion threshold: needs 145 MeV. At 100 MeV, proton is safe.

4Regime 3: The First Cracks (4 GeV)

The proton gets excited — and breaks

At 4 GeV the electron has 40 times more energy than in Regime 2. Its Lorentz factor γ = 4000/0.511 = 7,828. Its de Broglie wavelength:

λ = 2πħc / E = 2π × 197.3 / 4000 = 0.3099 fm (λ/rₚ = 0.368)

For the first time λ < rₚ. The electron can now see structure inside the proton. The centre-of-mass energy from equation (3):

√s = √(938.3² + 2×938.3×4000) = √(8,347,633) = 2890 MeV = 2.89 GeV

41 Reaction channels open at 4 GeV

Every channel whose threshold √sₜℎʳ ≤ 2.89 GeV is available:

42 The Δ(1232) resonance: the proton’s first excited state

The dominant process is excitation of the Δ⁺(1232) resonance. Just as hydrogen can absorb a photon and jump to a higher level, the proton absorbs energy and becomes the Δ. The mass difference:

(11)ΔE = MΔ − Mₚ = 1232 − 938.3 = 293.7 MeV

The Δ has spin 3/2 (compared to 1/2 for the proton), isospin 3/2, and decays in τ = 5.6 × 10⁻²⁴ s. At the speed of light, it travels only cτ = 1.69 fm — barely twice the proton radius — before decaying. It is born and dies within the proton’s volume. Decay modes: Δ⁺ → p + π⁰ (33%): Proton survives. Neutral pion flies out. π⁰ then decays: π⁰ → γγ (lifetime 8.4×10⁻¹⁷ s). Two photons hit the calorimeter. Δ⁺ → n + π⁺ (67%): Proton converts to neutron. Charged pion escapes. π⁺ is long-lived (τ = 26.0 ns) and traverses the detector.

43 A specific event: 4 GeV, θ = 10°

Let us trace one event in full. A 4 GeV electron hits a stationary proton and scatters at 10°. First compute the elastic scattered energy from equation (4):

E'ᵉˡ = 4000 / (1 + (2×4000/938.3)×sin²(5°)) = 4000 / (1 + 8.521×0.00760) = 4000/1.0648 = 3757 MeV

The momentum transfer squared from equations (5) and (6):

Q² = 4 × 4000 × 3757 × sin²(5°) = 60,112,000 × 0.00760 = 0.457 GeV²

If instead the proton gets excited into the Δ(1232), the virtual photon must carry exactly MΔ − Mₚ = 293.7 MeV of invariant mass. The electron comes out with E′ = 3757 − 293.7 = 3463 MeV, losing an extra 294 MeV compared to elastic scattering. This 294 MeV “missing energy” is the experimental signature of the resonance. In the Δ’s rest frame, the decay pion gets:

p_π* = √[ (MΔ² - (Mₚ+Mπ)²)(MΔ² - (Mₚ-Mπ)²) ] / (2MΔ) = 229 MeV/c

T_π = √(p_π² + Mπ²) - Mπ = √(52441 + 18225) - 135 = 266 - 135 = 131 MeV The Δ is boosted forward in the lab with βΔ = 0.571, γΔ = 1.22. Boosting the pion forward:

p_π(lab) ≈ γΔ(p_π* + βΔE_π*) = 1.22 × (229 + 0.571×266) = 1.22 × 381 = 465 MeV/c

Regime 3 summary: 4 GeV Wavelength: λ = 0.310 fm (λ/rₚ = 0.368) Centre-of-mass energy: √s = 2.89 GeV All pion channels open. Δ(1232) dominant. Elastic E′ at 10°: 3757 MeV | Inelastic (via Δ) E′: 3463 MeV Signature: 294 MeV missing energy + forward pion at ~465 MeV/c What comes out: e⁻ + (p or n) + (1-2 pions)

5Regime 4: Shattering the Proton (20 GeV)

Deep inelastic scattering — the electron sees quarks

In 1968, at SLAC, physicists fired 20 GeV electrons at stationary protons. The de Broglie wavelength:

λ = 2πħc / E = 2π × 197.3 / 20,000 = 0.0620 fm (λ/rₚ = 0.0737)

The electron now has 1/0.074 ≈ 14 “pixels” across the proton. It does not see a ball. It sees individual point-like objects inside. This was the experimental discovery of quarks.

51 The kinematic variables of deep inelastic scattering

In deep inelastic scattering (DIS), the electron exchanges a virtual photon with the proton. Four key variables:

(6)Q² = 4EE′ sin²(θ/2)

(7)ν = E − E′

(8)x = Q² / (2Mₚν)

(9)W² = Mₚ² + 2Mₚν − Q²

Q² is the momentum transfer squared: higher Q² means finer resolution (smaller wavelength). ν is the energy transferred to the proton. x (the Bjorken variable) is the fraction of the proton’s momentum carried by the struck quark. W is the invariant mass of the hadronic debris.

52 A single event in full detail: θ = 10°, E′ = 12 GeV

Before: electron E = 20 GeV rightward. Proton at rest E = 0.938 GeV. Total energy 20.938 GeV. Total forward momentum 20.000 GeV/c. Computing the kinematic variables from equations (5)–(8):

Q² = 4 × 20 × 12 × sin²(5°) = 960 × 0.00760 = 7.30 GeV²

ν = 20 − 12 = 8.0 GeV

x = 7.30 / (2 × 0.938 × 8.0) = 7.30 / 15.01 = 0.487

W = √(0.938² + 2×0.938×8.0 − 7.30) = √(0.880 + 15.008 − 7.30) = √8.59 = 2.93 GeV

Interpretation: the electron struck a quark carrying 48.7% of the proton’s momentum. At this x value, it is most likely an up quark (the proton contains two up quarks and one down quark; up quarks are favoured at high x). The hadronic debris has invariant mass W = 2.93 GeV and total energy Eℎʳˡ = ν + Mₚ = 8.938 GeV, flying forward at angle θℎʳˡ ≈ arctan(p⊥/p∥). Average particle multiplicity at W = 2.93 GeV is approximately 5–6 hadrons. Typical particle list from this event: 1 proton or neutron (2–6 GeV) The two “spectator” quarks that were not struck, carrying most of the forward momentum. 2–3 π⁺ pions (1–3 GeV each) Quark-antiquark pairs formed when the struck quark was ripped out. 1–2 π⁻ pions (0.5–2 GeV each) Charge conservation requires roughly equal π⁺ and π⁻ production. 1–2 π⁰ pions Decay instantly to two photons; hit the electromagnetic calorimeter. Occasionally 1 kaon (K⁺ or K⁰) If a strange quark pair was created from the vacuum. Conservation check at this event: Regime 4 summary: 20 GeV Wavelength: λ = 0.062 fm (λ/rₚ = 0.074) Resolution: 14 “pixels” across the proton Bjorken x = Q²/(2Mν): fraction of proton momentum carried by struck quark At x = 0.487: most likely hitting a valence up quark Hadronic debris: W = 2.93 GeV, 5-6 particles, all forward What comes out: e⁻ (deflected) + hadronic jet

ep_collision.py — ep_collision
runs in your browser
===========================================================================
ELECTRON-PROTON COLLISION: REAL NUMBERS
Electron beam hits stationary proton (fixed target)
===========================================================================

===========================================================================
SCENARIO A: 100 MeV electron on stationary proton
===========================================================================

--- Incoming electron ---
  Kinetic energy:  T = 100.0 MeV
  Total energy:    E = 100.51 MeV
  Momentum:        p = 100.51 MeV/c
  Lorentz gamma:   196.7
  Speed:           beta = 0.999987 c
  de Broglie lam:  12.34 fm
  lam / r_proton:  14.7  ->  cannot resolve proton interior

--- Centre-of-mass frame ---
  sqrt(s) = 1033.91 MeV = 1.0339 GeV
  Available energy above rest masses: 95.13 MeV

--- Pion production threshold ---
  Need sqrt(s) >= 1073.8 MeV to produce even one pi0
  That requires E_beam = 145.3 MeV (T = 144.8 MeV)
  We have sqrt(s) = 1033.9 MeV  ->  BELOW threshold

--- What happens (elastic scattering) ---
  The proton STAYS INTACT. Only two particles come out:
    1. Scattered electron (deflected)
    2. Recoiling proton

--- Scattered electron (lab frame) ---
  theta_lab    Ee' (MeV)    Te' (MeV)   dE (MeV)     E_recoil   Tp (MeV)
  --------------------------------------------------------------------
      10 deg       100.35        99.84       0.16       938.44       0.16
      20 deg        99.87        99.35       0.65       938.92       0.65
      30 deg        99.09        98.58       1.42       939.69       1.42
      45 deg        97.45        96.94       3.06       941.33       3.06
      60 deg        95.40        94.89       5.11       943.38       5.11
      90 deg        90.79        90.27       9.73       948.00       9.73
     120 deg        86.60        86.09      13.91       952.19      13.91
     150 deg        83.77        83.26      16.74       955.02      16.74
     180 deg        82.78        82.27      17.73       956.01      17.73

--- Recoiling proton ---
  At theta_e = 90 deg:
  Electron: E' = 90.79 MeV, theta = 90 deg
  Proton:   T  = 9.73 MeV, phi = 42.1 deg (forward)
  Proton speed: beta = 0.1429 c
  -> Proton barely moves. It is 1836x heavier than the electron.

--- Cross section (Mott, point proton) ---
  theta =  10 deg: dsig/dOmega = 0.8811 fm^2/sr = 8.8108 mbarn/sr
  theta =  30 deg: dsig/dOmega = 0.0105 fm^2/sr = 0.1052 mbarn/sr
  theta =  90 deg: dsig/dOmega = 0.0001 fm^2/sr = 0.0009 mbarn/sr


===========================================================================
SCENARIO B: 4 GeV electron on stationary proton
===========================================================================

--- Incoming electron ---
  Kinetic energy:  T = 4.0 GeV
  Momentum:        p = 4.0005 GeV/c  (ultra-relativistic: E ~ pc)
  Lorentz gamma:   7829
  de Broglie lam:  0.3099 fm
  lam / r_proton:  0.3683  ->  RESOLVES proton interior\!

--- Centre-of-mass ---
  sqrt(s) = 2.896 GeV
  Available hadronic mass: W = 2896 MeV

--- Production thresholds vs our sqrt(s) = 2896 MeV ---
  [OPEN  ]  e + p -> e + p (elastic)
           threshold: sqrt(s) = 939 MeV = 0.939 GeV
  [OPEN  ]  e + p -> e + p + pi0
           threshold: sqrt(s) = 1074 MeV = 1.074 GeV
  [OPEN  ]  e + p -> e + n + pi+
           threshold: sqrt(s) = 1080 MeV = 1.080 GeV
  [OPEN  ]  e + p -> e + Delta(1232) -> e + p + pi
           threshold: sqrt(s) = 1233 MeV = 1.233 GeV
  [OPEN  ]  e + p -> e + p + pi+ + pi-
           threshold: sqrt(s) = 1218 MeV = 1.218 GeV
  [OPEN  ]  e + p -> e + p + pi0 + pi0
           threshold: sqrt(s) = 1209 MeV = 1.209 GeV
  [OPEN  ]  e + p -> e + p + rho0(->pipi)
           threshold: sqrt(s) = 1714 MeV = 1.714 GeV
  [OPEN  ]  e + p -> e + p + 3pi
           threshold: sqrt(s) = 1357 MeV = 1.357 GeV
  [OPEN  ]  e + p -> e + p + p + pbar
           threshold: sqrt(s) = 2815 MeV = 2.815 GeV

--- Scattered electron at selected angles ---
  For ELASTIC scattering (proton stays intact):
  theta_lab    Ee' (MeV)  Q^2 (GeV^2) lam_probe (fm)
  --------------------------------------------------
       6 deg       3909.2        0.171          2.995
      10 deg       3757.1        0.457          1.835
      20 deg       3182.3        1.536          1.001
      30 deg       2546.1        2.729          0.750
      45 deg       1778.9        4.169          0.607

--- The Delta(1232) resonance ---
  The first thing the electron 'breaks off' the proton.
  p + gamma* -> Delta++, Delta+, Delta0, or Delta-
  Delta(1232) has spin 3/2, isospin 3/2
  Decays in ~5.6 x 10^-24 s (barely exists\!)
  Delta+ -> p + pi0  (BR ~33%%)
         -> n + pi+  (BR ~67%%)
  Lifetime: tau = hbar/Gamma, where Gamma ~ 117 MeV
  tau = hbar/Gamma = 5.63e-24 s
  Travels: c*tau = 1.69 fm before decaying
  -> It decays INSIDE the proton\! You never see the Delta directly.
     You see its decay products: a proton/neutron + pion(s).

--- What the detector sees at 4 GeV ---
  FORWARD (small angle):
    - Scattered electron (most of the beam energy)
    - Hadronic debris: proton or neutron + pions, mostly forward
  LARGE ANGLE (theta > 30 deg):
    - Hard-scattered electron (lost a LOT of energy)
    - Multiple pions spraying forward

  Typical event at theta = 10 deg, elastic:
    Electron out: E' = 3757 MeV at theta = 10 deg
    Proton out:   T  = 243 MeV at phi = 65.3 deg
    Proton speed: beta  = 0.6079 c

  Typical inelastic event (Delta production) at theta = 10 deg:
    Electron out: E' = 3438 MeV at theta = 10 deg (lost 562 MeV\!)
    Q^2 = 0.418 GeV^2
    The 562 MeV goes into creating the Delta(1232)
    Delta -> proton (938 MeV) + pi0 (135 MeV)
    or Delta -> neutron (940 MeV) + pi+ (140 MeV)
    In Delta rest frame: pion gets T = 131 MeV, proton gets T = 28 MeV
    Pion momentum (Delta frame): p = 229 MeV/c
    Delta boost: gamma = 1.22, beta = 0.5710
    Forward pion in lab: p ~ 464 MeV/c
    -> Pion flies FORWARD (same direction as beam)


===========================================================================
SCENARIO C: 20 GeV electron on stationary proton  (SLAC 1968)
===========================================================================

--- Incoming electron ---
  Energy:       E = 20.001 GeV
  Momentum:     p = 20.001 GeV/c
  de Broglie lam: 0.0620 fm
  lam / r_proton: 0.0737
  -> Resolves to 1/14 of proton radius\!
     This electron sees INSIDE the proton.

--- Centre-of-mass ---
  sqrt(s) = 6.20 GeV
  Enough energy for: ~38 pions

--- Deep inelastic scattering kinematics ---
  The electron hits a SINGLE QUARK inside the proton.

[output truncated at 160 lines — run the script for the rest]

6The Missing Momentum: Gluons

If you add up the momentum carried by all the quarks the electron can scatter from, you get a surprise. The momentum sum rule says:

(14)∫₀¹ x [ u(x) + ū(x) + d(x) + d̅(x) + g(x) ] dx = 1

But when you measure the quark contributions alone: ∫₀¹ x [ u(x) + d(x) ] dx ≈ 0.54 → neutral constituents carry ∼46% (gluons ≈ 41%, strange sea ≈ 5%) (15)

anchorThe quarks carry only about 54% of the proton’s total momentum. The other 46% is invisible to the electron because the electron interacts via the electromagnetic force, and whatever carries the missing momentum must be electrically neutral.
conjectureThese invisible carriers are the gluons — the particles that bind the quarks together. In octonion language this has a beautiful interpretation. The quarks are the components of the colour part of the octonion: red (·e₅), green (·e₆), blue (·e₇). The gluons are the non-associative couplings between these components — they are the algebra itself, not the elements within it. The electron, being a quaternion object, can interact with the quark components but not with the algebraic structure that binds them.

The proton’s momentum budget:

7Confinement: Why You Cannot Free a Quark

The most striking fact about the 20 GeV experiment is what does not come out: a free quark. The electron smashes a single quark with 8 GeV of energy transfer. The quark flies away. But it never emerges alone. The quark-quark potential has two terms. At short range, a weak attractive well (from gluon exchange). At long range, a linearly rising string tension:

(12)V(r) = −αₛ/r + κ r κ ≈ 0.18 GeV² ≈ 0.9 GeV/fm

The string tension κ ≈ 0.18 GeV² ≈ 0.9 GeV/fm means the force between separated quarks is approximately constant at about 15 tonnes — independent of distance. As the struck quark flies away, the energy stored in the stretched colour field grows at 0.9 GeV per femtometre. When the field energy exceeds the rest mass energy of a quark-antiquark pair (about 300 MeV for up/down quarks), the field snaps and a new pair is created from the vacuum. The new quark joins the departing quark to form a meson (pion). The process repeats. You put in 8 GeV trying to free a quark and all you get is more bound states. The distance at which the first string breaks:

rₛₜʳʳʳ = 2mᵠ / κ ≈ 300 MeV / (0.18 GeV²) = 1.67 GeV⁻¹ = 0.33 fm

After 0.33 fm of separation, a new quark-antiquark pair is created. How many pions does the snapping string produce? Hadron multiplicity grows logarithmically with W; empirically, W = 2.93 GeV yields 5–6 hadrons on average.

conjectureIn the octonion framework, confinement follows from a single premise — that observable quantities must be associative — and that premise is not proved here, so the argument is a conjecture with its assumption named rather than a theorem. Observable quantities must be formed from colour-singlet (associative) combinations. The colour components e₅, e₆, e₇ are individually non-associative. Only combinations where the colour vector sums to zero (white = red + green + blue, or quark + antiquark) restore associativity and therefore observability.

Confinement from the octonion algebra Observable states must be associative: (AB)C = A(BC) Single quark: carries colour e₅, e₆, or e₇ — non-associative — unobservable Meson (q q̅): colour × anticolour — associative — observable Baryon (qqq): R×G×B = white — associative — observable Granting associativity, this is not an extra postulate. It is a consequence of the octonion multiplication table.

8Bjorken Scaling: Proof That Quarks Are Points

If the proton were a continuous blob of charge, its structure function F₂(x, Q²) would change as you changed Q²: zooming in would reveal finer and finer texture. But SLAC found something different: F₂(x, Q²) → F₂(x) [Bjorken scaling: independent of Q²] (13) F₂ depends only on the ratio x = Q²/(2Mν), not on Q² alone. Whether you probe at Q² = 1 GeV² or Q² = 10 GeV², you see the same pattern. This is Bjorken scaling, and it means the objects inside the proton are point-like — they have no internal structure of their own. The scaling data from the original SLAC experiment (approximate): The ratio of maximum to minimum F₂ at any fixed x is less than 10% across a factor-of-five range in Q². This is scaling to within experimental precision. The objects inside the proton have no size. They are points. They are quarks.

9Summary: One System, Four Answers

The transition energy

The boundary between quaternion and octonion physics occurs where λ = rₚ:

pₜʳʳʳₛ = 2πħc / rₚ = 2π × 197.3 / 0.8414 = 1473 MeV/c ≈ 1.47 GeV/c

Below this momentum, the proton’s second quaternion is invisible. The proton looks like a point with a charge. Above this momentum, the second quaternion is revealed: confinement energy in e₄, colour charges in e₅, e₆, e₇, gluon couplings in the non-associative products. SLAC did not discover “quarks.” SLAC discovered the proton’s second quaternion.

Equation Index

(1) λ = 2πħc / p = h / p

(2) dσ/dΩ = (αħc / 2E)² × cos²(θ/2) / sin⁴(θ/2)

(3) √s = √( (E + Mₚ)² − pᵉ² ) = √( Mₚ² + 2MₚE ) [target at rest]

(4) √sₜℍ = Mₚ + Mπ = 938.3 + 135.0 = 1073.3 MeV

(5) E' = E / ( 1 + (2E/Mₚ) sin²(θ/2) ) [elastic, proton at rest]

(6) Q² = 4EE′ sin²(θ/2)

(7) ν = E − E′

(8) x = Q² / (2Mₚν)

(9) W² = Mₚ² + 2Mₚν − Q²

(10) Tₚ⁻⁺ˣ = 2Mₚpᵉ² / ( Mₚ² + 2MₚE + mᵉ² ) ≈ 2Mₚpᵉ² / Mₚ² for E ≫ Mₚ

(11) ΔE = MΔ − Mₚ = 1232 − 938.3 = 293.7 MeV

(12) V(r) = −αₛ/r + κ r κ ≈ 0.18 GeV² ≈ 0.9 GeV/fm

(13) F₂(x, Q²) → F₂(x) [Bjorken scaling: independent of Q²]

(14) ∫₀¹ x [ u(x) + ū(x) + d(x) + d̅(x) + g(x) ] dx = 1

(15) ∫₀¹ x [ u(x) + d(x) ] dx ≈ 0.54 → neutral constituents carry ∼46% (gluons ≈ 41%, strange sea ≈ 5%)

Symbols & Terms